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\markboth{Homological methods in Non-commutative Geometry -- Tokyo,
2007/2008}{Homological methods in Non-commutative Geometry -- Tokyo,
2007/2008}
\title{Homological methods in Non-commutative Geometry}
\author{D. Kaledin}
\begin{document}
\maketitle
These are lecture notes for a course I gave at the University of
Tokyo in the winter term of 2007/2008. The title of the course was
``Homological methods in Non-commutative geometry'', by which I mean
some assorted results about Hochschild and cyclic homology, on one
hand, and Hochschild cohomology and Kontsevich's Formality
Conjecture, on the other hand. The notes are essentially identical
to the hand-outs given out during the lectures -- I did not attempt
any serious revision.
There were eleven lectures in total. Every lecture is preceeded by a
brief abstract. The first seven deal with the homological part of
the story (cyclic homology, its various definitions, various
additional structures it possesses). Then there are four lecture
centered around Hochschild cohomology and the formality theorem. The
course ends rather abruptly, mostly because of the time
constraints. One thing which I regret omitting is an introduction to
the language of DG algebras and DG categories according to
B. To\"en, including his beautiful recent finiteness theorem
arXiv:math/0611546. Further possible topics include, for instance,
the homological structures associated to Calabi-Yau algebras and
Calabi-Yau categories, where we can identify Hochschild homology and
Hochschild cohomology, and consider the interplay between the
Connes-Tsygan differential on the former, and the Gerstenhaber
algebra structure on the later. The resulting notions --
Batalin-Vilkovisky algebras, non-commutative calculus of
Tsygan-Tamarkin, etc. -- are very beautiful and important, but I
don't feel ready to present them in introductory lectures. The same
goes for the more advanced parts of formality (for instance,
complete proof of Deligne conjecture, Etingof-Kazhdan quantization,
recent work of Dolgushev-Tsygan-Tamarkin and Calaque-Van den Bergh
on $G_\infty$-formality) and for deformation theory of abelian
categories.
So, in a nutshell, these lectures cover at most one third of a
reasonable textbook on the subject, and I cannot really extend them
because the other two thirds are still under active investigation by
many mathematicians around the world.
In addition, there is no bibliography (some references are included
in the text), and there are far fewer exact attributions than I
would like (which is due to my ignorance, and most certainly should
not be understood as claiming any original research).
Well, for what it's worth.
\subsubsection*{Acknowledgement.}
These lectures became possible only because of the hospitality and
encouragement of Prof. Yu. Kawamata, on whose invitation I spent
time in Tokyo, and to whom I am sincerely grateful. I am also
grateful to the people who attended the course, and especially to
Prof. T. Terasoma, for following the lectures and making numerous
comments and suggestions. I should also mention that the University
of Tokyo is one of the best places I know to be and do mathematics,
and I benefitted immensely from its superb infrastructure, great
people and warm atmosphere.
\lecture{}
\begin{center}
\begin{minipage}{0.7\linewidth}\small
The subject of Non-commutative geometry. Notions of a
non-commutative geometry. Dictionary between notions from calculus
and homological invariants. Hochschild Homology and Cohomology.
Hochschild-Kostant-Rosenberg Theorem. Bar-resolution and the
Hochschild complex. Cyclic homology (explicit definition).
\end{minipage}
\end{center}
\subsection{The subject of Non-commutative Geometry.}
It is an empirical fact that the idea of ``non-commutative
geometry'', when seen for the first time, is met with deep
scepticism (at least, this was my personal reaction for 10 years or
so). Let me start these lectures with a short justification of the
subject.
\medskip
Back in the nineteenth century, and today in high school math,
geometry was essentially set-theoretic: the subject of geometry was
points, lines (sets of points of special types), and so on. This
approach has been inherited by early algebraic geometry -- instead
of lines we maybe consider curves of higher degree, or
higher-dimensional algebraic varieties, but we still think of them
as sets of points with some additional structure.
However, starting from mid-twentieth century, and especially in the
work of Grothendieck, a new viewpoint appeared, which can be loosely
termed ``categorical'': one thinks of an algebraic variety simply as
an object of the category of algebraic varieties. The precise
``inner structure'' of an algebraic variety is not so important
anymore -- what is important is how it behaves with respect to other
varieties, what maps to or from other varieties does it admit, and
so on. ``Set of points'' is just one functor on the category of
algebraic varieties that we can use to study them; there are other
important functors, such as, for instance, various cohomology
theories.
\medskip
These two ``dual'' approaches to algebraic geometry are not mutually
exclusive, but rather complementary, and somewhat competing. To give
you a non-trivial example, let us consider the Minimal Model
Program. Here two methods of studying an algebraic variety $X$
proved to be very successful. One is to study rational curves on $X$,
their families, subvarieties they span etc. The other is to treat
$X$ as a whole and obtain results by considering its cohomology with
various coefficients and using Vanishing Theorems. For example, the
Cone Theorem claims that a certain part of the ample cone of $X$ is
polyhedral, with faces dual to certain classes in $H_2(X)$ called
``extremal rays''. If $X$ is smooth, the Theorem can be proved by
the ``bend-and-break'' techniques; extremal rays emerge as
fundamental classes of certain rational curves on $X$. On the other
hand, the Cone Theorem can be proved essentially by using
consistently the Kawamata-Viehweg Vanishing Theorem; this only gives
extremal rays as cohomology classes, with no generating rational
curves, but it works in larger generality (for instance, for a
singular $X$).
\bigskip
Now, the idea of ``non-commutative'' geometry is, in a nutshell, to
try to replace the notion of an affine algebraic variety $X = \Spec
A$ with something which would make sense for a non-commutative ring
$A$. The desire to do so came originally from physics -- one of the
ways to interpret the formalism of quantum mechanic is to say that
instead of the algebra of functions on a symplectic manifold $M$
(``the phase space''), we should consider a certain non-commutative
deformation of it. Mathematically, the procedure seems absurd. In
order to define a spectrum $\Spec A$ of a ring $A$, you need $A$ to
be commutative, otherwise you cannot even define ``points'' of
$\Spec A$ in any meaningful way. Thus the set-theoretic approach to
non-commutative geometry quickly leads nowhere.
However, and this is somewhat surprising, the categorical approach
does work: much more things can be generalized to the
non-commutative setting than one had any right to expect
beforehand. Let us list some of these things.
\begin{enumerate}
\item Algebraic K-theory.
\item Differential forms and polyvector fields.
\item De Rham differential and de Rham cohomology, Lie bracket of
vector fields, basic formalism of differential calculus.
\item Hodge theory (in its algebraic form given by Deligne).
\item Cartier isomorphisms and Frobenius action on cristalline
cohomology in positive characteristic.
\end{enumerate}
Of these, the example of K-theory is the most obvious one: Quillen's
definition of the K-theory of an algebraic variety $X = \Spec A$
involves only the abelian category $A\amod$ of $A$-modules, and it
works for a non-commutative ring $A$ without any changes
whatsoever. Before giving the non-commutative versions of the other
notions on the list, however, we need to discuss more precisely what
we mean by ``non-commutative setting''.
\subsection{The notion of a non-commutative variety.}
Actually, there are several levels of abstraction at which
non-commutative geometry can be built. Namely, we can take as our
definition of a ``non-commutative variety'' one of the following
four.
\begin{enumerate}
\renewcommand{\labelenumi}{{\normalfont(\arabic{enumi})}}
\item An associative ring $A$.
\item A differential graded (DG) algebra $A^\hdot$.
\item An abelian category $\CC$.
\item A triangulated category $\D$ ``with some enhancement''.
\end{enumerate}
The relation between these levels is not linear, but rather as
follows:
\begin{equation}\label{levels}
\begin{CD}
\text{\normalfont(1)} @>>> \text{\normalfont(2)}\\
@VVV @VVV\\
\text{\normalfont(3)} @>>> \text{\normalfont(4)}.
\end{CD}
\end{equation}
Given an associative ring $A$, we can treat it as a DG algebra
placed in degree $0$ -- this is the correspondence
$(1)\Rightarrow(2)$. Or else, we can consider the category $A\amod$
of left $A$-modules -- this is the correspondence
$(1)\Rightarrow(3)$. Given a DG algebra $A^\hdot$, we can construct
the derived category $\D(A^\hdot)$ of left DG $A^\hdot$-modules, and
given an abelian category $\CC$, we can consider its derived
category $\D(\CC)$ -- this is $(2)\Rightarrow(4)$ and
$(3)\Rightarrow(4)$.
Of course, in any meaningful formalism, the usual notion of a
(commutative) algebraic variety has to be included as a particular
case. In the list above, $(1)$ is the level of an affine algebraic
variety $X = \Spec A$. Passing from $(1)$ to $(3)$ gives the
category of $A$-modules, or, equivalently, the category of
quasicoherent sheaves on $X$. This makes sense for an arbitrary, not
necessarily affine scheme $X$ -- thus on level $(3)$, we can work
with any scheme $X$ by replacing it with its category of
quasicoherent sheaves. We can then pass to level $(4)$, and take the
derived category $\D(X)$.
What about $(2)$? As it turns out, an arbitrary scheme $X$ also
appears already on this level: the derived category $\D(X)$ of
quasicoherent sheaves on $X$ is equivalent to the derived category
$\D(A^\hdot)$ of a certain (non-canonical) DG algebra $A^\hdot$. The
rough slogan for this is that ``every scheme is derived-affine''.
Here are some other examples of non-commutative varieties that one
would like to consider.
\begin{enumerate}
\item Given a scheme $X$, one can consider a coherent sheaf $\A$ of
algebras on $X$ and the category of sheaves of $\A$. This is only
``slightly'' non-commutative, in the sense that we have an honest
commutative scheme, and the non-commutative algebra sheaf is of
finite rank over the commutative sheaf $\calo_X$ (e.g. if $X =
\Spec B$ is affine, then $\A$ comes from a non-commutative algebra
which has $B$ lying its center, and is of finite rank over this
center). However, there are examples where this is useful. For
instance, in the so-called {\em non-commutative resolutions}
introduced by M. Van den Bergh, $X$ is usually singular;
generically over $X$, $\A$ is a sheaf of matrix algebras, so that
its category of modules is equivalent to the category of coherent
sheaves on $X$, but near the singular locus of $X$, $\A$ is no
longer a matrix algebra, and it is ``better behaved'' than
$\calo_X$ -- e.g. it has finite homological dimension.
\item Many interesting categories come from representation theory --
representation of a Lie algebra, or of a quantum group, or
versions of these in finite characteristic, and so on. These have
appeared prominently, for examples, in the recent works of
R. Rouquier.
\item In sympletic geometry, there is the so-called {\em Fukaya
category} and its versions (e.g. the ``Fukaya-Seidel
category''). These only exist at level $(4)$ above, and they are
very hard to handle; still, the fully developed theory should
apply to these categories, too.
\end{enumerate}
Let us also mention that even if one is only interested in the usual
schemes $X$, looking at them non-commutatively is still non-trivial,
because {\em there are more maps between schemes $X$, $X'$ when they
are considered as non-commutative varieties}. E.g. on level $(4)$, a
map between triangulated categories is essentially a trinagulated
functor, or maybe a pair of adjoint triangulated functors, depending
on the specific formalism used -- but in any approach, a
Fourier-Mukai transform, for instance, gives a well-defined
non-commutative map. Flips and flops in the Minimal Model Program
are also expected to give non-commutative maps.
\medskip
Passing to a higher level of abstraction in \eqref{levels}, we lose
some information. A single abelian category can be equivalent to the
category of modules for different rings $A$ (this is known as Morita
equivalence -- e.g. a commutative algebra $A$ is Morita-equivalent
to its matrix algebra $M_n(A)$, for any $n \geq 2$). And a single
triangulated category can appear as the derived category of
quasicoherent sheaves on different schemes (e.g. related by the
Fourier-Mukai transform) and the derived category of DG modules over
different DG algebras (e.g. related by Koszul duality, the DG
version of Morita equivalence). However, it seems that the
information lost is inessential; especially if we think of various
homological invariants of a non-commutative variety, they all are
independent of the specifics lost when passing to $(4)$. While this
is not a self-evident first principle but rather an empirical
observation, it seems to hold -- again as a rough slogan,
``non-commutative geometry is derived Morita-invariant''. Thus it
would be highly desirable to develop the theory directly on level
$(4)$ and not bother with inrelevant data.
However, at present it is not possible to do this. The reason is the
well-known fact that the notion of triangulated category is ``too
weak''. Here are some instances of this.
\begin{enumerate}
\item ``Cones are not functorial''. Thus for a triangulated category
$\D$, the category of functors $\Fun(I,\D)$ for even the simplest
diagrams $I$ -- e.g. the category of arrows in $\D$ -- is not
triangulated.
\item Triangulated categories do not patch together well. For
instance, if we are given two triangulated categories $\D_1$,
$\D_2$ equipped with triangulated functors to a triangulated
category $\D$, the fibered product $\D_1 \times_\D \D_2$ is not
triangulated.
\item Given two triangulated categories $\D_1$, $\D_2$, the category
of triangulated functors $\Fun_{tr}(\D_1,\D_2)$ is not
triangulated.
\end{enumerate}
It is the consensus of all people working in the field that the
correct notion is that of a triangulated category with some
additional structure, called ``enhancement''; however, there is no
consensus as to what a convenient enhancement might be, exactly.
Popular candidates are ``DG-enhancement'',
``$A_\infty$-enhancement'' and ``derivator enhancement''. Within the
framework of these lectures, let me just say that the only
sufficiently developed notion of enhancement seems to be the DG
approach, but using it is not much different from simply working in
the context of DG algebras, that is, on our level $(2)$.
Thus is the present course, we will not attempt to work in the full
generality of $(4)$ -- we will start at $(1)$, and then maybe go to
$(2)$ and/or $(3)$.
However, it is important to keep in mind that $(4)$ is the correct
level. In particular, everything should and will be
``derived-Morita-invariant'' -- DG algebras or abelian categories
that have equivalent derived categories are indistinguishable from
the non-commutative point of view.
\subsection{A dictionary.}
Let us now give a brief dictionary between some notions of algebraic
geometry and their non-commutative counterparts. We will only do it
in the affine case (level $(1)$). For convenience, we have
summarized it in table form.
\begin{quote}
\begin{tabular}{|c|c|}
\hline
An affine scheme $X = \Spec A$ & An associative algebra $A$\\
\hline
$X$ is smooth & $A$ has finite homological dimension\\
\hline
Differential forms $\Omega^\hdot(X)$ & Hochschild homology classes
$HH_\idot(A)$ \\
\hline
Polyvector fields $\Lambda^\hdot\T(X)$ & Hochschild cohomology
classes $HH^\hdot(A)$\\
\hline
De Rham differential $d$ & Connes' differential $B$\\
\hline
De Rham cohomology $H^\hdot_{DR}(X)$ & Cyclic homology
$HC_\idot(A)$, $HP_\idot(A)$\\
\hline
Schouten bracket & Gerstenhaber bracket\\
\hline
Hodge-to-de Rham spectral sequence & Hochschild-to-cyclic spectral
sequence\\
\hline
Cartier isomorphisms & A
non-commutative version thereof\\
\hline
\end{tabular}
\end{quote}
Here are some comments on the table.
\begin{enumerate}
\item Polyvector fields are sections of the exterior algebra
$\Lambda^\hdot\T(X)$ generated by the tangent bundle $\T(X)$, and
Schouten bracket is a generalization of the Lie bracket of vector
field to polyvector fields. It seems that in non-commutative
geometry, it is not possible to just consider vector fields -- all
polyvector fields appear together as a package.
\item Similarly, {\em multiplication} in de Rham cohomology seems to
be a purely commutative phenomenon -- in the general
non-commutative setting, it does not exist.
\item The first line corresponds to a famous theorem of Serre which
claims that the category of coherent sheaves on a scheme $X$ has
finite homological dimension if and only if $X$ is regular. In the
literature, some alternative notions of smoothness for
non-commutative varieties are discussed; however, we will not use
them.
\item The last line takes place in positive characteristic, that is,
for schemes and algebras defined over a field $k$ with $p = \cchar
k > 0$.
\end{enumerate}
All the items in the left column are probably very familiar (expect
for maybe the last line, which we will explain in due course). The
notions in the right column probably are not familiar. In the first
few lectures of this course, we will explain them. We start with
Hochschild Homology and Cohomology.
\subsection{Hochschild Homology and Cohomology.}
Assume given an associative unital algebra $A$ over a field $k$.
\begin{defn}\label{hh.defn}
{\em Hochschild homology} $HH_\idot(A)$ of the algebra $A$ is given
by
\begin{equation}\label{hh.idot.eq}
HH_\idot(A) = \Tor_\idot^{A^{opp} \otimes A}(A,A).
\end{equation}
{\em Hochschild cohomology $HH^\hdot(A)$} of the algebra $A$ is given
by
\begin{equation}\label{hh.hdot.eq}
HH^\hdot(A) = \Ext^\hdot_{A^{opp} \otimes A}(A,A).
\end{equation}
\end{defn}
Here $A^{opp}$ is the opposite algebra to $A$ -- the same algebra
with multiplication written in the opposite direction (if $A$ is
commutative, then $A^{opp} \cong A$, but in general they might be
different). Left modules over $A^{opp} \otimes A$ are the same as
bimodules over $A$, and $A$ has a natural structure of $A$-bimodule,
called {\em the diagonal bimodule} -- this is the meaning of $A$ in
\eqref{hh.hdot.eq} and in the right-hand side of $\Tor_\idot(-,-)$
in \eqref{hh.idot.eq}. However, $A$ also has a natural structure of
a {\em right} module over $A^{opp} \otimes A$ -- and this is what we
use in the left-hand side of $\Tor_\idot(-,-)$ in
\eqref{hh.idot.eq}.
We note that by definition $HH^\hdot(A)$ is an algebra (take the
composition of $\Ext^\hdot$-s), and $HH_\idot(A)$ has a natural
structure of a right module over $HH^\hdot(A)$. In general, neither
of them has a structure of an $A$-module.
Given an $A$-bimodule $M$, we can also define Hochschild homology
and cohomology with coefficients in $M$ by setting
$$
HH_\idot(A,M) = \Tor_\idot^{A^{opp} \otimes A}(A,M),\qquad
HH^\hdot(A,M) = \Ext^\hdot_{A^{opp} \otimes A}(A,M).
$$
In particular, $HH_\idot(A,-)$ is the derived functor of the
left-exact functor $A\bimod \to k\Vect$ from $A$-bimodules to
$k$-vector spaces given by $M \mapsto A \otimes_{A^{opp} \otimes A}
M$. Equivalently, this functor can be defined as follows:
$$
M \mapsto M/\{am-ma\mid a \in A,m \in M\}.
$$
The reason Hochschild homology and cohomology is interesting -- and
indeed, the starting point for the whole brand of non-commutative
geometry which we discuss in these lecture -- is the following
classic theorem.
\begin{thm}[Hochschild-Kostant-Rosenberg, 1962]\label{hkr}
Assume that $A$ is commutative, and that $X = \Spec A$ is a smooth
algebraic variety of finite type over $k$. Then there exist
isomorphisms
$$
HH_\idot(A) \cong \Omega^\hdot(X),\qquad
HH^\hdot(A) \cong \Lambda^\hdot\T(X),
$$
where $\Omega^\hdot(A)$ are the spaces of differential forms on the
affine variery $X$, and $\Lambda^\hdot\T(A)$ are the spaces of
polyvector fields -- the sections of the exterior powers of the
tangent sheaf $\T(X)$.
\end{thm}
\proof{} To compute $HH_\idot(A)$ and $HH^\hdot(A)$, we need to find
a convenient projective resolution of the diagonal bimodule
$A$. Since $A$ is commutative, we can identify $A$ and $A^{opp}$, so
that $A$-bimodules are the same as $A \otimes A$-modules. Let $I
\subset A \otimes A$ be the kernel of the natural surjective map
$m:A \otimes A \to A$, $m(a_1 \otimes a_2) = a_1a_2$. Then $I$ is an
ideal in $A \otimes A$, and by definition, the module $\Omega^1(A)$
of $1$-forms on $A$ is equal to the quotient $I/I^2$. Thus we have a
canonical surjective map
$$
\eta:I \to \Omega^1(A).
$$
Since $X = \Spec A$ is smooth of finite type, $\Omega^1(A)$ is a
projective $A$-module. Therefore, if consider the $A$-bimodule $I$
as an $A$-module by restriction to one of the factors in $A \otimes
A$ -- say the second one -- then the map $\eta$ admits a splitting
map $\Omega^1(A) \to I$, which extends to a map
$$
s:A \otimes \Omega^1(A) \to I
$$
of $A$-bimodules. But the $A$-bimodule $A \otimes \Omega^1(A)$ is
projective; thus we can let $P_0 = A \otimes A$, $P_1 = A \otimes
\Omega^1(A)$, and we have a start of a projective resolution
$$
\begin{CD}
P_1 @>{s}>> P_0 @>{m}>> A
\end{CD}
$$
of the diagonal bimodule $A$. Extend it to a ``Koszul complex''
$P_\idot$ by setting $P_i = \Lambda^i_{A \otimes A}(P_1)$, $i \geq
0$, and extending $s$ to a derivation $d:P_{\idot+1} \to P_\idot$ of
this exterior algebra. This gives a certain complex $P_\idot$, and
it well-know that
\begin{quote}
$P_\idot$ is a resolution of $A$ outside of a certain Zariski-closed
subset $Z \subset X \times X$ which does not intersect the diagonal.
\end{quote}
Therefore the complex $P_\idot$ can be used to compute $HH_\idot(A)$
and $HH^\hdot(A)$; doing this gives the desired isomorphism.
\endproof
\begin{exc}
Show that the isomorphisms in Theorem~\ref{hkr} are canonical.
\end{exc}
We note that this proof does not need any assumptions on
characteristic (the original proof of Hochschild-Kostant-Rosenberg
was slightly different, and it only worked in characteristic $0$).
\subsection{The bar resolution and the Hochschild complex.}
The Koszul resolution is very convenient, but it only exists for a
smooth commutative algebra $A$. We will now introduce another
resolution for the diagonal bimodule called {\em the bar resulution}
which is much bigger, but exists in full generality. This gives a
certain large but canonical complex for computing $HH_\idot(A)$ and
$HH^\hdot(A)$.
The bar resolution $C_\idot(A)$ starts with the same free
$A$-bimodule $C_0(A) = A \otimes A$ as the Koszul resolution. Since
we want the resolution to exist for any $A$, there is not much we
can build upon to proceed to higher degrees -- we have to use $A$
itself. Thus for any $n \geq 1$, we let
$$
C_n(A) = A^{\otimes(n+2)} = A \otimes A^{\otimes n} \otimes A,
$$
the free $A$-bimodule generated by the $k$-vector space $A$. The
differential $C_{n+1}(A) \to C_n(A)$ is denoted $b'$ for historical
reasons, and it is given by
\begin{equation}\label{b.prime.eq}
b' = \sum_{i=1}^{n+2}(-1)^i\id^{\otimes i} \otimes m \otimes
\id^{\otimes n+2-i},
\end{equation}
where, as before, $m:A \otimes A \to A$ is the multiplication map.
We note that $b'$ is obviously an $A$-bimodule map.
There is also a version with coefficients: assume given an
$A$-bimodule $M$, and denote the $A$-action maps $A \otimes M \to
M$, $M \otimes A \to M$ by the same letter $m$. Then we let
$C_n(A,M) = A^{\otimes(n+1)} \otimes M$, $n \geq 0$, and we define
the map $b':C_{n+1}(A,M) \to C_n(A,M)$ by the same formula
\eqref{b.prime.eq}.
\begin{lemma}\label{bar.reso.1}
For any $A$, $M$, the complex $\langle C_\idot(A,M),b' \rangle$ is a
resolution of the bimodule $M$.
\end{lemma}
\proof{} The fact that $b'$ squares to $0$ is a standard computation
which we leave as an exersize (it also has an explanation in terms
of simplicial sets which we will give later). To prove that
$C_\idot(A,M)$ is a resolution, extend it to a complex
$C'_\idot(A,M)$ by shifting the degree by $1$ and adding the term
$A$ -- that is, we let
$$
C'_n(A,M) = A^{\otimes n} \otimes M
$$
for $n \geq 0$, with the differential $b'$ given by the same formula
\eqref{b.prime.eq}. Then we have to prove that $C'_\idot(A,M)$ is
acyclic. But indeed, the map $h:C'_\idot(A,M) \to C'_{\idot+1}(A,M)$
given by
$$
h(a_0 \otimes \dots \otimes a_n) = 1 \otimes a_0 \otimes \dots
\otimes a_n,
$$
obviously satisfies $h \circ b' + b' \circ h = \id$, thus gives a
contracting homotopy for $C'_\idot(A,M)$.
\endproof
\begin{exc}\label{A.t.M}
Show that for any $A$-bimodule $M$, the bimodule $A \otimes M$ is
acyclic for the Hochschild homology functor (that is, $HH_i(A,A
\otimes M) = 0$ for $i \geq 1$). Hint: compute $HH_i(A,A \otimes M)$
by using the bar resolution for the right $A^{opp} \otimes A$-module
$A$ in the left-hand side of $\Tor_\idot^{A^{opp} \otimes A}(A,A
\otimes M)$.
\end{exc}
By virtue of Exercise~\ref{A.t.M}, the resolution $C_\idot(A,M)$ can
be used for the computation of the Hochschild homology groups
$HH_\idot(A,M)$. This gives a complex whose terms are also given by
$A^{\otimes n} \otimes M$, $n \geq 0$, but the differential is given
by
\begin{equation}\label{b.eq}
b = b' + (-1)^{n+1}t,
\end{equation}
with the correction term $t$ being equal to
$$
t(a_0 \otimes \dots \otimes a_{n+1} \otimes m) = a_1 \otimes \dots
\otimes a_{n+1} \otimes ma_0
$$
for any $a_0,\dots,a_{n+1} \in A$, $m \in M$. This is known as the
{\em Hochschild homology complex}.
Geometrically, one can think of the components $a_0,\dots,a_{n-1},m$
of some tensor in $A^{\otimes n} \otimes M$ as having been placed at
$n+1$ points on the unit interval $[0,1]$, including the egde points
$0,1 \in [0,1]$; then each of the terms in the differential $b'$
corresponds to contracting an interval between two neighboring
points and multiplying the components sitting at its endpoints. To
visualize the differential $b$ in a similar way, one has to take
$n+1$ points placed on the unit circle $S^1$ instead of the unit
interval, including the point $1 \in S^1$, where we put the
component $m$.
\subsection{Cyclic homology -- explicit definition.}
In the case $M = A$, the terms in the Hochschild homology complex
are just $A^{\otimes n+1}$, $n \geq 0$, and they acquire an
additional symmetry: we let $\tau:A^{\otimes n+1} \to A^{\otimes
n+1}$ to be the cyclic permutation multiplied by $(-1)^n$. Note that
in spite of the sign change, we have $\tau^{n+1} = \id$, so that it
generates an action of the cyclic group $\Z/(n+1)\Z$ on every
$A^{\otimes n+1}$. The fundamental fact here is the following.
\begin{lemma}\label{cycl.lemma}
For any $n$, we have
\begin{align*}
(\id - \tau) \circ b' &= -b \circ (\id - \tau),\\
(\id + \tau + \dots + \tau^{n-1}) \circ b &= -b' \circ (\id +
\tau + \dots + \tau^n)
\end{align*}
as maps from $A^{\otimes n+1}$ to $A^{\otimes n}$.
\end{lemma}
\proof{} Denote $m_i = \id^i \otimes m \otimes \id^{n-i}:A^{\otimes
n+1} \to A^{\otimes n}$, $0 \leq i \leq n-1$, so that $b' = m_0 -
m_1 + \dots + (-1)^{n-1}m_{n-1}$, and let $m_n = t = (-1)^n(b -
b')$. Then we obviously have
$$
m_{i+1} \circ \tau = \tau \circ m_i
$$
for $0 \leq i \leq n-1$, and $m_0 \circ \tau = (-1)^nm_n$. Formally
applying these
identities, we conclude that
\begin{equation}\label{hc.bim.1}
\begin{aligned}
\sum_{0 \leq i \leq n}(-1)^im_i \circ (\id - \tau)
&= \sum_{0 \leq i \leq n}(-1)^im_i - m_0 - \sum_{1 \leq i \leq
n}(-1)^i\tau \circ m_{i-1} \\
&= -(\id - \tau) \circ \sum_{0 \leq i \leq n-1}(-1)^im_i,
\end{aligned}
\end{equation}
\begin{equation}\label{hc.bim.2}
\begin{aligned}
b' \circ (\id + \tau + \dots + \tau^n) &= \sum_{0 \leq i \leq n-1}
\sum_{0 \leq j \leq n} (-1)^im_i \circ \tau^j \\
&= \sum_{0 \leq j \leq i \leq n-1}(-1)^i\tau^j
\circ m_{i-j} + \sum_{1 \leq i \leq j \leq n}
(-1)^{i+n}\tau^{j-1} \circ m_{n+i-j} \\
&= -(\id + \tau + \dots + \tau^{n-1}) \circ b,
\end{aligned}
\end{equation}
which proves the claim.
\endproof
As a corollary, the following diagram is in fact a bicomplex.
\begin{equation}\label{hc.eq}
\begin{CD}
\dots @>>> A @>{\id}>> A @>{0}>> A\\
@. @AA{b}A @AA{b'}A @AA{b}A \\
\dots @>>> A \otimes A @>{\id + \tau}>> A \otimes A
@>{\id - \tau}>> A \otimes A\\
@. @AA{b}A @AA{b'}A @AA{b}A\\
\dots @. \dots @. \dots @. \dots\\
@. @AA{b}A @AA{b'}A @AA{b}A \\
\dots @>>> A^{\otimes n} @>{\id + \tau + \dots + \tau^{n-1}}>>
A^{\otimes n} @>{\id - \tau}>> A^{\otimes n}\\
@. @AA{b}A @AA{b'}A @AA{b}A
\end{CD}
\end{equation}
Here it is understood that the whole thing extends indefinitely to
the left, all the even-numbered columns are the same, all
odd-numbered columns are the same, and the bicomplex is invariant
with respect to the horizontal shift by $2$ columns.
\begin{defn}
The total homology of the bicomplex \eqref{hc.eq} is called the {\em
cyclic homology} of the algebra $A$, and denoted by $HC_\idot(A)$.
\end{defn}
We see right away that the first, the third, and so on column when
counting from the right is the Hochschild homology complex computing
$HH_\idot(A)$, and the second, the fourth, and so on column is the
acyclic complex $C_\idot'(A)$. (the top term is $A$, and the rest is
the bar resolution for $A$). Thus the spectral sequence for this
bicomplex has the form
\begin{equation}\label{h.dr.eq}
HH_\idot(A)[u^{-1}] \Rightarrow HC_\idot(A),
\end{equation}
where $u$ is a formal parameter of cohomological degree $2$, and
$HH_\idot(A)[u^{-1}]$ is shorthand for ``polynomials in $u^{-1}$ with
coefficients in $HH_\idot(A)$''. This is known as {\em
Hochschild-to-cyclic}, or {\em Hodge-to-de Rham} spectral sequence
(we will see in the next lecture that it reduces to the usual
Hodge-to-de Rham spectral sequence in the smooth commutative case).
Shifting \eqref{hc.eq} to the right by $2$ columns gives the {\em
periodicity map} $u:HC_{\idot+2}(A) \to HC_\idot(A)$, which fits
into an exact triangle
\begin{equation}\label{connes.ex}
\begin{CD}
HH_{\idot+2} @>>> HC_{\idot+2}(A) @>>> HC_\idot(A) @>>>,
\end{CD}
\end{equation}
known as the {\em Connes' exact sequence}. One can also invert the
periodicity map -- in other words, extend the bicomplex
\eqref{hc.eq} not only to the left, but also to the right. This
gives the {\em periodic cyclic homology} $HP_\idot(A)$. Since the
bicomplex for $HP_\idot(A)$ is infinite in both directions, there is
a choice involved in taking the total complex: we can take either
the product, or the sum of the terms. We take the product.
\begin{rem}
The $n$-th row of the complex \eqref{hc.eq} is the standard complex
which computes the homology $H_\idot(\Z/n\Z,A^{\otimes n})$ of the
cyclic group $\Z/n\Z$. In the periodic version, we have the
so-called {\em Tate homology} instead of the usual homology. It is
known that, $Z/n\Z$ being finite, Tate homology is always trivial
over a base field of characteristic $0$. Were we to take the sum of
terms of the periodic bicomplex instead of the product in the
definition of $HP_\idot(A)$, the corresponding spectral sequence
would have converged, and the resulting total complex would have
been {\em acyclic}. This is the first instance of an important
feature of the theory of cyclic homology: convergence or
non-convergence of various spectral sequences is often not
automatic, and, far from being just a technical nuissance, has a
real meaning.
\end{rem}
\lecture{}
\begin{center}
\begin{minipage}{0.7\linewidth}\small
Second bicomplex for cyclic homology. Connes' differential. Cyclic
homology and the de Rham cohomology in the HKR case. Homology of
small categories. Simplicial vector spaces and homology of the
category $\Delta^{opp}$.
\end{minipage}
\end{center}
\subsection{Second bicomplex for cyclic homology.}
Recall that in the end of the last lecture, we have defined cyclic
homology $HC_\idot(A)$ of an associative unital algebra $A$ over a
field $k$ as the homology of the total complex of a certain explicit
bicomplex \eqref{hc.eq} constructed from $A$ and its tensor
powers\footnote{By the way, a good reference for everything related
to cyclic homology is J.-L. Loday's book {\em Cyclic homology},
Springer, 1998. Personally, I find also very useful an old overview
article B. Feigin, B. Tsygan, {\em Additive $K$-theory}, in Lecture
Notes in Math, vol. 1289.}. This definition is very {\em ad
hoc}. Historically, it was arrived at as a result of a certain
computation of the homology of Lie algebras of matrices over $A$; it
is not clear at all what is the invariant meaning of this explicit
bicomplex. Next several lectures will be devoted mostly to various
alternative definitions of cyclic homology. Unfortunately, all of
them are {\em ad hoc} to some degree, and none is completely
satisfactory and should be regarded as final. No really good
explanation of what is going on exists to this day. But we can at
least do computations.
The first thing to do is to notice that not only we know that the
even-numbered columns $C_\idot'(A)$ of the cyclic bicomplex
\eqref{hc.eq} are acyclic, but we actually have a contracting
homotopy $h$ for them given by $h(a_0 \otimes \dots \otimes a_n) = 1
\otimes a_0 \otimes \dots \otimes a_n$. This can be used to remove
these acyclic columns entirely. The result is the {\em second
bicomplex for cyclic homology} which has the form
\begin{equation}\label{hc.B.eq}
\begin{CD}
@.@.@. A\\
@.@.@. @AA{b}A\\
@.@. A @>{B}>> A^{\otimes 2}\\
@.@. @AA{b}A @AA{b}A\\
@. A @>{B}>> A^{\otimes 2} @>{B}>> A^{\otimes 3}\\
@.@AA{b}A @AA{b}A @AA{b}A\\
A @>{B}>> A^{\otimes 2} @>{B}>> A^{\otimes 3} @>{B}>> A^{\otimes 4},\\
@AA{b}A @AA{b}A @AA{b}A @AA{b}A
\end{CD}
\end{equation}
with the horizontal differential $B:A^{\otimes n} \to
A^{\otimes(n+1)}$ given by
$$
B = (\id - \tau) \circ h \circ (\id + \tau + \dots + \tau^{n-1}).
$$
This differential $B$ is known as the {\em Connes' differential}, or
the {\em Connes-Tsygan differential}, or the {\em Rinehart
differential}. In the commutative case, it was discovered by
G. Rinehart back in the 1960es; then it was forgotten, and
rediscovered independently by A. Connes and B. Tsygan in about 1982
(in the general associative case).
\begin{lemma}
The diagram \eqref{hc.B.eq} is a bicomplex whose total complex is
quasiisomorphic to the total complex of \eqref{hc.eq}.
\end{lemma}
\proof{} This is a general fact from linear algebra which has
nothing to do with the specifics of the situation. Assume given a
bicomplex $K_{\idot,\idot}$ with differentials $d_{1,0}$, $d_{0,1}$,
and assume given a contracting homotopy $h$ for the complex $\langle
K_{i,\idot},d_{0,1} \rangle$ for every odd $i \geq 1$. Define the
diagram $\langle K'_{\idot,\idot}, d_{1,0}', d_{0,1}' \rangle$ by
$$
K'_{i,j} = K_{2i,j-i},\quad d'_{0,1} = d_{0,1}, \quad d'_{1,0} =
d_{1,0} \circ h \circ d_{1,0}.
$$
Then $d'_{1,0} \circ d'_{1,0} = d_{1,0} \circ h \circ d_{1,0}^2
\circ h \circ d_{1,0} = 0$, and
\begin{align*}
d'_{1,0} \circ d'_{0,1} + d'_{0,1} \circ d'_{1,0} &= d_{1,0} \circ h
\circ d_{1,0} \circ d_{0,1} + d_{0,1} \circ d_{1,0} \circ h \circ
d_{1,0}\\
&= - d_{1,0} \circ h \circ d_{0,1} \circ d_{1,0} - d_{1,0} \circ
d_{0,1} \circ h \circ d_{1,0}\\
&= - d_{1,0} \circ (h \circ d_{0,1} + d_{0,1} \circ h) \circ d_{1,0}
= -d_{1,0} \circ d_{1,0} = 0,
\end{align*}
so that $K'_{\idot,\idot}$ is indeed a bicomplex, and one checks
easily that the map
$$
\bigoplus_i(-1)^i\id \oplus (-1)^{i+1}(h \circ d_{1,0}):\bigoplus_i
K'_{i,\idot-i} = \bigoplus_i K_{2i,\idot-2i} \to \bigoplus_i
K_{i,\idot-i}
$$
is a chain homotopy equivalence between the total complexes of
$K_{\idot,\idot}$ and $K'_{\idot,\idot}$.
\endproof
\begin{exc}
Check this.
\end{exc}
\subsection{Comparison with de Rham cohomology.}
The main advantage of the complex \eqref{hc.B.eq} with respect to
\eqref{hc.eq} is that it allows the comparison with the usual de
Rham cohomology in the commutative case.
\begin{prop}\label{B=d}
In the assumptions of the Hochschild-Kostant-Rosenberg Theorem,
denote $n = \dim \Spec A$, and assume that $n!$ is invertible in the
base field $k$ (thus either $\cchar k = 0$, or $\cchar k > n$). Then
the HKR isomorphism $HH_\idot(A) \cong \Omega_A^\hdot$ extends to a
quasiisomorphism between the bicomplex \eqref{hc.B.eq} and the
bicomplex
$$
\begin{CD}
@.@.@. A\\
@.@.@. @AA{0}A\\
@.@. A @>{d}>> \Omega_A^2\\
@.@. @AA{0}A @AA{0}A\\
@. A @>{d}>> \Omega_A^2 @>{d}>> \Omega_A^3\\
@.@AA{0}A @AA{0}A @AA{0}A\\
A @>{d}>> \Omega_A^2 @>{d}>> \Omega_A^3 @>{d}>> \Omega_A^4,\\
@AA{0}A @AA{0}A @AA{0}A @AA{0}A
\end{CD}
$$
where the vertical differential is $0$, and the horizontal
differential is the de Rham differential $d$.
\end{prop}
\proof{} First we show that under the additional assumption of the
Proposition, the HKR isomorphism extends to a canonical
quasiisomorphism $P$ between the Hochschild complex and the complex
$\langle \Omega^\hdot_A, 0 \rangle$. This quasiisomorphism $P$ is
given by
$$
P(a_0 \otimes a_1 \otimes \dots \otimes a_i) = \frac{1}{i!}a_0da_1
\wedge \dots \wedge a_i.
$$
This is obviously a map of complexes: indeed, since $d(a_1a_2) = a_1
da_2 + a_2 da_1$ by the Leibnitz rule, the expression for $P(b(a_0
\otimes \dots \otimes a_i))$ consists of terms of the form
$$
a_0a_j da_1 \wedge \dots \wedge da_{j-1} \wedge da_{j+1} \wedge \dots
\wedge da_i,
$$
every such term appears exactly twice, and with opposite signs. Thus
$P$ induces a map $p:HH_\idot(A) \to \Omega^\hdot_A$. By HKR, both
sides are isomorphic flat finitely generated $A$-modules; by
Nakayama Lemma, to prove that $p$ an isomorphism, it suffices to
prove that it is surjective. This is clear -- since $A$ is
commutative, the alternating sum
$$
\sum_\sigma \sgn(\sigma)a_0 \otimes \sigma(a_1 \otimes \dots \otimes
a_i)
$$
over all the permutations $\sigma$ of the indices $1,\dots,i$ is a
Hochschild cycle for any $a_0,\dots,a_i \in A$, and we have
$$
P\left(\sum_\sigma \sgn(\sigma)a_0 \otimes \sigma(a_1 \otimes \dots \otimes
a_i)\right) = a_0da_1 \wedge \dots \wedge a_i.
$$
So, $p$ is an isomorphism, and $P$ is indeed a quasiisomorphism. It
remains to prove that it sends the Connes-Tsygan differential $B$ to
the de Rham differential $d$ -- that is, we have $P \circ B = d
\circ P$. This is also very easy to see. Indeed, every term in
$B(a_0 \otimes \dots a_i)$ contains $1$ as one of the factors. Since
$1$ is annihilated by the de Rham differential $d$, the only
non-trivial contribution to $P(B(a_0 \otimes \dots a_i))$ comes from
the terms which contain $1$ as the first factor, so that we have
\begin{align*}
P(B(a_0 \otimes \dots a_i)) &= \sum_{j=0}^{i-1}P(h(\tau^j(a_0 \otimes
\dots \otimes a_i))) = \frac{1}{i!}\sum_{j=0}^{i-1}\tau^j(da_0 \wedge
\dots \wedge da_i)\\
& = \frac{1}{(i-1)!}da_0 \wedge \dots \wedge da_i,
\end{align*}
which is exactly $d(P(a_0 \otimes \dots \otimes a_i))$.
\endproof
\begin{corr}
In the assumptions of Proposition~\ref{B=d}, we have a natural
isomorphism
$$
HP_\idot(A) \cong H^\hdot_{DR}(\Spec A)((u)).
$$
\end{corr}
\proof{} Clear.\endproof
\begin{rem}
For example, the Connes-Tsygan differential $B$ in the lowest
degree, $B:A \to A^{\otimes 2}$, is given by
$$
B(a) = 1 \otimes a + a \otimes 1,
$$
which is very close to the formula $a \otimes 1 - 1 \otimes a$ which
gives the universal differential $A \to \Omega^1(A)$ into the module
of K\"ahler differentials $\Omega^1(A)$ for a commutative algebra
$A$. The difference in the sign is irrelevant because of the HKR
identification of $HH_1(A)$ and $\Omega^1(A)$ -- if one works out
explicitly the identification given in Lecture 1, one checks that $1
\otimes a$ goes to $0$, so that it does not matter with which sign
we take it. The comparison map $P$ in the lowest degree just sends
$a \otimes b$ to $adb$, so that $P(B(a)) = da$.
\end{rem}
\subsection{Generalities on small categories.}
Our next goal is to give a slightly less {\em ad hoc} definition of
cyclic homology also introduced by A. Connes. This is based on the
techniques of the so-called homology of small categories. Let us
describe it.
For any small category $\Gamma$ and any base field $k$, the category
$\Fun(\Gamma,k)$ of functors from $\Gamma$ to $k$-vector spaces is
an abelian category, and the direct limit functor
$\lim_{\overset{\to}{\Gamma}}$ is right-exact. Its derived functors
are called {\em homology functors} of the category $\Gamma$ and
denoted by $H_\idot(\Gamma,E)$ for any $E \in \Fun(\Gamma,k)$. For
instance, if $\Gamma$ is a groupoid with one object with
automorphism group $G$, then $\Fun(\Gamma,k)$ is the category of
$k$-representations of the group $G$; the homology
$H_\idot(\Gamma,-)$ is then tautologically the same as the group
homology $H_\idot(G,-)$. Analogously, the inverse limit functor
$\lim_{\overset{\gets}{\Gamma}}$ is left-exact, and its derived
functors $H^\hdot(\Gamma,-)$ are the cohomology functors of the
category $\Gamma$. In the group case, this corresponds to the usual
cohomology of the group. By definition of the inverse limit, we have
$$
H^\hdot(\Gamma,E) = \Ext^\hdot(k^\Gamma,E),
$$
where $k^\Gamma$ denotes the constant functor from $\Gamma$ to
$k\Vect$. In particular, $H^\hdot(\Gamma,k^\Gamma) =
\Ext^\hdot(k^\Gamma,k^\Gamma)$ is an algebra, and the homology
$H_\idot(\Gamma,k^\Gamma)$ with constant coefficients is a module
over this algebra.
In general, it is not easy to compute the homology of a small
category $\Gamma$ with arbitrary coefficients $E \in
\Fun(\Gamma,k)$. One way to do it is to use resolutions by the {\em
representable functors} $k_{[a]}$, $[a] \in \Gamma$ -- these are by
definition given by
$$
k_{[a]}([b]) = k[\Gamma([a],[b])]
$$
for any $[b] \in \Gamma$, where $\Gamma([a],[b])$ is the set of maps
from $[a]$ to $[b]$ in $\Gamma$, and $k[-]$ denotes the $k$-linear
span. By Yoneda Lemma, we have $\Hom(k_{[a]},E') = E'([a])$ for any
$E' \in \Fun(\Gamma,k)$; therefore $k_{[a]}$ is a projective object
in $\Fun(\Gamma,k)$, higher homology groups $H_i(\Gamma,k_{[a]})$,
$i \geq 1$ vanish, and again by Yoneda Lemma, we have
\begin{equation}\label{lim.repr}
\Hom(\lim_{\overset{\to}{\Gamma}}k_{[a]},k) \cong
\Hom(k_{[a]},k^\Gamma) \cong k^\Gamma([a]) = k,
\end{equation}
so that $H_0(\Gamma,k_{[a]}) = k$. Every functor $E \in
\Fun(\Gamma,k)$ admits a resolution by sums of representable
functors --- for example, we have a natural adjunction map
$$
\bigoplus_{[a] \in \Gamma}E([a]) \otimes k_{[a]} \to E,
$$
and this map is obviously surjective. Analogously, for cohomology,
we can use {\em co-representable functors} $k^{[a]}$ given by
$$
k^{[a]}([b]) = k[\Gamma([b],[a])]^*;
$$
they are injective, $H^0(\Gamma,k^{[a]}) \cong k$, and every $E \in
\Fun(\Gamma,k)$ has a resolution by products of functors of this
type.
\medskip
One can also think of functors in $\Fun(\Gamma,k)$ as ``presheaves
of $k$-vector spaces on $\Gamma^{opp}$''. This is of course a very
complicated name for a very simple thing, but it is useful because
it brings to mind familiar facts about sheaves on topological spaces
or \'etale sheaves on schemes. Most of these facts hold for functor
categories as well, and the proofs are actually much
easier. Specifically, it is convenient to use a version of
Grothendieck's ``formalism of six functors''. Namely, if we are
given two small categories $\Gamma$, $\Gamma'$, and a functor
$\gamma:\Gamma \to \Gamma'$, then we have an obvious restriction
functor $\gamma^*:\Fun(\Gamma',k) \to \Fun(\Gamma,k)$. This functor
has a left-adjoint $\gamma_!$ and a right-adjoint $f_*$, called the
left and right {\em Kan extensions}. (If you cannot remember which
is left and which is right, but are familiar with sheaves, then the
notation $\gamma_!$, $\gamma_*$ will be helpful.)
The direct and inverse limit over a small category $\Gamma$ are
special cases of this construction -- they are Kan extensions with
respect to the projection $\Gamma \to \ppt$ onto the point category
$\ppt$. The representable and co-representable functors $k_{[a]}$,
$k^{[a]}$ are obtained by Kan extensions with respect to the
embedding $\ppt \to \Gamma$ of the object $[a] \in \Gamma$. Given
three categories $\Gamma$, $\Gamma'$, $\Gamma''$, and two functors
$\gamma:\Gamma \to \Gamma'$, $\gamma':\Gamma' \to \Gamma''$, we
obviously have $(\gamma' \circ \gamma)^* \cong \gamma^* \circ
\gamma^{'*}$, which implies by adjunction $\gamma'_! \circ \gamma_!
\cong (\gamma' \circ \gamma)_!$ and $\gamma'_* \circ \gamma_* \cong
(\gamma' \circ \gamma)_*$. In general, the Kan extensions
$\gamma_!$, $\gamma_*$ have derived functors $L^\hdot f_!$, $R^\hdot
f_*$; just as in the case of homology and cohomology, one can
compute them by using resolutions by representable
resp. corepresentable functors.
\subsection{Homology of the category $\Delta^{opp}$.}
Probably the first useful fact about homology of small categories is
a description of the homology of the category $\Delta^{opp}$, the
opposite to the category $\Delta$ of finite non-empty totally
ordered sets. We denote by $[n] \in \Delta^{opp}$ the set of
cardinality $n$. Objects $E \in \Fun(\Delta^{opp},k)$ are known as
{\em simplicial $k$-vector spaces}. Explicitly, such an object is
given by $k$-vector spaces $E([n])$, $n \geq 1$, and various maps
between them, among which one traditionally distinguishes the {\em
face maps} $d_n^i:E([n+1]) \to E_{[n]}$, $0 \leq i \leq n$ -- the
face map $d_n^i$ corresponds to the injective map $[n] \to [n+1]$
whose image does not contain the $(i+1)$-st element in $[n+1]$.
\begin{lemma}\label{delta.comp}
For any simplicial vector space $E \in \Fun(\Delta^{opp},k)$, the
homology $H_\idot(\Delta^{opp},E)$ can be computed by the {\em
standard complex} $E_\idot$ given by $E_n = E([n+1])$, $n \geq 0$,
with differential $d:E_n \to E_{n-1}$, $n \geq 1$, equal to
$$
d = \sum_{0 \leq i \leq n} (-1)^id_n^i.
$$
\end{lemma}
\proof{} By definition, we have a map $E_0 = E([1]) \to
H_0(\Delta^{opp},E)$, which obviously factors through the cokernel
of the differential $d$, and this is functorial in $E$.
Denote by $H'_\idot(\Delta^{opp},E)$ the homology groups of the
standard complex $E_\idot$. Then every short exact sequence of
simplicial vector spaces induces a long exact sequence of
$H_\idot'(\Delta^{opp},-)$, so that $H_\idot'(\Delta^{opp},-)$ form
a $\delta$-functor. Moreover, $H'_0(\Delta^{opp},E)$ is by
definition the cokernel of the map $d = d_1^0 - d_1^1:E([2]) \to
E([1])$. This is the same as the direct limit of the diagram
$$
E([2])
\begin{array}{c}\overset{d_1^0}{\longrightarrow}\\[-3.5mm]
\underset{d_1^1}{\longrightarrow}
\end{array}
E([1])
$$
of two $k$-vector spaces $E([2])$, $E([1])$ and two maps $d_1^0$,
$d_1^1$ between them (a direct limit of this type is called a {\em
coequalizer}). Since this diagram has an obvious map to
$\Delta^{opp}$, we have a natural map
$$
H_0'(\Delta^{opp},E) \to
\lim_{\overset{\Delta^{opp}}{\longrightarrow}}E =
H_0(\Delta^{opp},E),
$$
and by the universal property of derived functors, it extends to a
canonical map
\begin{equation}\label{can.delta}
H'_\idot(\Delta^{opp},E) \to H_\idot(\Delta^{opp},E)
\end{equation}
of $\delta$-functors. We have to prove that it is an
isomorphism. Since every $E \in \Fun(\Delta^{opp},k)$ admits a
resolution by sums of representable functors $k_{[n]}$, $[n] \in
\Delta^{opp}$, it suffices to prove that the map \eqref{can.delta}
is an isomorphism for all $E = k_{[n]}$ (this is known as {\em the
method of acyclic models}). This is clear:
$H_i(\Delta^{opp},k_{[n]})$ is $k$ for $i=0$ and $0$ otherwise, and
the left-hand side of \eqref{can.delta} is the homology of the
standard complex of an $n$-simplex, which is also $k$ in degree $0$
and $0$ in higher degrees.
\endproof
\begin{exc}
Compute the cohomology $H^\hdot(\Delta^{opp},E)$. Hint: compute
$k^{[1]}$.
\end{exc}
\lecture{}
\begin{center}
\begin{minipage}{0.7\linewidth}\small
Connes' cyclic category $\Lambda$. Cyclic homology as homology of
the category $\Lambda$. Yet another bicomplex, and a definition of
cyclic homology using arbitrary resolutions.
\end{minipage}
\end{center}
\subsection{Connes' category $\Lambda$.}
For applications to cyclic homology, A. Connes introduced a special
small category known as {\em the cyclic category} and denoted by
$\Lambda$. Objects $[n]$ of $\Lambda$ are indexed by positive
integers $n$, just as for $\Delta^{opp}$. Maps between $[n]$ and
$[m]$ can be defined in various equivalent ways; we give two of
them.
\bigskip
\noindent
{\em Topological description}. The object $[n]$ is thought of as a
``wheel'' -- the circle $S^1$ with $n$ distinct marked points,
called {\em vertices}. A continuous map $f:[n] \to [m]$ is {\em
good} if it sends marked points to marked points, has degree $1$,
and is {\em monotonous} in the following sense: for any connected
interval $[a,b] \subset S^1$, the preimage $f^{-1}([a,b]) \subset
S^1$ is connected. Morphisms from $[n]$ to $[m]$ in the category
$\Lambda$ are homotopy classes of good maps $f:[n] \to [m]$.
\bigskip
\noindent
{\em Combinatorial description}. Consider the category $\Cycl$ of
linearly ordered sets equipped with an order-preserving endomorphism
$\sigma$. Let $[n] \in \Cycl$ be the set $\Z$ with the natural
linear order and endomorphism $\sigma:\Z \to \Z$, $\sigma(a) = a +
n$. Let $\Lambda_\infty \subset \Cycl$ be the full subcategory
spanned by $[n]$, $n \geq 1$ -- in other words, for any $n$, $m$,
let $\Lambda_\infty([n],[m])$ be the set of all maps $f:\Z \to \Z$
such that
\begin{equation}\label{lambda.defn.eq}
\begin{aligned}
f(a) \leq f(b) \quad \text{ whenever }a \leq b,\qquad
f(a+n) = f(a)+m,
\end{aligned}
\end{equation}
for any $a,b \in \Z$. For any $[n],[m] \in \Lambda_\infty$, the set
$\Lambda_\infty([n],[m])$ is acted upon by the endomorphism $\sigma$
(on the left, or on the right, by definition it does not matter). We
define the set of maps $\Lambda([n],[m])$ in the category $\Lambda$
by $\Lambda([n],[m])=\Lambda_\infty([n],[m])/\sigma$.
\bigskip
Here is the correspondence between the two definitions. First of
all, we note that homotopy classes of continuous monotonous maps
from $\R$ to itself whcih send integral points into integral points
are obviously in one-to-one correspondence with non-descreaing maps
from $\Z$ to itself. Now, in the topological description above, we
may assume that if we consider $S^1$ as the unit disc in the complex
plane $\C$, then the marked points are placed at the roots of
unity. Then the universal cover of $S^1$ is $\R$, and after
rescaling, we may assume that exactly the integral points are
marked. Thus any good map $f:S^1 \to S^1$ induces a map $\R \to \R$
which sends integral points into integral points, or in other words,
a non-decreasing map from $\Z$ to itself. Such a map $\R \to \R$
comes from a map $S^1 \to S^1$ if and only if the corresponding map
$\Z \to \Z$ commutes with $\sigma$.
There is also an explicit description of maps in $\Lambda$ by
generators and relations which we will not need; an interested
reader can find it, for instance, in Chapter 6 of Loday's book.
Given an object $[n] \in \Lambda$, it will be convenient to denote
by $V([n])$ the set of vertices of the wheel $[n]$ (in the
topological description), and it will be also convenient to denote
by $E([n])$ the set of {\em edges} of the wheel -- that is, the
clock-wise intervals $(s,s') \subset S^1$ between the two
neighboring vertices $s,s' \in V([n])$.
\begin{lemma}
The category $\Lambda$ is self-dual: we have $\Lambda \cong
\Lambda^{opp}$.
\end{lemma}
\proof{} In the combinatorial description, define a map
$\Lambda_\infty([m],[n]) \to \Lambda_\infty([n],[m])$ by $f \mapsto
f^o$, $f^o:\Z \to \Z$ given by
$$
f^o(a) = \min\{b \in \Z|f(b) \geq a\}.
$$
This is obviously compatible with compositions and bijective, so
that we get an isomorphism $\Lambda_\infty \cong
\Lambda_\infty^{opp}$. Being compatible with $\sigma$, it descends
to $\Lambda$.
In the topological description, note that for any map $f:[n] \to [n]$
and any edge $e=(s,s') \in E([m])$, the preimage $f^{-1}(e) \subset
S^1$ with respect to the corresponding good map $f:S^1 \to S^1$
lies entirely within a single edge $e' \in E([m'])$, Thus we get a
natural map $f^o:E([m]) \to E([n])$. We leave it to the reader to
check that this extends to a duality functor $\Lambda \to
\Lambda^o$ which interchanges $V([n])$ and $E([n])$.
\endproof
If we only consider those maps in \eqref{lambda.defn.eq} which send
$0 \in \Z$ to $0$, then the resulting subcategory in
$\Lambda_\infty$ is equivalent to $\Delta^{opp}$.
\begin{exc}
Check this. Hint: use the duality $\Lambda \cong \Lambda^o$.
\end{exc}
This gives a canonical embedding $j:\Delta^{opp} \to
\Lambda_\infty$, and consequently, an embedding $j:\Delta^{opp} \to
\Lambda$ (this is injective on maps). Functors in $\Fun(\Lambda,k)$
are called {\em cyclic $k$-vector spaces}. Any cyclic $k$-vector
space $E$ defines by restriction a simplicial $k$-vector space $j^*E
\in \Fun(\Delta^{opp},k)$.
\subsection{Homology of the category $\Lambda$.}
The category $\Lambda$ conveniently encodes the maps $m_i$ and
$\tau$ between various tensor powers $A^{\otimes n}$ used in the
complex \eqref{hc.eq}: $m_i$ corresponds to the map $f \in
\Lambda([n+1],[n])$ given by
$$
f(a(n+1)+b) =
\begin{cases}
an+b, &\quad b \leq i,\\
an+b-1, &\quad b > i,
\end{cases}
$$
where $0 \leq b \leq n$, and $\tau$ is the map $a \mapsto a+1$,
twisted by the sign (alternatively, one can say that $m_i$ are
obtained from face maps in $\Delta^{opp}$ under the embedding
$\Delta^{opp} \subset \Lambda_p$). The relations $m_{i+1} \circ \tau
= \tau \circ m_i$, $0 \leq i \leq n-1$, and $m_0 \circ \tau =
(-1)^nm_n$ between these maps which we used in the proof of
Lemma~\ref{cycl.lemma} are encoded in the composition laws of the
category $\Lambda$. Thus for any object $E \in \Fun(\Lambda,k)$ --
they are called {\em cyclic vector spaces} -- one can form the
bicomplex of the type \eqref{hc.eq}:
\begin{equation}\label{hc.lambda}
\begin{CD}
\dots @>>> E([1]) @>{\id }>>
E([1]) @>{\id - \tau}>> E([1])\\
@. @AA{b}A @AA{b'}A @AA{b}A \\
\dots @>>> E([2]) @>{\id + \tau}>> E([2])
@>{\id - \tau}>> E([2])\\
@. @AA{b}A @AA{b'}A @AA{b}A\\
\dots @. \dots @. \dots @. \dots\\
@. @AA{b}A @AA{b'}A @AA{b}A \\
\dots @>>> E([n]) @>{\id + \tau + \dots + \tau^{n-1}}>>
E([n]) @>{\id - \tau}>> E([n])\\
@. @AA{b}A @AA{b'}A @AA{b}A
\end{CD}
\end{equation}
(where $b$ and $b'$ are obtained from $m_i$ and $\tau$ by the same
formulas as in \eqref{hc.eq}). We also have the periodic version,
the Connes' exact sequence and the Hodge-to-de Rham spectral
sequence (where the role of Hochschild homology is played by the
homology $H_\idot(\Delta^{opp},j^*E)$).
\begin{lemma}\label{cycl.comp}
For any $E \in \Fun(\Lambda,k)$, the homology $H_\idot(\Lambda,E)$
can be computed by the bicomplex \eqref{hc.eq}.
\end{lemma}
\proof{} As in Lemma~\ref{delta.comp}, we use the method of acyclic
models. We denote by $H'_\idot(\Lambda,E)$ the homology of the total
complex of the bicomplex \eqref{hc.lambda}. Just as in
Lemma~\ref{delta.comp}, we have a natural map $H'_0(\Lambda,E) \to
H_0(\Lambda,E)$, we obtain an induced functorial map
$$
H'_\idot(\Lambda,E) \to H_\idot(\Lambda,E),
$$
and we have to prove that it is an isomorphism for $E=k_{[n]}$, $[n]
\in \Lambda$. We know that for such $E$, in the right-hand side we
have $k$ in degree $0$ and $0$ in higher degrees. On the other hand,
the action of the cyclic group $\Z/m\Z$ generated by $\tau \in
\Lambda([m],[m])$ on $\Lambda([n],[m])$ is obviously free, and we
have
$$
\Lambda([n],[m])/\tau \cong \Delta^{opp}([n],[m])
$$
-- every $f:\Z \to \Z$ can be uniquely decomposed as $f = \tau^j
\circ f_0$, where $0 \leq j < m$, and $f_0$ sends $0$ to
$0$. The rows of the complex \eqref{hc.eq} compute
$$
H_\idot(\Z/m\Z,k_{[n]}([m])) \cong k\left[\Delta^{opp}([n],[m])\right],
$$
and the first term in the corresponding spectral sequence is the
standard complex for the simplicial vector space $k_{[n]}^\Delta \in
\Fun(\Delta^{opp},k)$ represented by $[n] \in
\Delta^{opp}$. Therefore this complex computes
$H_\idot(\Delta^{opp},k_{[n]}^\Delta)$, and we are done by
Lemma~\ref{delta.comp}.
\endproof
There is one useful special case where the computation of
$H_\idot(\Lambda,E)$ is even easier.
\begin{defn}\label{clean.defn}
A cyclic vector space $E \in \Fun(\Lambda,k)$ is {\em clean} if for
any $[n] \in \Lambda$, the homology $H_i(\Z/n\Z,E([n]))$ with
respect to the $\Z/n\Z$-action on $E([n])$ given by $\tau$ is
trivial for all $i \geq 1$.
\end{defn}
In practice, a cyclic vector space can be clean for two
reasons. First, $E([n])$ might be a free $k[\Z/n\Z]$-module for any
$n$. Second, the base field $k$ might have characteristic $0$, so
that finite groups have no higher homology with any coeffients. In
any case, for a clean $E \in \Fun(\Lambda,k)$, computing the
homology of the rows of the bicomplex \eqref{hc.lambda} reduces to
taking the coinvariants $E([n])_\tau$ with respect to the
autmorphism $\tau$, and the whole \eqref{hc.lambda} reduces to a
complex
\begin{equation}\label{hc.clean}
\begin{CD}
\dots @>{b}>> E([n])_\tau @>{b}>> \dots @>{b}>> E([2])_\tau @>{b}>>
E([1])_\tau,
\end{CD}
\end{equation}
with the differential induced by the differential $b$ of
\eqref{hc.lambda}. We note that the coinvariants $E([n])_\tau$, $n
\geq 1$, do not form a simplicial vector space; nevertheless, the
differential $b$ is well-defined.
\subsection{The small category definition of cyclic homology.}
Assume now again given an associative unital algebra $A$ over a
field $k$. To define cyclic homology $HC_\idot(A)$ as homology of
the cyclic category $\Lambda$, one constructs a cyclic $k$-vector
space $A_\#$ in the following way: for any $[n] \in \Lambda$,
$A_\#([n]) = A^{\otimes n}$, where we think of the factors $A$ in
the tensor product as being numbered by vertices of the wheel $[n]$,
and for any map $f:[n] \to [m]$, the corresponding map
$A_\#(f):A^{\otimes n} \to A^{\otimes m}$ is given by
\begin{equation}\label{hash.map}
A_\#(f)\left(\bigotimes_{i \in V([n])}a_i\right)=
\bigotimes_{j \in V([m])} \prod_{i \in f^{-1}(j)}a_i,
\end{equation}
where $V([n])$, $V([m])$ are the sets of vertices of the wheel
$[n],[m] \in \Lambda$. We note that for any $j \in V([m])$, the
finite set $f^{-1}(j)$ has a natural total order given by the
clockwise order on the circle $S^1$. Thus, although $A$ need not be
commutative, the product in the right-hand side is well-defined. If
$f^{-1}(j)$ is empty for some $j \in V([m])$, then the right-hand
side involves a product numbered by the empty set; this is defined
to be the unity element $1 \in A$.
As an immediate corollary of Lemma~\ref{cycl.comp}, we obtain the
following.
\begin{prop}\label{hc.lambda.prop}
We have a natural isomorphism $HC_\idot(A) \cong
H_\idot(\Lambda,A_\#)$.\endproof
\end{prop}
This isomorphism is also obviously compatible with the periodicity,
the Connes' exact sequence, and the Hodge-to-de Rham spectral
sequence. In particular, the standard complex for the simplicial
$k$-vector space $j^*A_\#$ is precisely the Hochschild homology
complex, so that we have $HH_\idot(A) =
H_\idot(\Delta^{opp},j^*A_\#)$.
\begin{exc}
Show that the Hochschild homology complex which computes
$HH_\idot(A,M)$ for an $A$-bimodule $M$ also is the standard complex
for a simplicial $k$-vector space. Does it extend to a cyclic vector
space?
\end{exc}
\subsection{Example: yet another bicomplex for cyclic homology.}
The definition of cyclic homology using small categories may seem
too abstract at first, but this is actually a very convenient
technical tool: it allows to control the combinatorics of various
complexes in a quite efficient way. As an illustration of this, let
me sketch, in $\cchar 0$, yet one more description of cyclic
homology by an explicit complex (this definition has certain
advantages explained in the next subsection).
\bigskip
For any $k$-vector space $V$ equipped with a non-zero covector $\eta
\in V^*$, $\eta:V \to k$, contraction with $\eta$ defines a
differential $\delta:\Lambda^{\hdot + 1}V \to \Lambda^\hdot V$ on
the exterior algebra $\Lambda^\hdot V$, and the complex $\langle
\Lambda^\hdot V,\delta \rangle$ is acyclic, so that $\Lambda^{\geq
1}V$ is a resolution of $k = \Lambda^0V$. This construction depends
functorially on the pair $\langle V,\eta \rangle$, so that it can be
applied poinwise to the representable functor $k_{[1]} \in
\Fun(\Lambda,k)$ equipped with the natural map $\eta:k_{[1]} \to
k^\Lambda$. The result is a resolution $\Lambda^\hdot k_{[1]}$ of
the constant cyclic vector space $k^\Lambda \in \Fun(\Lambda,k)$.
Here is another description of the exterior powers $\Lambda^\hdot
k_{[1]}$. Consider a representable functor $k_{[i]}$ for some $i
\geq 1$, and let $\ok_{[i]} \in \Fun(\Lambda,k)$ be its quotient
given by
$$
\ok_{[i]}([n]) = k\left[\Lambda([i],[n])\right]/\{f \in
\Lambda([i],[n])\mid f \text{ not injective}\};
$$
in other words, $\ok_{[i]}([n])$ is spanned by injective maps from
$[i]$ to $[n]$. Then $k_{[i]}$ is acted upon by the cyclic group
$\Z/i\Z$ of automorphisms of $[i] \in \Lambda$, this action descends
to the quotient $\ok_{[i]}$, and we have
$$
\Lambda^ik_{[i]} = \left(\ok_{[i]}\right)_\tau,
$$
where $\tau:\ok_{[i]} \to \ok_{[i]}$ is the generator of $\Z/i\Z$
twisted by $(-1)^{i+1}$. The differential $\delta$ lifts
to a differential $\delta:\ok_{[i]} \to \ok_{[i-1]}$ given by the
alternating sum of the maps $\ok_{[i]} \to \ok_{[i-1]}$ induced by
the $i$ injective maps $[i-1] \to [i]$. We note, however, that the
complex $\ok_{[\idot]}$ is no longer a resolution of $k^\Lambda$.
\begin{lemma}\label{sign}
For any $i \geq 1$, we have $H_j(\Lambda,\ok_{[i]})=0$ if $j \neq
i-1$, and $k$ if $j = i-1$. The $\Z/i\Z$-action on
$k=H_{i-1}(\Lambda,\ok_{[i]})$ by the $\Z/i\Z$-action on $\ok_{[i]}$
is given by the sign representation. Moreover, for any $E \in
\Fun(\Lambda,k)$, we have $H_\idot(\Lambda,\ok_{[i]} \otimes E)
\cong H_{\idot+i-1}(\Delta^{opp},E)$, with the sign action of $\Z/i\Z$.
\end{lemma}
\proof{} The cyclic object $\ok_{[i]} \in \Fun(\Lambda,k)$ is clean,
and the corresponding complex \eqref{hc.clean} is the quotient of
the standard complex of the elementary $(i-1)$-simplex by the
subcomplex spanned by all faces of dimension less than $i-1$. In
other words, $H_\idot(\Lambda,\ok_{[i]})$ is the reduced homology of
the $(i-1)$-sphere. This proves the first claim. The second claim is
obvious: the term of degree $i-1$ in the complex \eqref{hc.clean} is
isomorphic to $(k[\Z/i\Z])_\tau$, and this is the sign
representation by the definition of $\tau$. The third claim now
follows immediately from the well-known K\"unneth formula, which says
that for any simplicial vector spaces $V,W \in
\Fun(\Delta^{opp},k)$, the standard complex of the product $V
\otimes W$ is naturally quasiisomorphic to the product of the
standard complexes for $V$ and $W$.
\endproof
Now, for any cyclic $k$-vector space $E \in \Fun(\Lambda,k)$, and
any $[i] \in \Lambda$, the product $E \otimes \ok_{[i]} \in
\Fun(\Lambda,k)$ is clean, so that it makes sense to consider the
complex \eqref{hc.clean}. Then the differential $\delta:\ok_{[i]}
\to \ok{[i-1]}$ induces a map between these complexes, and we can
form a bicomplex $K_{\idot,\idot}(E)$ given by
\begin{equation}\label{hc.eq.3}
K_{i,j}(E) = \left(\overline{k}_i([j+1]) \otimes E([j+1])\right)_\tau,
\end{equation}
where the horizontal differential $K_{\idot+1,\idot}(E) \to
K_{\idot,\idot}(E)$, henceforth denoted by $\wt{B}$, is induced by
$\delta$, and the vertical differential is the Hochschild
differential $b$, as in \eqref{hc.clean}.
\begin{lemma}\label{hc.eq.3.lem}
Assume that $\cchar k = 0$. Then for any $E \in \Fun(\Lambda,k)$,
the total complex of the bicomplex \eqref{hc.eq.3} computes the
homology $H_\idot(\Lambda,k)$.
\end{lemma}
\proof{} Since $\cchar k = 0$, every cyclic vector space is clean,
and we can compute cyclic homology by using the complex
\eqref{hc.clean}. Since $\langle\Lambda^{\geq
1}k_{[1]},\delta\rangle$ is a resolution of the constant cyclic
vector space $k^\Lambda$, we have
$$
H_\idot(\Lambda,E) \cong \Hh_\idot(\Lambda,K_\idot \otimes E),
$$
where $K_\idot \cong \Lambda^{\hdot+1}k_{[1]}$, and the differential
in $K_\idot \otimes E$ is induced by $\delta$. Applying
\eqref{hc.clean} to the right-hand side {\em almost} gives the
bicomplex \eqref{hc.eq.3} -- the difference is that we take $K_i =
(\ok_{[i+1]})_\tau$ instead of $\ok_{[i+1]}$. Thus it suffices to
prove that the natural map
$$
H_\idot(\Lambda,\ok_{[i]} \otimes E) \to
H_\idot\left(\Lambda,\left(\ok_{[i]}\right)_\tau \otimes E\right)
$$
is an isomorphism for any $[i] \in \Lambda$. But since $\cchar k =
0$, the cyclic groups have no homology, so that the right-hand side
is isomorphic to
$$
H_\idot(\Lambda,\ok_{[i]} \otimes E)_\tau.
$$
And by Lemma~\ref{sign}, $\tau$ on $H_\idot(\Lambda,\ok_{[i]}
\otimes E)$ is the identity map.
\endproof
Assume now that $E = A_\#$ for some associative unital $A$-algebra
$A$. Then the bicomplex \eqref{hc.eq.3} is similar to the second
bicomplex \eqref{hc.B.eq} for cyclic homology in the following
sense: for any $i \geq 0$, the column $K_{i,\idot}(A_\#)$ of
\eqref{hc.eq.3} computes the Hochschild homology $HH_\idot(A)$, with
the same degree shift as in \eqref{hc.B.eq}.
What happens is the following. Recall that to obtain the Hochschild
homology complex, one uses the bar resolution $C_\idot(A)$. However,
to compute the Hochschild homology $HH_\idot(A)$, any other
resolution would do. In particular, we can take any integer $n \geq
2$, and consider the $n$-fold tensor product
$$
C^n_\idot(A) = C_\idot(A) \otimes_A C_\idot(A) \otimes_A \dots
\otimes_A C_\idot(A).
$$
This is obviously a complex of free $A$-bimodules, and it is
quasiisomorphic to $A \otimes_A A \otimes_A \dots \otimes_A A \cong
A$, so that it is a good resolution. Using this resolution to
compute $HH_\idot(A)$, we obtain a complex $CH_\idot^n(A)$ whose
$l$-th term $CH_l^n(A)$ is the sum of several copies of
$A^{\otimes(n+l)}$, and these copies are numbered by elements in the
set
$$
M_l^n = \Lambda_{inj}([n],[l+n])/\tau
$$
of injective maps $[n] \to [l+n]$ considered modulo the action of
the cyclic permutation $\tau:[l+n] \to [l+n]$. In other words, the
terms of the complex $CH^n_\idot(A)$ are numbered by wheels $[n+m]$,
$m \geq 0$, with $n$ marked points considered modulo cyclic
permutation. These $n$ points cut the wheel into $n$ intervals of
lengths $l_1,l_2,\dots,l_n$ with $l_1 + l_2 + \dots + l_n = m+n$,
and the corresponding term in $CH^n_l(A)$ computes the summand
$$
A \otimes_{A^{opp} \otimes A} \left(C_{l_1-1}(A) \otimes_A C_{l_2-1}(A)
\otimes_A \dots \otimes_A C_{l_1-1}(A)\right)
$$
in
$$
CH_\idot^n(A) = A \otimes_{A^{opp} \otimes A} \left(C_\idot(A) \otimes_A
\dots \otimes_A C_\idot(A)\right).
$$
The differential $\wt{b}:CH_{\idot+1}^n(A) \to CH_\idot^n(A)$
restricted to the term which corresponds to some injective $f:[n]
\to [n+l+1]$ is the alternating sum of the maps $m_i$ corresponding
to surjective maps $[n+l+1] \to [n+l]$ such that the composition
$[n] \to [n+l+1] \to [n+l]$ is still injective -- in other words, we
allow to contract edges of the marked wheel $[n+l+1]$ {\em unless an
edge connects two marked points}. Of course, $CH^1_\idot(A)$ is the
usual Hochschild homology complex, and $\wt{b} = b$ is the usual
Hochschild differential (since there is only one marked point, every
edge can be contracted).
We leave it to the reader to check that the complex $CH^n_\idot(A)$
is precisely isomorphic to the complex $K_{n,\idot+n}(A_\#)$.
One can also show that the periodicity in $HC_\idot(A)$ corresponds
to shifting the bicomplex \eqref{hc.eq.3} by one column to the left,
just as in \eqref{hc.B.eq}, so that the Hodge filtration on
$HC_\idot(A)$ is also induced by the stupid filtration on
\eqref{hc.eq.3} in the horizontal direction. Thus {\em a postriori},
\eqref{hc.eq.3} and \eqref{hc.B.eq} are even quasiisomorphic as
bicomplexes, and the horizontal differential $\wt{B}$ in
\eqref{hc.eq.3} can be identified with the Connes-Tsygan
differential $B$. However, this is not at all easy to see by a
direct computation.
\subsection{Cyclic homology computed by arbitrary resolution.}
To show why \eqref{hc.eq.3} is useful, let me show how it can be
modified so that the bar resolution $C_\idot(A)$ is replaced with an
arbitrary projective resolution $P_\idot$ of the diagonal bimodule
$S$ (I follow the exposition in my paper {\em Cyclic homology with
coefficients}, {\tt math.KT/0702068}, which is based on ideas of
B. Tsygan).
For simplicity, I will only explain how to do this for the first two
columns of \eqref{hc.eq.3}. This gives a resolution-independent
description of the Connes-Tsygan differential $B = \wt{B}$, but says
nothing about possible higher differentials in the Hodge-to-de Rham
spectral sequence.
Fix a projective resolution $P_\idot$ with the augmentation map
$r:P_\idot \to A$. Consider the resolution $P_\idot^2 = P_\idot
\otimes_A P_\idot$ of the same diagonal bimodule $A$. Note that the
augmentation map $r$ induces {\em two} quasiisomorphisms
$r_0,r_1:P_\idot^2 \to P_\idot$ given by
$$
r_0 = r \otimes_A \id, \qquad r_1 = \id \otimes_A r.
$$
In general, there is no reason why these two maps should be
equal. However, being two maps of projective resolutions of $A$ which
induce the same identity map on $A$ itself, they should be
chain-homotopic. Choose a chain homotopy $\iota:P_\idot^2 \to
P_{\idot+1}$.
Now consider the complexes
$$
\overline{P}_\idot = A \otimes_{A^{opp} \otimes A} P_\idot,
\quad \overline{P}^2_\idot = A \otimes_{A^{opp} \otimes A} P_\idot^2
$$
which compute $HH_\idot(A)$, and the maps $\overline{r}_0$,
$\overline{r}_0$, $\overline{\iota}$ between them induces by $r_0$,
$r_1$ and $\iota$. Notice that the complex $\overline{P}_\idot^2$
has another description: we have
$$
\overline{P}^2_\idot = \bigoplus_{l,\idot-l} A \otimes_{A^{opp}
\otimes A} P_l \otimes_A P_{\idot-l},
$$
and for any two $A$-bimodules $M$, $N$, we have
$$
A \otimes_{A^{opp} \otimes A} (M \otimes_A N) = M \otimes N/\left\{ma
\otimes n - m \otimes an,am \otimes n - m \otimes na\mid a \in A, m
\in M, n \in N \right\},
$$
which is manifestly symmetric in $m$ and $n$. Thus we have a natural
involution $\tau:\overline{P}^2_\idot \to
\overline{P}^2_\idot$. This involution obviously interchanges
$\overline{r}_0$ and $\overline{r}_1$, but there is no reason why it
should be in any way compatible with the map $\overline{\iota}$ --
all we can say is that $\tau \circ \overline{\iota}$ is another
chain homotopy between $\overline{r}_0$ and $\overline{r}_1$. Thus
the map
$$
\wt{B} = \overline{\iota} - \tau \circ
\overline{\iota}:\overline{P}^2_\idot \to \overline{P}_{\idot+1}
$$
commutes with the differentials.
\begin{lemma}
The map $\wt{B}$ induces the same map on the Hochschild homology
$HH_\idot(A)$ as the Connes-Tsygan differential $B$.
\end{lemma}
\proof[Sketch of a proof.] One checks that the map we need to
describe does not depend on choices: neither of a projective
resolution $P_\idot$, since any two such resolutions are
chain-homotopy equivalent, nor of the map $\iota$, since any two
such are chain-homotopic to each other. Thus to compute it, we can
take any $P_\idot$ and any $\iota$. If we take $P_\idot =
C_\idot(A)$, the bar-resolution, and let $\iota$ be the sum of
tautological maps $A^{\otimes l} \otimes A^{\otimes l'} \to
A^{\otimes l + l'}$, then $\wt{B}$ is precisely the same as in the
bicomplex \eqref{hc.eq.3}.
\endproof
\begin{rem}
In the assumptions of the Hochschild-Kostant-Rosenberg Theorem, it
would be very interesting to try to work out explicitly the map
$\wt{B}$ for the Koszul resolution.
\end{rem}
\lecture{}
\begin{center}
\begin{minipage}{0.7\linewidth}\small
Combinatorics of the category $\Lambda$: cohomology of $\Lambda$ and
$\Lambda_{\leq n}$, periodicity, classifying spaces. Fibrations and
cofibrations of small categories; $\Lambda_\infty$ as a fibered
category over $\Lambda$.
\end{minipage}
\end{center}
\subsection{Cohomology of the category $\Lambda$ and periodicity.}
In the last lecture, we have shown that the homology
$H_\idot(\Lambda,E)$ with coefficients in some cyclic vector space
$E \in \Fun(\Lambda,E)$ can be computed by the standard complex
\eqref{hc.lambda}; in particular, we have the periodicity map
$u:H_{\idot+2}(\Lambda,E) \to H_\idot(\Lambda,E)$ and the Connes'
exact triangle
$$
\begin{CD}
H_\idot(\Delta^{opp},j^*E) @>>> H_\idot(\Lambda,E) @>{u}>> H_{\idot
- 2}(\Lambda,E) @>>>,
\end{CD}
$$
where $j:\Delta^{opp} \to \Lambda$ is the embedding defined in the
last lecture. Today, we want to give a more invariant description of
the periodicity map. That such a description should exist is
more-or-less clear. Indeed, homology $H_\idot(\Lambda,-)$ --- or
rather, hyperhomology $\Hh(\Lambda,-)$ --- is a functor from the
derived category $\D(\Lambda,k)$ of the abelian category
$\Fun(\Lambda,k)$ to the derived category $\D(k\Vect)$. By
definition, this functor is adjoint to the tautological embedding
$\D(k\Vect) \to \D(\Lambda,k)$, $k \mapsto k^\Lambda$, so that by
Yoneda Lemma, every natural transformation $H_{\idot+2}(\Lambda,-)
\to H_\idot(\Lambda,-)$ should be induced by an element in
$$
\Ext^2(k^\Lambda,k^\Lambda) = H^2(\Lambda,k).
$$
Thus to describe periodicity, we have to compute the cohomology
$H^\hdot(\Lambda,k)$ of the category $\Lambda$ with constant
coefficients $k = k^\Lambda \in \Fun(\Lambda,k)$.
The computation itself is not difficult: since the category
$\Lambda$ is self-dual, the complex \eqref{hc.lambda} has an obvious
dualization, and exactly the same argument as in the proof of
Lemma~\ref{cycl.comp} shows that dualized complex computes
$H^\hdot(\Lambda,E)$ for any $E \in \Fun(\Lambda,k)$. For the
constant functor $k$, this gives
\begin{equation}\label{lambda.coho}
H^\hdot(\Lambda,k) \cong k[u],
\end{equation}
where, as before, $k[u]$ means ``the space of polynomials in one
formal variable $u$ of degree $2$''. It is only slightly more
difficult to see that the isomorphism \eqref{lambda.coho} is an
algebra isomorphism, and the action of the generator $u \in
H^2(\Lambda,k)$ on homology $H_\idot(\Lambda,-)$ is the periodicity
map. One can argue, for instance, as follows. The same operation of
``shifting the bicomplex by two columns'' induces a periodicity map
$H^\hdot(\Lambda,E) \to H^{\hdot+2}(\Lambda,E)$; this map is
functorial, thus (1) induced by an element in $H^2(\Lambda,k)$, and
obviously the same one, and (2) compatible with the algebra
structure on
$$
H^\hdot(\Lambda,k) = \Ext^\hdot(k^\Lambda,k^\Lambda),
$$
so that $H^\hdot(\Lambda,k)$ must be a unital algebra over the
polynomial algebra $k[u]$ generated by the periodicity map. Since
by \eqref{lambda.coho}, it is isomorphic to $k[u]$ as a
$k[u]$-module, it must also be isomorphic to $k[u]$ as an algebra.
\medskip
\noindent
However, it will be useful to have a more explicit description of
the generator $u \in H^2(\Lambda,k)$.
\medskip
\noindent
To obtain such a description, we use the topological interpretation
of the category $\Lambda$ --- in other words, we treat $[n] \in
\Lambda$ as a wheel formed by marking $n$ points on the circle
$S^1$. Note that this defines a cellular decomposition of the
circle: its $0$-cells are vertices $v \in V([n])$, and its $1$-cells
are edges $e \in E([n])$. Denote by $C_\idot([n])$ the corresponding
complex of length $2$ which computes the homology
$H_\idot(S^1,k)$. Any map $f \in \Lambda([n],[m])$ induces a
cellular map $S^1 \to S^1$, or at any rate, a map which sends
$0$-skeleton into $0$-skeleton, and thus induces a map $C_\idot([n])
\to C_\idot([m])$. In this way, $C_\idot([n])$ becomes a length-$2$
complex of cyclic vector spaces. Since the homology of the circle
$H_i(S^1,k)$ is equal to $k$ for $i=0,1$ and $0$ otherwise, and does
not depend on the cellular decomposition, the homology of the
complex $C_\idot \in \Fun(\Lambda,k)$ is $k^\Lambda$ in degree $0$
and $1$, and $0$ in other degrees. Thus we have an exact sequence
\begin{equation}\label{dg.exa.1}
\begin{CD}
0 @>>> k^\Lambda @>>> C_1 @>>> C_0 @>>> k @>>> 0
\end{CD}
\end{equation}
of cyclic vector spaces. Explicitly, $V([n]) \cong
\Lambda([1],[n])$, so that $C_0([n]) = k[V([n])] =
k[\Lambda([1],[n])]$, and $C_0$ is canonically isomorphic to the
representable functor $k_{[1]}$. As for $C_1$, we have by definition
$$
C_1([n]) = k[E([n])] = k[\Lambda([n],[1])],
$$
and the map $C_1(f):C_1([n]) \to C_1([m])$ corresponding to a map
$f:[n] \to [m]$ is given by
\begin{equation}\label{k^1}
C_1(f)(e) = \sum_{e' \in f^{o-1}(e)}e' \in k[E([m])]
\end{equation}
for any edge $e \in E([n])$, so that $C_1$ is canonically identified
with the corepresentable functor $k^{[1]}$. All in all, the exact
sequence \eqref{dg.exa.1} can be rewritten as
\begin{equation}\label{dg.exa}
\begin{CD}
0 @>>> k^\Lambda @>>> k^{[1]} @>>> k_{[1]} @>>> k^\Lambda @>>> 0.
\end{CD}
\end{equation}
This represents by Yoneda a certain class in $H^2(\Lambda,k) =
\Ext^2(k^\Lambda,k^\Lambda)$.
\begin{lemma}\label{perio}
The class $u' \in H^2(\Lambda,k)$ represented by \eqref{dg.exa} is
equal to the periodicity generator $u$.
\end{lemma}
\proof{} Let us first prove the equality up to an invertible
constant. To do this, it suffices to prove that the cone of the map
$H_{\idot+2}(\Lambda,k) \to H_\idot(\Lambda,k)$ induced by $u'$ is
isomorphic to $k$ in degree $0$ and trivial in other degrees. This
cone is the hyperhomology $\Hh(\Lambda,C_\idot)$. Since
$C_0=k_{[1]}$ is representable, it already has all the homology we
want from the cone, so that we have to prove that
$$
H_\idot(\Lambda,C_1) = H_\idot(\Lambda,k^{[1]})=0
$$
(in all degrees). Denote by $M$ the kernel of the natural map
$k_{[1]} \to k^\Lambda$, so that we have short exact sequences
$$
\begin{CD}
0 @>>> M @>>> k_{[1]} @>>> k^\Lambda @>>> 0,\\
0 @>>> k^\Lambda @>>> k^{[1]} @>>> M @>>> 0.
\end{CD}
$$
Computing the homology long exact sequence for the first of these
exact sequences, we see that the boundary differential
$\delta_1:H_i(\Lambda,M) \to H_{i+1}(\Lambda,k)$ is non-trivial, so
that the first short exact sequence is not split, and that in fact
$\delta_1$ is an isomorphism for all $i \geq 0$. To prove the claim,
it suffices to check that the boundary differential
$\delta_2:H_{i+1}(\Lambda,M) \to H_i(\Lambda,k)$ in the second long
exact sequence also is an isomorphism for all $i$. Since everything
is compatible with with $k[u']$-action, it suffices to prove it for
$i=0$ -- in other words, we have to prove that the generator of
$H_0(\Lambda,k) = k$ goes to $0$ under the map $k^\Lambda \to
k^{[1]}$. But if not, this means by definition that the second short
exact sequence is split. This is not possible: the duality $\Lambda
\cong \Lambda^{opp}$ together with the usual duality $k\Vect^{opp}
\to k\Vect$, $V \mapsto V^*$ induce a fully faithfull duality
functor $\Fun(\Lambda,k)^o \to \Fun(\Lambda,k)$, and this functor
sends our short exact sequences into each other.
As for the constant, we note that it obviously must be universal,
thus invertible in any field, thus either $1$ or $-1$. On the other
hand, in the definition of \eqref{dg.exa} there is a choice: we have
to choose an orientation of the cirle $S^1$. Switching the
orientation changes the sign of $u'$, so that we can always achieve
$u=u'$. We leave it at that.
\endproof
\subsection{Canonical resolution.}
We can extend the exact sequence \eqref{dg.exa} to a resolution of
the constant functor $k^\Lambda$ by iterating it -- the result is a
complex of the form
$$
\begin{CD}
\dots @>>> k^{[1]} @>>> k_{[1]} @>>> k^{[1]} @>>> k_{[1]},
\end{CD}
$$
where the maps $k^{[1]} \to k_{[1]}$ are as in \eqref{dg.exa}, and
the maps $k_{[1]} \to k^{[1]}$ are the composition maps $k_{[1]} \to
k^\Lambda \to k^{[1]}$. Moreover, for any cyclic vector space $E
\in \Fun(\Lambda,k)$, we have a canonical resolution
\begin{equation}\label{per.res}
\begin{CD}
\dots @>>> k^{[1]} \otimes E @>>> k_{[1]} \otimes E @>>> k^{[1]}
\otimes E @>>> k_{[1]} \otimes E.
\end{CD}
\end{equation}
The periodicity map for $E$ is induced by $\id \otimes u \in
\Ext^2(E,E)$, and it can be represented explicitly by the obvious
periodicity endomorphism of \eqref{per.res} which shift everything
to the left by two terms.
It is instructive to see what happens if compute
$H_\idot(\Lambda,E)$ by replacing $E$ with \eqref{per.res}, as in
Lemma~\ref{hc.eq.3.lem} in the last Lecture. Both $k_{[1]} \otimes
E$ and $k^{[1]} \otimes E$ are clean in the sense of
Definition~\ref{clean.defn}, so that we can compute
$H_\idot(\Lambda,-)$ by the complex \eqref{hc.clean}. Applying it to
\eqref{per.res} gives a double complex $M_{i,j}(E)$ with terms
$$
M_{i,j}(E) =
\begin{cases}
(k_{[1]}([j+1]) \otimes E([j+1]))_\tau, &\quad i \text{ even},\\
(k^{[1]}([j+1]) \otimes E([j+1]))_\tau, &\quad i \text{ odd}.
\end{cases}
$$
To identify further $M_{0,j}(E) = E([j+1])$, we need to choose a
vertex $v \in V([j+1])$ (for instance, we may fix the embedding
$j:\Delta^{opp} \to \Lambda$), and to identify $M_{1,j}(E) =
E([j+1])$, we need to to choose an edge $e \in E([j+1])$ (for
instance, since choosing $v \in V([j+1])$ cuts the wheel and defines
a total order on $E([j+1])$, we can take the last edge with respect
to this order). To compute the differential $b:M_{i,j}(E) \to
M_{i,j-1}(E)$, we note that for any contraction $[j+1] \to [j]$ of
an edge $e' \in E([j+1])$, the corresponding face map
$m_e:k_{[1]}([j+1]) \to k_{[1]}([j])$ sends the chosen vertex $v \in
k[V([j+1])] = k_{[1]}([j+1])$ to the chosen vertex $v \in
k[V([j])]$. On the other hand, it immediately follows from
\eqref{k^1} that the face map $m'_{e'}:k^{[1]}([j+1]) \to
k^{[1]}([j])$ sends the chosen last edge $e \in k[E([j+1])]$ to $e
\in k[E([j])]$ if $e \neq e'$, and to $0$ otherwise. Thus the
diferential $b:M_{i,j}(E) \to M_{i,j-1}(E)$ is given by
$$
b = \sum_{0 \leq l \leq j}(-1)^jr_lm_l,
$$
where $r_l=0$ if $i$ is odd and $l=j$, and $r_l=1$ otherwise. Thus
$M_{\idot,\idot}(E)$ becomes exactly isomorphic to the original
bicomplex \eqref{hc.lambda} for the cyclic vector space $E$. We also
have $H_\idot(\Lambda,E \otimes k^{[1]}) = 0$, and
$H_\idot(\Lambda,E \otimes k_{[1]})=H_\idot(\Delta^{opp},j^*E)$.
\subsection{Nerves and geometric realizations.}
To anyone who studied algebraic topology, the cohomology algebra
$H^\hdot(\Lambda,k) = k[u]$ of the category $\Lambda$ will seem
familiar: the same algebra appears as the cohomology algebra
$H^\hdot(\C P^\infty,k)$ of the infinite-dimensional complex
projective space $\C P^\infty$, the classifying space $BU(1)$ for
the unit circle group $U(1)=S^1$. This is not a simple coincidence.
The relation between $\Lambda$ and $\C P^\infty$ has been one of the
recurring themes of the whole theory of cyclic homology from its
very beginning.
The relation occurs at various levels, and while the most advanced
ones are not properly understood even today, we do understand the
picture up to a certain point. The next level after the cohomology
isomorphism is that of the so-called {\em geometric realizations}.
Unfortunately, we do not have time to present the notion of the
geometric realization in full detail (it is easily available in the
literature; my personal favourite is the exposition in Chapter I of
Gelfand-Manin's book, also Quillen has a nice and concise exposition
in his paper on higher K-theory in Lecture Notes in Math.,
vol. 341). Let us just briefly remind the reader that to any small
category $\Gamma$, one associated a simplicial set $N(\Gamma)$
called {\em the nerve} of the category $\Gamma$. By definition,
$0$-simplices in $N(\Gamma)$ are objects of $\Gamma$, $1$-simplices
are morphisms, $2$-simplices are pairs of composable morphisms
$[a_1] \to [a_2] \to [a_3]$, and so on -- $n$-simplices in
$N(\Gamma)$ are functors to $\Gamma$ from the totally ordered set
$[n+1]$ considered as a category in the usual way. Given a
simplicial set $X \in \Fun(\Delta^{opp},\Sets)$, one forms a
topological space $|X|$ called the {\em geometric realization} of
$X$ by gluing together the elementary simplices $\Delta^n$, one for
each $n$-simplex in $X([n+1])$. Given a small category $\Gamma$, we
will call $|N(\Gamma)|$ its geometric realization, and we will
denote it simply by $|\Gamma|$.
Here are some simple properties of the geometric realization.
\begin{enumerate}
\item We have $|\Gamma| \cong |\Gamma^{opp}|$.
\item A functor $\gamma:\Gamma \to \Gamma'$ induces a map $|\gamma|:|\Gamma|
\to |\Gamma'|$, and a map $\gamma_1 \to \gamma_2$ between functors
$\gamma_1$, $\gamma_2$ induces a homotopy between $|\gamma_1|$ and
$|\gamma_2|$.
\item Consequently, if a functor $\gamma:\Gamma \to \Gamma'$ has an
adjoint, then $|\gamma|$ is a homotopy equivalence. In
particular, if $\Gamma$ has a final, or an initial object, then
$|\Gamma|$ is contractible.
\item If $\Gamma$ is a connected groupoid, and an object $[a] \in
\Gamma$ has automorphism group is $G$, then up to homotopy,
$|\Gamma|$ is the classifying space $BG$.
\end{enumerate}
To any functor $E \in \Fun(\Gamma,k)$, one associates a constructible
sheaf $\E$ of $k$-vector spaces on $|\Gamma|$ by the following rule:
for any $n$-simplex $[a_0] \to \dots \to [a_n]$ of $N(\Gamma)$, the
restriction of $\E$ to the corresponding simplex $\Delta^n \subset
|\Gamma|$ is the constant sheaf with fiber $E([a_0])$, and the
gluing maps are either identical or induced by the action of
morphisms in $\Gamma$. The gives an exact comparison functor
$\Fun(\Gamma,k) \to \Shv(|\Gamma|,k)$. This functor is fully
faithful, and it is even fully faithful on the level of derived
categories: for any $E,E' \in \Fun(\Gamma,k)$ with corresponding
sheaves $\E,\E' \in \Shv(|\Gamma|,k)$, the natural map
$$
\Ext^\hdot(E,E') \to \Ext^\hdot(\E,\E')
$$
is an isomorphism in all degrees (to prove it, one can, for
instanse, use the Godement resolution of $E \in \Fun(\Gamma,k)$ by
representable sheaves, as in Lecture 2). Of course, the comparison
functor is not an equivalence: in general, the category
$\Shv(|\Gamma|,k)$ is much larger. However, we have the following
obvious fact.
\begin{defn}
A functor $E \in \Fun(\Gamma,k)$ is {\em locally constant} if for
any morphism $f:[a] \to [a']$ in $\Gamma$, the corresponding map
$E([a]) \to E([a'])$ is invertible.
\end{defn}
\begin{lemma}
The comparison functor induces an equivalence between the derived
category $\D_{lc}(\Gamma,k)$ of complexes with locally constant
homology and the derived category $\Shv_{lc}(|\Gamma|,k)$ of
complexes of sheaves on $|\Gamma|$ whose homology sheaves are
locally constant.\endproof
\end{lemma}
\begin{corr}\label{contr}
Assume that for any field $k$ and for any locally constant $E \in
\Fun(\Gamma,k)$, we have $H_\idot(\Gamma,E)=E([a])$, where $[a] \in
\Gamma$ is s fixed object. Then $|\Gamma|$ is contractible.
\end{corr}
\proof{} By the well-known Whitehead Theorem, a map $f:X \to Y$ of
$CW$-complexes is a homotopy equivalence if for any local systems
$A$ on $Y$, $B$ on $X$, the induced maps $H_\idot(X,f^*A) \to
H_\idot(Y,A)$, $H_\idot(X,B) \to H_\idot(Y,f_*B)$ are isomorphisms.
\endproof
Going back to the cyclic category $\Lambda$: our goal is to prove
that $|\Lambda|$ is homotopically equivalent to $\C P^\infty$. We
will do it indirectly, in two steps: first, we prove that the
realization $|\Lambda_\infty|$ of the category $\Lambda_\infty$ is
contractible, then we prove that the projection functor
$\Lambda_\infty \to \Lambda$ induces a fibration $|\Lambda_\infty|
\to |\Lambda|$ whose fiber is the circle $S^1 = U(1)$ -- thus
$|\Lambda_\infty|$ can be taken as the contractible space $EU(1)$,
and $|\Lambda|$ is homotopy equivalent to the classifying space
$EU(1)/U(1) = BU(1) \cong \C P^\infty$. For the first step, we only
need Corollary~\ref{contr}, but for the second step, we need to
develop some machinery of fibrations for small categories.
\subsection{Fibrations and cofibration of small categories.}
The notion of a fibered and cofibered category was introduced by
Grothendieck in SGA1, Ch.VI, which is perhaps still the best
reference for those who can read French; nowadays, this machinery is
usually called {\em Grothendieck construction}. Let me give the
basic definitions.
Assume given a functor $\gamma:\Gamma' \to \Gamma$ between small
categories $\Gamma$, $\Gamma'$. By the {\em fiber} $\Gamma'_{[a]}$
over an object $[a] \in \Gamma$ we understand the subcategory
$\Gamma'_{[a]} \to \Gamma'$ of objects $[a'] \subset \Gamma'$ such
that $\gamma([a'])=[a]$, and morphisms $f$ such that $\gamma(f) =
\id$. A morphism $f:[a] \to [b]$ in $\Gamma'$ is called {\em
Cartesian} if it has the following universal property:
\begin{itemize}
\item any morphism $f':[a'] \to [b]$ such that $\gamma(f') =
\gamma(f)$ factors through $f$ by means of a unique map $[a'] \to
[a]$ in $\Gamma'_{\gamma([a])}$.
\end{itemize}
\begin{defn}\label{fib.defn}
A functor $\gamma:\Gamma' \to \Gamma$ is called a {\em fibration} if
\begin{enumerate}
\item for any $f:[a] \to [b]$ in $\Gamma$, and any $b' \in
\Gamma'_{[b]}$, there exists a Cartesian morphism $f':[a'] \to
[b']$ such that $\gamma(f')=f$, and
\item the composition of two Cartesian morphisms is Cartesian.
\end{enumerate}
\end{defn}
Condition \thetag{i} here mimics the ``covering homotopy'' condition
in the definition of a fibration in algebraic topology, but it is in
fact much more precise --- indeed, the Cartesian covering morphism
$f'$, having the universal property, is uniquely
defined. Grothendieck also introduced ``cofibrations'' as functors
$\gamma:\Gamma' \to \Gamma$ such that $\gamma^{opp}:\Gamma^{'opp}
\to \Gamma^{opp}$ is a fibration. This terminology is slightly
unfortunate because the topological analogy is still a fibration --
``cofibration'' in topology means something completely
different. For this reason, now the term ``op-fibration'' is
sometimes used. However, we will stick to Grothendieck's original
terminology.
Assume given a fibration $\gamma:\Gamma' \to \Gamma$ and a morphism
$f:[a] \to [b]$ in $\Gamma$. Then for any $[b'] \in \Gamma'_{[b]}$,
we by definition have a Cartesian morphism $f':[a'] \to [b']$, and
using the universal property of the Cartesian morphism, one checks
that the correspondence $[b'] \mapsto [a']$ is functorial: we have a
functor $f^*:\Gamma_{[b]}' \to \Gamma'_{[a]}$, $[b'] \mapsto
[a']$. Using condition \thetag{ii} of Definition~\ref{fib.defn}, one
checks that for any composable pair of maps $f$, $g$, we have a
natural isomorphism $(f \circ g)^* \cong g^* \circ f^*$, and there
is a compatibility constraint for these isomorphisms when we are
given a composable triple $f$, $g$, $h$. All in all, the
correspondence $[a] \mapsto \Gamma'_{[a]}$, $f \mapsto f^*$ defines
a contravriant ``weak functor'' from $\Gamma$ to the category of
small categories. Conversely, every such ``weak functor'',
appropriately defined, arises in this way. This was the main reason
for Grothendieck's definition of a fibration -- it gives a nice and
short replacement for the cumbersome notion of a weak functor, with
all its higher isomorphisms and compatibility constraints.
\medskip
Today, we will only need one basic fact about fibrations, and we
will use it without a proof.
\begin{defn}
A fibration $\gamma:\Gamma' \to \Gamma$ is {\em locally constant} if
for any $f$ in $\Gamma$, the functor $f^*$ is an equivalence.
\end{defn}
\begin{prop}\label{fibr}
Assume given a connected small category $\Gamma$ and a locally
constant fibration $\gamma:\Gamma' \to \Gamma$. Then the homotopy
fiber of the induced map $|\gamma|:|\Gamma'| \to |\Gamma|$ is
naturally homotopy equivalent to the realization $|\Gamma'_{[a]}|$
of the fiber over any object $[a] \in \Gamma$.\endproof
\end{prop}
\subsection{Computation of $|\Lambda|$.}
We can now prove that the realization $|\Lambda|$ is equivalent to
$\C P^\infty$. We start with the following.
\begin{lemma}
The realization $|\Lambda_\infty|$ is contractible.
\end{lemma}
\proof{} By Corollary~\ref{contr}, it suffices to prove that
$H_\idot(\Lambda_\infty,E) \cong E([1])$ for any locally constant $E
\in \Fun(\Lambda_\infty,k)$. The homology of the category
$\Lambda_\infty$ can be computed by a complex similar to
\eqref{hc.lambda}: we take \eqref{hc.lambda} and remove everything
except for the two right-most columns. We leave it to the reader to
check that this indeed computes $H_\idot(\Lambda_\infty,E)$ (while
the rows of the complex now have only length $2$, they still compute
the homology of the infinite cyclic group $\Z = \Aut([n])$, and the
same proof as in Lemma~\ref{cycl.comp} works). Since we now only
have two columns, and one of them is contractible, the Connes' exact
sequence reduces to an isomorphism
$$
H_\idot(\Delta^{opp},j^*E) \cong H_\idot(\Lambda_\infty,E).
$$
Since $j^*E$ is obviously locally constant, it suffices to check
that the realization $|\Delta^{opp}|$ of the category $\Delta^{opp}$
is contractible. This is clear --- $\Delta^{opp}$ has an
initial object.
\endproof
\begin{lemma}\label{infty.fib}
The natural functor $\Lambda_\infty \to \Lambda$ is a locally
constant fibration whose fiber is the groupoid $\ppt_\Z$ with one
object whose automorphisms group is $\Z$.
\end{lemma}
\proof{} We use the combinatorial description of the category
$\Lambda$. Then for any $[n],[m] \in \Lambda$, the map
$\Lambda_\infty([n],[m]) \to \Lambda([n],[m])$ is surjective by
definition, and one checks easily that {\em any} map $f \in
\Lambda_\infty([n],[m])$ is Cartesian. The fiber, again by
definition, has one object, and its automorphism group is freely
generated by the automorphism $\sigma$.
\endproof
\begin{prop}
We have a homotopy equivalence $|\Lambda| \cong \C P^\infty \cong
BU(1)$.
\end{prop}
\proof{} By Proposition~\ref{fibr} and Lemma~\ref{infty.fib}, the
homotopy fiber of the map $|\Lambda_\infty| \to |\Lambda|$ is
homotopy equivalent to $|\ppt_\Z|$, and since $|\Lambda_\infty|$ is
contractible, this implies that $|\ppt_\Z|$ is homotopy equivalent
to the loop space of $|\Lambda|$. But $\ppt_\Z$ is a groupoid, so
that $|\ppt_\Z|$ is equivalent to the classifying space $B\Z \cong
S^1$. This means that $|\Lambda|$ has only one non-trivial homotopy
group, namely $\pi_2(|\Lambda|) = \Z$, so that it must be the
Eilenberg-MacLane space $K(\Z,2) = \C P^\infty$.
\endproof
As a corollary, we see that the derived category
$\D_{lc}(\Lambda,k)$ of complexes of cyclic objects with locally
constant homology objects is equivalent to the derived category of
complexes of sheaves of $\C P^\infty$ with locally constant homology
sheaves. The objects in this latter category are also know as
$U(1)$-equivariant constructible sheaves on the point $\ppt$.
\lecture{}
\begin{center}
\begin{minipage}{0.7\linewidth}\small
The structure of $\D_{lc}(\Lambda,k)$ and $\Fun(\Lambda,k)$;
Dold-Kan equivalence, mixed complexes. Cyclic bimodules. Cyclic
homology as a derived functor.
\end{minipage}
\end{center}
\subsection{The structure of the category $\Fun(\Lambda,k)$.}
In the last lecture, we have proved that the geometric realization
$|\Lambda|$ of the Connes' cyclic category is homotopy equivalent to
the infinite projective space $\C P^\infty$. In particular, we have
an equivalence
$$
\D_{lc}(\Lambda,k) \cong \D_{lc}(\Shv(\C P^\infty,k)),
$$
where $\D_{lc}$ means ``the full subcategory in the derived category
$\D(\Lambda,k)$ spanned by complexes with locally constant
homology'', and similarly in the right-hand side. The category in
the right-hand side is also equivalent to the derived category of
$S^1$-equivariant sheaves on a point. Besides these topological
descriptions, there is also the following very simple combinatorial
description.
Let $\D^{per}(k\Vect)$ be the periodic derived category of the
category $k\Vect$ -- namely, $\D^{per}(k)$ is the triangulated
category obtained by considering the category of quadruples $\langle
V_+,V_-,d_+,d_-\rangle$ of two vector spaces $V_+$, $V_-$ and two
maps $d_+:V_+ \to V_-$, $d_-:V_- \to V_+$ such that $d_+ \circ d_- =
d_- \circ d_+ = 0$, and inverting quasiisomorphisms. Equivalently,
$\D^{per}(k)$ is the homotopy category of $2$-periodic complexes
$V_\idot$ of $k$-vector spaces (with $V_+ = V_{2\idot}$,
$V_-=V_{2\idot+1}$, and $d_+$, $d_-$ being the components of the
differential). Just as the usual derived category $\D(k\Vect)$ has
filtered version $\DF(k\Vect)$, we define the filtered periodic
category $\DF^{per}(k\Vect)$ by considering $2$-periodic filtered
complexes $F^\hdot V_\idot$ such that $F^\idot V_\idot \cong
F^{\idot+1}V_{\idot+2}$ -- note the shift in the filtration! Then
for any cyclic vector space $E \in \Fun(\Lambda,k)$, the periodic
cyclic homology $HP_\idot(E)$ equipped with the Hodge filtration is
an object in $\DF^{per}(k\Vect)$, so that we have a natural functor
$$
HP_\idot(-):\D(\Lambda,k) \to \DF^{per}(k\Vect).
$$
\begin{exc}
Show that the induced functor $\D_{lc}(\Lambda,k) \to
\DF^{per}(k\Vect)$ is an equivalence of categories. Hint: both
categories are generated by $k$, so that it suffices to compare
$\Ext^\hdot(k,k)$.
\end{exc}
Thus an object $\D_{lc}(\Lambda,k)$, when compared to its periodic
cyclic homology equipped with the Hodge filtration, contains exactly
the same amount of information, we lose nothing by taking
$HP_\idot(-)$. What can be said about non-constant cyclic vector
spaces --- in other words, how complicated is the category
$\Fun(\Lambda,k)$? Unfortunately, the answer is ``very
complicated''.
This might not seem surprising, because the category $\Lambda$
contains so many maps. However, so does the category
$\Delta$. Nevertheless, there is the following surprising fact,
discovered about 50 years ago independently by A. Dold and D. Kan.
\begin{thm}[Dold,Kan]
The abelian category $\Fun(\Delta^{opp},k)$ of simplicial $k$-vector
spaces is equivalent to the category $C^{\leq 0}(k)$ of complexes of
$k$-vector spaces cocentrated in non-positive degrees.
\end{thm}
\proof{} There are many proofs, but they all involve either
non-trivial computations, or non-trivial combinatorics. We will not
give any of them, but we will indicate what the equivalence
is. Given a simplicial vector space $E \in \Fun(\Delta^{opp},k)$, we
take its stadrd complex $E_\idot$, and we replace it with its {\em
normalized quotient} $N(E)_\idot$ given by
$$
N(E)_i = E_i/\sum\Im s_j,
$$
where $s_j:E_{i-1} \to E_i$ are the degeneration maps (induced by
surjective maps $[i] \to [i-1]$).
\endproof
There exist also various generalizations of the Dold-Kan
equivalence. First, the category $\Fun(\Delta,k)$ of co-simplicial
vector spaces is equivalent to the category $C^{\geq 0}(k)$ of
complexes concentrated in non-negative degrees (this is not
surprising, since $\Fun(\Delta,k)$ and $\Fun(\Delta^{opp},k)$ are
more-or-less dual to each other). One can also consider the
subcategory $\Delta_+ \subset \Delta$ with the same objects, and
only those maps $[n] \to [m]$ which send the first element to the
first element. Then $\Fun(\Delta_+^{opp},k)$ is equivalent to the
category of $k$-vector spaces graded by non-positive integers
(restriction to $\Delta^{opp}_+ \subset \Delta^{opp}$ corresponds to
forgetting the differential in the complex). Finally, if one
``truncates'' $\Delta$ and considers the full subcategory
$\Delta_{\leq n} \subset \Delta$ spanned by objects $[1],\dots,[n]$,
then $\Fun(\Delta_{\leq n}^{opp},k)$ is equivalent to the category
$C^{[1-n,0]}(k)$ of complexes concentrated in degrees from $1-n$ to
$0$, and similarly for $\Fun(\Delta_{\leq n},k)$ and for $\Delta_+$.
Now, we have a natural embedding $\Delta^{opp} \subset \Lambda$, so
that we have a flag of subcategories $\Delta_+^{opp} \subset
\Delta^{opp} \subset \Lambda$. We know that $\Lambda$ is self-dual,
$\Lambda \cong \Lambda^{opp}$. One checks easily that
$\Delta_+^{opp}$ is preserved by this self-duality --- we have
$\Delta^{opp}_+ \cong \Delta_+$. The intermediate category
$\Delta^{opp} \subset \Lambda$ is not preserved, so that by duality,
we get an embedding $\Delta \subset \Lambda$. All in all, we have
the following diagram.
$$
\begin{CD}
\Delta_+ \cong \Delta_+^{opp} @>>> \Delta\\
@VVV @VVV\\
\Delta^{opp} @>>> \Lambda \cong \Lambda^{opp}.
\end{CD}
$$
Applying restrictions and the Dold-Kan equivalence, we associate to
any cyclic vector space $E \in \Fun(\Lambda,k)$ a complex $E_\idot
\in C^{\leq 0}(k)$ and a complex $E^\hdot \in C^{\geq 0}(k)$, and
since the diagram of categories commutes, we also have natural
identifications $E_i \cong E^i$ as $k$-vector spaces. In other
words, we have a collection $E_i$, $i \geq 0$ of $k$-vector spaces
and {\em two} differentials $b:E_i \to E_{i-1}$, $B:E_i \to
E_{i+1}$. One can check that these differentials anti-commute,
$bB+Bb=0$. The result is what is known in the literature as a {\em
mixed complex}.
\begin{defn}
A {\em mixed complex} $E_\idot$ is a collection $E_i$, $i \geq 0$ of
$k$-vector spaces and two maps $b:E_i \to E_{i-1}$, $B:E_i \to
E_{i+1}$ such that $b^2 = B^2 = bB+Bb=0$.
\end{defn}
Mixed complexes form a nice abelian category $M^{\leq 0}(k)$ which
is not much more complicated than the category of complexes $C^{\leq
0}(k)$, and we have a comparison functor $\Fun(\Lambda,k) \to
M(k)$. But the obvious analog of the Dold-Kan Theorem is {\em wrong}
--- the comparison functor is not an equivalence.
The only fact which is true is the following: define the derived
category $\D(M(k))$ of mixed complexes by inverting the maps which
are quasiisomorphisms with respect to the differential $b$. Then the
comparison functor $\D_{lc}(\Lambda,k) \to \D(M(k))$ is an
equivalence (and $\D(M(k))$ is equivalent to $\DF^{per}(k)$ -- this
is an instance of the so-called {\em Koszul}, or $S-\Lambda$ {\em
duality}). However, even when we pass to the restricted categories
$\Fun(\Lambda_{\leq n},k)$, $M^{[1-n,0]}(k)$, with the obvious
notation, the comparison functor probably is an equivalence only if
$k$ has characteristic $0$. And for non-locally constant functors,
things only get worse.
To sum up: while in the literature on cyclic homology people often
use mixed complexes as a basic object, especially in characteristic
$0$, in reality, cyclic vector spaces contain strictly more
information. And we will see later at least one example where this
extra information is crucially important.
\subsection{Projecting to $\D_{lc}(\Lambda,k)$.}
One moral of the above story is that it is much preferable to work
with the locally constant subcategory $\D_{lc}(\Lambda,k)$ rather
then with the whole category $\D(\Lambda,k)$. An immediate problem
is that the cyclic vector space $A_\# \in \Fun(\Lambda,k)$ defined
for an associative unital $k$-algebra $A$ is not locally constant
unless $A = k$. However, we can force it to be locally
constant. Namely, the embedding $\D_{lc}(\Lambda,k) \subset
\D(\Lambda,k)$ admits a left-adjoint functor $\LC:\D(\Lambda,k) \to
\D_{lc}(\Lambda,k)$. Since $\D_{lc}(\Lambda,k) \subset
\D(\Lambda,k)$ is a full subcategory, $\LC$ is identical on
$\D(\Lambda,k)$, and since it contains the constant cyclic vector
space $k^\Lambda$ which corepresents the homology functor, the
homology functor factors through $\LC$, so that for any $E \in
\D(\Lambda,k)$, we have a canonical isomorphism $H_\idot(\Lambda,E)
\cong H_\idot(\Lambda,\LC(E))$.
The existence of the adjoint functor $\LC$ is easy to prove by
general nonsense, but it is perhaps more interesting to use the
following explicit construction.
Consider the category $\Lambda^{opp} \times \Lambda$, and consider
the functor $\I \in \Fun(\Lambda^{opp} \times \Lambda,k)$ spanned by
the $\Hom$-functor: we set
$$
\I([n] \times [m]) = k \left[\Lambda([m],[n])\right].
$$
Denote by $\pi$, $\pi^o$ the natural projections $\pi:\Lambda^{opp}
\times \Lambda \to \Lambda$, $\pi:\Lambda^{opp}
\times \Lambda \to \Lambda^{opp}$. We claim that for any $E \in
\Fun(\Lambda,k)$, we have a natural isomorphism
\begin{equation}\label{lambda.times}
H_\idot(\Lambda^{opp} \times \Lambda,\I \otimes \pi^*E) \cong
H_\idot(\Lambda,k).
\end{equation}
Indeed, by an obvious version of the K\"unneth formula, we can
compute the homology in the left-hand side first along
$\Lambda^{opp}$, and then along $\Lambda$. Then it suffices to show
that for any $[n] \in \Lambda$, we have a functorial isomorphism
$$
H_\idot(\Lambda^{opp},E([n]) \otimes \I|_{\Lambda^{opp} \times [n]})
\cong E([n]).
$$
But here we can take $E([n])$ out of the brackets, so that it
suffices to consider the case $E([n]) = k$, and the restriction
$\I_{\Lambda^{opp} \times [n]}$ is nothing but the representable
functor $k_{[n]}^{\Lambda^{opp}}$, so that its homology is indeed
isomorphic to $k$ concentrated in degree $0$.
But on the other hand, we can compute the left-hand side of
\eqref{lambda.times} by first using the projection $\pi^o$. By
general nonsence, we have
$$
H_\idot(\Lambda^{opp} \times \Lambda,\I \otimes \pi^*E) \cong
H_\idot(\Lambda^{opp},L^\hdot\pi^o_!(\I \otimes \pi^*E)),
$$
and since $\Lambda \cong \Lambda^{opp}$, we can define $\LC(E) =
L^\hdot\pi^o_!(\I \otimes \pi^*E))$. All we have to do is to prove
that it is locally constant. Indeed, by the K\"unneth formula, for
any $[n] \in \Lambda^{opp} \cong \Lambda$ we have
$$
\LC(E)([n]) = H_\idot(\Lambda,k_{[n]} \otimes E),
$$
where the representable functor $k_{[n]}$ is the restriction of $\I$
to $[n] \times \Lambda \subset \Lambda^{opp} \times \Lambda$. But
$k_{[n]}$ is clean in the sense of Definition~\ref{clean.defn}, so
that
$$
H_\idot(\Lambda,k_{[n]} \otimes E) \cong
H_\idot(\Delta^{opp},k^{\Delta^{opp}}_{[n]} \otimes j^*E).
$$
By the well-known K\"unneth formula for simplicial vector spaces, the
right-hand side is canonically isomorphic to
$$
H_\idot(\Delta^{opp},k^{\Delta^{opp}}_{[n]}) \cong
H_\idot(\Delta^{opp},E) \otimes H_\idot(\Delta^{opp},E),
$$
which is manifestly independent of $[n]$.
\subsection{Cyclic bimodules.}
If one writes down explicitly $\LC(E)$ by using the complex
\eqref{hc.clean}, the result is very similar to the ``third
bicomplex'' \eqref{hc.eq.3} for cyclic homology which we defined in
Lecture 3. One can also clearly see why that construction only
worked in characteristic $0$. The columns in \eqref{hc.eq.3} are
naturally assembled into a cyclic object, not in a simplicial one;
when we simply imposed the differential $\wt{B}$ on them, we in
effect forgot the cyclic structure and only considered the
underlying simplicial structure. In $\cchar 0$, this did not matter
-- the cyclic group action on each column is actually trivial, so
that we we compute $H_\idot(\Lambda,\LC(E))$ by \eqref{hc.clean},
taking coinvariants with respect to $\tau$ can be omitted. In the
general case, we do need to compute honestly the cyclic homology
$H_\idot(\Lambda,\LC(E))$.
However, the bicomplex \eqref{hc.eq.3}, although it only worked in
$\cchar 0$, was very interesting for the computation of the cyclic
homology $HC_\idot(A)$ of an associative algebra $A$, because it had
a version where the bar resolution $C_\idot(A)$ of the diagonal
$A$-bimodule could be replaced with an arbitrary resolution
$P_\idot$ (at least for the two rightmost columns). Now that we know
the full truth, can we perhaps give a version of that construction
which is valid in any characteristic and for all columns, not only
the two rightmost ones?
It turns out that we can do even better --- it is possible to obtain
the whole $\LC_\idot(A_\#)$ as an object of $\D_{lc}(\Lambda,k)$
completely canonically, without any explicit choice at all, neither
of a resolution $P_\idot$, nor of the homotopy $\iota$, as in
Lecture 3, part 3.5. Or rather, the choices do occur, but they are
all packed into a single choice of a projective resolution in some
appropriate abelian category, and cyclic homology is obtained as a
derived functor on this abelian category (just as Hochschild
homology is the derived functor on the abelian category $A\bimod$ of
$A$-bimodules).
\medskip
To construct this new category, which we call the category of {\em
cyclic $A$-bimodules}, we use the technique of fibered and cofibered
categories explained in Lecture 4.
\medskip
Assume given a small category $\Gamma$ and a category $\Cc$ equipped
with a cofibration $\pi:\Cc \to \Gamma$. Thus for any $[a] \in
\Gamma$, we have the fiber $\Cc_{[a]}$, and for any map $f:[a] \to
[b]$, we have a transition functor $f_!:\Cc_{[a]} \to \Cc_{[b]}$.
Denote by $\Sec(\Cc)$ the category of sections $\Gamma \to \Cc$ of
the projection $\pi:\Cc \to \Gamma$. Explicitly, an object $M \in
\Sec(\Cc)$ is given by a collection of objects $M_{[a]} \in
\Cc_{[a]}$ for all $[a] \in \Gamma$, and of transtion maps
$\iota_f:f_!M_{[a]} \to M_{[b]}$ for all $f:[a] \to [b]$, subject ot
natural compatibilities.
\begin{prop}\label{abe.sec}
Assume that all the fibers $\Cc_{[a]}$ of the cofibration $\pi:\Cc
\to \Gamma$ are abelian, and all the transition functors
$f_!:\Cc_{[a]} \to \Cc_{[b]}$ are left-exact. Then the category
$\Sec(\Cc)$ is abelian.
\end{prop}
\proof[Sketch of a proof.] To prove that an additive category is
abelian, one has to show that it has kernels and cokernels, and they
satisfy some additional conditions (such as ``the cokernel of the
kernel is isomorphic to the kernel of the cokernel). The kernel and
cokernel of a map $\phi:M \to N$ in $\Sec(\Cc)$ are taken fiberwise,
$\Coker\phi_{[a]} = \Coker\phi_{[a]}$, $\Ker\phi_{[a]} =
\Ker\phi_{[a]}$. The transition maps of the kernel are induced from
those of $M$, and to construct the transtion maps for the cokernel,
one uses the fact that the transition functors $f_!$ are
right-exact. All the extra conditions can be checked fiberwise,
where they follow from the assumption that all fibers are abelian.
\endproof
As we can see from its explicit description, the category
$\Sec(\Cc)$ is rather large. One can define a smaller subcategory by
only considering those sections that are {\em Cartesian} -- that is,
any $f:[a] \to [b]$ goes to a Cartesian map in $\Cc$. Equivalently,
in the explicit description above, all the transition maps
$\iota_f:f_!M_{[a]} \to M_{[b]}$ must be isomorphisms. This is often
a much smaller category, but it need not be abelian (unless all the
transition functors are exact, not just right-exact, which rarely
happens in practice). A reasonable thing to do is to consider the
derived category $\D(\Sec(\Cc))$ and the full subcategory
$\D_{cart}(\Sec(\Cc)) \subset \D(\Sec(\Cc))$ of complexes with
Cartesian cohomology.
\medskip
Assume now given an associative unital algebra $A$, and consider the
category $A\bimod$ of $A$-bimodules. This is a unital tensor
category: we have the (non-symmetric) associative tensor product
functor $m:A\bimod \times A\bimod \to A\bimod$, $M_1 \times M_2
\mapsto M_1 \otimes_A M_2$. Moreover, we can also consider the
category $A^{\otimes 2}\bimod$ of $A^{\otimes 2}$-bimodules, and the
exterior product functor $A\bimod \times \A\bimod \to A^{\otimes
2}\bimod$, $M_1 \otimes M_2 \mapsto M_1 \boxtimes M_2$ is a fully
faithful embedding. The tensor product functor then obviously
extends to a right-exact functor $m:A^{\otimes 2}\bimod \to
A\bimod$. Since the tensor product on $A\bimod$ is associative, we
can iterate this and obtain the right-exact tensor product functors
$m_n:\A^{\otimes n}\bimod \to A\bimod$ for any $n \geq 1$. For $n =
0$, we take $A^{\otimes 0}$ to be $k$, and $m_0:k\Vect \to A\bimod$
is the functor which sends $k$ to the unit object of $A\bimod$.
What we want to do now is to take the construction of the cyclic
vector space $A_\#$, and replace the unital associative algebra $A$
with the unital associative tensor category $A\bimod$. The result is
a category cofibered over $\Lambda$ which we denote by
$A\bimod_\#$. The fibers are given by
$$
A\bimod_\#([n]) = A^{\otimes n}\bimod,
$$
and the transition functors $f_!$ are induced by the multiplication
functors $m_n$ by the same formula \eqref{hash.map} as in the
definition of the cyclic vector space $A_\#$.
\begin{defn}
A {\em cyclic $A$-bimodule} $M$ is a Cartesian section of the
cofibration $A\bimod_\# \to \Lambda$.
\end{defn}
Explicitly, a cyclic $A$-bimodule $M$ is given a collection of
$M_{[n]} \in A^{\otimes n}\bimod$, $n \geq 1$, and transition maps
between them. However, because all transition maps are isomorphisms,
the bimodules $M_{[n]}$, $n \geq 2$ can be computed from the first
bimodule $M_1 = M_{[1]}$ --- it suffices to apply the transition
functor $f_!$ for some map $f:[1] \to [n]$. Since such a map is not
unique, extending a given $M_1 \in A\bimod$ to a cyclic bimodule
requires extra data. It is enough, for instance, to specify an
$A^{\otimes 2}$-bimodule isomorphism $\tau:A \boxtimes M \to M
\boxtimes A$ such that the induced maps $\tau_{23}:A \boxtimes A
\boxtimes M \to A \boxtimes M \boxtimes A$, $\tau_{12}:A \boxtimes M
\boxtimes A \to M \boxtimes A \boxtimes A$, $\tau_{23}:M \boxtimes A
\boxtimes A \to A \boxtimes A \boxtimes M$ satisfy
\begin{equation}\label{tr.eq}
\tau_{31} \circ \tau_{12} \circ \tau_{23} = \id.
\end{equation}
The category of cyclic $A$-bimodules is abelian, but this is an
accident: the category that must be abelian for general reasons is
the category $\Sec(A\bimod_\#)$ of all sections of the cofibration
$A\bimod_\# \to A$. Thus we consider the derived category
$\DL(A\bimod)= \D(\Sec(A\bimod_\#))$, and we define the {\em derived
category of cyclic bimodules} $\DL_{lc}(A\bimod)$ as the full
subcategory
$$
\DL_{lc}(A\bimod) = \D_{cart}(\Sec(A\bimod_\#)) \subset
\D(\Sec(A\bimod_\#)) = \DL(A\bimod)
$$
of complexes with Cartesian cohomology.
We note that even though the category $\Sec_{cart}(A\bimod_\#)$ of
cyclic bimodules {\em per se} happens to be abelian, its derived
category is smaller than $\DL_{lc}(A\bimod)$. For instance, if $A =
k$, so that $A^{\otimes n} = k$ for any $n \geq 0$, with identical
transition funcotrs, then $\Sec(A\bimod)$ is exactly equivalent to
$\Fun(\Lambda,k)$, and the Cartesian sections correspond to locally
constant functors. But every locally constant cyclic vector space is
constant, while $\DL_{lc}(k\bimod) \cong \D_{lc}(\Lambda,k)$ is a
non-trivial category.
\subsection{Cyclic homology as a derived functor.}
We now recall that the Hochschild homology $H_\idot(A,M)$ with
coefficients in an $A$-bimodule $M$ is by definition the derived
functor of the functor $M \mapsto A \otimes_{A^{opp} \otimes A} M$,
which can be equivalently described as the following right-exact
{\em trace functor}
$$
\tr(M) = M/ \{am-ma \mid a \in A, m \in M\}.
$$
We prefer this description because it clearly has the following
``trace property'': for any two $A$-bimodules $M$, $N$, there exists
a canonical isomorphism $\tr(M \otimes_A N) \cong \tr(N \otimes
M)$. Even more generally, for any $A^{\otimes n}$-module $M_n$, we
can define
$$
\tr(M_n) = M/ \{am-m\sigma(a) \mid a \in A^{\otimes n},m \in M_n\},
$$
where $\sigma:A^{\otimes n} \to A^{\otimes n}$ is the cyclic
permutation. These trace functors obviously commute with the transition
functors of the cofibered category $A\bimod_\#$, so that $\tr$
extends to a functor $\tr:A\bimod_\# \to k\Vect$ which sends every
Cartesian map to an isomorphism of vector spaces.
We can now apply the trace functor $\tr$ fiberwise, to obtain a
Cartesian functor $\tr:A\bimod_\# \to k\Vect^\Lambda$, where
$k\Vect^\Lambda = k\Vect \times \Lambda$ is the constant cofibration
with fiber $k\Vect$. This induces a right-exact functor
$$
\tr:\Sec(A\bimod_\#) \to \Fun(\Lambda,k),
$$
and since $\tr:A\bimod_\# \to k\Vect^\Lambda$ is Cartesian, the
derived functor $L^\hdot\tr:\DL(A\bimod) \to \D(\Lambda,k)$ sends
$\DL_{lc}(A\bimod)$ into $\D_{lc}(\Lambda,k)$.
\begin{defn}
The {\em cyclic homology $HC_\idot(A,M)$} of the algebra $A$ with
coefficients in some $M \in \DL(A\bimod)$ is given by
$$
HC_\idot(A,M) = H_\idot(\Lambda,L^\hdot\tr(M)).
$$
\end{defn}
In general, it is not easy to construct cyclic bimodules. However,
one cyclic bimodule manifestly exists for any algebra $A$ --- this
is $A_\#$, with the diagonal $A^{\otimes n}$-bimodule structure on
every $A_\#([n]) = A^{\otimes n}$.
\begin{prop}\label{coeff.hc}
For any algebra $A$, we have $HC_\idot(A,A_\#) \cong HC_\idot(A)$.
\end{prop}
\proof{} Notice that we can define a simpler notion of cyclic
homology with coefficients in some $M \in \DL(A\bimod)$ -- we can
forget the $A^{\otimes n}$-bimodule structure on $M([n])$, and treat
$M$ simply as a complex of cyclic vector spaces. Denote
$H_\idot(\Lambda,M)$ by $HC_\idot'(A,M)$. We have obvious projection
maps $M([n]) \to \tr(M([n]))$ which induce a functorial map
\begin{equation}\label{hc.prime}
HC'_\idot(A,M) \to HC_\idot(A,M).
\end{equation}
We have to show that this map is an isomorphism for $M = A_\#$. It
suffices to prove that it is an isomorphism for any $M \in
\DL(A\bimod)$, or even for any $M \in \Sec(A\bimod_\#)$. We note
that the evaluation at $[n] \in \Lambda$ induces a functor
$\Sec(A\bimod) \to A^{\otimes n}\bimod$, which has a left-adjoint
$i^{[n]}_!:A^{\otimes n} \to \Sec(A\bimod)$. Explicitly, for any
$A^{\otimes n}$-bimodule $P$, we have
\begin{equation}\label{i.n}
i^{[n]}_! P([m]) = \bigoplus_{f:[n] \to [m]}f_!P.
\end{equation}
If $P$ is projective, then $i^{[n]}_!P$ is projective in
$\Sec(A\bimod)$ by adjunction, and $\Sec(A\bimod)$ obviously has
enough projectives of this type, so it is enough to prove that
\eqref{hc.prime} is an isomorphism for $M = i^{[n]}_!P$. Even
further, it is enough to consider objects $P^n \in \Sec(A\bimod)$
given by
$$
P^n = i^{[n]}_! A^{\otimes n} \otimes A^{\otimes n},
$$
where on the right-hand side we have the free $A^{\otimes
n}$-bimodule with one generator.
Since $P^n$ is projective, we have $L^i\tr(P^n)=0$ for $i \geq 1$,
and $\tr(P^n) \in \Fun(\Lambda,k)$ is isomorphic to $A^{\otimes n}
\otimes k_{[n]}$; thus the right-hand side of \eqref{hc.prime} with
$M = P^n$ is canonically isomorphic to $A^{\otimes n}$ in degree
$0$, and trivial otherwise. As for the left-hand side, we see from
\eqref{i.n} that
$$
P^n([m]) = \bigoplus_{f:[n] \to [m]} A^{\otimes (n+m)}.
$$
In particular, it is clean, so that $H_\idot(\Lambda,P^n)$ can be
computed by the complex \eqref{hc.clean}. We leave it to the reader
to check that the resulting complex $P^n_\idot$ can be described as
follows: if we take the augmented bar resolution $C_\idot(A)$,
$C_i(A) = A^{\otimes i+1}$ and consider the $n$-fold tensor power
$C^n_\idot = C_\idot(A)^{\otimes n}$, then
$$
P^n_i = C^n_{i+1}
$$
for any $i \geq 0$. Since the whole complex $C^n_\idot$, being the
$n$-fold tensor power of the acyclic complex $C_\idot(A)$, is itself
acyclic, the complex $P^n_\idot$ is a resolution for the $0$-th term
$C^n_0$, which is again $A^{\otimes n}$.
\endproof
\lecture{}
\begin{center}
\begin{minipage}{0.7\linewidth}\small
Cyclic homology for general tensor
categories. Morita-invariance. Example: cyclic homology of a group
algebra. Regulator map.
\end{minipage}
\end{center}
\subsection{Cyclic homology for general tensor categories.}
In the last lecture, we have constructed the derived category
$\DL_{lc}(A\bimod)$ of cyclic bimodules over an associative algebra
$A$, and we have re-defined cyclic homology by means of a trace
functor $\tr:\DL_{lc}(A\bimod) \to \D_{lc}(\Lambda,k)$. The algebra
$A$ itself essentially only appeared in the construction though the
tensor category $A\bimod$ of $A$-bimodules. A natural question is,
can we do the same construction for a more general tensor category
$\Cc$?
To start with, we need to construct a category $\Cc_\#$ cofibered
over $\Lambda$. Here there is one problem: there is no well-defined
tensor product for general abelian categories. Namely, we can
introduce the following.
\begin{defn}
Assume given two abelian $k$-linear categories $\Cc_1$, $\Cc_2$.
The {\em tensor product} $\Cc_1 \otimes \Cc_2$ is a $k$-linear
abelian category equipped with a functor $\Cc_1 \times \Cc_2 \to
\Cc_1 \otimes \Cc_2$ which is $k$-linear and right-exact in each
variable, and has the following universal property:
\begin{itemize}
\item for any $k$-linear abelian category $\Cc'$, any functor $\Cc_1
\times \Cc_2 \to \Cc'$ which is $k$-linear and right-exact in each
variable factors through $\Cc_1 \times \Cc_2 \to \Cc_1 \otimes
\Cc_2$, and the facrorization is unique up to an isomorphism.
\end{itemize}
\end{defn}
The problem is, while the tensor product in this sense is obviously
unique up to an equivalence, it does not always exist. However, it
does exist for categories of modules or bimodules: one can show that
for any $k$-algebras $A$, $B$, we have $A\amod \otimes B\amod \cong
(A \otimes B)\amod$, $A\bimod \otimes B\bimod \cong (A \otimes
B)\bimod$ -- thus the category $A^{\otimes n}\bimod$ which we used
in the last lecture is actually $A\bimod^{\otimes n}$ in the sense
of Definition~\ref{prod.cat}. There are other interesting cases,
too. Thus we simply impose this as an assumption.
\begin{defn}\label{prod.cat}
A $k$-linear abelian tensor category $\Cc$ is {\em good} if it has
arbitrary sums, the tensor product functor is right-exact in each
variable, and for any $n$, there exists a tensor product
$\Cc^{\otimes n}$ in the sense of Definition~\ref{prod.cat}.
\end{defn}
\begin{rem}
Sometimes in the representation-theoretic literature, ``tensor
category'' means ``symmetric tensor category'' -- that is, the
tensor product is not only bilinear, but also symmetric -- and
tensor categories in the normal sense are called ``monoidal''. The
reason for this is completely unclear to me, and this is bad
terminology -- in the standard language of category theory,
``monoidal'' does not imply that the tensor product is a bilinear
functor.
\end{rem}
Given a good $k$-linear unital tensor category $\Cc$, we can
literally repeat the construction of the last lecture and obtain a
category $\Cc_\#$ which is cofibered over $\Lambda$ -- the fiber
$(\Cc_\#)_{[n]}$ is the category $\Cc^{\otimes n}$, and the
transition functors $f_!$ are obtained from the tensor product
functors $m_n:\Cc^{\otimes n} \to \Cc$ (for $n=0$, we take
$\Cc^{\otimes n} = k\Vect$, and $m_0:k\Vect \to \Cc$ is the functor
which sends $k$ to the unit object in $\Cc$). Again, the category
$\Sec(\Cc_\#)$ of sections of the cofibration $\Cc_\# \to \Lambda$
is abelian by Proposition~\ref{abe.sec}, so that we can consider the
derived category $\DL(\Cc)=\D(\Sec(\Cc_\#))$ and the full
triangulated subcategory $\DL_{lc}(\Cc)=\D_{cart}(\Sec(\Cc_\#))$
spanned by Cartesian sections. We will call $\DL_{lc}(\Cc_\#)$ the
{\em cyclic envelope} of $\Cc_\#$.
Cyclic envelope only depends on the tensor category $\Cc$. However,
already to define Hochschild homology $HH_\idot(\Cc)$ of the
category $\Cc$, we need an extra datum -- a right-exact ``trace
functor''.
\begin{defn}
Assume given a good $k$-linear tensor category $\Cc$. A {\em trace
functor} on $\Cc$ is a functor $\tr:\Cc \to k\Vect$ which is
extended to a functor $\tr:\Cc_\#\to k\Vect$ in such a way that
$\tr(f)$ is invertible for any Cartesian map $f$ in
$\Cc_\#/\Lambda$.
\end{defn}
Explicitly, a trace functor is given by a functor $\tr:\Cc \to
k\Vect$ and an isomorphism
\begin{equation}\label{tau.eq}
\tau: \tr(M \otimes N) \cong \tr(N \otimes M)
\end{equation}
for any two objects $M,N \in \Cc$. The isomorphism $\tau$ should be
functorial in both $M$ and $N$, and satisfy the condition $\tau_{31}
\circ \tau_{12} \circ \tau_{23} = \id$, as in \eqref{tr.eq}. We
leave it to the reader to check that such an isomorphism $\tau$
uniquely defines an extension of $\tr$ to the whole category
$\Cc_\#$.
Given a good $k$-linear tensor category $\Cc$ equipped with a trace
functor $\tr$, we can repeat the construction of the last lecture:
we extend $\tr$ to a functor $\tr:\Sec(\Cc_\#) \to \Fun(\Lambda,k)$,
and consider the corresponding dervied functor $L^\hdot\tr:\DL(\Cc)
\to \D(\Lambda,k)$. As before, it sends $\DL_{lc}(\Cc) \subset
\DL(\Cc)$ into $\D_{lc}(\Lambda,k)$.
\begin{defn}
{\em Hochschild homology} $HH_\idot(\Cc,\tr)$ of the pair $\langle
\Cc,\tr \rangle$ is given by
$$
HH_\idot(\Cc,\tr) = L^\hdot\tr (\I),
$$
where $\I \subset \Cc$ is the unit object. {\em Cyclic homology}
$HC_\idot(\Cc,\tr)$ of the pair $\langle \Cc,\tr \rangle$ is given
by
$$
HC_\idot(\Cc,\tr) = H_\idot(\Lambda,L^\hdot\tr \I_\#),
$$
where $\I_\# \in \Sec_{cart}(\Cc_\#)$ is the Cartesian section of
$\Cc_\# \to \Lambda$ which sends an object $[n] \in \Lambda$ to
$\I^{\boxtimes n} \in \Cc^{\otimes n}$, the $n$-th power of the unit
object $\I \in \Cc$.
\end{defn}
Of course, in the case $\Cc = A\bimod$, $\tr$ as in the last
lecture, we have $HC_\idot(A\bimod,\tr) = HC_\idot(A,A_\#) =
HC_\idot(A)$ by virtue of Proposition~\ref{coeff.hc}.
\subsection{Morita-invariance of cyclic homology.}
As an application of the general formalism developed above, we prove
that Hochschild and cyclic homology of an associative algebra $A$
only depends on the category $A\amod$ of left $A$-modules. This is
known as {\em Morita invariance}.
A typical situation is the following. Assume given two $k$-algebras
$A$, $B$, and a $k$-linear functor $F:A\amod \to B\amod$. Assume
that $F$ is right-exact and commutes with infinite direct
sums. Consider the $B$-module $P = F(A)$. Since $\End_A(A) =
A^{opp}$, $P$ is not only a left $B$-module, but also a right
$A$-module -- in other words, an $A-B$-bimodule. By definition, we
have $F(A) = A \otimes_A P$; since $F$ is right-exact and commutes
with arbitrary sums, the same is true for any $M \in A\amod$ -- the
bimodule $P$ represents the functor $F$ in the sense that we have a
functorial isomorphism
$$
F(M) \cong M \otimes_A P.
$$
If $F$ is an equivalence of categories, then the inverse equivalence
$F^{-1}$ is of course also right-exact and commutes with sums; thus
we have a $B-A$-bimodule $P^o$ representing $F^{-1}$, and since $F
\circ F^{-1} \cong \Id$, $F^{-1} \circ F \cong \Id$, we have
isomorphisms
\begin{equation}\label{morita.ppo}
A \cong P \otimes_B P^o \in A\bimod, \qquad B \cong P^o \otimes_A P \in
B\bimod.
\end{equation}
\begin{prop}
Assume given two associative $k$-algebras $A$, $B$, and an
equivalence $A\amod \cong B\amod$. Then there exist natural
isomorphisms $HH_\idot(A) \cong HH_\idot(B)$, $HC_\idot(A) \cong
HC_\idot(B)$.
\end{prop}
\proof{} As we have already proved, every right-exact $k$-linear
functor $G:A\amod \to A\amod$ which commutes with sums is
represented by an $A$-bimodule $Q$. Conversely, every $Q \in
A\bimod$ represents such a functor. Tensor product of bimodules
corresponds to the composition of functor. Therefore the $k$-linear
tensor category $A\bimod$ only depends on the $k$-linear abelian
category $A\amod$, and can be recovered as the category of
endofunctors of $A\amod$ of a certain kind ($k$-linear, right-exact,
preserving sums). Thus in our situation, we have a natural
equivalence $F:A\bimod \cong B\bimod$ of $k$-linear abelian tensor
categories. It induces an equivalence of the corresponding
categories of cyclic bimodules. To finish the proof, it suffices to
prove that the equivalence $A\bimod \cong B\bimod$ is compatible
with the natural trace functors on both side. This is obvious: for
any $M \in A\bimod$, we have
$$
\tr(M) = A \otimes_{A^{opp} \otimes A} M \cong B
\otimes_{B^{opp} \otimes B} (P \otimes P^o)
\otimes_{A^{opp} \otimes A} M \cong B \otimes_{B^{opp} \otimes B}
F(M) = \tr(F(M)),
$$
where $P$ and $P^o$ are as in \eqref{morita.ppo}.
\endproof
\subsection{Example: group algebras}
Traditionally, in every exposition of cyclic homology, the authors
devote some time to one very special case, that of a group
algebra. I don't really know why --- whether it's because this is
needed to construct the regulator map from higher algebraic
$K$-theory, or because there are interesting new things special for
the group algebra case, or for some other reason. But let me follow
the tradition. This will also give us an example where the general
theory of cyclic homology for tensor categories is applied to a
tensor category which is not a category of bimodules.
Assume given a group $G$, and consider the group algebra
$k[G]$. This is an associative unital algebra, so it has Hochschild
and cyclic homology, and the category of $k[G]$-bimodules is a
tensor category. However, since $G$ is a group, the category $G\amod
= k[G]\amod$ of representation of $G$ a.k.a. left $k[G]$-modules is
a tensor category in its own right. Moreover, there is an obvious
functor $\gamma:G\amod \to k[G]\bimod$ which sends a representation
$V \in G\amod$ to a functor $G\bimod \to G\bimod$ given by $M
\mapsto M \otimes V$ (here we use the interpretation of
$k[G]$-bimodules as endofunctors of the category $G\amod$). This
functor is obviously exact and obviously tensor. Explicitly, it is
given by
$$
\gamma(V) = V \otimes R,
$$
where we denote $R = k[G]$, the left $k[G]$-action on $V \otimes R$
is through $V$ and $R$, and the right action is through $R$: we have
$g_1(v \otimes g)g_2 = g_1v \otimes g_1gg_2$. If we have two
representations $V_1,V_2 \in G\amod$, the natural isomorphism
$\gamma(V_1 \otimes V_2) \cong \gamma(V_1) \otimes_{k[G]}
\gamma(V_2)$ is given by the map
\begin{equation}\label{tens.eq}
V_1 \otimes V_2 \otimes R \to (V_1 \otimes R) \otimes_{k[G]} (V_2
\otimes R)
\end{equation}
which sends $v_1 \otimes v_2 \otimes g$ to $(v_1 \otimes 1) \otimes
(v_2 \otimes g)$, where $1 \in G$ is the unity element.
Since the functor $\gamma$ is tensor, the usual trace functor $\tr$
on $k[G]\bimod$ gives by restriction a trace functor $\tr^R = \tr
\circ \gamma$ on $G\amod$. Explicitly, it is given by
$$
\tr^R(V) = (V \otimes R)/\{g_1v \otimes g_1g - v \otimes gg_1\mid
g,g_1 \in G,v \in V\},
$$
and since the quotient is over all $g$ and all $g_1$, we might as
well replace $g$ with $gg_1^{-1}$. Then we have
$$
\tr^R(V) = (V \otimes R)/\{g_1v \otimes g_1gg_1^{-1} - v \otimes g\}
= (V \otimes R)_G,
$$
the $G$-coinvariants in the $G$-representation $V \otimes R$, where
$R$ is equipped with the adjoint $G$-action. One can also check, and
this is important, that the identification survives on the level of
derived functors --- the natural map
\begin{equation}\label{der.tr}
L^\hdot\tr (\gamma(V)) \to L^\hdot\tr^R(V) = H_\idot(G,V \otimes R)
\end{equation}
is an isomorphism in all degrees. For example, for the trivial
representation $V=k$, we obtain an isomorphism $HH_\idot(k[G]) \cong
H_\idot(G,R)$. The isomorphism $\tau^R:\tr^R(V_1 \otimes V_2) \cong
\tr^R(V_2 \otimes V_1)$ of \eqref{tau.eq} is induced by the usual
symmetry isomorphism $V_1 \otimes V_2 \to V_2 \otimes V_1$ and the
isomorphism \eqref{tens.eq}; explicitly, $\tau^R$ is the map on the
spaces of coinvariants induced by the map
\begin{equation}\label{wt.tau.eq}
\wt{\tau}^R: V_1 \otimes V_2 \otimes R \to V_2 \otimes V_1 \otimes
R, \qquad \wt{\tau}^R(v_1 \otimes v_2 \otimes g) = gv_2 \otimes v_1
\otimes g.
\end{equation}
One easily checks that the map $\wt{\tau}^R$ defined in this way is
actually a map of $G$-representations.
Applying the general theory of cyclic homology with
coefficients, we extend this isomorphism to an isomorphism
$$
HC_\idot(k[G]) \cong HC_\idot(G\amod,\tr^R).
$$
We now note that the adjoint representation $R = k[G]$ canonically
splits into a direct sum $R = \bigoplus_{\langle g \rangle} R^g$
over the conjugacy classes $\langle g \rangle \subset G$, $R^g =
k[\langle g \rangle]$, and this induces a canonical direct sum
decomposition
\begin{equation}\label{cnj.eq}
\tr^R = \bigoplus_{\langle g \rangle} \tr^g
\end{equation}
of the trace functor $\tr^R$: we set $\tr^g(V) = (V \otimes R^g)_G$,
and since the isomorphism $\wt{\tr}^R$ of \eqref{wt.tau.eq}
obviously respects the direct sum decomposition, the isomorphism
$\tau^R$ induces isomorphisms \eqref{tau.eq} for every component
$\tr^g$. Therefore we actually have a canonical direct sum
decomposition of cyclic homology:
\begin{equation}\label{cnj.classes}
HC_\idot(k[G]) = \bigoplus_{\langle g
\rangle}HC_\idot(G\amod,\tr^g),
\end{equation}
and a corresponding decomposition for $HH_\idot(k[G])$.
However, we can say more. Consider the component $\tr^1$ in the
decomposition \eqref{cnj.eq} which corresponds to the unity element
$1 \in G$. Then we have $\tr^1(V) = V_G$, the space of
$G$-coinvariants, and
$$
HH_\idot(G\amod,\tr^1) \cong H_\idot(G,k).
$$
What can we say about the cyclic homology $HC_\idot(G\amod,\tr^1)$?
Looking at \eqref{wt.tau.eq}, we see that the isomorphism $\tr^1(V_1
\otimes V_2) \cong \tr^1(V_2 \otimes V_1)$ for the trace functor
$\tr^1$ is induced by the symmetry isomorphism $v_1 \otimes v_2 \to
v_2 \otimes v_1$. We can rephrase this in the following way: since
the tensor category $G\amod$ is symmetric, {\em any} right-exact
functor $F:G\amod \to k\Vect$ canonically extends to a trace functor
$F_\#:G\amod_\# \to k\Vect$, and it is this trace functor structure
that $\tr^1$ has --- we have $\tr^1 \cong \Coinv_\#$, where
$\Coinv:G\amod \to k\Vect$ is the coinvariants functor, $V \mapsto
V_G$.
In other words, the identity functor $\Id:G\amod \to G\amod$ {\em
can also be considered as a trace functor}, albeit with values in
$G\amod$ rather than $k\Vect$, so that we have a functor
$$
L^\hdot\Id:\DL(G\amod) \to \D(\Lambda,G\amod) = \D(\Lambda \times
\ppt_G,k),
$$
where $\ppt_G$ is the category with one object with automorphism
group $G$, and the trace functor $L^\hdot\tr^1$ factors through
$L^\hdot\Id$, so that we have
$$
HC_\idot(G\amod,\tr^1) = H_\idot(\Lambda \times
\ppt_G,L^\hdot\Id(\I_\#)).
$$
Moreover, $\Id$ is exact, so that there is no need to take its
derived functor, and we simply have $L^\hdot\Id(\I_\#) = \Id(\I_\#) =
k^{\Lambda \times \ppt_G}$, the constant cyclic $k$-vector space
with the trivial action of $G$. Thus by the K\"unneth formule, we
have
$$
HC_\idot(G\amod,\tr^1) \cong H_\idot(\Lambda \times \ppt_G),k) =
H_\idot(\Lambda,k) \otimes H_\idot(\ppt_G,k).
$$
Since $H_\idot(\ppt_G,k) = H_\idot(G,k) = HH_\idot(G\amod,\tr^1)$,
we conclude that {\em the Hodge-to-de Rham spectral sequence for the
cyclic homology $HC_\idot(G\amod,\tr^1)$ canonically degenerates}:
we have a canonical isomorphism
\begin{equation}\label{hdr.uni}
HC_\idot(G\amod,\tr^1) \cong HH_\idot(G\amod,\tr^1)[u^{-1}]
\end{equation}
for the unity component in the direct sum decomposition
\eqref{cnj.classes}.
\subsection{The regulator map}
To finish today's lecture, let me give the standard application of
the above computation of groups algebras: I will construct the {\em
higher regulator} a.k.a. {\em higher Chern character map} from
Quillen's higher $K$-theory to cyclic homology.
Recall that to define higher $K$-theory of an algebra $A$, one
considers the group $GL_\infty(A) = \lim_{\to}GL_N(A)$ of infinite
matrices over $A$ and its classifying space $BGL_\infty(A)$. This is
of course an Eilenberg-MacLane space of type $K(\pi,1)$. However,
Quillen defined a certain very non-trivial operation called {\em the
plus-construction} with replaces a topological space $X$ with
another topological space $X^+$ so that the homology is the same,
$H_\idot(X,\Z) \cong H_\idot(X^+,\Z)$, but $X^+$ has an abelian
fundamental group. Then by definition, higher $K$-groups of $A$ are
given by
$$
K^\hdot(A) = \pi_\idot(BGL_\infty(A)^+),
$$
the homotopy groups of the plus-construction $BGL_\infty(A)$.
These groups are very hard to compute (not surprisingly, since
homotopy groups in general are hard to compute). Fortunately, to
construct the regulator, we do not need to do it. Namely, for any
topological space $X$, there exists a canonical Hurewitz map
$\pi_\idot(X) \to H_\idot(X)$. The regulator map factors through the
Hurewitz map for $BGL_\infty^+$, so that the source of the map we
will construct is actually the homology $H_\idot(BGL_\infty^+)$. At
this point, we can also get rid of the plus-construction: by its
very definition, $H_\idot(X) = H_\idot(X^+)$, so that
$H_\idot(BGL_\infty^+) = H_\idot(BGL_\infty) =
H_\idot(GL_\infty,\Z)$, the homology of the group $GL_\infty(A)$
with trivial coefficients. In fact, our map will further factor
through $H_\idot(GL_\infty(A),k)$.
What is the natural target of the regulator map? Comparison with the
Chern character map in algebraic geometry suggests at first that
this should the de Rham cohomology groups $H_{DR}^\hdot(-)$ --- in
our situation, these correspond to the periodic cyclic homology
groups $HP_\idot(A)$. However, it is known that the Chern character
actually behaves nicely with respect to the Hodge filtration --- the
Chern character map $K_0(X) \to \bigoplus_iH_{DR}^{2i}(X)$ for an
smooth algebraic variety $X$ actually factors through $\bigoplus_i
F^iH_{DR}^{2i}(X)$. In the non-commutative situation, this
corresponds to taking the $0$-th graded piece of the Hodge
filtration on $HP_\idot(A)$. This has its own name.
\begin{defn}
The {\em negative cyclic homology} $HC^-_\idot(A)$ of an algebra $A$
is the $0$-th term $F^0HP_\idot(A)$ of the Hodge filtration on
$HP_\idot(A)$.
\end{defn}
If we compute $HP_\idot(A)$ by the standard periodic bicomplex, then
computing $HC^-_\idot(A)$ amount to removing all the columns {\em to
the left} of the $0$-th one --- as opposed to the usual
$HC_\idot(A)$, where we remove everything to the right. This
explains the adjective ``negative''.
So, what we want to do is to construct a map
$H_\idot(GL_\infty(A),k) \to HC^-_\idot(A)$. This is done in three
steps.
First, fix some integer $N$, and consider the group algebra
$k[GL_N(A)]$. This has a natural map into the algebra $\Mat_N(A)$ of
$N \times N$-matrices in $A$ -- an element $g \in GL_N(A)$ goes to
itself considered as an element in $\Mat_N(A)$. The map of algebras
induces a map of negative cyclic homology; passing to the limit, we
obtain a map
$$
\lim_{\to}HC_\idot^-(k[GL_N(A)]) \to \lim_{\to} HC_\idot^-(\Mat_N(A)).
$$
Second, we observe that by Morita-invariance of cyclic homology, the
directed system in the right-hand side is actually constant --- we
have $HC_\idot^-(\Mat_N(A)) \cong HC_\idot^-(A)$ for any $N$. Thus
we have constructed a map
$$
\lim_{\to}HC_\idot^-(k[GL_N(A)]) \to HC_\idot^-(A).
$$
Finally, we use the direct sum decomposition \eqref{cnj.classes} ---
we take the graded piece of \eqref{cnj.classes} corresponding to the
unity element $1 \in GL_N(A)$, and apply the canonical Hodge-to-de
Rham degeneration \eqref{hdr.uni}. This gives a canonical map
$$
H_\idot(GL_\infty(A),k) = \lim_{\to}H_\idot(GL_N(A),k) \to
\lim_{\to}HC_\idot^-(k[GL_N(A)]).
$$
Composing the two maps above, and plugging in the Hurewitz map, we
obtain the higher regulator map $K_\idot(A) \to HC_\idot^-(A)$.
\lecture{}
\begin{center}
\begin{minipage}{0.7\linewidth}\small
Cartier isomorphism in the commutative case. The categories
$\Lambda_p$. Frobenius and quasi-Frobenius maps. Non-commutative
case: the Cartier isomorphism for algebras with a quasi-Frobenius
map. Remarks on the general case.
\end{minipage}
\end{center}
\subsection{Cartier isomorphism in the commutative case.}
The goal of this lecture is to explain the construction of the
so-called {\em Cartier isomorphism} for algebras over a finite field
$k$. We start by recalling what happens in the commutative case.
Fix a finite field $k$ of characteristic $p = \cchar k$, and
consider a smooth affine variety $X = \Spec A$ over $k$. Assume that
$p > \dim X$, and consider the de Rham complex
$\Omega^\hdot_X$. This complex behaves very differently from what we
have in characteristic $0$. For instance, in characteristic $0$, a
function $f$ is closed with respect to the de Rham differential if
and only if it is locally constant. In our situation, however, the
$p$-th power $a^p$ of any $a \in A$ is closed: we have $df^p =
pf^{p-1}df = 0$. About fifty years ago, P. Cartier has shown that
this gives all the closed functions, and moreover, the situation in
higher degrees is similar --- for any $n \geq 0$, there exists a
canonical {\em Cartier isomorphism}
$$
C:H_{DR}^n(X) \cong \Omega_{X^\tw}^n
$$
between the de Rham cohomology group $H_{DR}^n(X)$ and the space
$\Omega^n_{X^\tw}$ of $n$-forms on the so-called ``Frobenius twist''
$X^\tw = \Spec A^\tw$ of the variety $X$ --- $A^\tw$ coincides with
$A$ as a ring, but the $k$-algebra structure is twisted by the
Frobenius automorphism of the field $k$.
Let us briefly sketch the construction of the inverse isomorphism
$C^{-1}:\Omega_{X^\tw}^n \to H_{DR}^n(X)$ (this is
simpler). Consider the ring $W(k)$ of Witt vectors of the field $k$
--- that is, the unramified extension of $\Z_p$ whose residue field
is $k$. Since $X$ is an affine variety, we can lift it to a smooth
variety $\wt{X}$ over $W(k)$ so that $X = \wt{X} \otimes_{W(k)}
k$. Moreover, we can lift the Frobenius map $F:X \to X^\tw$ to a map
$\wt{F}:\wt{X} \to \wt{X}^\tw$, where $\wt{X}^\tw$ means the twist
with respect to the Frobenius automorphism of $W(k)$. For any
$1$-form $fdg \in \Omega^1_{\wt{X}}$, we have
$$
\wt{F}^*(fdg) = f^pdg^p \mod p,
$$
so that the pullback map $\wt{F}^*:\Omega^1_{\wt{X}^\tw} \to
\Omega^1_{\wt{X}}$ is divisible by $p$, and cosequently, $\wt{F}^*$ on
$\Omega^n_{\wt{X}^\tw}$ is divisible by $p^n$. Let us make this
division and consider the map
$$
\overline{F}:\wt{\Omega}^\hdot_{\wt{X}^\tw} \to \Omega^\hdot_{\wt{X}}
$$
given by $\overline{F} = \frac{1}{p^n}\wt{F}^*$ in degree $n$, where
$\wt{\Omega}^\hdot_{\wt{X}^\tw}$ is the de Rham complex of the
variety $\wt{X}^\tw$ with differential multiplied by $p$. Then it is
not difficult to check --- for instance, by a computation in local
coordinates --- that the map $\overline{F}$ is a
quasiisomorphism. Reducing it modulo $p$, we obtain a
quasiisomorphism
$$
\bigoplus_n\Omega^n_{X^\tw} \cong \Omega^\hdot_X,
$$
where the differential in the left-hand side, being divisible by $p$,
reduces to $0$. One then checks that the components of this
quasiisomorphisms in individual degrees do not depend on our choices
--- neither of the lifting $\wt{X}$, nor on the lifting
$\wt{F}$. These are the inverse Cartier maps.
We note that the Cartier maps are not easy to write down by an
explicit formula even when $X$ is a curve, expect for one especially
simple case --- and contrary to the expectations, the simple case is
not the affine line $X = \Spec k[t]$, but the multiplicative group
$X = \Spec k[t,t^{-1}]$. In this case, we have
$$
C^{-1}\left(f\frac{dt}{t}\right) = f^p\frac{dt}{t}
$$
for any $f \in k[t,t^{-1}]$. Analogously, in dimension $n$, we have
a similar explicit formula for the torus $X = T = \Spec k[L]$, the
group algebra of a lattice $L = \Z^n$.
\subsection{The categories $\Lambda_p$.}
To generalize this construction to the non-commutative case, we need
one piece of linear algebra which we now describe.
Recall that in the combinatorial description, the cyclic category
$\Lambda$ was obtained as a quotient of the category
$\Lambda_\infty$: for any $[m],[n] \in \Lambda$, we have
$\Lambda([n],[m]) = \Lambda_\infty([n],[m])/\sigma$. For any
positive integer $l$, we can define a category $\Lambda_l$ by a
similar procedure: $\Lambda_l$ has the same objects as $\Lambda$,
and we set
$$
\Lambda_l([n],[m]) = \Lambda_\infty([n],[m])/\sigma^l
$$
for any $[n],[m] \in \Lambda_l$. We have an obvious projection
$\pi:\Lambda_l \to \Lambda$; just as the projection $\Lambda_\infty
\to \Lambda$, this is a connected bifibration whose fiber is the
groupoid $\ppt_l$ with one object and $\Z/l\Z$ as its automorphism
group. One the other hand, we also have an embedding $i:\Lambda_l
\to \Lambda$ which sends $[n] \in \Lambda_l$ to $[nl] \in
\Lambda$. Just as for $\Lambda$, the embedding $j:\Delta^{opp} \to
\Lambda_\infty$ induces an embedding $j_l:\Delta^{opp} \to \Lambda_l$.
It turns out that most of the facts about the homology of the
category $\Lambda$ immediately generalize to $\Lambda_l$, with the
same proofs. In particular, for any $E \in \Fun(\Lambda_l,k)$, the
homology $H_\idot(\Lambda_l,E)$ can be computed by a bicomplex
\begin{equation}\label{hc.lambda.l}
\begin{CD}
\dots @>>> E([1]) @>{\id }>>
E([1]) @>{\id - \tau}>> E([1])\\
@. @AA{b}A @AA{b'}A @AA{b}A \\
\dots @>>> E([2]) @>{\id + \dots + \tau^{l-1}}>> E([2])
@>{\id - \tau}>> E([2])\\
@. @AA{b}A @AA{b'}A @AA{b}A\\
\dots @. \dots @. \dots @. \dots\\
@. @AA{b}A @AA{b'}A @AA{b}A \\
\dots @>>> E([n]) @>{\id + \tau + \dots + \tau^{ln-1}}>>
E([n]) @>{\id - \tau}>> E([n]),\\
@. @AA{b}A @AA{b'}A @AA{b}A
\end{CD}
\end{equation}
we have a periodicity map $H_{\idot+2}(\Lambda_l,E) \to
H_\idot(\Lambda_l,E)$ which fits into a Connes' exact sequence
$$
\begin{CD}
H_\idot(\Delta^{opp},j_l^*E) @>>> H_\idot(\Lambda_l,E) @>{u}>>
H_{\idot-2}(\Lambda_l,E) @>>>,
\end{CD}
$$
and the periodicity map $u$ is induced by the action of the
generator $u$ of the cohomology algebra $H^\hdot(\Lambda_l,k) \cong
k[u]$. As in Lecture 4, this generator admits an explicit Yoneda
representation by a length-$2$ complex $j_{l*}^{opp}j_l^{opp*}k \to
j_{l!}j_l^*k$. Moreover it is easy to check that this complex
coincides with the pullback of the analogous complex in
$\Fun(\Lambda,k)$ with respect to $i:\Lambda_l \to \Lambda$, so that
$i$ induces an isomorphism
$$
i^*:H^\hdot(\Lambda,k) \to H^\hdot(\Lambda_l,k)
$$
sending the periodicity generator to the periodicity
generator. However, there is also one new and slightly surprising
fact.
\begin{lemma}\label{i.compa}
For any associative unital algebra $A$ over $k$, the natural map
$$
M:H_\idot(\Lambda_l,i^*A_\#) \to H_\idot(\Lambda_l,A_\#)
$$
is an isomorphism.
\end{lemma}
\proof{} Since $i^*$ is compatible with the periodicity maps, it
suffices to prove that the natural map
$$
H_\idot(\Delta^{opp},j_l^*i^*A_\#) \to H_\idot(\Delta^{opp},j^*A_\#)
$$
on Hochschild homology is an isomorphism. By definition, we have
$$
j^*A_\# \cong C_\idot(A) \otimes_{A \otimes A^{opp}} A,
$$
where $C_\idot(A)$ is the bar resolution considered as a simplicial
set. Writing down explicitly the definition of $i:\Lambda_l \to
\Lambda$, one deduces that
$$
j_l^*i^*A_\# \cong (C_\idot(A) \otimes_A \dots \otimes_A C_\idot(A))
\otimes_{A \otimes A^{opp}} A,
$$
with $l$ factors $C_\idot(A)$. But since $C_\idot(A)$ is a
resolution of $A$, so is the product in the right-hand side. We
conclude that $H_\idot(\Delta^{opp},j_l^*i^*A_\#)$ is just the
Hochschild homology $HH_\idot(A)$ computed by a diffewrent
resolution, and $M$ is indeed an isomorphism.
\endproof
\begin{exc}
Prove that the map $M$ is an isomorphism for any cyclic vector space
$E \in \Fun(\Lambda,k)$, not just for $A_\#$. Hint: use the acyclic
models method, and show that $\Fun(\Lambda,k)$ has a generator of
the form $A_\#$.
\end{exc}
\subsection{Frobenius and quasi-Frobenius maps.}
Assume now given an associative unital algebra $A$ over $k$;
motivated by comparison theorems of Lecture 2, we want to construct
a Cartier isomorphism of the form
\begin{equation}\label{car}
HH_\idot(A^\tw)((u)) \cong HP_\idot(A).
\end{equation}
Unfortunately, the procedure that we have used in the commutative
case breaks down right away: there is no Frobenius map in the
non-commutative case. The endomorphism $F:A \to A$ given by $a \mapsto
a^p$ is neither additive nor multiplicative for a general
non-commutative algebra $A$.
To analyze the difficulty, split $F$ into the composition
$$
\begin{CD}
A^\tw @>{\phi}>> A^{\otimes p} @>{m}>> A
\end{CD}
$$
of the map $\phi$ given by $\phi(a) = a^{\otimes p}$, and the
multiplication map $m:A^{\otimes p} \to A$, $m(a_1 \otimes \dots
\otimes a_p) = a_1 \dots a_p$. The map $\phi$ is not additive, nor
multiplicative, but this is always so, be $A$ commutative or not. It
is the map $m$ that causes the problem: if $A$ is commutative, it is
an algebra map, and in general it is not.
This is where the $p$-cyclic category $\Lambda$ helps. Although the
map $m$ is not an algebra map, so that no Frobenius map acts on $A$,
we still can get an action of this nonexisting Frobenius on
Hochschild and cyclic homology by extending $m$ to the isomorphism
$$
M:H_\idot(\Lambda_p,i^*A_\#) \to H_\idot(\Lambda,A_\#)
$$
of Lemma~\ref{i.compa}. As for the map $\phi$, which behaves very
badly in all cases, it turns out that it can be replaced by a
different map within a certain large class of them. Namely, the only
important property of the map $\phi$ is the following.
\begin{lemma}\label{V.p}
For any vector space $V$ over $k$, the map $\phi:V^\tw \to
V^{\otimes p}$ induces an isomorphism
$$
H_i(\Z/p\Z,V^\tw) \cong H_i(\Z/p\Z,V^{\otimes p})
$$
for any $i \geq 1$, where the cyclic group $\Z/p\Z$ acts trivially
on $V^\tw$, and by the cyclic permutation on $V^{\otimes p}$.
\end{lemma}
\proof{} The homology of the cyclic group $\Z/p\Z$ with coefficients
in some representation $M$ can be computed by the standard periodic
complex $M_\idot$ with terms $M_i = M$, $i \geq 0$, and the
differentials $d_- = 1-\sigma$ in odd degrees and $d_+ = 1 + \sigma
+ \dots + \sigma^{p-1} = (1-\sigma)^{p-1}$ in even degrees, where
$\sigma$ is the generator of $\Z/p\Z$. For the trivial
representation $V^\tw$, $d_+ = d_- = 0$. The map $\phi$ obviously
sends $V^\tw$ into the $\sigma$-invariant subspace in $V^{\otimes
p}$, thus into the kernel of both $d_+$ and $d_-$. We have to show
that (1) $\phi$ becomes additive modulo the image of the
corresponding differential $d_-$, $d_+$, and (2) it actually becomes
an isomorphism. Choose a basis in $V$, so that $V
= k[S]$ is the $k$-vector space generated by a set $S$. Then
$V^{\otimes p} = k[S^p]$. Decompose $S^p = S \coprod (S \setminus
S)$, where $S \subset S^p$ is embedded as the diagonal, and consider
the corresponding decomposition $V^{\otimes p} = V \oplus V'$, where
$V' = k[S^p \setminus S]$. This decomposition is $\Z/p\Z$-invariant,
thus compatible with $d_+$ and $d_-$; morever, $\phi$ obviously
becomes an additive isomorphism if we replace $V^{\otimes p}$ with
its quotient $V^{\otimes p}/V' = V$. Thus it suffices to prove that
the complex which computes $H_\idot(\Z/p\Z,V')$ is acyclic in
degrees $\geq 1$. This is obvious --- the $\Z/p\Z$-action on $S^p
\setminus S$ is free.
\endproof
For a more natural formulation of Lemma~\ref{V.p}, one can invert
the periodicity endomorphism of the homology functor
$H_\idot(\Z/p\Z,-)$ to obtain the so-called {\em Tate homology}
$\vH_\idot(\Z/p\Z,-)$ (this is the same procedure that we used in
passing from $HC_\idot(-)$ to $HP_\idot(-)$). Then Lemma~\ref{V.p}
claims that $\phi$ induces a canonical isomorphism
$$
\vH_\idot(\Z/p\Z,V^\tw) \cong \vH_\idot(\Z/p\Z,V^{\otimes p})
$$
in all degrees. We will call it {\em the standard isomorphism}.
\begin{defn}
A {\em quasi-Frobenius map} for an associative unital algebra $A$
over $k$ is a $\Z/p\Z$-equivariant algebra map $\Phi:A^\tw \to
A^{\otimes p}$ which induces the standard isomorphism on Tate
homology $\vH_\idot(-)$.
\end{defn}
Given an algebra $A$ with a quasi-Frobenius map $\Phi$, we can
construct an inverse Cartier map \eqref{car} right away. Namely,
comparing the bicomplex \eqref{hc.lambda.l} with the usual cyclic
bicomplex \eqref{hc.lambda}, we see that the only difference is that
the differential $1 + \tau + \dots + \tau^{n-1}$ is replaced with
$$
1 + \tau + \dots + \tau^{np-1} = (1 + \sigma + \circ + \sigma^{p-1})
\circ (1 + \tau + \dots + \tau^{n-1}),
$$
where we have used the fact that $\sigma = \tau^n$. But for some $E'
\in \Fun(\Lambda_p,k)$ of the form $E' = \pi^*E$, $E \in
\Fun(\Lambda,p)$, we have $\sigma = 1$, so that $1 + \sigma + \circ
+ \sigma^{p-1} = p = 0$. Therefore we have a natural identification
\begin{equation}\label{hh.p}
HP_\idot(\pi^*A^\tw_\#) \cong HH_\idot(A^\tw)((u)),
\end{equation}
where on the left-hand side we have a periodic version of the
homology $H_\idot(\Lambda_p,\pi^*A^\tw_\#)$. On the other hand, the
quasi-Frobenius map $\Phi$ induces a map $\Phi:\pi^*A_\# \to
i^*A_\#$, which induces a map on periodic homology. We define the
inverse Cartier map $C^{-1}$ as the map
\begin{equation}\label{car.defn.eq}
C^{-1} = M \circ \Phi: HH_\idot(A^\tw)((u)) \cong
HP_\idot(\pi^*A^\tw_\#) \to HP_\idot(i^*A_\#) \to HP_\idot(A).
\end{equation}
We must say that this comparatively easy situation is quite rare ---
in fact, the only situation where I know that a quasi-Frobenius map
exists is the case of a group algebra $A = k[G]$ of some group $G$
(one can take, for instance, the map $\Phi:k[G] \to k[G^p] =
k[G]^{\otimes p}$ given by $\Phi(g) = g^{\otimes p}$, $g \in
G$). This is perhaps not surprising, since in the commutative case,
the situation was also explicit and simple only for the torus $A =
k[L]$. It remains to do three things.
\begin{enumerate}
\item Prove that the map $C^{-1}$ is an isomorphism.
\item Compare it to the usual inverse Cartier isomoprhism in the
commutative case.
\item Explain what to do when no quasi-Frobenius map is available.
\end{enumerate}
I will give a sketch of \thetag{i} next, under an additional
assumption that the algebra $A$ has finite homological dimension ---
it seems that this is a necessary assumption. I will leave
\thetag{ii} as a not very difficult but tedious exersize. As for
\thetag{iii}, this is unfortunately quite involved, and I cannot
really present the procedure here in any detail, however sketchy;
let me just mention that the only new thing in the general case is a
certain generalization of the notion of a quasi-Frobenius map, while
everything that concerns cyclic homology {\em per se} remains
more-or-less the same as in the simple case. I refer the reader to
Section 5 of my paper {\tt arXiv.math/0708.1574} for an introductory
exposition, with the detailed proofs given in {\tt
arXiv.math/0611623}.
\subsection{Cartier isomorphism for algebras with a quasi-Frobenius
map.}
We assume given an associative algebra $A/k$ with a quasi-Frobenius
map $\Phi$, and we want to prove that the corresponding inverse
Cartier map \eqref{car.defn.eq} is an isomorphism. We note that the
map $M$ induces an isomorphism by Lemma~\ref{i.compa}, so that what
we have to prove is that $\Phi$ also induces an isomorphism on
periodic cyclic homology.
We will need one technical notion. Note that the embedding
$j:\Delta^{opp} \to \Lambda_p$ extends to an embedding
$\wt{j}:\Delta^{opp} \times \ppt_p \to \Lambda_p$. Thus every $E \in
\Fun(\Lambda_p,k)$ gives by restriction a simplicial
$\Z/p\Z$-representation $\wt{j}^*E \in \Fun(\Delta^{opp} \times
\ppt_p,k) \cong \Fun(\Delta^{opp},\Z/p\Z\amod)$. By the Dold-Kan
equivalence, $\wt{j}^*E$ can treated as a complex of
$\Z/p\Z$-representations.
\begin{defn}\label{lambda.small}
An object $E \in \Fun(\Lambda_p,k)$ is {\em small} if $\wt{j}^*E$ is
chain-homotopic to a complex of $\Z/p\Z$-modules which is of finite
length.
\end{defn}
\begin{lemma}\label{no.hp}
Assume given a small $E \in \Fun(\Lambda_p,k)$ such that $E([n])$ is
a free $\Z/p\Z$-module for any $[n] \in \Lambda_p$ (the action of
$\Z/p\Z$ is generated by $\sigma = \tau^n$). Then we have
$$
HP_\idot(E) = 0.
$$
\end{lemma}
\proof[Sktech of a proof.] We have $H_\idot(\Lambda_p,E) =
H_\idot(\Lambda,L^\hdot\pi_!E)$, and the Connes' exact sequence for
$L^\hdot\pi_!E$ gives an exact triangle
$$
\begin{CD}
H_{\idot-1}(\Lambda_p,E) @>>> H_\idot(\Delta^{opp},\wt{j}^*E) @>>>
H_\idot(\Lambda_p,E) @>{\pi^*u}>>,
\end{CD}
$$
where the connecting map is induced by the pullback $\pi^*u \in
H^2(\Lambda,k)$ of the generator $u \in H^2(\Lambda,k)$. Computing
$H^2(\Lambda_p,k)$ by a cohomological version of the bicomplex
\eqref{hc.lambda.l}, as in Lecture 4, we find that $\pi^*u = 0$
(this is the same computation as in \eqref{hh.p}). Therefore, to
prove that
$$
HP_\idot(E) = \lim_{\overset{u}{\gets}}H_\idot(\Lambda_p,E)
$$
vanishes, it suffices to prove the vanishing of
$$
\lim_{\overset{u}{\gets}}H_\idot(\Delta^{opp} \times
\ppt_p,\wt{j}^*E),
$$
where $\wt{j}^*u \in H^2(\Delta^{opp} \times \ppt_p,k)$ is the
restriction of the periodicity generator $u \in
H^2(\Lambda_p,k)$. Using the Yoneda representation of $u$, we see
that with respect to the K\"unneth isomorphism $H^2(\Delta^{opp}
\times \ppt_p,k) \cong H^\hdot(\Z/p\Z,k)$, the class $\wt{j}^*u$
corresponds to the periodicity generator of
$H^2(\Z/p\Z,k)$. Therefore in the spectral sequence
$$
H_\idot(\Delta^{opp},\lim_{\overset{u}{\gets}}H_\idot(\Z/p\Z,\wt{j}^*E))
\Rightarrow \lim_{\overset{u}{\gets}}H_\idot(\Delta^{opp} \times
\ppt_p,\wt{j}^*E),
$$
the limit in the left-hand side is the Tate homology
$\vH_\idot(\Z/p\Z,\wt{j}^*E)$. Since $E$ is small, the spectral
sequence converges, and since $E([n])$ is a free
$\Z/p\Z$-representation for every $[n]$, the Tate homology in
question is equal to $0$.
\endproof
\begin{prop}
Assume given an associative algebra $A$ equipped with a
quasi-Frobenius map $\Phi:A^\tw \to A^{\otimes p}$, and assume that
the category $A\bimod$ of $A$-bimodules has finite homological
dimension. Then the Cartier map \eqref{car.defn.eq} for the algebra
$A$ is an isomorphism.
\end{prop}
\proof{} We first note that the object $i^*A_\# \in
\Fun(\Lambda_p,k)$ is small in the sense of
Definition~\ref{lambda.small}. Indeed, by assumption, the diagonal
$A$-bimodule $A$ admits a finite projective resolution
$P_\idot$. Therefore the bar resolution $C_\idot(A)$ is
chain-homotopic to a finite complex $P_\idot$, its $p$-th power
$C_\idot(A) \otimes_A \dots \otimes_A C_\idot(A)$ is chain-homotopic
to the finite complex $P_\idot \otimes_A \dots \otimes_A P_\idot$,
and the induced chain homotopy equivalence between $i^*A_\#$ and the
finite complex
$$
(P_\idot \otimes_A \dots \otimes_A P_\idot) \otimes_{A^{opp} \otimes
A} A
$$
is obviously compatible with the $\Z/p\Z$-action. Moreover,
$\pi^*A_\#$ is also small. It remains to notice that any
quasi-Frobenius map $\Phi$ must be injective (otherwise it sends
some element $a \in A^\tw = \vH_0(\Z/p\Z,A^\tw)$ to $0$), and its
cokernel $A^{\otimes p}/\Phi(A^{\otimes p})$ by definition has no
Tate homology.
\begin{exc}
Prove that for some $k[\Z/p\Z]$-module $V$, $\vH_\idot(\Z/p\Z,V)=0$
if and only if $V$ is free. Hint: identifying $k[\Z/p\Z] =
k[t]/t^p$, $\sigma \mapsto 1+t$, show that $V$ decomposes into a
direct sum of modules of the form $k[t]/t^l$, $0 < l \leq p$, and
check the statement for all $l$.
\end{exc}
We conclude that $A^{\otimes p}/\Phi(A^{\otimes p})$ is a free
$k[\Z/p\Z]$-module. Therefore for every $n$, the module $A^{\otimes
pn}/\Phi^{\otimes n}(A^{\tw\otimes n})$ is free, and has no Tate
homology. This means that the cokernel $i^*A_\#/\Phi(\pi^*A_\#)$
satisfies the assumptions of Lemma~\ref{no.hp}, and $\Phi$ indeed
induces an isomorphism between $HP_\idot(\pi^*A_\#)$ and
$HP_\idot(i^*A_\#)$.
\endproof
\begin{rem}
In the smooth commutative case, the assumption that $A\bimod$ has
finite homological dimension just means that $A$ is of finite type
over $k$. In the general case of the theorem, when no
quasi-Frobenius map is available, one actually needs to assume that
the homological dimension is less than $2p-1$. In the commutative
case, this reduces to $p > \dim\Spec A$.
\end{rem}
\lecture{}
\begin{center}
\begin{minipage}{0.7\linewidth}\small
Hochschild cohomology of an associative algebra and its
Morita-invariance. Hochschild cohomology complex. Multiplication and
the Eckman-Hilton argument. Derivations of the tensor algebra and
the Gerstenhaber bracket on Hochschild cohomology. Hochschild
cohomology and deformations. Quantizations. Kontsevich formality
(statements).
\end{minipage}
\end{center}
\subsection{Generalities on Hochschild cohomology.}
Up to now, we were studying Hochschild homology of associative
algebras and related concepts --- cyclic homology, regulator maps,
and so on. We will now turn to the other half of the story:
Hochschild cohomology.
We recall (see Definition~\ref{hh.defn}) that the {\em Hochschild
cohomology} $HH^\hdot(A,M)$ of an associative unital algebra $A$
over a field $k$ with coefficients in an $A$-bimodule $M \in
A\bimod$ is given by
$$
HH^\hdot(A,M) = \Ext^\hdot_{A\bimod}(A,M),
$$
where $A$ in the right-hand side is the diagonal bimodule $A \in
A\bimod$. {\em Hochschild cohomology of an algebra $A$} is its
cohomology with coefficients in the diagonal bimodule, $HH^\hdot(A)
= HH^\hdot(A,A)$.
We note right away that the Hochschild cohomology groups
$HH^\hdot(A)$ are Morita-invariant --- that is, they only depend on
the category $A\amod$ of left $A$-modules. Indeed, all we need to
compute $HH^\hdot(A)$ is the tensor abelian category $A\bimod$ with
its unit object $A \in A\bimod$; as we have seen already in Lecture
6, these only depend on $A\amod$.
When $A$ is commutative and $X = \Spec A$ is smooth, the
Hochschild-Kostant-Rosenberg Theorem (Theorem~\ref{hkr}) provides a
canonical identification
$$
HH^\hdot(A) \cong H^0(X,\Lambda^\hdot\T_X),
$$
where $\T_X$ is the tangent bundle to $X$. Roughly speaking,
Hochschild cohomology is in the same relation to Hochschild homology
as vector fields are to differential forms. We note, however, that
to describe $HH^\hdot(A)$, we need not only the tangent bundle
$\T_X$, but all its exterior powers $\Lambda^\hdot\T_X$, so that
Hochschild cohomology contains not only vector fields, but all the
polyvector fields, too. In the non-commutative setting, there is no
reasonable way to work only with vector fields, we have to treat all
the polyvector fields as a single package.
\medskip
Just as in the case of Hochschild homology, we can compute
Hochschild cohomology $HH^\hdot(A)$ of an algebra $A$ by using the
canonical bar resolution $C_\idot(A)$ of the diagonal bimodule
$A$. This gives the {\em Hochschild cohomology complex} with terms
$$
\Hom(A^{\otimes n},A), \quad n \geq 0,
$$
where $\Hom$ means the space of all $k$-linear maps. Maps $f \in
\Hom(A^{\otimes n},A)$ are called {\em Hochschild cochains}; we can
treat an $n$-cochain as an $n$-linear $A$-valued form on $A$. The
differential $\delta$ in the Hochschild cohomology complex is given
by
\begin{equation}\label{delta.hh}
\delta(f)(a_0,\dots,a_n) = a_0f(a_1,\dots,a_n) - \sum_{0 \leq j <
n}(-1)^jf(a_0,\dots,a_ja_{j+1},\dots,a_n) +
(-1)^{n+1}f(a_0,\dots,a_{n-1})a_n.
\end{equation}
For example, if $f = a \in A$ is a $0$-cochain, then $\delta(f)$ is
given by $\delta(f)(b) = ab - ba$; if $f:A \to A$ is a $1$-cochain,
then we have
$$
\delta(f)(a,b) = af(b)+f(a)b-f(ab).
$$
We conclude that the space $HH^0(A) \subset A$ of Hochschild
$0$-cocycles is the center of the algebra $A$; the space of
Hochschild $1$-cocycles is the space of all {\em derivations} $f:A
\to A$ (that is, maps that satisfy the Leibnitz rule
$f(ab)=af(b)+f(a)b$). The Hochschild cohomology group $HH^1(A)$ is
the space of all derivations $A \to A$ considered modulo the {\em
inner derivations} given by $b \mapsto ab-ba$.
\subsection{Multiplication and the Eckman-Hilton argument.}
By definition, Hochschild cohomology $HH^\hdot(A) = \Ext^\hdot(A,A)$
of an associative unital algebra $A$ is equipped with an additional
structure: an associative multiplication, given by the Yoneda
product on $\Ext$-groups.
However, the abelian category $A\bimod$ is a tensor category, and
the diagonal bimodule $A \in A\bimod$ is its unit object. This
defines a second multiplication operation on $HH^\hdot(A)$: given
two elements $\alpha,\beta \in \Ext^\hdot(A,A)$, we can consider
their tensor product $\alpha \otimes_A \beta \in
\Ext^\hdot(A,A)$.
Both multiplications are obviously associative, and it seems that
this is all we can claim. However, a moment's reflection shows that
more is true.
\begin{lemma}\label{eck}
The two multiplications on Hochschild cohomology $HH^\hdot(A)$ are
the same, and moreover, this canonical multiplication is
(graded)commutative.
\end{lemma}
\proof{} It is easy to see that the two multiplications we have
defined obey the following distribution law:
\begin{equation}\label{distro}
(\alpha_1 \otimes_A \alpha_2) \cdot (\beta_1 \otimes_A \beta_2) =
(-1)^{\deg\alpha_2\deg\beta_1} \alpha_1\beta_1 \otimes_A
\alpha_2\beta_2,
\end{equation}
for any $\alpha_1,\alpha_2,\beta_1,\beta_2 \in HH^\hdot(A)$. This
formally implies the claim:
$$
\alpha\beta = (\alpha \otimes_A 1) \cdot (1 \otimes_A \beta) =
(\alpha \cdot 1) \otimes_A (1 \cdot \beta) = \alpha \otimes_A \beta,
$$
and similar for the commutativity, which we leave to the reader.
\endproof
This observation is known as the {\em Eckman-Hilton argument}: two
associative multiplications which commute according to
\eqref{distro} are commutative and equal. It first appeared in
algebraic topology --- essentially the same argument shows that the
homotopy groups $\pi_i(X)$ of a topological space $X$ are abelian
when $i \geq 2$. Although the Eckman-Hilton argument is very
elementary, it captures an essential feature of the whole story: in
fact, all the results about Hochschild cohomology can be deduced
from an elaboration of this semi-trivial observation. A good
reference for this is a paper by M. Batanin, {\tt
arXiv:math/0207281}. In these lectures, we will not attempt such an
extreme treatment and follow a more conventional path, only refering
to the Eckman-Hilton argument when it simplifies the exposition.
One example of this is an explicit description of the product in
$HH^\hdot(A)$ in terms of Hochschild cochains. Writing down the
Yoneda product in terms of $\Ext$'s computed by an explicit
resolution is usually rather cumbersome, and the resulting formulas
are not nice. However, the tensor product $f \otimes_A g$ of two
Hochschild cochains $f:A^{\otimes n} \to A$, $g:A^{\otimes m} \to A$
is very easy to write down: it is given by
\begin{equation}\label{mult}
(f \otimes_A g)(a_1,\dots,a_{n+m}) =
f(a_1,\dots,a_n)g(a_{n+1},\dots,a_{n+m}).
\end{equation}
By Lemma~\ref{eck}, the Yoneda product is given by the same formula.
\subsection{The Gerstenhaber bracket.}
Recall now that the space of vector fields on a smooth algebraic
variety has an additional structure: the Lie bracket. It turns out
that such a bracket, known as the {\em Gerstenhaber bracket}, also
exists for an arbitrary associative unital algebra $A$. To define
it, we need to introduce a completely different construction of the
Hochschild cohomology complex.
Assume given a $k$-vector space $V$, and consider the free
graded associative coalgebra $T_\idot(V)$ generated by $V$ placed in
degree $1$ --- explicitly, we have
$$
T_nV = V^{\otimes n}, \quad n \geq 0.
$$
Consider the graded Lie algebra $DT^\hdot(V)$ of all {\em coderivations}
of the coalgebra $T_\idot(V)$ --- the notion of a coderivation of a
coalgebra is dual to that of a derivation of an algebra, and we
leave it to the reader to write down a formal definition. Then since
the coalgebra $T_\idot(V)$ is freely generated by $V$, every $\delta
\in DT^\hdot(V)$ is uniquely determined by its composition with the
projection $T_\idot(V) \to V$, so that we have
\begin{equation}\label{dl}
DT^{n+1}(V) \cong \Hom(V^{\otimes n},V), \quad n \geq 0.
\end{equation}
\begin{lemma}\label{ass.del}
Assume that $\cchar k \neq 2$. A coderivation $\mu \in DT^1(V) =
\Hom(V^{\otimes 2},V)$ satisfies $\mu^2=0$ if and only if the
corresponding binary operation $V^{\otimes 2} \to V$ is associative.
\end{lemma}
\proof{} Since $\mu$ is an odd derivation, $\mu^2 =
\frac{1}{2}\{\mu,\mu\}:T_{\idot+2}(V) \to T_\idot(V)$ is also a
derivation; thus it suffices to prove that the map $\mu^2:V^{\otimes
3} \to V$ is equal to $0$ if and only if the map $\mu:V^{\otimes 2}
\to V$ is associative. This is obvious: by the Leibnitz rule, we
have
$$
\mu^2(v_1,v_2,v_3) = \mu(\mu(v_1,v_2),v_3) -
\mu(v_1,\mu(v_2,v_3))
$$
for any $v_1,v_2,v_3 \in V$.
\endproof
Thus if we are given an associative algebra $A$, the product in $A$
defines an element $\mu \in DT^1(A) = \Hom(A^{\otimes 2},A)$ such
that $\{m,m\}=0$. Then setting $\delta(a) = \{\mu,a\}$ for any $a
\in DT^\hdot(A)$ defines a differential $\delta:DT^\hdot(A) \to
DT^{\hdot+1}(A)$ and turns $DT^\hdot(A)$ into a graded Lie
algebra. But as we can see from \eqref{dl}, the space $DT^n(A)$ is
exactly the space of Hochschild $(n+1)$-cochains of the algebra $A$.
\begin{exc}
Check that under the identification \eqref{dl}, the differential
$\delta$ in $DT^\hdot(A)$ becomes equal to the differential in the
Hochschild cohomology complex.
\end{exc}
Thus the Hochschild complex for the algebra $A$ becomes a graded Lie
algebra, with a Lie bracket of degree $-1$, and we get an induced
graded Lie bracket on Hochschild cohomology $HH^\hdot(A)$. This is
known as the {\em Gerstenhaber bracket}. Explicitly, the
Gerstenhaber bracket $\{f,g\}$ of two cochains $f:A^{\otimes n} \to
A$, $g:A^{\otimes m} \to A$ is given by
\begin{equation}\label{ger.bra}
\begin{aligned}
\{f,g\}(a_1,\dots,a_{n+m-1}) &= \sum_{1 \leq i <
n}(-1)^if(a_1,\dots,g(a_i,\dots,a_{i+m-1}),\dots,a_{n+m-1})\\
&\quad - \sum_{1 \leq i <
m}(-1)^ig(a_1,\dots,f(a_i,\dots,a_{i+n-1}),\dots,a_{n+m-1}).
\end{aligned}
\end{equation}
\begin{exc}
Prove this. Hint: use the Leinitz rule.
\end{exc}
We note that if we take $g = \mu$, \eqref{ger.bra} recovers the
formula \eqref{delta.hh} for the differential $\delta$ in the
Hochschild cohomology complex. On the other hand, if both $f$ and
$g$ are $1$-cochains --- that is, $k$-linear maps from $A$ to itself
--- then $\{f,g\}:A \to A$ is their commutator, $\{f,g\} = fg -
gf$. If $f$ and $g$ are also $1$-cocycles, that is, derivations of
the algebra $A$, then so is their commutator $\{f,g\}$: the
Gerstenhaber bracket on $HH^1(A)$ is given by the commutator of
derivations.
\medskip
Thus we have two completely different interpretation of the
Hochschild complex, and two natural structures on it: the
multiplication and the Lie bracket. These days, the corresponding
structure on $HH^\hdot(A)$ is usually axiomatized under the name of
a {\em Gerstenhaber algebra}.
\begin{defn}\label{ger.a.defn}
A {\em Gerstenhaber algebra} is a graded-commutative algebra
$B^\hdot$ equipped with a graded Lie bracket $\{-,-\}$ of degree
$-1$ such that
\begin{equation}\label{poi.ger}
\{a,bc\} = \{a,b\}c + (-1)^{\deg b\deg c}\{a,c\}b
\end{equation}
for any $a,b,c \in B^\hdot$.
\end{defn}
\begin{exc}
Check that the Hochschild cohomology algebra $HH^\hdot(A)$ equipped
with its Gerstenhaber bracket satisfies \eqref{poi.ger}, so that
$HH^\hdot(A)$ is a Gerstenhaber algebra in the sense of
Definition~\ref{ger.a.defn}.
\end{exc}
We note that the definition of a Gerstenhaber algebra is very close
to that of a {\em Poisson algebra} --- the difference is that the
bracket has degree $-1$, and \eqref{poi.ger} acquires a sign. We
will discuss this analogy in more detail at a later time.
\subsection{Hochschild cohomology and deformations.}
By far the most common application of Hochschild cohomology is its
relation to deformations of associative algebras. We will explain
this in the form of the so-called {\em Maurer-Cartan} formalism
popularized by M. Kontsevich.
Assume given an Artin local algebra $S$ with maximal ideal $\m \in
S$ and residue field $k = A/\m$. By an {\em $S$-deformation}
$\wt{A}$ of an associative unital $k$-algebra $A$ we will understand
a flat $S$-algebra $\wt{A}$ equipped with an isomorphism $\wt{A}/\m
\cong A$.
Assume given such a deformation $\wt{A}$, choose a $k$-linear
splitting $A \to \wt{A}$ of the projection $\wt{A} \to \wt{A}/\m
\cong A$, and extend it to an $S$-module map $\wt{A} \cong A
\otimes_k S$ --- since $\wt{A}$ is flat, this map is an isomorphism.
We leave it to the reader to check that Lemma~\ref{ass.del} extends
to flat $S$-modules, with the same statement and proof. Then the
multiplication map $\mu:\wt{A} \otimes_S \wt{A} \to \wt{A}$ can be
rewritten as
\begin{equation}\label{mu.gamma}
\mu = \mu_0 + \gamma \in \Hom(A^{\otimes 2},A) \otimes S,
\end{equation}
where $\mu_0$ is the multiplication map in $A$. If the splitting map
$A \to \wt{A}$ is compatible with the multiplication, then
$\gamma=0$; but in general, it is a non-trivial correction term with
values in $\Hom(A^{\otimes 2},A) \otimes \m \subset \Hom(A^{\otimes
2},A) \otimes S$. All we can say is that, since both $\mu_0$ and
$\mu$ are associative, by Lemma~\ref{ass.del} we have $\{\mu,\mu\} =
0$ and $\{\mu_0,\mu_0\}=0$. This can be rewritten as the {\em
Maurer-Cartan equation}
\begin{equation}\label{mc.eq}
\delta(\gamma) + \frac{1}{2}\{\gamma,\gamma\} = 0,
\end{equation}
where $\delta$ is the Hochschild differential of the algebra
$A$. Conversely, every solution $\gamma$ of the Maurer-Cartan
equation defines by \eqref{mu.gamma} an associative product
structure on the $S$-module $A \otimes_k S$.
This establishes the correspondence between $S$-deformations of the
algebra $A$ and $\m$-valued degree-$1$ solutions of the
Maurer-Cartan equation in the differential graded Lie algebra
$DT^\hdot(A)$. We denote the set of these solutions by
$MC(DT^\hdot(A),\m)$; by definition, it only depends on the
differential graded Lie algebra $DT^\hdot(A)$ and the local Artin
algebra $S$ with its maximal ideal $\m \subset S$.
How canonical is this correspondence? There is one choice: that of
an $S$-module identification $\wt{A} \cong A \otimes S$. The set of
all such identifications is a torsor over the algebraic group
$GL_{S,\m}(A)$ of all $S$-linear invertible maps $A \otimes S \to A
\otimes S$ which are equal to identity modulo $\m$. Assume now that
$\cchar k = 0$. Then we note that since $S$ is local and Artin, this
algebraic group is unipotent, and therefore it is completely
determined by its Lie algebra $\Hom(A,A) \otimes \m \cong DT^0(A)
\otimes \m$. Changing an identification $\wt{A} \cong A \otimes S$
changes the solution $\gamma \in MC(DT^\hdot(A),\m)$, so that we
have an action of the group $GL_{S,\m}(A)$ on
$MC(DT^\hdot(A),\m)$. The corresponding action of its Lie algebra
$DT^0(A) \otimes \m$ is easy to describe: an element $l \in DT^0(A)
\otimes \m$ sends $\mu$ to $\{\mu,l\}$, which in terms of $\gamma$
is given by
$$
\gamma \mapsto \{\mu_0,l\} + \{\gamma,l\} = \delta(l) +
\{\gamma,l\},
$$
where $\delta:DT^0(A) \to DT^1(A)$ is the differential in
$DT^\hdot(A)$.
This is the general pattern of deformation theory in the
Maurer-Cartan formalism. To a deformation problem, one associates a
differential graded Lie algebra $L^\hdot$, which ``controls'' the
problem in the following sense: isomorphism classes of deformations
over a local Artin base $\langle S,\m \rangle$ are in one-to-one
correspondence with solutions of the Maurer-Cartan equation in $L^1
\otimes \m$, considered modulo the natural action of the unipotent
algebraic group corresponding to the nilpotent Lie algebra $L^0
\otimes \m$ (because of this passage from a Lie algebra to a
unipotent group, the formalism only works well in characteristic
$0$). In the case of deformations of an associative algebra $A$, we
have just shown that the controlling differential graded Lie algebra
is the Hochschild cohomology complex $DT^\hdot(A)$.
As an interesting special case, one can consider the so-called
first-order deformations --- that is, one takes $S = k[h]/h^2$, the
algebra of dual numbers. Then $\m = k$ and $\m^2 = 0$, so that the
Lie algebra $L^0 \otimes \m \cong L^0$ is abelian, the corresponding
unipotent group is simply the vector space $L^0 \otimes \m$, and its
action is given by $\gamma \mapsto \gamma + dl$, $l \in L^0$. On the
other hand, the term $\{\gamma,\gamma\}$ in the Maurer-Cartan
equation vanishes. Thus the set of isomorphism classes of
deformations is naturally identified with the degree-$1$ cohomology
classes of the complex $L^\hdot$. We note that this special case
does not require the assumption $\cchar k = 0$ --- indeed,
integrating an {\em abelian} Lie algebra to a unipotent group does
not require exponentiation, so that no denomitators occurs.
In particular, the first-order deformations of an associative
algebra $A$ are classified, up to an isomorphism, by elements in the
second Hochschild cohomology group $HH^2(A)$.
We also note that while we have introduce the Maurer-Cartan
formalism in the case of a local Artin base $S$, it immediately
extends to complete deformations over a complete local Noetherian
base: the only difference is that the Lie algebra $L^\hdot \otimes
\m$ should be replaced with its $\m$-adic completion, and its
degree-$0$ term $L^\hdot \otimes \m$ becomes not nilpotent but
pro-nilpotent.
\subsection{Example: quantizations.}
A useful particular case of the deformation formalism described
above is that of a commutative algebra $A$: assume given a
commutative algebra $A$, and assume that $X = \Spec A$ is a smooth
algebraic variety. Under the Hochschild-Kostant-Rosenberg
isomorphism
$$
HH^\hdot(A) = H^0(X,\Lambda^\hdot\T_X),
$$
the group $HH^1(A)$ corresponds to the space of vector fields on
$X$, and the Gerstenhaber bracket is the usual Lie bracket of vector
fields. The bracket between $HH^1(A) = H^0(X,\T_X)$ and $HH^0(A) =
H^0(X,\calo_X)$ is given by the action of a vector field on the
space of functions. The bracket on $HH^i(A)$, $i \geq 2$ is uniquely
defined by \eqref{poi.ger}; it is known as the {\em Schouten
bracket} of polyvector fields.
Deformations of the algebra $A$ are classified by $HH^2(A) =
H^0(X,\Lambda^2\T_X)$, the space of bivector fields on $X$. Such a
field $\Theta \in H^0(X,\Lambda^2\T_X)$ defines a bracket operation
$\{-,-\}$ on $\calo_X$ by the rule
$$
\{f,g\} = \langle df \wedge dg,\Theta \rangle.
$$
This bracket is obviously a derivation with respect to either of the
arguments: we have $\{f_1f_2,g\} = f_2\{f_1,g\} +
f_1\{f_2,g\}$. Moreover, it satisfies the Jacobi identity if and
only if $[\Theta,\Theta] = 0$ with respect to the Schouten
bracket. In this case, $\Theta$ is called a {\em Poisson bivector},
and $A$ acquires a structure of a {\em Poisson algebra}.
\begin{defn}
A {\em Poisson algebra} is a commutative algebra $A$ equipped with a
Lie bracket $\{-,-\}$ such that $\{f_1f_2,g\} = f_2\{f_1,g\} +
f_1\{f_2,g\}$ for any $f_1,f_2,g \in A$.
\end{defn}
A natural source of Poisson algebra structures on $A$ is given by
its {\em quantizations}.
\begin{defn}
A {\em quantization} $\wt{A}$ of the algebra $A$ is a flat complete
associative unital $k[[h]]$-algebra $\wt{A}$ equipped with an
isomorphism $\wt{A}/h \cong A$.
\end{defn}
For any quantization $\wt{A}$, there obviously exists a unique
bracket $\{-,-\}$ on $A$ such that
\begin{equation}\label{bra}
\wt{f}\wt{g}-\wt{g}\wt{f} = h\{f,g\} \mod h^2
\end{equation}
for any $f,g \in A$ and arbitrary $\wt{f},\wt{g} \in \wt{A}$ such
that $\wt{f} = f \mod h$, $\wt{g} = g \mod h$. It is easy to check
that this bracket defines a Poisson algebra structure on $A$. On the
other hand, $\wt{A}$ can be treated as a $k[[h]]$-deformation of
$A$, so that we have a solution $\gamma \in \Hom(A^{\otimes
2},A)[[h]]$ of the Maurer-Cartan equation. Its leading term $\Theta
\in \Hom(A^{\otimes 2},A)$ is a Hochschild cocycle, thus gives a
bivector on $X$.
\begin{exc}
Check that the bracket on $A$ defined by the bivector $\Theta$ is
equal to the bracket given by \eqref{bra}.
\end{exc}
The equation $[\Theta,\Theta] = 0$ also immediately follows from the
Maurer-Cartain equation.
\subsection{Kontsevich formality: the statement.}
For some time, an important open question was whether the above
construction can be reversed: given a Poisson algebra structure on a
commutative smooth algebra $A$, can we extend it to a quantization
$\wt{A}$? Or, equivalently: given an element $\Theta \in HH^2(A)$
such that $\{\Theta,\Theta\} = 0$, can we extend it to a solution of
the Maurer-Cartan equation in $DT^\hdot(A)[[h]]$? A positive answer
to this was first conjectured and then proved by M. Kontsevich. In
fact, he proved the following stronger fact.
\begin{thm}[Kontsevich Formality Theorem]
Let $A = k[x_1,\dots,x_n]$ be a polynomial algebra over a field $k$
of characteristic $0$. Then the DG Lie algebra $DT^\hdot(A)$ is {\em
formal} --- that is $DT^\hdot(A)$ is quasiisomorphic to its
cohomology $HH^{\hdot+1}(A)$ (the DG Lie algebra formed by the
Hochschild cohomology groups of $A$, with trivial differential).
\end{thm}
Here the precise meaning of ``quasiisomorphic'' is the following:
there exists a chain of DG Lie algebras $L_i^\hdot$ and DG Lie
algebra maps $DT^\hdot(A) \gets L_1^\hdot \to L_2^\hdot \gets \dots
\to HH^{\hdot+1}(A)$ such that all the maps induces isomorphisms on
cohomology of the complexes. Unfortunately, in general there does
not exist a single DG Lie algebra quasiisomorphism $HH^{\hdot+1}(A)
\to DT^\hdot(A)$ (in particular, the canonical
Hochschild-Kostant-Rosenberg map is not compatible with the
bracket). However, this is not important for the deformation theory.
\begin{exc}
Check that for any local Artin $\langle S,\m \rangle$, a DG Lie
algebra quasiisorphism $L_1^\hdot \to L_2^\hdot$ between two DG Lie
algebras $L_1^\hdot$, $L_2^\hdot$ induces a map between the solution
sets $MC(L_1^\hdot,\m)$ and $MC(L_2^\hdot,\m)$ of the Maurer-Cartan
equation which identified the sets of equivalence classes of the
solutions.
\end{exc}
This together with the Formality Theorem implies that quantizations
of the algebra $A$ are in one-two one correspondence with
equivalence classes of the solutions of the Maurer-Cartan equations
in the DG Lie algebra $HH^{\hdot+1}(A)$. However, since the
differential in this algebra is trivial, the Maurer-Cartan equation
simply reads $\{\Theta,\Theta\} = 0$. In particular, any Poisson
bivector on $A$ canonically gives such a solution.
There are two proofs of the Kontsevich Formality Theorem: the
original proof of Kontsevich, which is largely combinatorial, and a
second proof by D. Tamarkin --- this is more conceptual, but it
requires a much more detailed study of the the Hochschild cohomology
complex $DT^\hdot(A)$. Roughly speaking, one proves an even stronger
theorem: $DT^\hdot(A)$ and $HH^{\hdot+1}(A)$ are quasiisomorphic not
only as DG Lie algebras, but as Gerstenhaber algebras. This stronger
statement is actually easier; in fact, Tamarkin shows without much
difficulty that any Gerstenhaber algebra which has cohomology
algebra $HH^\hdot(A)$ must be formal. The real difficulty in the
proof is the following: {\em a priori}, the Hochschild cohomology
complex $DT^\hdot(A)$ {\em is not a Gerstenhaber algebra} ---
indeed, while it does have a Lie bracket and a multiplication, the
multiplication \eqref{mult} is commutative only on the level of
cohomology, not on the nose. What precise structure does exist on
$DT^\hdot(A)$ is a subject of the so-called {\em Deligne
Conjecture}. We will return to this later, after introducing some
appropriate machinery.
\lecture{}
\begin{center}
\begin{minipage}{0.7\linewidth}\small
The language of operads. Poisson and associative
operad. Gerstenhaber operad and small discs. Braided
algebras. Deligne Conjecture.
\end{minipage}
\end{center}
\subsection{The language of operads.}
These days, it has become common practice to use the language of the
so-called {\em operads} to describe various non-trivial algebraic
structures such as that of a Gerstenhabe algebra. It must be
mentioned that the notion of an operad has been introduced 35 years
ago by P. May essentially as a quick hack; it is not very natural,
and in many cases it is not quite what one needs, so that
descriptions using operads tend to be somewhat ugly and somewhat
artificial. But at least, from the formal point of view, everything
is well-defined. We will only sketch most proofs. For a complete
exposition which covers much if not all the material in this
lecture, I refer the reader, for instance, to the paper {\tt
arXiv:0709.1228} by V. Ginzburg and M. Kapranov which is now
considered one of the standard references on the subject (the paper
was published in 1994, and I am grateful to V. Ginzburg who finally
put it on arxiv in 2007). Another reference is the foundational
paper {\tt arXiv:hep-th/9403055} by E. Getzler and J.D.S. Jones, but
this has to be used with care, since some advanced parts of it were
later found to be wrong.
To define an operad, let $\Gamma$ be the category of finite sets,
and let $\tGamma$ be the category of arrows in $\Gamma$ (objects are
morphisms $f:S' \to S$ between $S',S \in \Gamma$, morphisms are
commutative squares). Then $\Gamma$ has a natural embedding into
$\tGamma$: every finite set $S$ has a canonical morphism $p^S:S \to
\ppt$ into the finite set $\ppt \in \Gamma$ with a single
element. We note that every $f \in \tGamma$, $f:S' \to S$
canonically decomposes into a coproduct
\begin{equation}\label{copr}
f = \coprod_{s \in S}f^s,
\end{equation}
where $f^s \in \tGamma$ is the canonical map $p^{f^{-1}(s)}f^{-1}(s)
\to \ppt$ corresponding to the preimage $f^{-1}(s) \subset S'$.
\begin{defn}\label{opera.defn}
An {\em operad} $O_\idot$ of $k$-vector spaces is a rule which
assigns a vector space $O_f$ to any $f \in \tGamma$ together with
the following operations:
\begin{enumerate}
\item for any pair $f:S' \to S$, $g:S'' \to S'$ of composable maps,
a map $\mu_{f,g}:O_f \otimes O_g \to O_{f \circ g}$,
\item for any $f \in \tGamma$, $f:S' \to S$, an isomorphism
$$
O_f \cong \bigotimes_sO_{f^s},
$$
where $f^s = p^{f^{-1}(s)}$ are as in \eqref{copr}.
\end{enumerate}
Moreover, the assignment $f \mapsto O_f$ should be functorial with
respect to isomorphisms in $\tGamma$, the maps in \thetag{i} and
\thetag{ii} should be functorial maps, and for any triple $f,g,h \in
\tGamma$ of composable maps, the square
$$
\begin{CD}
O_f \otimes O_g \otimes O_h @>>> O_{f \circ g} \otimes O_h\\
@VVV @VVV\\
O_f \otimes O_{g \circ h} @>>> O_{f \circ g \circ h}
\end{CD}
$$
should be commutative.
\end{defn}
It is useful to require also that $O_{\id} \cong k$ for an identity
map $\id:S \to S$, and we shall do so. We note that by virtue of
\thetag{ii}, it is sufficient to specify only the vector spaces
$O_{p^S}$ for the canonical maps $p_S:S \to \ppt$ (these are usually
denoted $O_S$, or simply $O_n$, where $n$ is the cardinality of
$S$). However, the way we have formulated the definition makes it
slightly more natural, and slightly easier to generalize.
\begin{defn}\label{alg.defn}
An {\em algebra} $A$ over an operad $O_\idot$ of $k$-vector spaces
is a $k$-vector space $A$, together with an action map
$$
a_f:O_f \otimes A^{\otimes S_1} \to A^{\otimes S_2}
$$
for any $f \in \tGamma$, $f:S_1 \to S_2$, where for any finite set
$S \in \Gamma$, we denote by $A^{\otimes S}$ the tensor product of
copies of $A$ numbered by elements $s \in S$. The maps $a_f$ should
be functorial with respect to isomoprhisms in $\tGamma$ and satisfy
the following rules:
\begin{enumerate}
\item For a pair $f,g \in \tGamma$ of composable maps, we should
have $a_f \circ a_g = a_{f \circ g} \circ \mu_{f,g}$.
\item For any $f \in \tGamma$, $f:S' \to S$, we should have
$$
a_f = \bigotimes_{s \in S} a_{f^s}.
$$
\end{enumerate}
\end{defn}
As in the definition of an operad, \thetag{ii} insures that it is
sufficient to specify the action maps $a_n=a_S=a_{p^S}:O_{p^S}
\otimes A^{\otimes S} \to A$ for all $S \in \Gamma$, but our
formulation is slightly more natural. We also note that algebras
over a fixed operad $O$ form a category, which has a forgetfull
functor into the category of $k$-vector spaces. The left-adjoint
functor associates to a $k$-vector space $V$ the free $O$-algebra
$F_OV$ generated by $V$, which is explicitly given by
\begin{equation}\label{free.alg}
F_OV = \bigoplus_{S \in \Gamma}\left(O_S \otimes V^{\otimes
S}\right)_{\Aut(S)},
\end{equation}
where the sum is over all the isomorphism classes of finite sets ---
in other words, over all integers --- and $\Aut(S)$ is the symmetric
group of all automorphisms of a finite set $S$.
\medskip
The reasoning behind these definitions is the following. We want to
describe algebras of a certain kind --- associative algebras,
commutative algebras, Lie algebras, Poisson algebras, etc. To do so,
one usually says that an algebra is a vector space $A$ equipped with
some multilinear structural maps which satisfy some axioms
(associativy, the Jacobi identity, and so forth). However, this is
not always convenient --- just as describing a concrete algebra by
its generators and relations is usually too cumbersome. An operad
$O$ encodes {\em all the polylinear operations} we want our algebra
to have. More precisely, given some $f:S_1 \to S_2$, we collect in
the vector space $O_f$ all the operations from $A^{\otimes S_1}$ to
$A^{S_2}$ which can be obtained from the structural maps by
composing them and substituting one into the other; and we take the
quotient by all the relations our concrete type of algebraic
structure imposes on these compositions. Moreover, we only want to
consider those algebraic structures which are defined by operations
with values in $A$ itself, not its tensor powers. This is the reason
for the condition \thetag{ii} in Definition~\ref{opera.defn} and
Definition~\ref{alg.defn}.
\subsection{Examples.}
Probably the simplest example of an operad is obtained by taking
$O_f = k$, the $1$-dimensional vector space, for any $f \in
\tGamma$. This operad is denoted by $\Com$. A moment's reflection
shows that algebras over $\Com$ are nothing but {\em commutative
associative unital algebras}. Indeed, by definition, we must have a
unique action map
$$
a_S:A^{\otimes S} \to A
$$
for any $S \in \Gamma$, and moreover, this map should be functorial
with respect to isomorphisms in $\Gamma$ --- in other words,
$a_{p^S}$ must the equivariant with respect to the natural action of
the symmetric group $\Aut(S)$. Thus first, we must have a
commutative multiplication $\mu:A^{\otimes 2} \to A$ corresponding
to the generator of $\Com_2 = k$, and second, any way to compose
this operation to obtain an operation $A^{\otimes n} \to A$ for any
$n$ must give the same result --- which for $n=3$ implies
associativity,
$$
\mu \circ (\mu \otimes \id) = \mu \circ (\id \otimes \mu).
$$
One checks easily that conversely, associativity implies the
uniqueness for any $n \geq 3$. The free $\Com$-algebra $F_{\Com}V$
generated by a vector space $V$ is given by \eqref{free.alg} and
coincides with the symmetric algebra $S^\hdot V$.
\begin{exc}
Check that for a $\Com$-algebra $A$, the action map
$a_0:k = A^{\otimes 0} \to A$ provides a unity in the
commutative associative algebra $A$.
\end{exc}
A slightly more difficult example is the operad $\Ass$ which encodes
the structure of an associative unital algebra: it is usually
described by setting
$$
\Ass_S = k[\Aut(S)],
$$
the regular representation of the symmetric group $\Aut(S)$. To
define the operadic composition, one can, for example, consider the
so-called {\em category $\Sigma$ of non-commutative sets}: objects
are finite sets, morphisms from $S'$ to $S$ are pairs of a map $f:S'
\to S$ of finite sets and a total ordering on every preimage
$f^{-1}(s)$, $s \in S$. The composition is obvious, and we obviously
have the forgetfull functor $\gamma:\Sigma \to \Gamma$ which forgets
the total orders. Then we set
\begin{equation}\label{ass.descr}
\Ass_f = k[\{f' \in \Sigma(S',S)\mid \gamma(f')=f\}]
\end{equation}
for any $f \in \Gamma$, $f:S' \to S$, and the composition in
$\Sigma$ induces the composition maps $\Ass_f \otimes \Ass_g \to
\Ass_{f \circ g}$. The free algebra $F_{\Ass}V$ generated by a
vector space $V$ is the tensor algebra $T^\hdot V$.
Let us assume from now on that the base field $k$ has characteristic
$0$, $\cchar k = 0$. For any vector space $V$, the diagonal map $V
\to V \oplus V$ induces a coproduct $T^\hdot V \to T^\hdot V \otimes
T^\hdot V$ which turns the tensor algebra $T^\hdot V$ into a
cocommutative Hopf algebra. Since $\cchar k = 0$, this means that
$T^\hdot V$ is the universal envelopping algebra of some Lie algebra
$L^\hdot V$. In fact, by the universality property of a universal
envelopping algebra, $L^\hdot V$ is the free Lie algebra generated
by $V$. The universal envelopping algebra $T^\hdot V$ acquires a
Poincar\'e-Birkhoff-Witt increasing filtration $K_\idot T^\hdot V$,
and the associated graded quotient with respect to this filtration
is the symmetric algebra generated by $L^\hdot V$ --- we have a
canonical identification
$$
\gr^F_\idot T^\hdot V \cong S^\hdot L^\hdot V.
$$
This graded quotient is a Poisson algebra, and it is easy to see by
spelling out the universal properties that $P_\idot V = \gr^F_\idot
T^\hdot V$ is actually the free Poisson algebra generated by $V$.
Now, both the PBW filtration and the isomorphism $\gr^F_\idot
T^\hdot V \cong P_\idot V$ are functorial in $V$; this implies that
what we actually have is a decreasing filtration $F^\hdot\Ass$ on
the associative operad $\Ass$, and an identification
$\gr_F^\hdot\Ass \cong \Poi$ between the associated graded quotient
of $\Ass$ and an operad $\Poi$ which encodes the structure of a
Poisson algebra (in particular, the PBW filtration on $\Ass$ is
compatible with the operadic structure). We see that $\Poi$ is in
fact an operad of graded vector spaces. This is also obvious from
the definition: if we assign degree $0$ to multiplication and degree
$1$ to the Poisson bracket, then all the axioms of a Poisson algebra
are compatible with these degrees.
The highest degree term of the PBW filtration on $\Ass$ --- or
equivalently, the highest term in the associated graded quotient
$\gr_F^\hdot\Ass \cong \Poi$ --- is the Lie operad $\Lie$; the
natural maps $\Lie \to \Ass$, $\Lie \to \Poi$ encode the fact that
both a Poisson algebra and an associative algebra are Lie algebras
in a canonical way (in the associative case, the bracket is given by
the commutator, $[a,b] = ab-ba$). We note that it is not trivial to
describe $\Lie$ explicitly. For example, the dimension of $\Lie_n$
is $(n-1)!$. If the base field $k$ is algebraically closed, then
$\Lie_n$ can be described as the representation of the symmetric
group $\Sigma_n$ induced from the non-trivial character of the
cyclic subgroup $\Z/n\Z \subset \Sigma_n$ spanned by the long
cycle. It is a pleasant exersize to check that this representation
is actually defined over $k$ even when $k$ is not algebraically
closed.
\medskip
Finally, the example that interest us most is that of Gerstenhaber
algebras. Since the definition of a Gerstenhaber algebra differs
from that of a Poisson algebra only in the degree assigned to the
bracket, one might expect that Gerstenhaber algebras are controlled
by an operad $\Gerst^\hdot$ essentially isomorphic to
$\Poi^\hdot$. This is true, but there is the following
subtlety. Both $\Poi^\hdot$ and $\Gerst^\hdot$ are operads of graded
$k$-vector spaces, but this can means one of two distinct things:
either we define the product of graded vector spaces simply as their
product with induced grading, or we treat the degree as a
homological degree. The difference is in the symmetry isomorphism
$\sigma:V_\idot \otimes W_\idot \to W_\idot \otimes V_\idot$ of the
tensor product of graded vector spacee $V_\idot$, $W_\idot$: if the
degree is homological, then by convention we introduce the sign and
define $\sigma$ by
$$
\sigma(a \otimes b) = (-1)^{\deg a\deg b}b \otimes a.
$$
Now, $\Gerst^\hdot$ and $\Poi^\hdot$ are both operads of graded
vector spaces, and the difference between them is the following: the
action of the symmetric group $\Aut(S)$ on $\Gerst_S$ is twisted by
the sign representation --- for any $n$, $S$, we have
\begin{equation}\label{g=p}
\Gerst^n_S \cong \Poi^n_S \otimes \eps^{\otimes n},
\end{equation}
where $\eps$ is the one-dimensional sign representation of
$\Aut(S)$. But while $\Poi^\hdot$ is a graded operad in the usual
naive sense, the degree in $\Gerst^\hdot$ is homological, and
because of this, the isomorphisms \eqref{g=p} are still compatible
with the operadic structure.
\subsection{Little cubes operad.}
It turns out, however, that there is a different, more conceptual
construction of the Gerstenhaber operad $\Gerst$.
One immediately notes that in the definition of an operad, one can
use any symmetric monoidal category instead of the category of
$k$-vector spaces. Thus we can speak not only about operads of
vector spaces, or graded vector spaces, but also abouts operads of
sets and operads of topological spaces. And historically, it was the
operads of topological spaces which appeared first --- specifically,
the so-called {\em operad of little $n$-cubes}.
Let $I$ be the unit interval $[0,1]$. Fix a positive integer $n$,
and consider the cube $I^n$ of size $1$ of dimension $n$. For any
finite set $S$, say that an {\em $S$-cube configuration in $I^n$} is
an open subset in $I^n$ whose complement is the union connected
components numbered by elements of $S$, each being a subcube in $I$
of smaller size, whose faces are parallel to faces of $I^n$. Let
$O_S^n$ be the set of all such configurations. A configuration is
completely determined by the centers and the sizes of all the cubes,
so that $O_S^n$ is naturally an open subset in $(I^{(n+1)})^S$. This
turns it into a topological space.
We now note that the collection $O_S^n$ with a fixed $n$ naturally
defines an operad of topological spaces. The composition is given by
the following procedure: take an $S_1$-cube configuration in $I^n$,
rescale it to a smaller size, and plug it into an $S_2$-cube
configuration by filling in one of the connected components of its
complement. When the sizes fit, the result is obviously an $(S_1
\cup S_2 \setminus \{s\})$-cube configuration, where $s \in S_2$ is
the point which we used for the operation. We leave it to the reader
to check that this procedure indeed gives a well-defined operad, and
that all the structure maps of this operad are continuous maps.
\begin{defn}
The operad $O_\idot^n$ is called the {\em operad of little
$n$-cubes}.
\end{defn}
What one is interested in is not the topological spaces $O_S^n$ but
their homotopy types, and these have a simpler
description. Forgetting the size of a cube defines a projection
$O_S^n \to (I^n)^S \setminus \Diag$, the complement to all the
diagonals in the power $(I^n)^S$, and this projection is a homotopy
equivalence --- in other words, $O_S^n$ is homotopy-equivalent to
the {\em configuration space} of injective maps from $S$ to
$I^n$. Equivalently, one can take $\R^n$ instead of the cube
$I^n$. Unfortunately, the structure of the operad is not visible in
this model.
If $n=1$, we can go even further: the configuration space of
injective maps from $S$ to the interval $I$ has $|\Aut(S)|$
connected components, numbered by the induced total order on the set
$S$, and each connected component is a simplex, thus
contractible. We conclude that $O_S^1$ is homotopy-equivalent to the
(discrete finite) set of total orders on $S$.
Now, taking the homology with coefficients in $k$ turns any operad
of topological spaces into an operad of graded $k$-vector
spacers. In particular, for any $n \geq 1$ we have an operad formed
by $H_\idot(O^n_S,k)$.
\begin{exc}
Check that for $n=1$, $H_\idot(O^n_\idot,k)$ is the operad
$\Ass_\idot$. Hint: use the description \eqref{ass.descr}.
\end{exc}
\begin{prop}\label{grst}
Algebras over the homology operad $H_\idot(O^2_S,k)$ of the operad
$O^2_\idot$ of little squares are the same as Gerstenhaber algebras,
and $H_\idot(O^2_S,k)$ is isomorphic to the Gerstenhaber operad
$\Gerst^\hdot$.
\end{prop}
\proof{} This is an essentially well-known but rather non-trivial
fact; for example, it implies that $H_n(O^2_n,k)$ is the $n$-th
space $\Lie_n$ of the Lie operad --- as far as I know, this was
first proved by V. Arnold back in the late 60-es.
Let us first construct a map of operads $a_\idot:\Gerst^\hdot_\idot
\cong H_\idot(O^2_\idot,k)$. The component $\Gerst^\hdot_2$ is
spanned by the product and the bracket, and $O^2_2$ is the
complement to the diagonal in the product $I^2 \times I^2$, which is
homotopy-equivalent to the circle $S^1$. We define $a_2$ by sending
the product in $\Gerst^0_2$ to the class of a point in $H_0(S^1,k)
\cong k$, and the bracket in $\Gerst^1_2$ to the fundamental class
in $H_1(S^1,k) \cong k$.
\begin{exc}
Check that this extends to a map of operads. Hint: since all the
relations in $\Gerst^\hdot$ invlove only three indeterminates, it is
sufficient to consider $O^2_3$.
\end{exc}
Now assume by induction that $a_i$ is an isomorphism for all $i \leq
n$. By definition, $\Gerst^\hdot_{n+1}$ is spanned by all
expressions involving the product and the bracket in $n+1$
indeterminates $x_1,\dots,x_{n+1}$. Substituting the unity instead
of $x_{n+1}$ gives a map $\Gerst^\hdot_{n+1} \to \Gerst_n$; this map
is obviously surjective. Substituting $\{x_{n+1},x_i\}$ instead of
$x_i$ gives a map $\Gerst^\hdot_n \otimes k[S] \to
\Gerst^{\hdot+1}_{n+1}$, where $S$ is the set of indeterminates
$x_1,\dots,x_n$. Since $\{1,x_i\}$ is by definition equal to $0$, we
have a sequence
\begin{equation}\label{es.1}
\begin{CD}
\Gerst^{\hdot-1}_n \otimes k[S] @>>> \Gerst^\hdot_{n+1} @>>>
\Gerst^\hdot_n @>>> 0.
\end{CD}
\end{equation}
which is exact on the right.
On the geometric side, filling in the $(n+1)$-st cube in a cube
configuration --- or equivalently, forgetting the $(n+1)$-st point
in a configuration of points in $\R^2$ --- defines a projection
$O_{n+1}^2 \to O_n^2$, and this is a fibration with fiber $E^n =
\R^2 \setminus S$, where $S \subset \R^2$ is the configuration of
the remaining $n$ distinct points. We have the Leray spectral
sequence
$$
H_\idot(O^2_n,H_\idot(E^2_n,k)) \Rightarrow H_\idot(O^2_{n+1},k).
$$
The homology $H_\idot(E^2_n,k)$ is only non-trivial in degrees $0$
and $1$; the group $H_1(E^2_n,k)$ can be naturally identified with
$k[S]$ by sending $s \in S$ to a small circle around its image in
$\R^2$. The fundamental group of the base $O^2_n$ is the pure braid
group, and it is easy to check that it acts trivially on
$H_\idot(E^2_n,k)$, so that the spectral sequence reads
$$
H_\idot(O^2_n,k) \otimes H_\idot(E^2_n,k) \Rightarrow
H_\idot(O^2_{n+1},k).
$$
Moreover, replacing $\R^2$ with $\C$, we can treat $O^2_S = \C^S
\setminus \Diag$ as a complex algebraic variety whose homology
groups have Hodge structures, and in particular, weights. One
checks easily that $H_n(O^2_S,k)$ is pure Hodge-Tate of weight
$2n$. Therefore the Leray spectral sequence degenerates, so that,
taking in account the isomorphism $H_1(E^2_N,k) \cong k[S]$, we have
a short exact sequence
\begin{equation}\label{es.2}
\begin{CD}
0 @>>> H_{\idot-1}(O^2_n,k) \otimes k[S] @>>> H_\idot(O^2_{n+1},k)
@>>> H_\idot(O^2_n,k) @>>> 0.
\end{CD}
\end{equation}
Now, it is obvious from the construction of the map $a_\idot$ that
it is a map between \eqref{es.1} and \eqref{es.2}, so that we have a
commutative diagram
\begin{equation}\label{es.3}
\begin{CD}
@. \Gerst^{\hdot-1}_n \otimes k[S] @>{f}>> \Gerst^\hdot_{n+1} @>>>
\Gerst^\hdot_n @>>> 0\\
@. @V{a_n}VV @VV{a_{n+1}}V @VV{a_n}V\\
0 @>>> H_{\idot-1}(O^2_n,k) \otimes k[S] @>>> H_\idot(O^2_{n+1},k)
@>>> H_\idot(O^2_n,k) @>>> 0.
\end{CD}
\end{equation}
Moreover, we now that $a_n$ is an isomorphism, which implies in
particular that the map $f$ in \eqref{es.3} is injective. To prove
that $a_{n+1}$ is also an isomorphism, it suffices to prove that the
top row forms a short exact sequence. But we also have the
projection $O^1_{n+1} \to O^1_n$, and it induces a short exact
sequence
$$
\begin{CD}
0 @>>> \Ass_n \otimes k[S] @>>> \Ass_{n+1} @>>> \Ass_n @>>> 0
\end{CD}
$$
which gives \eqref{es.1} under taking the associated graded with
respect to the Poincar\'e-Birkhoff-Witt filtration and using the
isomorphism $\Gerst^\hdot \cong \Poi^\hdot$. Since this sequence is
exact, and its associated graded is exact on the left and on the
right, it must also be exact in the middle term for dimension
reasons.
\endproof
\subsection{Braided algebras and Tamarkin's proof.}
What we did in Proposition~\ref{grst} was to take two different
operads, that of $1$-cubes and that of $2$-cubes, and identify, up
to a sign twist, $H_\idot(O^2_\idot,k)$ with a certain associated
graded quotient of $H_\idot(O^1_\idot,k)$ (which reduces to
$H_0(O^1_\idot,k)$). We now note that $H_\idot(O^2_\idot,k)$ can
also be treated as an associated graded quotient. Namely, given a
topological space $X$, one can consider its singular chain complex
$C_\idot(X,k)$. Every complex $E_\idot$ has a ``canonical
filtration'' $F^\hdot E_\idot$ given by
$$
F^iE_j =
\begin{cases}
0, &\quad j < i,\\
\Ker d, &\quad j=i,\\
E_j, &\quad j > i,
\end{cases}
$$
where $d$ is the differential. The associated graded quotient
$\gr_F^\hdot E_\idot$ is canonically quasiisomorphic to the sum of
homology of the complex $E_\idot$. In particular, we have
$$
\gr_F^\hdot C_\idot(X,k) \cong H_\idot(X,k).
$$
Thus passing to homology is, up to quasiisomorphism, equivalent to
taking the associated graded quotient with respect to the canonical
filtration.
Given an operad $X_\idot$ of topological spaces, we can consider the
DG operad formed by $C_\idot(X_\idot,k)$. The canonical filtration,
being canonical, is automatically compatible with the operadic
structure, and the associated graded quotient $\gr_F^\hdot
C_\idot(X_\idot,k)$ is quasiisomorphic to $H_\idot(X_\idot,k)$.
In particular, we can consider the operad
$C_\idot(O^2_\idot,k)$. Its canonical filtration in fact behaves
similarly to the PBW filtration on $\Ass = H_0(O^1_\idot,k)$,
although to define it, we do not need to use the structure of an
operad. The associated graded quotient $\gr_F^\hdot
C_\idot(O^2_\idot,k)$ is quasiisomorphic to the Gerstenhaber operad
$\Gerst$.
\begin{defn}\label{br.defn}
A {\em braided algebra} is a DG algebra over the DG operad
$C_\idot(O^2_\idot,k)$.
\end{defn}
The term ``braided algebra'' comes from the relation between $O^2_n$
and the pure braid group $B_n$ of $n$ braids: we have $\pi_1(O^2_n)
= B_n$, and one can show that $O^2_n$ has no higher homotopy groups,
so that it is homotopy-equivalent to the classifying space of
$B_n$.
We note that as stated, Definition~\ref{br.defn} is almost useless,
since the singular chain complex $C_\idot(X)$ of a topological space
is huge --- one cannot expect the DG operad $C_\idot(O^2_\idot,k)$
to act on anything reasonable. However, what one can do is to invert
quasiisomorphisms and consider DG algebras over some DG operad
$O_\idot$ ``up to quasiisomorphism'', in the same way as we did for
DG Lie algebras. A convenient formalism for this is provided by the
so-called {\em closed model categories} originally introduced by
Quillen (a modern reference is the book ``Model categories'' by
M. Hovey). This gives a certain well-defined category
$\Ho(O_\idot)$, and, what is important, it only depends on the
defining operad ``up to a quasiisomorphism'' --- a quasiisomorphism
$O'_\idot \to O_\idot$ between DG operads induces an equivalence
$\Ho(O'_\idot) \cong \Ho(O_\idot)$. In practice, one is only
interested in braided algebras up to a quasiisomorphism, that is, in
objects of the category $\Ho(C_\idot(O^2_\idot,k))$; and to
construct such an algebra, it is sufficient to have a DG algebra
over some DG operad quasiisomorphic to $C_\idot(O^2_\idot,k)$. It is
this structure which one has on the Hochschild cohomology complex of
an associative unital algebra $A$.
\begin{thm}[Deligne Conjecture]
For any unital associative $k$-algebra $A$, its Hochschild
cohomology complex is a DG algebra over a DG operad which is
quasiisomorphic to $C_\idot(O^2_\idot,k)$.
\end{thm}
This statement has an interesting history. Originally it was a
question, not even a conjecture, asked in 1993 by P. Deligne. Almost
immediately it was wrongly proved by Getzler and Jones, and
independently, also wrongly, by A. Voronov. But in 1998, Tamarkin
has discovered his amazingly short proof of the Kontsevich Formality
Theorem, which used Deligne conjecture; under close scrutiny, the
mistakes were found, and new complete proofs by several groups of
people were available by 2000 (among those people I should mention
at least Tamarkin, Voronov, J. McClure-J. Smith, and
M. Kontsevich-Y. Soibelman). In almost all the proofs, the authors
actually construct a single DG operad which works for all
associative algebras, but all of them are rather complicated and
unnatural. The real reason for this is that what acts naturally on
Hochschild cohomology is not an operad but a more complicated
object, and this is currently under investigation. However, for
practical purposes such as Formality Theorem, any solution is good,
since it can used as a black box.
Assuming Deligne Conjecture, Tamarkin's proof of the Formality
Theorem is a combination of the following two results.
\begin{thm}[Tamarkin,Kontsevich]\label{grst.form}
The DG operad $C_\idot(O^2_\idot,k)$ itself is formal, that is,
there exists a chain of quasiisomorphisms connecting it to the
Gerstenhaber operad $\Gerst^\hdot = H_\idot(O^2_\idot,k)$.
\end{thm}
\begin{thm}[Tamarkin]\label{tama.thm}
Let $A$ be the polynomial algebra $k[x_1,\dots,x_n]$ in $n$
variables, equipped with the natural action of the group $GL(n,k)$
which interchanges the variables. Any DG algebra over $\Gerst$ which
is equipped with a $GL(n,k)$-action and whose cohomology is
isomorphic to $HH^\hdot(A)$ as a $GL(n,k)$-equivariant Gerstenhaber
algebra is formal.
\end{thm}
It is the second result that was the original discovery of Tamarkin,
and its proof was very simple. But then the problems with Deligne
Conjecture appeared... in the course of fixing them, Kontsevich
suggested that the operad $C_\idot(O^2_\idot,k)$ itself should be
formal, and Tamarkin promptly proved it (but this proof was
combinatorial and not simple at all). Later on, Kontsevich gave a
different proof, also combinatorial. There is also a very simple
argument in folklore which deduces Theorem~\ref{grst.form} from
Hodge Theorey, similarly to the classic formality result of
Deligne-Griffits-Morgan-Sullivan, but this, to the best of my
knowledge, has never been written down. In any case, one thing is
very important: the quasiisomorphisms in Theorem~\ref{grst.form}, no
matter how one produces them, are very non-trivial, and they usually
depend on transcedental things like periods of differential forms or
the so-called ``Drinfeld associator''. In addition, there is no
canonical choice of these quasiisomorphisms --- one expects that the
conjectural ``motivic Galois group'', or even the usual Galois group
$\Gal(\overline{\Q}/\Q)$, acts on the set of these quasiisomorphisms
in a very non-trivial way. On the other hand, the DG operads which
appear in the solutions to Deligne Conjecture are quite canonical,
and their action on Hochschild cohomology is elementary and defined
over $\Q$.
\lecture{}
\begin{center}
\begin{minipage}{0.7\linewidth}\small
Combinatorics of planar trees. Comparison theorem. Brace operad and
its action on the Hochschild cohomology complex.
\end{minipage}
\end{center}
\subsection{Planar trees.}
The topic of today's lecture is Deligne Conjecture --- we want to
construct an operad $O_\idot$ quasiisomorphic to the chain complex
operad $C_\idot(O^2_\idot,k)$ of the operad of little squares so
that $O_\idot$ acts in a natural way on the Hochschild cohomology
complex of an associative algebra $A$.
We start by introducing a certain combinatorial model of the operad
of little squares (or rather, it will be more convenient for us to
work with little discs).
By a {\em planar tree} we will understand an unoriented connected
graph with no cycles and one distiguished vertex of valency $1$
called {\em the root}, equipped with a cyclic order on the set of
edges attached to each vertex. Given such a tree $T$, we will denote
by $V(T)$ the set of all non-root vertices of $T$, and we will
denote by $E(T)$ the set of all edges of $T$ not adjacent to the
root.
More generally, by an {\em $n$-planar tree} we will understand an
unoriented connected graph with no cycles and $n$ distiguished
vertices of valency $1$, called {\em external vertices}, one of
which is additionally distinguished and called the root; again, the
graph should be equipped with a cyclic order on the set of edges
attached to each vertex. We note that this automatically induces a
cyclic order on the set of external vertices, so that $n$-planar
trees are naturally numbered by an object $[n]$ of the cyclic
category $\Lambda$. For an $n$-planar tree $T$, $V(T)$ denotes the
set of all non-external vertices, and $E(T)$ denotes the set of of
edges not adjacent to external vertices.
Given a tree $T$, we denote by $|T|$ its geometric realization, that
is, a CW complex with vertices of $T$ as $0$-cells and edges of $T$
as $1$-cells. We note that for every planar tree $T$, $|T|$ can be
continuously embedded into the unit disc $D$ so that the root of $T$
goes to $1 \in D$, the external vertices, if any, go to points on
the boundary $S^1 \subset D$ and split it into a wheel graph, the
rest of $|T|$ is mapped into the interior of the disc, and for every
vertex $v \in V(T)$, the given cyclic order on the edges adjacent to
$v$ is the clockwise order. Moreover, the set of all such embeddings
with a natural topology is contractible, so that the embedding is
unique up to a homotopy, and the homotopy is also unique up to a
homotopy of higher order, and so on.
Given a tree $T$ and an edge $e \in E(T)$, we may contract $e$ to a
vertex and obtain a new tree $T^e$. The contrations of different
edges obviously commute, so that for any $n$ edges $e_1,\dots,e_n
\in E(T)$, we have a unique tree $T^{e_1,\dots,e_n}$ obtained by
contracting $e_1,\dots,e_n$. By construction, we have a natural map
$V(T) \to V(T^{e_1,\dots,e_n})$ and a natural map of realizations
$|T| \to |T^{e_1,\dots,e_n}|$.
Assume given a finite set $S$. By a {\em tree marked by $S$} we will
understand a planar tree or an $n$-planar tree $T$ together with an
injective map $S \to V(T)$. The vertices in the image of this map
are called marked, the other ones are unmarked. A marked tree $T$ is
{\em stable} if every unmarked vertex $v \in V(T) \setminus S$ has
valency at least $3$. Given a marked tree $T$ which is unstable, we
can canonically produce a stable tree $T'$ by first recursively
removing all unmarked vertices of valency $1$ and edges leading to
them, and them removing unmarked vertices of valency $2$ and gluing
together the corresponding edges. We will call this $T'$ the {\em
stabilization} of $T$.
Given a stable marked tree $T$ and some edges $e_1,\dots,e_n \in
E(T)$, we mark the contraction by composing the map $S \to V(T)$
with the natural map $V(T) \to V(T^{e_1,\dots,e_n})$. If the
resulting map is injective, then this is again a stable marked tree.
\begin{exc}
Check that for any two trees $T$, $T'$ stably marked by the same set
$S$, there exists at most one subset $\{e_1,\dots,e_n\} \subset
E(T)$ such that $T^{e_1,\dots,e_n} \cong T$. Hint: removing an edge
splits a tree $T$ into two connected components; first prove that an
edge is uniquely defined by the corresponding partition of the set
$V(T)$.
\end{exc}
By virtue of this exercize, for every $[n] \in \Lambda$ the
collection of all $n$-planar trees stably marked by the same finite
set $S$ acquires a partial order: we say that $T \geq T'$ if and
only if $T'$ can be obtained from $T$ by contraction. We will denote
this partially ordered set by $\Tt^{[n]}_S$, or simply by $\Tt_S$ if
$[n] = [1]$. This is our combinatorial model for the configuration
space.
\begin{thm}\label{comp.thm}
For any $[n] \in \Lambda$, the classifying space $|\Tt^{[n]}_S|$ of
the partially ordered set $\Tt_S$ is homotopy equivalent to the
configuration space $D^S \setminus \Diag$ of injective maps $S \to
D$ to the unit disc $D$.
\end{thm}
As a first step of proving this, let us construct a map $|\Tt_S| \to
D$.
Let us denote by $\B_S$ the fundamental groupoid of the
configuration space $D^S \setminus \Diag$: objects are points, that
is, injective maps $f:S \to D$, morphisms from $f$ to $f'$ are
homotopy classes of paths, that is, homotopy classes of continuous
maps $S \times I \to D$, whose restriction to $S \times \{0\}$,
resp. $S \times \{1\}$ is equal to $f$, resp. $f'$ (here $I = [0,1]$
is the unit interval, and $S \times I$ is equipped with the product
topology --- it is the disjoint union of $S$ copies of $I$). Since
$D^S \setminus \Diag$ is an Eilenberg-MacLane space of type
$K(\pi,1)$, we have the homotopy equivalence $D^S \setminus \Diag
\cong |\B_S|$.
Now consider the following category $\wt{\Tt}_S$. Objects are stable
marked trees $T$ together with an embedding $f:|T| \to D$. Maps from
$f:|T| \to D$ to $f':|T'| \to D$ exist only if $T \geq T'$, and they
are homotopy classes of continous maps $\gamma:|T| \times I \to D$
such that the restriction $\gamma:|T| \times \{x\} \to D$ is
injective for any $x \in [0,1[$, the restriction $\gamma:|T| \times
\{0\} \to D$ is equal to the map $f$, and the restriction
$\gamma:|T| \times \{1\} \to D$ is the composition of the natural
map $|T| \to |T'|$ and the map $f':|T'| \to D$.
Then on one hand, we have a forgetfull functor $\wt{\Tt}_S \to
\Tt_S$ which forgets the embedding, and since the space of
embeddings is contractible, this is an equivalence of categories.
On the other hand, we have a comparison functor $\wt{\Tt}_S \to
\B_S$ which sends an embedded stable marked tree $|T| \subset D$ to
the subset of its marked points $S \subset |T| \subset D$, and
forgets the rest. Then Theorem~\ref{comp.thm} for $n = 1$ follows
immediately from the following.
\begin{prop}\label{comp}
The comparison functor $\wt{\Tt}_S \to \B_S$ induces a homotopy
equivalence $|\Tt_S| \cong |\wt{\Tt}_S| \cong |\B_S|$.
\end{prop}
\subsection{Stratified spaces and homology equivalences.}
Our strategy of proving Proposition~\ref{comp} is the same as in the
study of the Gerstenhaber operad in the last lecture: we want to
apply induction on the cardinality of $S$ by forgetting points
one-by-one and considering the corresponding projections of the
configuration spaces.
Thus we assume given a finite set $S'$ and an element $v \in S'$,
and we denote $S = S' \setminus \{v\}$. Then forgetting $v$ defines
a projection $\B_{S'} \to \B_S$. On the other hand, unmarking $v$
and applying stabilization defines a projection $\Tt_{S'} \to
\Tt_S$. This is obviously compatible with the comparison functors,
so that we have a commutative diagram
\begin{equation}\label{ind.comp}
\begin{CD}
\Tt_{S'} @>>> \B_{S'}\\
@VVV @VVV\\
\Tt_S @>>> \B_S.
\end{CD}
\end{equation}
\begin{defn}
An {\em abelian fibration} $\Cc$ over a small category $\Gamma$ is a
fibration $\Cc/\Gamma$ such that all fibers $\Cc_{[a]}$, $[a] \in
\Gamma$ are abelian categories, and all the transition functors
$f^*:\Cc_{[b]} \to \Cc_{[a]}$, $f:[a] \to [b]$ are left-exact.
\end{defn}
Just as in Proposition~\ref{abe.sec}, one shows easily that the
category of sections $\Sec(\Cc)$ of an abelian fibration $\Cc/\Gamma$
is an abelian category.
\begin{defn}\label{hom.eq}
A functor $\gamma:\Gamma \to \Gamma'$ between small categories is
said to be a {\em homological equivalence} if
\begin{enumerate}
\item for any abelian fibration $\Cc/\Gamma'$, the pullback functor
$\gamma^*::\D(\Sec(\Cc)) \to \D(\Sec(\gamma^*\Cc))$ is a fully
faithful embedding, and
\item the essential image $\gamma^*(\Sec(\Cc)) \subset
\Sec(\gamma^*\Cc)$ consists of such $E \in \Sec(\gamma^*\Cc)$ that
for any map $f:[a] \to [b]$ in $\Gamma$ with invertible
$\gamma(f)$, the transition map $E_{[a]} \to f^*E_{[b]}$ is
invertible.
\end{enumerate}
\end{defn}
Here $\gamma^*\Cc = \Cc \times_{\Gamma'} \Gamma$ is the pullback of
the abelian fibration $\Cc/\Gamma'$, $\D(-)$ stand for the derived
category, $E|_{[a]}$ is the restriction of $E$ to the fiber
$(\gamma^*\Cc)_{[a]} \cong \Cc_{\gamma([a])}$, and similarly for
$E_{[b]}$. For example, if $\Gamma' = \ppt$, $\gamma:\Gamma \to
\ppt$ is the projection to the point, and $\Cc = k\Vect$, the
conditions of the definition say that $\D_{lc}(\Gamma,k)$ is
equivalent to the derived category $\D(k\Vect)$. As we saw in
Corollary~\ref{contr}, this implies that the geometric realization
$|\Gamma|$ is contractible.
\begin{exc}\label{hom=>hot}
Prove that if $\gamma:\Gamma \to \Gamma'$ is a homological
equivalence, then the induced map $|\gamma|:|\Gamma| \to |\Gamma'|$
is a homotopy equivalence.
\end{exc}
The reason we have put the additional abelian fibration $\Cc$ in
Definition~\ref{hom.eq} is that this way, it becomes recursive: we
have the following.
\begin{lemma}\label{hom.rec}
Assume given cofibrations $\Gamma'_1/\Gamma_1$,
$\Gamma'_2/\Gamma_2$, a functor $\gamma:\Gamma_1 \to \Gamma_2$, and
a Cartesian functor $\gamma':\Gamma_1' \to \Gamma_1
\times_{\Gamma_2} \Gamma_2' \to \Gamma_2'$. Then if $\gamma$ is a
homological equivalence, and $\gamma'$ restricts to a homological
equivalence on all the fibers, then $\gamma'$ itself is a
homological equivalence.
\end{lemma}
\begin{exc}
Prove this. Hint: first show that for any cofibration $\pi:\Gamma'
\to \Gamma$ and any abelian fibration $\Cc/\Gamma'$, there exists an
abelian fibration $\pi_*\Cc$ whose fibers are given by
$$
(\pi_*\Cc)_{[a]} = \Sec(\Cc|_{\Gamma'_{[a]}}), \qquad [a] \in \Gamma,
$$
where $\Cc|_{\Gamma'_{[a]}}$ means the restriction to the fiber
$\Gamma_{[a]}' \subset \Gamma'$, and that $\Sec(\Cc) \cong
\Sec(\pi_*\Cc)$.
\end{exc}
\begin{exc}\label{cocarto}
Assume given diagrams of categories and functors
$$
\begin{CD}
\Gamma_1 @>>> \Gamma_{12}\\
@AAA @AAA\\
\Gamma_0 @>>> \Gamma_2
\end{CD}
\qquad\qquad\qquad
\begin{CD}
\Gamma'_1 @>>> \Gamma'_{12}\\
@AAA @AAA\\
\Gamma'_0 @>>> \Gamma'_2
\end{CD}
$$
which are cocartesian in the sense that for any category $\Cc$, we
have
$$
\Fun(\Gamma_{12},\Cc) \cong \Fun(\Gamma_1,\Cc)
\times_{\Fun(\Gamma_0,\Cc)} \Fun(\Gamma_2,\Cc),
$$
and similarly for $\Gamma'$. Assume given a functor $\gamma=\langle
\gamma_0,\gamma_1,\gamma_2,\gamma_{12}\rangle$ between them. Prove
that if $\gamma_0$, $\gamma_1$ and $\gamma_2$ are homological
equivalences, then so is $\gamma_{12}$.
\end{exc}
Unfortunately, Lemma~\ref{hom.rec} cannot be used to analyze
\eqref{ind.comp} directly, since the projection functor $\Tt_{S'}
\to \Tt_S$ is not a cofibration. To correct this, we have to
``compactify'' the categories $\Tt_S$ by allowing non-injective
markings $S \to V(T)$ --- geometrically, this corresponds to adding
the diagonals $\Diag \subset D^S$ to the configuration space $D^S
\setminus \Diag$.
So, first, for every finite set $S$ we define the category $\tT_S$
whose objects are trees $T$ equipped with a map $f:S \to V(T)$ such
that the induced embedding $\Im(f) \subset V(T)$ is a stable
marking, with maps given by contractions of edges.
Second, we consider the topological space $D^S$ as a space
stratified by the diagonals, and we define the category $\bB_S$ as
the its ``stratified fundamental groupoid'' in the following sense.
\begin{defn}
The {\em stratified fundamental groupoid} of a topological space $X$
stratified by strata $X_i \subset X$ is the category whose objects
are points $x \in X$, and whose maps from $x_1 \in X_1$ to $x_2 \in
X_2$ exist only when $X_2 \subset X_1$, and are given by homotopy
classes of paths $f:I \to X_1$ from $x_1$ to $x_2$ such that $f(I)
\cap X_2 = f(1) = p_2$, and $f(I) \cap X_3 = \emptyset$ for any
proper substratum $X_3 \subset X_2$.
\end{defn}
Explicitly, an object in $\bB_S$ is given by a not necessarily
injective map $f:S \to D$, and maps from $f_0$ to $f_1$ are given by
homotopy classes of maps $\gamma:f_0(S) \times I \to D$ such that
$\gamma:f_0(S) \times \{x\} \to D$ is injective for any $x \in
[0,1[$, $\gamma:f_0 \times \{1\} \to D$ is a map onto $f_1(S)
\subset D$, and the composition $\gamma \circ f_0:S \to f_0(S) \to
f_1(S)$ is equal to $f_1$.
\begin{exc}\label{strat}
Let $\langle X,X_i \subset X\rangle$ be a stratified topological
space, and let $\pi_1(X)$ be its stratified fundamental
groupoid. Prove that the category $\Fun(\pi_1(X)^{opp},k)$ is
equivalent to the category of constructible sheaves of $k$-vector
spaces on $X$ which are locally constant along the open
strata. Hint: consider first the case $X = I$, with a single proper
stratum $X_1 = \{1\} \subset I$.
\end{exc}
We leave it to the reader to check that the comparison functor
\eqref{ind.comp} extends to a functor $\tT_S \to \bB_S$, and we have
a commutative diagram
\begin{equation}\label{ind.comp.1}
\begin{CD}
\tT_{S'} @>>> \bB_{S'}\\
@VVV @VVV\\
\tT_S @>>> \bB_S.
\end{CD}
\end{equation}
\subsection{The comparison theorem.}
We can now prove the comparison theorem between $\tT_S$ and $\bB_S$.
\begin{prop}\label{hom.eq.comp}
The comparison functor $\tT_{S} \to \bB_S$ is a homological
equivalence for any finite set $S$.
\end{prop}
\proof{} One checks easily that the vertical projections in
\eqref{ind.comp.1} are cofibrations; thus by induction, it suffices
to check that the comparison functor induces a homological
equivalence on all the fibers.
Fix a tree $T \in \tT_S$, and consider a tree $T' \in
\left(\tT_{S'}\right)_{T}$. When we remove the mark $v \in S'$ from
$T'$, one of the following four things might happen:
\begin{enumerate}
\item The tree remains stable, with the vertex $v \in V(T') = V(T)$
possibly becoming unmarked.
\item An unmarked vertex of valency $2$ appears; under
stabilization, it is removed, and adjacent edges are glued
together to give an edge $e \in E(T)$.
\item An unmarked vertex of valency $1$ appears; under
stabilization, we remove this vertex and the adjacent edge.
\item An unmarked vertex of valency $1$ appears; under
stabilization, we remove it with its edge, and then an unmarked
vertex of valency $2$ appears, which also has to be removed.
\end{enumerate}
In the case \thetag{i}, $T'$ is completely determined by specifying
$v \in V(T)$, and in the case \thetag{ii}, by specifying $e \in
E(T)$. To describe the combinatorial invariants in \thetag{iii}, it
is convenient to embed the tree $T$ into the disc $D$ and draw a
small disc around each vertex $v \in V(T)$. The boundary of this
disc is a wheel graph $[n] \in \Lambda$ whose vertices correspond to
edges adjacent to $v$. Edges of these graphs are called {\em angles}
of $T$, and the set of all angles of $T$ is denoted by $A(T)$. Then
in the case \thetag{iii}, to determine $T'$ we need to specify the
other vertex $v \in V(T)$ of the removed edge, and the (unique)
angle $a \in A(T)$ which this removed edge intersects. Finally, in
the case \thetag{iv}, $T'$ is determined by the new edge $e \in
E(T)$ containing the removed vertex of valency $2$, and the ``side''
of this edge at which the removed edge was attached. The set of
these sides is denoted by $S(T)$ (it is of course a $2$-fold cover
of the set $E(T)$). We note that every side defines an angle
attached to each of the two vertices of the corresponding edge.
To sum up: the fiber of the projection $\tT_{S'} \to \tT_S$ over a
tree $T \in \tT_S$ is the set
$$
F_T = V(T) \cup E(T) \cup A(T) \cup S(T),
$$
with some partial order.
\begin{exc} Check that $F_T$ has the following order: an edge
$e \in E(T)$ is less than either of its vertices, an angle $a \in
A(T)$ is less than the vertex where it lives, and a side $s \in
S(T)$ is less than the corresponding edge, and less than the two
angles it defines.
\end{exc}
On the other hand, the fiber $F_p$ of the projection $\bB_{S'} \to
\bB_S$ over an object represented by $p:S \to D$ is the stratified
fundamental groupoid of the pair $f(S) \subset D$. To finish the
proof, it suffices to prove the following.
\begin{lemma}
Assume given a possibly unstable marked tree $T$ embedded into the
disc $D$, $|T| \subset D$, and let $p:S \to D$ be the corresponding
embedding of the set of markings $S \subset V(T)$. Then the
comparison functor $F_T \to F_p$ is a homological equivalence.
\end{lemma}
\proof{} Choose a vertex $v \in T$ of valency $1$, let $T'$ be the
tree obtained by removing $v$ and the adjacent edge $e \in E(T)$,
and let $p':S' \to D$ be the embedding of its set of markings $S'
\subset V(T')$. Then we have a cocartesian diagram
$$
\begin{CD}
F_{T'} @>>> F_T\\
@AAA @AAA\\
F_e @>>> F_v,
\end{CD}
$$
where $F_e \subset F_T$ is the subset consisting of $e$ and its two
sides, and $F_v \subset F_T$ is the subset consisting of $v$, all
its adjacent edges, all its angles, and all their sides. On the
other hand, we can shrink $D$ to a small neighborhood of $|T|
\subset D$ and then decompose it into the union of a small disc
$D_v$ centered at $v$ and a neighborhood $D_{T'}$ of $|T'| \subset
D$ so that the intersection $D_v \cap D_{T'}$ is contractible with
no stratification. This gives a cocartesian diagram
$$
\begin{CD}
F_{p'} @>>> F_p\\
@AAA @AAA\\
\ppt @>>> F_v,
\end{CD}
$$
where $F_v$ is the fundamental groupoid of $D_v$ if $v$ is unmarked,
and the stratified fundamental groupoid of $\{v\} \subset D_v$ if
$v$ is marked. By virtue of Exercize~\ref{cocarto}, we can apply
induction. Thus it suffices to prove that $F_e$ is homologically
equivalent to a point, and the comparison functor $F_v \to F_v$ is a
homological equivalence (both if $v$ is marked and if it is not). We
leave it as an exercize. Hint: in the marked case, show first that
the partially ordered set $F_v \setminus \{v\}$ is homologically
equivalent to the fundamental groupoid of a circle $S^1$.
\endproof
\proof[Proof of Proposition~\ref{comp}.] An immediate corollary of
Proposition~\ref{hom.eq.comp}: every abelian fibration $\Cc/\B_S$,
resp. $\Cc/\Tt_S$ can obviously be extended to $\bB_S$, resp. $\tT_S$
by setting $\Cc_{[a]}=0$ for any $[a] \in \bB_S \setminus \B_S$,
resp. $[a] \in \tT_S \setminus \Tt_S$, and this does not change the
category of sections; therefore the comparison functor $\Tt_S \to
\B_S$ is also a homological equivalence, and this implies the claim
by Exercize~\ref{hom=>hot}.
\endproof
To finish the proof of Theorem~\ref{comp.thm}, it remains to
consider $n$-planar trees for $n \geq 2$. Note that for any $[n] \in
\Lambda$ and a fixed embedding $f:[n-1] \to [n]$, we have a natural
projection $\pi^f:\Tt_S^{[n]} \to \Tt_S^{[n-1]}$ obtained by the
removing the external vertex not contained in the image of $f$ and
applying stabilization.
\begin{exc}\label{n1}
Check that $\pi^f$ is a cofibration whose fiber $E_T$ over a tree $T
\in \Tt_S^{[n-1]}$ is the partially ordered set of cells of a
certain cell decomposition of the open interval $]0,1[$, with the
order by given adjacency (the decomposition may depend on
$T$). Deduce that $\pi^f$ is a homological equivalence.
\end{exc}
This Exercize together with Exercize~\ref{hom=>hot} finish the proof
of Theorem~\ref{comp.thm}.
\subsection{Regular partially ordered sets.}
By virtue of Theorem~\ref{comp.thm}, instead of studying the chain
complex $C_\idot(D^S \setminus \Diag)$ directly, we may study
complexes which compute the homology of the partially ordered set
$\T_S$ (considered as a small category). This turns out to be easy,
since the partially ordered set $\T_S$ is well-behaved.
Assume given a partially ordered set $P$. For any $p \in P$, denote
by $\delta_p \in \Fun(P^{opp},k)$ the functor given by
$\delta_p(p)=k$, $\delta_p(p') = 0$ if $p \neq p'$.
\begin{defn}
A finite partially ordered set $P$ is called {\em regular} if for
any $p \in P$, we have
\begin{equation}\label{reg.delta}
H_i(P^{opp},\delta_p) \cong
\begin{cases}
k, &\quad i = n,\\
0, \quad \text{otherwise}
\end{cases}
\end{equation}
where $n$ is some integer $n \geq 0$ called the {\em index} of $p$
and denoted $\ind(p)$.
\end{defn}
\begin{exc}\label{prodoo}
Prove that the product $P_1 \times P_2$ of two regular partially
ordered sets is regular.
\end{exc}
\begin{exc}\label{subso}
Prove that $P$ is regular if and only if for any $p \in P$, so the
set $U_p = \{p' \in P| p' \leq q\}$.
\end{exc}
\begin{prop}
The partially ordered set $\Tt^{[n]}_S$ is regular for any finite
set $S$ and any $[n] \in \Lambda$, and the index of a tree $T \in
\Tt^{[n]}_S$ is equal to $\ind(T) = n - 2 - v(T)$, where $v(T)$ is
the cardinality of $V(T)$.
\end{prop}
\proof{} For any tree $T \in \Tt^{[n]}_S$, the partially ordered set
$U_T$ of Exercize~\ref{subso} is isomorphic to
\begin{equation}\label{prdd}
U_T \cong \prod_{v \in V(T)} \Tt^{[n_v]}_{s(v)},
\end{equation}
where $[n_v]$ is the set of edges adjacent to the vertex $v$ with
its given cyclic order, and $s(v)$ is $\ppt$ if $v$ is marked and
$\emptyset$ if $v$ is unmarked. Thus by Exercize~\ref{prodoo}, it
suffices to consider the cases $S=\ppt$ and $S = \emptyset$. In
either of these cases, we use induction on $n$. The sets
$\Tt^{[1]}_{\emptyset}$ and $\Tt^{[2]}_{\emptyset}$ are empty; the
sets $\Tt_{\ppt}^{[1]}$ and $\Tt^{[3]}_{\emptyset}$ both consist of
one point, thus giving the induction base. For the induction step,
choose an embedding $[n-1] \to [n]$, and consider the corresponding
projection $\Tt_S^{[n]} \to \Tt_S^{[n-1]}$. This is a
cofibration. Its fibers $E_T$ have been described in
Exercize~\ref{n1}, and it is easy to check that they are
regular. Moreover, for any $T \leq T' \in \Tt_S^{[n-1]}$, the
corresponding transition map $E_T \to E_{T'}$ is obviously a
homological equivalence. To finish the proof of the inductive step
and the Proposition, it suffices to apply the following to every
$\delta_T \in \Fun(\Tt^{[n]opp}_S,k)$.
\begin{lemma}
Assume given a fibration $\gamma:\Gamma' \to \Gamma$ of small
categories, and assume that the transistion functor $f^*$ is a
homological equivalence for any map $f:[a] \to [b]$ in
$\Gamma$. Then for any $E \in \Fun(\Gamma',k)$ and any $[a] \in
\Gamma$, there exists an isomorphism
\begin{equation}\label{bc}
(L^\hdot\gamma_!E)([a]) \cong H_\idot(\Gamma'_{[a]},E_{[a]}),
\end{equation}
where $E_{[a]} \in \Fun(\Gamma'_{[a]},k)$ is the restriction of $E$
to the fiber $\Gamma'_{[a]} \subset \Gamma'$.
\end{lemma}
\proof{} Let $i:\ppt \to \Gamma$ be the embedding of the object
$[a]$, and let $i':\Gamma'_{[a]} \to \Gamma'$ be the embedding of
the fiber. Then we have the adjunction map
$$
i_! \circ \gamma_! \circ i^{'*} \cong \gamma_! \circ i'_! \circ
i^{'*} \to \gamma_!,
$$
which by adjunction induces a base change map $\gamma_! \circ i^{'*}
\to i^* \circ \gamma_!$. Taking derived functors, we obtain a map
\eqref{bc} functorially for any $E$. To prove that it is an
isomorphism, it suffices to consider the case of a representable
$E$, $E = k_{[b']}$ for some $[b'] \in \Gamma'$. Then the left-hand
side of \eqref{bc} is canonically isomorphic to
$k[\Gamma([b],[a])]$, where $[b] = \gamma([b']) \in \Gamma$. On the
other hand, since $\gamma$ is a fibration, we have a canonical
identification
$$
k_{[b']}|_{\Gamma'_{[a]}} \cong \bigoplus_{f \in
\Gamma([b],[a])}(f^*)_!k_{[b']}|_{\Gamma'_{[b]}}.
$$
But since $f^*$ is a homological equivalence for any $f \in
\Gamma([b],[a])$, we have
$$
H_\idot(\Gamma_{[a]}',(f^*)_!k_{[b']}|_{\Gamma'_{[b]}}) \cong
H_\idot(\Gamma_{[b]}',k_{[b']}|_{\Gamma'_{[b]}}) \cong k,
$$
which finishes the proof.
\endproof
\begin{exc}
Prove that the homology $H_\idot(P,k) = H_\idot(P^{opp},k)$ of a
finite partially ordered set $P$ can be computed by a complex
$C_\idot(P,k)$ with terms $C_i(P,k) = \bigoplus_{\ind(p)=i}k$. Hint:
take a maximal element $p \in P$, let $P' = P \setminus \{p\}$, and
consider the short exact sequence
$$
\begin{CD}
0 @>>> j^{opp}_!k^{P'} @>>> k^P @>>> \delta_p @>>> 0,
\end{CD}
$$
where $k^P \in \Fun(P^{opp},k)$, $k^{P'} \in \Fun(P^{'opp},k)$ are
the constant functors, and $j:P' \to P$ is the embedding.
\end{exc}
\subsection{The brace operad.}
Now consider the partially ordered set $\Tt_S$. It is regular, so
its cohomology can be computed by a complex $C_\idot(\Tt_S,k)$,
which we denote by $C_\idot(S)$ to simplify notation. Unfortunately,
there is some abguity in the differentials of the complexes
$C_\idot(S)$ (for a discussion, see the Kontsevich-Soibelman paper
arxiv:math/0001151). As it turns out, with the appropriate choice of
the differentials, the complexes $C_\idot(S)$ form a DG operad,
called the {\em brace operad}, and this operad acts naturally on the
Hochschild cohomology complex of any associative algebra $A$.
Namely, assume given an associative unital algebra $A$, and assume
given an $m$-cochain $f \in \Hom(A^{\otimes m},A)$ and $l$ other
cochains $g_j \in \Hom(A^{\otimes n_j},A)$, $1 \leq j \leq l$ of
degrees $n_1,\dots,n_l$. Then the {\em brace} $f\{g_1,\dots,g_l\}$
is the cochain of degree $M=m + n_1 + \dots + n_l - l$ given by
\begin{multline*}
f\{g_1,\dots,g_l\}(a_1,\dots,a_M) = \sum_I (-1)^\eps_I
f(a_1, \dots, a_{i_1-1}, g_1(a_{i_1}, \dots, a_{i_1+n_1-1}),
a_{i_1+n_1}, \dots,\\ a_{M-m-n+1}, g_l(a_{M-m-n_l+1}, \dots,
a_{M-m+i_l}), a_{M-m+i_l+1}, \dots, a_M),
\end{multline*}
where the sum is taken over all the multiindices $1 \leq i_1 < \dots
< i_l \leq m$, and $\eps_I$ is given by
$$
\eps_I = \sum_{1 \leq j \leq l}n_j(i_j-1).
$$
If $m < l$, then the set of multiindices is empty, and the brace is
set to be $0$.
In other words, the brace is obtained by substituting
$g_1,\dots,g_l$ into $f$ in all possible ways, and taking the
alternating sum.
Then every tree $T \in \Tt_S$ defines an $S$-linear operation
$\alpha_T$ on the Hochschild cohomology complex of $A$ by the
following inductive rule.
\begin{enumerate}
\item If $T$ is the tree with exactly one vertex of valency $\geq
1$, and this vertex is marked, then
$$
\alpha_T(f_1,\dots,f_n) = f_1(f_2,\dots,f_n),
$$
where $f_1,\dots,f_n$ are cochains numbered by elements in $S =
V(T)$, and $f_1$ corresponds to the marked vertex.
\item If in the situation above the vertex is unmarked,
$$
\alpha_T(f_1,\dots,f_n) = f_1 \cdot f_2 \cdot\dots\cdot f_n.
$$
\item In the general case, split $T$ into two trees $T_1$, $T_2$ by
cutting an edge $e \in E(T)$, marking one of the resulting new
vertices, and declaring the other one the new root vertex, and let
$$
\alpha_T(f_1,\dots,f_n) =
\alpha_{T_1}(\alpha_{T_2}(f_1,\dots,f_l),f_{l+1},\dots,f_n),
$$
where $T_1$ is the subtree which contains the original root, and
$\alpha_{T_2}$ corresponds to the new marked vertex of $T_1$.
\end{enumerate}
It is not too difficult to check that the brace operation is
associative in the appropriate sense, so that the operation in
\thetag{iii} does not depend on the choice of the edge $e \in
E(T)$. To make \thetag{ii} similar to \thetag{i}, we note that since
$A$ is associative, we have a preferred cochain $\mu \in
\Hom(A^{\otimes n},A)$ for any $n \geq 0$ given by the product, and
$$
\mu\{f_1,\dots,f_n\} = f_1 \cdot f_2 \cdot\dots\cdot f_n.
$$
Moreover, it is clear that the collection of the operations
$\alpha_T$ is closed under substitution --- more precisely,
$\alpha_T$ span a suboperad in the endomorphism operad of the
Hochschild cohomology complex of the algebra $A$. This defines an
operad structure on the graded vector spaces $C_\idot(S) =
k[\Tt_S]$.
\begin{thm}\label{deligne}
With the appropriate choice of the differentials in the complexes
$C_\idot(S)$, the operad structure on $C_\idot(S)$ defined by the
brace operation and the action of this operad on the Hochschild
cohomology complexes is compatible with the differential, so that we
have a DG operad, and for any associative unital algebra $A$, its
Hochschild cohomology complex is a DG algebra over the DG operad
$C_\idot(S)$.
\end{thm}
I do not give the exact differentials, since I will not prove this
result anyway (see the quoted paper of Kontsevich-Soibelman, and
also arxiv:math/9910126 of McClure and Smith, where a closely
related result is proved). Rather, to finish the lecture, I want to
discuss what the result means, and what would a conceptually clear
proof look like.
\subsection{Discussion.}
First of all, we note that Theorem~\ref{deligne} {\em does not prove
the Deligne Conjecture}.
\bigskip
Indeed, while we have constructed quasiisomorphisms between the
chain complexes of configuration spaces $D^S \setminus \Diag$ and
the complexes $C_\idot(S)$ which act on Hochschild cohomology, we
did not prove that they are compatible with the operadic
structure. A natural way to do this would be to extend
Theorem~\ref{comp.thm} to a comparison theorem between operads; this
would also take care of all the signs. But it is completely
impossible to do this: while the groupoids $\bB_S$ do form an operad
in an approppriate $2$-categorical sense, the partially ordered sets
$\Tt_S$ {\em do not}.
Namely, the operadic structure would allow one to replace a marked
vertex $v$ in a tree $T$ with another tree $T'$. But this is only
possible if $v$ has valency $1$ --- otherwise it is not clear what
to do with the extra edges coming into $v$. We can only replace $v$
with an $n$-planar tree, where $n$ is the valency of $v$.
This is why there is a sum in our definition of the brace operation
--- essentially this is an averaging over all possible ways to take
care of the extra edges; and this becomes possible only after we
pass to the chain complex. What happens is that we consider the
canonical quasiisomorphism $C_\idot(\Tt_S^{[n]},k) \to
C_\idot(\Tt_S,k) = C_\idot(S)$ obtained by projection, and forcibly
invert it.
\medskip
Considering all the $n$-planar trees together does not help much:
they do not an operad either, because they can be sustituted one
into the other only if the valencies match.
\medskip
To me, the best way to prove Deligne Conjecture would be not to
force the pieces into submission, but rather, to formalize the
structure that the partially ordered sets $\Tt_S$ and $\Tt_S^{[n]}$
do possess; this amounts to generalizing the notion of an operad by
replacing the category $\Gamma$ of finite sets with something else
--- for example, an appropriately defined category of trees, with
\eqref{prdd} playing the role of the product decomposition
\eqref{copr}. However, as far as I know, this has not been
done. M. Batanin has realized a similar plan, but a different
replacement for $\Gamma$ --- he introduces a notion of a
``non-$\Sigma$ $2$-operad'' which is encoded by the ``category of
$2$-ordinals''; this category is not directly related to trees, but
rather, gives another model of the configuration spaces of points on
a disc. Recently D. Tamarkin has shown in {\tt arXiv:math/0606553}
how to prove the Deligne Conjecture in this language. The other
existing approaches to Deligne Conjecture (for example in the papers
by Voronov, McClure-Smith, Kontsevich-Soibelman, in fact also in the
original paper by Getzler-Jones) are more indirect. What these
authors do is the following: they construct a different and much
larger DG operad which maps both onto the brace operad and onto the
chain complex operad of small discs, and show that both maps are
quasiisomorphisms. The construction usually involve doing some very
intricate cellular subdivisions of the configuration spaces and a
lot of combinatorics. My feeling is that the ``final solution'' of
the Deligne Conjecture is not yet known.
Finally, some bibliographical notes. I have borrowed the formula for
the brace operation from the paper {\tt arXiv:math/9910126} of
McClure and Smith, together with the signs. The brace operad also
appeared there, or rather, a version of it slightly different from
the one presented here (I note that the authors use ``formulas''
instead of planar trees, but these objects are in fact
identical). Exactly the same complex as above appears in {\tt
arXiv:math/0001151} by Konstevich and Soibelman, and also in other
places in the literature. So does the partially ordered set of
planar trees. But our proof of the comparison theorem seems to be
new. The usual approach is to take a certain cellular subdivision of
the configuration space and quote the general theorem which says
that if the subdivision is nice enough, then the geometric
realization of the partially ordered set of cells in a space is
homotopy equivalent to the space itself. An exact subdivision which
corresponds to trees also appears in Kontsevich-Soibelman, but
without proof. The other references that I know use different
subdivisions which give different partially ordered sets, and then
use combinatorics of varying degrees of difficulty to identify the
result with trees.
\lecture{}
\begin{center}
\begin{minipage}{0.7\linewidth}\small
Deformations of DG algebras and $A_\infty$ algebras. Deformations in
the Poisson and the Gerstenhaber case. Formality and
deformations. Tamarkin's Theorem.
\end{minipage}
\end{center}
\subsection{The language of $A_\infty$-maps.}
In this last lecture, I will try to sketch the proof of
D. Tamarkin's theorem which I have already formulated as
Theorem~\ref{tama.thm}. I start with a discussion of associative DG
algebras.
Assume given an associative unital DG algebra $A^\hdot$ over a field
$k$. To define Hochschild cohomology $HH^\hdot(A^\hdot)$, one can
naively write down the Hochschild cohomology complex, just as in the
case of usual associative algebras, and obtain a bicomplex;
Hochschild cohomology $HH^\hdot(A^\hdot)$ is the cohomology of the
total complex of this bicomplex. We note that since the complex
$\Hom(A^{\hdot \otimes n},A^\hdot)$ for every $n \geq 1$ has terms
both of positive and of negative degrees, there is an ambiguity in
taking the total complex of a bicomplex: one can take either the
sum, or the product of the diagonal terms. For the definition of
Hochschild cohomology, one needs to take the product: the degree-$n$
term of the resulting total complex is given by
$$
\prod_{i \geq 0} \Hom^{n-i}(A^{\hdot\otimes i},A^\hdot),
$$
where $\Hom^n(-)$ stands for the term of degree $n$ in the complex
of $\Hom$'s. More invariantly, one can consider the category of DG
modules over $A^\hdot$, and formally invert quasiisomorphisms. The
result is a triangulated category $\D(A^\hdot\amod)$ known as {\em
derived category of DG-modules over $A^\hdot$}. Analogously, one
defines the triangulated category $\D(A^\hdot\bimod)$ of DG
$A^\hdot$-bimodules. Then we have
$$
HH^\hdot(A^\hdot) =
\Rhom^\hdot_{\D(A^\hdot\bimod)}(A^\hdot,A^\hdot),
$$
where $A^\hdot$ in the right-hands side is the diagonal bimodule.
Recall now that for ordinary associative algebras, Hochschild
cohomology could be also used to describe deformations. What is the
situation with DG algebras? It turns out that a similar theory
exists, but it describes deformations of DG algebras ``up to a
quasiisomorphism'', as in Lecture 8.
To explain how this works, we first describe briefly a convenient
technical tool --- the notion of an {\em $A_\infty$-map}. For a very
good overview of this subject with detailed references, I refer the
reader to a paper {\tt arXiv:math/0510508} by B. Keller.
Assume given an associative DG algebra $A^\hdot$, with or without
unit, and consider the free coalgebra $T_\idot(A^\hdot)$ generated
by $A^\hdot$. Then by Lemma~\ref{ass.del}, $T_\idot(A^\hdot)$ has a
natural structure of a bicomplex, with one differential induced by
the differential in $A^\hdot$, and the other induced by
multiplication. Its total complex is then a DG coalgebra with
counit. For technical reasons, we need to remove the counit, and we
denote the corresponding coalgebra by
$\ttt_\idot(A^\hdot)$. Explicitly,
\begin{equation}\label{a.inf.eq}
\ttt_\idot(A^\hdot) = \bigoplus_{i \geq 1}A^{\hdot\otimes i}[i]
\end{equation}
as a graded vector space.
\begin{exc}
Prove that if $A^\hdot$ is a DG algebra with unit, then the complex
$\ttt_\idot(A^\hdot)$ is acyclic. Hint: show that
$\ttt_\idot(A^\hdot)$ is exactly the acyclic bar complex
$C'_\idot(A^\hdot)$ of Lecture 1 (Lemma~\ref{bar.reso.1}).
\end{exc}
\begin{exc}\label{doub}
Assume that $A^\hdot$ itself is a free associative DG algebra
without unit generated by a complex $V^\hdot$, $A^\hdot =
\ttt^\hdot(V^\hdot) = \bigoplus_{i \geq 1} V^{\hdot\otimes
i}[i]$. Prove that the natural map
$$
V^\hdot[1] \to \ttt^\hdot(V^\hdot)[1] = A^\hdot[1] \to \ttt_\idot(A^\hdot)
$$
is a quasiisomorphism. Hint: using the previous exercize, first show
that the complex $T_\idot(A^\hdot)$ computes
$\Tor_\idot^{A^\hdot}(k,k)$, where $k$ is the trivial left,
resp. right $A^\hdot$-module.
\end{exc}
\begin{defn}
An {\em $A_\infty$-map} between associative DG algebras $A^\hdot_1$,
$A^\hdot_2$ is a DG coalgebra map $\phi:\ttt_\idot(A^\hdot_1) \to
\ttt_\idot(A^\hdot_2)$.
\end{defn}
Since the coalgebra $\ttt_\idot(A^\hdot_2)$ is free, an
$A_\infty$-map $\phi$ is completely defined by the induced map
$\phi:\ttt_\idot(A^\hdot_1) \to A^\hdot_2$, and this can be
decomposed as
$$
\phi = \phi_0 + \phi_1 + \dots + \phi_i + \dots
$$
according to \eqref{a.inf.eq}. Here $\phi_0$ is simply a map of
complexes $\phi_0:A^\hdot_1 \to A^\hdot_2$. If all the components
$\phi_i$, $i \geq 1$ are equal to zero, then $\phi_0:A^\hdot_1 \to
A^\hdot_2$ is just a map which commutes with multiplication --- that
is, a DG algebra map in the usual sense. In general, however,
$\phi_0$ commutes with multiplication only up to a homotopy, and
this homotopy is $\phi_1:A^{\hdot\otimes 2}_1 \to
A^\hdot_2[-1]$. This in turn commutes with multiplication in an
appropriate sense, but only up to a homotopy given by $\phi_2$, and
so on.
\begin{defn}
An $A_\infty$-map $\phi$ is {\em a quasiisomorphism} if so is its
component $\phi_0$.
\end{defn}
Of course, a quasiisomorphism $\phi:A^\hdot_1 \to A^\hdot_2$ between
two DG algebras is also an
$A_\infty$-quasi\-is\-o\-mor\-phism. However, while it is often not
invertible in any sense as a DG algebra map, the resulting
$A_\infty$-map admits an inverse, in the following sense.
\begin{lemma}\label{a.inf.qi}
Assume given an $A_\infty$-quasiisomorphism $\phi$ from a DG algebra
$A_1^\hdot$ to a DG algebra $A_2^\hdot$. Then there exists an
$A_\infty$-quasiisomorphism $\phi^{-1}$ from $A^\hdot_2$ to
$A^\hdot_1$ such that both $\phi \circ \phi^{-1}$ and $\phi^{-1}
\circ \phi$ induce identity maps on cohomology.
\end{lemma}
\proof{} Since $\phi$ is a quasiisomorphism, there exists a map
$\phi^{-1}_0:A^\hdot_2 \to A^\hdot_1$ of the underlying complexes
which induces an iverse map on cohomology. We extend it to an
$A_\infty$-map by induction. Namely, for any DG algebra $A^\hdot$,
denote by $\ttt_{< i}(A^\hdot) \subset \ttt_\idot(A^\hdot)$ the
subcoalgebra consisting of components $A^{\hdot\otimes j}[j]$ with
$j \leq i$, and assume given a DG coalgebra map $\phi^{-1}_{<
i}:\ttt_{< i}(A^\hdot_2) \to \ttt_{< i}(A^\hdot_1)$ which induces a
map on cohomology inverse to that induced by $\phi$. Extend
$\phi_{*>> DT^\hdot(A) @>>> A^\hdot @>>>.
\end{CD}
$$
We note that strictly speaking, we had to consider the reduced
Hochschild cohomology even in the deformation theory of the usual
associative algebras. However, there it made no difference: if
$A^\hdot = A$ is concentrated in degree $0$, we have $HH^i(A) =
\HH^i(A)$ for any $i \geq 2$, and the spaces of the Maurer-Cartan
solutions are also isomorphic. For a DG algebra $A^\hdot$ which has
non-trivial terms in positive degrees, they might be different.
In the interests of full disclosure, let me mention that in some
situations, one can also consider $A_\infty$-maps which have
non-trivial $(-1)$-component; these correspond, roughly speaking, to
functors between categories of DG modules which are not induced by a
map of DG algebra. One can also develop a deformation theory which
is controlled by the full Hochschild cohomology complex
$DT^\hdot(A^\hdot)$; this is ``the deformation theory of the
category of DG modules'', in some appropriate sense. Deformations of
the category of DG modules which do not come from deformation of a
DG algebra do exist, and they are sometimes known as ``deformations
in the gerby direction''. However, this lies outside of the scope of
the present course.
\subsection{Poisson cohomology.}
What we really need to study for Theorem~\ref{tama.thm} is
Gerstenhaber algebras, not associative ones; thus we need to extend
the above formalism to the Gerstenhaber case. For simplicity, we
start with the Poisson case. The reference here is, for instance,
the Appendix to my joint paper with V. Ginzburg {\tt
arXiv:math/0212279}.
Assume given a vector space $V$. The free Poisson coalgebra
$P_\idot(V)$ generated by $V$ is the associated graded quotient of
the free associative coalgebra $T_\idot(V)$ with respect to the
Poincar\'e-Birkhoff-Witt filtration. It turns out that an analog of
Lemma~\ref{ass.del} holds in the Poisson situation, too.
\begin{lemma}\label{poi.del}
Poisson algebra structures on $V$ are in one-to-one correspondence
with coderivations $\delta:P_\idot(V) \to P_{\idot-1}(V)$ of degree
$1$ such that $\{\delta,\delta\}=0$.
\end{lemma}
\proof[Sketch of a proof.] By definition, we have $P_2(V) =
\gr_{PBW} V^{\otimes 2} = S^2(V) \oplus \Lambda^2(V)$, the sum of
the symmetric and the exterior square of the vector space $V$. Thus
a coderivation $\delta$ consists of two components, $\delta_0:S^2(V)
\to V$ and $\delta_1:\Lambda^2(V) \to V$. The component $\delta_0$
defines the multiplication, and $\delta_1$ defines the Poisson
bracket. The commutator $\{\delta,\delta\}$ has three components,
$\{\delta_0,\delta_0\}$, $\{\delta_1,\delta_1\}$ and
$2\{\delta_1,\delta_0\}$; their vanishing means, respectively, that
the multiplication is associative, the bracket satisfies the Jacobi
identity, and that the bracket satisfies the Leibnitz rule with
respect to the multiplication. The proof is a direct computation
which I leave as an exercize (or see the quoted paper {\tt
arXiv:math/0212279}).
\endproof
Thus given a Poisson algebra $A$, we have a canonical differential
on the free Poisson coalgebra $P_\idot(A)$, and we can consider the
DG Lie algebra $DP^\hdot(A)$ of all coderivations of $P_\idot(A)$.
\begin{defn}
{\em Poisson cohomology} $HP^\hdot(A)$ of the Poisson algebra $A$ is
the cohomology of the complex $DP^\hdot(A)$.
\end{defn}
As in Lecture 8, we can also consider the DG Lie algebra
$DT^\hdot(A)$ of coderivations of the tensor coalgebra $T_\idot(A)$,
and this is nothing but the Hochschild cohomology complex of the
algebra $A$. The PBW filtration on $T_\idot(A)$ induces a filtration
on $DT^\hdot(A)$, and we have $\gr^\hdot_{PBW}DT^\hdot(A) \cong
DP^\hdot(A)$. The component $DL^\hdot(A) = \gr^0_{PBW}DT^\hdot(A)$
is particularly important; it depends only on the multiplcation in
$A$, and it coincides with the DG Lie algebra of coderivation of the
free Lie coalgebra $L^\hdot(A)$ generated by $A$. This is known as
the {\em tangent complex} of the commutative algebra $A$, and it
computes the so-called {\em Harrison cohomology} of $A$. We note
that the differential in $DL^\hdot(A)$ is $A$-linear, so that it is
a DG Lie algebra of $A$-modules (in fact, free $A$-modules). As
such, it is quasiisomorphic to the complex
$$
\Rhom^\hdot_A(\Omega_\idot(A),A),
$$
where $\Omega_\idot(A)$ is the {\em cotangent complex} of $A$ first
constructed by L. Illusie. The whole Lie algebra $DP^\hdot(A)$ also
has the structure of an $A$-module, and coincides with the total
complex of the bicomplex
$$
\gr^\hdot_{PBW}DT^\hdot(A) \cong \Lambda^\hdot_ADL^\hdot(A).
$$
One differential in this bicomplex is induced by the differential in
$DL^\hdot(A)$, thus by multiplication in $A$ --- explicitly, the
multiplication gives a class $\mu \in DL^1(A)$, and the
differential is given by $\alpha \mapsto \{\mu,\alpha\}$. The other
differential in the bicomplex comes from the Poisson bracket in $A$
--- the bracket gives a class
\begin{equation}\label{poi.class}
\Theta \in \Hom(\Lambda^2A,A) \subset \Lambda^2_A(DL^0(A)),
\end{equation}
and the differential is given by $\alpha \mapsto \{\Theta,\alpha\}$.
In general, it is very diffucult to compute $DP^\hdot(A)$ and the
Harrison complex $DL^\hdot(A)$. However, the situation becomes much
simpler when the algebra $A$ is smooth --- that is, in the situaton
of the Hochschild-Kostant-Rosenberg Theorem. In this case, the
cotangent complex $\Omega_\idot(A)$ reduces to the module
$\Omega(A)$ of K\"ahler differentials of $A/k$, and this module is
flat. Therefore $DL^\hdot(A)$ has non-trivial cohomology only in
degree $1$, and this cohomology is canonically identified with the
module $\T(A)$ of derivations of the algebra $A$ (that is, vector
fields on $X = \Spec A$). The higher quotients
$\gr^\hdot_{PBW}DT^\hdot(A)$ are then isomorphic to modules of
polyvector fields on $X$, so that the PBW filtration is in fact
split --- $\gr^\hdot_{PBW}DT^\hdot(A)$ is quasiisomorphic to the
same space of polyvector fields $H^0(X,\Lambda^\hdot\T_X)$ as the
full Hochschild cohomology complex $DT^\hdot(A)$. Under this
identification, the class $\Theta$ from \eqref{poi.class}
corresponds to the Poisson bivector $\Theta \in
H^0(X,\Lambda^2\T_X)$. To sum up:
\begin{prop}\label{poi.ho}
Assume given a smooth Poisson algebra $A$ of finite type over a
characteristic-$0$ field $k$. Then the Poisson cohomology complex
$DP^\hdot(A)$ is quasiisomorphic to the complex with terms
$$
H^0(X,\Lambda^\hdot\T_X)
$$
and with differential given by $a \mapsto [\Theta,a]$, where $\Theta
\in H^0(X,\Lambda^2\T_X)$ is the Poisson bivector.\endproof
\end{prop}
I will not prove this Proposition. Let me just mention that it is
rather easy to reduce the statement to the case when $A =
S^\hdot(V)$ is the symmetric algebra generated by a $k$-vector space
$V$ --- in other words, a polynomial algebra --- and then the
crucial fact is the quasiisomorphism $L^\hdot(\overline{S}_\idot(V))
\cong V$, analogous to the quasiisomorphism of Exercize~\ref{doub}
(here $\overline{S}_\idot(-)$ means the free commutative coalgebra
without unit).
One way to establish this quasiisomorphism uses a more careful
analysis of the Hochschild-Kostant-Rosenberg map of Lecture 2, which
shows how it interacts with the symmetric group actions on the terms
$A^{\otimes n}$ of the Hochschild complex; the reader can find such
a proof, for instance, in Loday's book.
Another and slightly more conceptual proof uses the notion of
``Koszul duality of operads'' introduced in Ginzburg-Kapranov {\tt
arXiv:0709.1228}. One of the statements there is that the Lie and
the commutative operad are ``Koszul dual'', and this includes, as a
part of the package, canonical quasiisomorphisms
$L^\hdot(\overline{S}^\idot(V)) \cong V$ and
$\overline{S}^\hdot(L_\idot(V)) \cong V$. The second
quasiisomorphism is semi-obvious, since the left-hand side
$\overline{S}^\hdot(L_\idot(V))$, with the degree-$0$ term
$S^0(L_\idot(V))$ added, is nothing but the standard Chevalley
complex which computes Lie algebra homology
$H_\idot(L^\hdot(V),k)$. Then the first quasiisomorphism, which we
actually need, can be deduced by the general formalism of Koszul
duality. I refer the reader to {\tt arXiv:0709.1228} for further
details.
\medskip
Assuming Proposition~\ref{poi.ho}, we see that for a smooth algebra
$A$ --- in particular, for a polynomial algebra $S^\hdot(V)$ --- the
Poisson cohomology can be computed by the very explicit complex
whose terms are polyvector fields. This complex was first discovered
by J.-L. Brylinski in the early 80es, so that it is sometimes called
the {\em Brylinski complex}. But when the Poisson bivector $\Theta$
is non-degenerate, so that the smooth Poisson variety $X = \Spec A$
is actually symplectic, the Poisson cohomology becomes even simpler.
\begin{exc}
Prove that for any smooth Poisson variety $X$, the map $\Omega^1_X
\to \T_X$ given by contraction with the Poisson bivector $\Theta$
extends to a multiplicative map
$$
\Omega^\hdot_X \to \Lambda^\hdot\T_X
$$
from the de Rham complex of $X$ to the Brylinski complex $\langle
\Lambda^\hdot\T_X,[-,\Theta]\rangle$.
\end{exc}
Applying this in the affine symplectic case $X = \Spec A$, we see
that $\Omega^1_X \to \T_X$ is actually an isomorphism, so that the
Brylinski complex is quasiisomorphic to the de Rham complex, and the
Poisson cohomology $HP^\hdot(A)$ is isomorphic to the de Rham
cohomology $H^\hdot_{DR}(X)$ (in particular, it does not depend on
the symplectic/Poisson structure at all). When $A=S^\hdot(V)$ is the
polynomial algebra generated by a symplectic vector space $V$, with
the Poisson structure induced by the symplectic form on $V$, we have
$$
HP^i(A) = H^i_{DR}(X) =
\begin{cases}
k, &\quad i =0,\\
0, &\quad i \geq 1,
\end{cases}
$$
where $X = \Spec A$ is the affine space.
\medskip
The Gerstenhaber case works in exactly the same way, except that we
now have to care of the gradings, and use reduced cohomology.
\begin{defn}
The {\em Gerstenhaber cohomology complex} $DG^\hdot(A^\hdot)$,
resp. the {\em reduced Gerstenhabe cohomology complex}
$\DG^\hdot(A^\hdot)$ of a Gerstenhaber DG algebra $A^\hdot$ is the
DG Lie algebra of coderivations of the free Gerstenhaber coalgebra
with, resp. without unit generated by $A^\hdot[1]$.
\end{defn}
There is also a version of the $A_\infty$-formalism for DG
Gerstenhaber algebra, and the classification theorem for
deformations of DG Gerstenhaber algebras up to a quasiisomorphism;
this is completely parallel to the associative case and left to the
reader. The end result is that deformations ``up to a
quasiisomorphims'' of a DG Gerstenhaber algebra $A^\hdot$ are
controlled by the DG Lie algebra $\DG^\hdot(A^\hdot)$.
\begin{exc}\label{odd.sympl}
Let $A^\hdot = S^\hdot(V^\hdot)$ be graded polynomial algebra
generated by a graded vector space $V$, with the Gerstenhaber
structure induced by a non-degenerate graded sympletic form
$\Lambda^2(V^\hdot) \to k[-1]$. Show that the reduced Gerstenhaber
cohomology complex $\DG^\hdot(A^\hdot)$ is quasiisomorphic to the
quotient $A^\hdot/k$, where $k \to A^\hdot$ is the unit map $\lambda
\mapsto \lambda \cdot 1$.
\end{exc}
\subsection{Tamarkin's Theorem.}
We can now explain how to prove Tamarkin's Theorem, ar rather, the
following version of it.
\begin{prop}\label{tama.prop}
Let $A^\hdot = S^\hdot(V)$ be the polynomial algebra generated by a
vector space $V$, and assume given a DG Gerstenhaber algebra
$B^\hdot$ whose cohomology is isomorphic to the Hochschild
cohomology Gerstenhaber algebra $HH^\hdot(A)$. Aswsume in addition
that $B^\hdot$ admits an action of the group $GL(V)$ such that the
isomorphism $H^\hdot(B^\hdot) \cong HH^\hdot(A)$ is
$GL(V)$-equivariant. Then the DG Gerstenhaber algebra $B^\hdot$ is
formal, that is, quasiisomorphic to $HH^\hdot(A)$.
\end{prop}
\proof{} Consider the canonical filtration $F_\idot B^\hdot$ on the
Gerstenhaber algebra $B^\hdot$. Then we have a canonical
quasiisomorphism $\gr_\idot^F B^\hdot \cong HH^\hdot(A)$, and this
quasiisomorphism, being canonical, is compatible with the
Gerstenhaber algebra structure and with the $GL(V)$-action. There is
a standard way to interpret the associated graded quotient
$\gr_\idot^FB^\hdot$ as a special fiber of a certain deformation of
the algebra $B^\hdot$ (known as ``the deformation to the normal
cone''). Namely, consider the Rees algebra
$$
\wt{B}^\hdot = \bigoplus_i F_iB^\hdot
$$
defined by the canonical filtration. This is also a Gerstenhaber
algebra which has an additional grading by $i$. Moreover, the
embeddings $F_iB^\hdot \subset F_{i+1}B^\hdot$ give a certain
endomorphism of $\wt{B}^\hdot$ of degree $1$ which we denote by
$h$. Then $\wt{B}^\hdot$ is a graded Gerstenhaber algebra over $S =
k[h]$. Its generic fiber $\wt{B}^\hdot \otimes_S k[h,h^{-1}]$ is
isomorphic to $B^\hdot \otimes k[h,h^{-1}]$, while its special fiber
$\wt{B}^\hdot/h$ is isomorphic to $\gr_F^\hdot B^\hdot$. Thus we
have a $GL(V)$-equivariant $S$-deformation of the Gerstenhaber
algebra $\gr_F^\hdot B^\hdot \cong HH^\hdot(A)$, and we have to show
that this deformation is trivial up to a quasiisomorphism. But by
the Hochschild-Kostant-Rosenberg Theorem, we have $HH^\hdot(A) =
S^\hdot(V \oplus V^*[-1])$, and it is easy to check that the
Gerstenhaber structure is induced by the natural pairing $V \otimes
(V^*[-1]) \to k[-1])$ (it suffices to check this on the generators
$V \oplus V^*[-1]$, and this is a trivial exercize). Applying
Exercize~\ref{odd.sympl}, we conclude that
$$
\DG^\hdot(HH^\hdot(A)) \cong HH^\hdot(A)/k.
$$
In the right-hand side, the $GL(V)$-invariant part is trivial in
degrees $\geq 2$, so that {\em every} $GL(V)$-equivariant
deformation of the DG Gerstenhaber algebra $HH^\hdot(A)$ is trivial
up to a quasiisomorphism.
\endproof
As we have noted already in Lecture 9, this reduces Kontsevich
Formality Theorem to Theorem~\ref{grst.form}, the formality of the
chain operad of little discs (and the Deligne Conjecture). Indeed,
once these both are established, we know that the Hochschild
cohomology complex $DT^\hdot(A^\hdot)$ is a DG Gerstenhaber algebra.
It is obviously $GL(V)$-equivariant, thus formal by
Proposition~\ref{tama.prop}.
\end{document}
*