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Hamiltonian systems, symplectic flows, classical integrable systems
6
votes
Question about Hodge number
For a Kaehler surface, the Hodge numbers are topological invariants.
By Hodge Index Theorem, the signature of the Poincare pairing is equal
to 2h^{2,0} +2 - h^{1,1}, hence h^{2,0} is a topological inv …
0
votes
Examples of symplectic non-Kahler classes.
This is probably the simplest example.
Take the Fubini-Study form $\omega$ on $CP^2$.
Then $-\omega$ is symplectic, but never Kaehler,
because by Yau's theorem $CP^2$ admits a
unique (standard) comp …
4
votes
Accepted
Relation between kahler potential and Hermitian metric
The function $\psi:=\log(fh^{-1})$ satisfies $\partial\bar\partial \psi=0$, because $ \partial\bar\partial\log f=\partial\bar\partial\log h =\omega$.
Such functions are called pluriharmonic. Locally a …
3
votes
Holomorphic version of Darboux's theorem
To put it differently, is it true that X locally looks like a cotangent bundle?
This is false. Indeed, take an elliptic curve $C$ inside an elliptic
K3 surface. If it had a neighbourhood $U$ which is …
2
votes
Isomorphism of cotangent bundles..
The answer is "never"
(except for the trivial case of
a multiplication by a constant). Suppose
that you have coordinates $x_1, ..., x_n$ on M.
Then $x_i dx_i$ is closed, hence
$x_i \phi(dx_i)$ is als …
6
votes
Spaces of symplectic embeddings: Bundle? Smoothness?
It should not be hard to produce a Frechet manifold structure
on the space of symplectic submanifolds. A symplectic submanifold of a symplectic
manifold has a neighbourhood which is symplectomorphic t …
2
votes
Do holomorphic symplectic manifolds admit (high codimension) embeddings in some standard space?
You can ask a similar question for
holomorphic contact manifolds: when
such a manifold can be embedded to
a projectivization of the cotangent bundle.
The answer is known (for projective holomorphicall …
7
votes
Accepted
Is the generic deformation of a symplectic variety affine?
Being "affine" in this case does not make much sense,
because the hyperkaehler deformation is a complex manifold, without
a fixed algebraic structure. Simpson produced an example of a
hyperkaehler …
2
votes
Holonomy group of a non-compact Kaehler manifold
Yes, the holonomy of this manifold is in $SU(n)$. Indeed, the Chern connection on the canonical bundle is flat and its holonomy preserves $\Omega$, because its curvature is $\partial\bar\partial |\Ome …
6
votes
0
answers
191
views
Isotopy classes of $CP^1$ in 4-manifolds
Let $S_1$, $S_2$ be homologous embedded 2-spheres
in a compact smooth 4-manifold. Under which additional
conditions are they smoothly isotopic? I am interested
in the state of the art picture when $S_ …