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Hamiltonian systems, symplectic flows, classical integrable systems

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_ …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar
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 …
Misha Verbitsky's user avatar