Misha Verbitsky ([info]tiphareth) wrote,
@ 2004-05-09 04:35:00
Current mood: accomplished

LCK manifolds with LCK potential
вот вкратце запись того, что мы придумали за эту неделю
совместно с румынским другом.

Локально конформно кэлеровы (LCK) многообразия суть
многообразия, которые накрываются кэлеровыми, с монодромией,
действующей на накрытии умножая кэлерову форму на константу.
Они вайсмановы, если допускают голоморфный поток, действующий
конформно на метрику.

1. Motivation. A deformation of an LCK manifold is not LCK;
a deformation of a Vaisman manifold is not Vaisman (Belgun).
However, a deformation of a Vaisman manifold is LCK, because
the Kaehler potential on a covering deforms to a potential
for LCK metric. This suggests a need for a new class of LCK

\newcommand{\calo}{{\cal O}}
\newcommand{\Z}{{\Bbb Z}}
\newcommand{\C}{{\Bbb C}}

2. Definition. An LCK manifold with a potential is one
which admits a Kaehler covering $\tilde M, \omega$, and a positive
smooth function $\phi:\; \tilde M \arrow \R$ (LCK potential)
which is proper (level sets are compact), is mapped to
$const \phi$ by monodromy and serves as a Kaehler potential:
$\6\bar\6 \phi = \omega$.

3. A small deformation of LCK with a potential is
LCK with a potential. All Vaisman manifolds admit an
LCK potential. Not all LCK manifolds admit an LCK potential.

4. Given an LCK manifold with a potential, consider
the set $\tilde M(a) = \{x\in \tilde M | \phi(x) >a\}$. This set
is holomorphically concave. By Rossi-Andreotti-Siu
(see Marinescu: math.CV/0403044, math.CV/0210485),
$\tilde M(a)$ is an open subset of
a Stein variety $\tilde M_c$, with
(at most) isolated singularities. The ring
of holomorphic functions on the "filled in" part
of $\tilde M(a)$ is identified with the ring
of CR-functions on the level set $\phi^_{1}(a)$.

We should assume, most likely, that $\dim M >2$,
otherwise Rossi-Andreotti-Siu will not work (a careful
reading of Marinescu might help).

5. Therefore, one could extend the embedding
$\tilde M(a)\arrow \tilde M_c$ to
$\tilde M\arrow \tilde M_c$, filling all $\tilde M$
to a Stein variety with al most isolated singularities.

6. Moreover, the covering group $\Gamma$ of $\tilde M$ over $M$
acts on $\tilde M_c$ by holomorphic automorphisms.

7. Now we assume that $\Z \cong \Gamma$ acting on $\tilde M_c$
by non-trivial conformal automorphisms, and let
$\tau$ be its generator, which is a contraction
(that is, maps $\omega$ to $c\omega$ with $c<1$).
Let $f$ be a holomorphic function on $\tilde M_c$.
The variety $\tilde M_c$ is covered by a sequence
of compact subsets $B_r= \tilde M_c\backslash $\tilde M(r)$.
Since $\tau$ maps $B_r$ to $B_{rc}$, the sequence
$\tau^n f$ in uniformly bounded on compact sets.
Therefore, it has a subsequence converging
to a holomorphic function. It is easy
to check that the limit reaches its maximum
on any $B_r$, hence by maximum principle
this limit is constant. On the other hand,
$\sup \tau^n f$ stays the same, hence
the limit is the same for all subsequences
of $\tau^n f$. We obtain that $\tau^n f$
converges to a constant.

8. This gives a homomorphism from $\calo \tilde M_c$
to $\C$. The corresponding ideal sheaf has support
on $\tilde M_c\backslash \tilde M$. Therefore,
$\tilde M_c$ is a one-point compactification
of $\tilde M$. Denote the added point by $z$

9. Let $m$ be the ideal of $z$.
Using a version of Schwarz lemma, we need to show
that $\tau$ acts with eigenvalues $<1$ on $m/m^2$.

10. Therefore, the formal logarithm of $\tau$
converges, and $\tau$ is obtained by integrating
a holomorphic flow.

11. Using an argument similar to alg-geom/9612013,
we show that $\tau$ acts with finite Jordan blocks on
the formal completion $\hat \calo_z$ of the local ring $\calo_z$
of germs of analytic functions in $z\in \tilde M_c$

12. Estimating the Taylor coefficients of
formal solution of ODE, we find that any
formal solution of ODE is in fact analytic. Therefore,
$\tau$ acts with finite Jordan blocks on $\calo_z$.

13. Picking enough holomorphic functions in these
finite-dimensional eigenspaces, we find an embedding
$\tilde M_c\hookrightarrow \C^n$, such that the
monodromy $\Gamma$ acts on $\C^n$ equivariantly.
Then $M$ is embedded to $\C^n\backslash 0 / \Gamma$,
where $\Gamma$ acts linearly with eigenvalues $<1$.

14. Such quotient should be called "Hopf manifold";
to avoid confusion with class 0 Hopf manifolds of Kodaira -
"linear Hopf manifold". As an end result, we obtain that any
LCK with potential is embedded to a linear Hopf manifold.
The converse is also true, of course.

15. This gives a new proof of our "Kodaira theorem"
on immersion of Vaisman manifolds to Hopf manifolds.

16. There is a tradeoff - we used to have an immersion
from Vaisman to Hopf, now we have an embedding from
Vaisman to a linear Hopf, which might be non-Vaisman.

17. I suspect that this target Hopf manifold
is in fact Vaisman, assuming that $M$ is Vaisman.
We should check.

All the best

(Post a new comment)

2004-05-09 12:57 (link)
Вам определенно пора на костер!

(Reply to this)

2004-05-09 16:39 (link)
Миша, Вы лучшее, что у нас есть.

(Reply to this)

2004-05-09 17:27 (link)
What's the most direct list of books/articles I would have to read to be able to understand this?

(Reply to this) (Thread)

2004-05-09 21:14 (link)
Basically - Griffiths/Harris "Algebraic Geometry"
Henry Cartan's "Complex Analysis", Dragomur/Ornea
"Locally conformally kaehler manifolds".

All the best

(Reply to this) (Parent) (Thread)

2004-05-10 22:06 (link)
Запостил бы эту хрень в мат. конфу или mailing list, там где тебя поймут. А проповедовать это широкой публике - все равно что метать свиней перед бисером.

Lick my manifold, блин, потенциально.

(Reply to this) (Parent)

2004-05-10 08:17 (link)
The variety $\tilde M_c$ is covered by a sequence
of compact subsets $B_r= \tilde M_c\backslash $\tilde M(r)$.
Since $\tau$ maps $B_r$ to $B_{rc}$, the sequence
$\tau^n f$ in uniformly bounded on compact sets.
Therefore, it has a subsequence converging
to a holomorphic function.

такими заговорами надо из людей всякую дерриду изгонять

(Reply to this) (Thread)

2004-05-14 06:20 (link)
Миша, какого черта вы рассказываете про свои мат. исследования?
Из тех, кто общается с вами понимает что-нибудь, возможно, только

(Reply to this) (Parent)

(Post a new comment)

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