When does contractible space of almost complex structures taming a given symplectic form $omega$ contain an...
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Given a symplectic form $omega$ on a compact symplectic manifold $X$, we know there is a contractible homotopy class $mathcal{J}_{omega}$ of almost complex structures that tame $omega$. A subset of these is also compatible with $omega$, in that $omega(cdot, Jcdot cdot)$ defines a Riemannian metric on the manifold. How do know, other than things like odd Betti numbers being even, if $omega$ has an integrable member $J_{omega, int}$ of $mathcal{J}_{omega}$, so that $(X,omega, J_{omega, int}, omega(cdot, J_{omega, int}cdot cdot))$ is a Kaehler manifold?
general-topology differential-geometry symplectic-geometry kahler-manifolds almost-complex
asked Mar 3 '16 at 20:53
Sinister Cutlass
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Given a symplectic form $omega$ on a compact symplectic manifold $X$, we know there is a contractible homotopy class $mathcal{J}_{omega}$ of almost complex structures that tame $omega$. A subset of these is also compatible with $omega$, in that $omega(cdot, Jcdot cdot)$ defines a Riemannian metric on the manifold. How do know, other than things like odd Betti numbers being even, if $omega$ has an integrable member $J_{omega, int}$ of $mathcal{J}_{omega}$, so that $(X,omega, J_{omega, int}, omega(cdot, J_{omega, int}cdot cdot))$ is a Kaehler manifold?
general-topology differential-geometry symplectic-geometry kahler-manifolds almost-complex
asked Mar 3 '16 at 20:53
Sinister Cutlass
1,054510
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Given a symplectic form $omega$ on a compact symplectic manifold $X$, we know there is a contractible homotopy class $mathcal{J}_{omega}$ of almost complex structures that tame $omega$. A subset of these is also compatible with $omega$, in that $omega(cdot, Jcdot cdot)$ defines a Riemannian metric on the manifold. How do know, other than things like odd Betti numbers being even, if $omega$ has an integrable member $J_{omega, int}$ of $mathcal{J}_{omega}$, so that $(X,omega, J_{omega, int}, omega(cdot, J_{omega, int}cdot cdot))$ is a Kaehler manifold?
general-topology differential-geometry symplectic-geometry kahler-manifolds almost-complex
asked Mar 3 '16 at 20:53
Sinister Cutlass
1,054510
Given a symplectic form $omega$ on a compact symplectic manifold $X$, we know there is a contractible homotopy class $mathcal{J}_{omega}$ of almost complex structures that tame $omega$. A subset of these is also compatible with $omega$, in that $omega(cdot, Jcdot cdot)$ defines a Riemannian metric on the manifold. How do know, other than things like odd Betti numbers being even, if $omega$ has an integrable member $J_{omega, int}$ of $mathcal{J}_{omega}$, so that $(X,omega, J_{omega, int}, omega(cdot, J_{omega, int}cdot cdot))$ is a Kaehler manifold?
general-topology differential-geometry symplectic-geometry kahler-manifolds almost-complex
general-topology differential-geometry symplectic-geometry kahler-manifolds almost-complex
asked Mar 3 '16 at 20:53
Sinister Cutlass
1,054510
asked Mar 3 '16 at 20:53
Sinister Cutlass
1,054510
asked Mar 3 '16 at 20:53
Sinister Cutlass
1,054510
asked Mar 3 '16 at 20:53
Sinister Cutlass
1,054510
asked Mar 3 '16 at 20:53
Sinister Cutlass
1,054510
1,054510
add a comment |
add a comment |
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There are LOTS of additional obstructions (for example, not every finitely presented group whose abelianization has even rank is a fundamental group of a Kaehler manifold). Most of the known obstructions are listed, for example, here.
answered Nov 16 at 9:41
guest_1213
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
There are LOTS of additional obstructions (for example, not every finitely presented group whose abelianization has even rank is a fundamental group of a Kaehler manifold). Most of the known obstructions are listed, for example, here.
answered Nov 16 at 9:41
guest_1213
111
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guest_1213 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
1
down vote
There are LOTS of additional obstructions (for example, not every finitely presented group whose abelianization has even rank is a fundamental group of a Kaehler manifold). Most of the known obstructions are listed, for example, here.
answered Nov 16 at 9:41
guest_1213
111
New contributor
guest_1213 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
1
down vote
up vote
1
down vote
There are LOTS of additional obstructions (for example, not every finitely presented group whose abelianization has even rank is a fundamental group of a Kaehler manifold). Most of the known obstructions are listed, for example, here.
answered Nov 16 at 9:41
guest_1213
111
New contributor
guest_1213 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
There are LOTS of additional obstructions (for example, not every finitely presented group whose abelianization has even rank is a fundamental group of a Kaehler manifold). Most of the known obstructions are listed, for example, here.
answered Nov 16 at 9:41
guest_1213
111
New contributor
guest_1213 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 16 at 9:41
guest_1213
111
New contributor
guest_1213 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered Nov 16 at 9:41
guest_1213
111
answered Nov 16 at 9:41
guest_1213
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111
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New contributor
guest_1213 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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