Mixing
Chern classes of symplectic manifolds
$begingroup$
I have seen that people assign chern classes to the tangent bundle of symplectic manifolds. This confuses me, because to my knowledge chern classes detect differences in the complex structures of vector bundles.
I know that there is a canonical way to assign almost complex structures $J$ to symplectic manifolds $(M,omega)$. However, this mechnism seems to depend on a choice of metric $g$.
(This is because locally there exists a matrix $A$ such that $omega(v,w)=g(Av,w)$ and we can define a complex structure $J=Q^{-1} A$ where $Q^2=-A^2$.)
So why is this well defined?
differential-geometry definition complex-geometry symplectic-geometry
asked Jan 24 at 14:12
klirkklirk
1
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add a comment |
$begingroup$
I have seen that people assign chern classes to the tangent bundle of symplectic manifolds. This confuses me, because to my knowledge chern classes detect differences in the complex structures of vector bundles.
I know that there is a canonical way to assign almost complex structures $J$ to symplectic manifolds $(M,omega)$. However, this mechnism seems to depend on a choice of metric $g$.
(This is because locally there exists a matrix $A$ such that $omega(v,w)=g(Av,w)$ and we can define a complex structure $J=Q^{-1} A$ where $Q^2=-A^2$.)
So why is this well defined?
differential-geometry definition complex-geometry symplectic-geometry
asked Jan 24 at 14:12
klirkklirk
1
$endgroup$
add a comment |
$begingroup$
I have seen that people assign chern classes to the tangent bundle of symplectic manifolds. This confuses me, because to my knowledge chern classes detect differences in the complex structures of vector bundles.
I know that there is a canonical way to assign almost complex structures $J$ to symplectic manifolds $(M,omega)$. However, this mechnism seems to depend on a choice of metric $g$.
(This is because locally there exists a matrix $A$ such that $omega(v,w)=g(Av,w)$ and we can define a complex structure $J=Q^{-1} A$ where $Q^2=-A^2$.)
So why is this well defined?
differential-geometry definition complex-geometry symplectic-geometry
asked Jan 24 at 14:12
klirkklirk
1
$endgroup$
I have seen that people assign chern classes to the tangent bundle of symplectic manifolds. This confuses me, because to my knowledge chern classes detect differences in the complex structures of vector bundles.
I know that there is a canonical way to assign almost complex structures $J$ to symplectic manifolds $(M,omega)$. However, this mechnism seems to depend on a choice of metric $g$.
(This is because locally there exists a matrix $A$ such that $omega(v,w)=g(Av,w)$ and we can define a complex structure $J=Q^{-1} A$ where $Q^2=-A^2$.)
So why is this well defined?
differential-geometry definition complex-geometry symplectic-geometry
differential-geometry definition complex-geometry symplectic-geometry
asked Jan 24 at 14:12
klirkklirk
1
asked Jan 24 at 14:12
klirkklirk
1
asked Jan 24 at 14:12
klirkklirk
1
asked Jan 24 at 14:12
klirkklirk
1
asked Jan 24 at 14:12
klirkklirk
1
1
add a comment |
add a comment |
1 Answer
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The inclusion $U(n) to Sp(2n,mathbb R)$ is a maximal compact subgroup, hence a homotopy equivalence. This means principal $U(n)$-bundles are equivalent to principal $Sp(2n,mathbb R)$-bundles. (Here is a reference, also see nLab page for symplectic group.)
As for your question about the metric - any two compatible almost complex structures lead to isomorphic complex vector bundles, because the space of CACS is path-connected, while the space of topological vector bundles on a manifold is discrete.
In more detail, given two CACS $J_0$ and $J_1$, you can build a path of almost complex structures $J_t$ (cf. Auroux's notes for path-connectedness of space of CACS), which leads to a structure of complex vector bundle on $p_M^*(TM) to M times I$. Then use the fact that any family of (topological) vector bundles over a (paracompact) manifold is constant (cf. Prop 1.7 of Hatcher's K-theory).
answered Jan 24 at 15:03
BenBen
4,283617
$endgroup$
1
$begingroup$
So you are saying that if I take two different metrics and construct the associated almost complex structure, then the resulting bundles are isomorphic, because the almost complex structures are both compatible with $omega$?
$endgroup$
– klirk
Jan 24 at 15:26
$begingroup$
@klirk Sorry - I'm not carefully analyzing your description of how to produce the almost complex structure from the metric. However, if it's really canonical then it will be the same canonical assignment as mentioned in Auroux's note I linked to, which does produce a compatible (see the proposition there) ACS. I'll leave it to you to make sure that's really the construction you are talking about.
$endgroup$
– Ben
Jan 24 at 15:33
$begingroup$
thx for taking your time
$endgroup$
– klirk
Jan 24 at 15:44
$begingroup$
@klirk No problem. You might want to see the previous lecture in Auroux's course (ocw.mit.edu/courses/mathematics/…) for more notes about the polar decomposition. Actually if you follow the notes there, he uses pd to construct $sqrt{-A^2}$ just as you are doing, so I'm pretty sure this is the construction you are referring to.
$endgroup$
– Ben
Jan 24 at 15:52
1
$begingroup$
@klirk Yes that's right.
$endgroup$
– Ben
Jan 24 at 16:02
|
show 1 more comment
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1 Answer
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$begingroup$
The inclusion $U(n) to Sp(2n,mathbb R)$ is a maximal compact subgroup, hence a homotopy equivalence. This means principal $U(n)$-bundles are equivalent to principal $Sp(2n,mathbb R)$-bundles. (Here is a reference, also see nLab page for symplectic group.)
As for your question about the metric - any two compatible almost complex structures lead to isomorphic complex vector bundles, because the space of CACS is path-connected, while the space of topological vector bundles on a manifold is discrete.
In more detail, given two CACS $J_0$ and $J_1$, you can build a path of almost complex structures $J_t$ (cf. Auroux's notes for path-connectedness of space of CACS), which leads to a structure of complex vector bundle on $p_M^*(TM) to M times I$. Then use the fact that any family of (topological) vector bundles over a (paracompact) manifold is constant (cf. Prop 1.7 of Hatcher's K-theory).
answered Jan 24 at 15:03
BenBen
4,283617
$endgroup$
1
$begingroup$
So you are saying that if I take two different metrics and construct the associated almost complex structure, then the resulting bundles are isomorphic, because the almost complex structures are both compatible with $omega$?
$endgroup$
– klirk
Jan 24 at 15:26
$begingroup$
@klirk Sorry - I'm not carefully analyzing your description of how to produce the almost complex structure from the metric. However, if it's really canonical then it will be the same canonical assignment as mentioned in Auroux's note I linked to, which does produce a compatible (see the proposition there) ACS. I'll leave it to you to make sure that's really the construction you are talking about.
$endgroup$
– Ben
Jan 24 at 15:33
$begingroup$
thx for taking your time
$endgroup$
– klirk
Jan 24 at 15:44
$begingroup$
@klirk No problem. You might want to see the previous lecture in Auroux's course (ocw.mit.edu/courses/mathematics/…) for more notes about the polar decomposition. Actually if you follow the notes there, he uses pd to construct $sqrt{-A^2}$ just as you are doing, so I'm pretty sure this is the construction you are referring to.
$endgroup$
– Ben
Jan 24 at 15:52
1
$begingroup$
@klirk Yes that's right.
$endgroup$
– Ben
Jan 24 at 16:02
|
show 1 more comment
$begingroup$
The inclusion $U(n) to Sp(2n,mathbb R)$ is a maximal compact subgroup, hence a homotopy equivalence. This means principal $U(n)$-bundles are equivalent to principal $Sp(2n,mathbb R)$-bundles. (Here is a reference, also see nLab page for symplectic group.)
As for your question about the metric - any two compatible almost complex structures lead to isomorphic complex vector bundles, because the space of CACS is path-connected, while the space of topological vector bundles on a manifold is discrete.
In more detail, given two CACS $J_0$ and $J_1$, you can build a path of almost complex structures $J_t$ (cf. Auroux's notes for path-connectedness of space of CACS), which leads to a structure of complex vector bundle on $p_M^*(TM) to M times I$. Then use the fact that any family of (topological) vector bundles over a (paracompact) manifold is constant (cf. Prop 1.7 of Hatcher's K-theory).
answered Jan 24 at 15:03
BenBen
4,283617
$endgroup$
1
$begingroup$
So you are saying that if I take two different metrics and construct the associated almost complex structure, then the resulting bundles are isomorphic, because the almost complex structures are both compatible with $omega$?
$endgroup$
– klirk
Jan 24 at 15:26
$begingroup$
@klirk Sorry - I'm not carefully analyzing your description of how to produce the almost complex structure from the metric. However, if it's really canonical then it will be the same canonical assignment as mentioned in Auroux's note I linked to, which does produce a compatible (see the proposition there) ACS. I'll leave it to you to make sure that's really the construction you are talking about.
$endgroup$
– Ben
Jan 24 at 15:33
$begingroup$
thx for taking your time
$endgroup$
– klirk
Jan 24 at 15:44
$begingroup$
@klirk No problem. You might want to see the previous lecture in Auroux's course (ocw.mit.edu/courses/mathematics/…) for more notes about the polar decomposition. Actually if you follow the notes there, he uses pd to construct $sqrt{-A^2}$ just as you are doing, so I'm pretty sure this is the construction you are referring to.
$endgroup$
– Ben
Jan 24 at 15:52
1
$begingroup$
@klirk Yes that's right.
$endgroup$
– Ben
Jan 24 at 16:02
|
show 1 more comment
$begingroup$
The inclusion $U(n) to Sp(2n,mathbb R)$ is a maximal compact subgroup, hence a homotopy equivalence. This means principal $U(n)$-bundles are equivalent to principal $Sp(2n,mathbb R)$-bundles. (Here is a reference, also see nLab page for symplectic group.)
As for your question about the metric - any two compatible almost complex structures lead to isomorphic complex vector bundles, because the space of CACS is path-connected, while the space of topological vector bundles on a manifold is discrete.
In more detail, given two CACS $J_0$ and $J_1$, you can build a path of almost complex structures $J_t$ (cf. Auroux's notes for path-connectedness of space of CACS), which leads to a structure of complex vector bundle on $p_M^*(TM) to M times I$. Then use the fact that any family of (topological) vector bundles over a (paracompact) manifold is constant (cf. Prop 1.7 of Hatcher's K-theory).
answered Jan 24 at 15:03
BenBen
4,283617
$endgroup$
The inclusion $U(n) to Sp(2n,mathbb R)$ is a maximal compact subgroup, hence a homotopy equivalence. This means principal $U(n)$-bundles are equivalent to principal $Sp(2n,mathbb R)$-bundles. (Here is a reference, also see nLab page for symplectic group.)
As for your question about the metric - any two compatible almost complex structures lead to isomorphic complex vector bundles, because the space of CACS is path-connected, while the space of topological vector bundles on a manifold is discrete.
In more detail, given two CACS $J_0$ and $J_1$, you can build a path of almost complex structures $J_t$ (cf. Auroux's notes for path-connectedness of space of CACS), which leads to a structure of complex vector bundle on $p_M^*(TM) to M times I$. Then use the fact that any family of (topological) vector bundles over a (paracompact) manifold is constant (cf. Prop 1.7 of Hatcher's K-theory).
answered Jan 24 at 15:03
BenBen
4,283617
edited Jan 24 at 15:34
answered Jan 24 at 15:03
BenBen
4,283617
answered Jan 24 at 15:03
BenBen
4,283617
answered Jan 24 at 15:03
BenBen
4,283617
4,283617
1
$begingroup$
So you are saying that if I take two different metrics and construct the associated almost complex structure, then the resulting bundles are isomorphic, because the almost complex structures are both compatible with $omega$?
$endgroup$
– klirk
Jan 24 at 15:26
$begingroup$
@klirk Sorry - I'm not carefully analyzing your description of how to produce the almost complex structure from the metric. However, if it's really canonical then it will be the same canonical assignment as mentioned in Auroux's note I linked to, which does produce a compatible (see the proposition there) ACS. I'll leave it to you to make sure that's really the construction you are talking about.
$endgroup$
– Ben
Jan 24 at 15:33
$begingroup$
thx for taking your time
$endgroup$
– klirk
Jan 24 at 15:44
$begingroup$
@klirk No problem. You might want to see the previous lecture in Auroux's course (ocw.mit.edu/courses/mathematics/…) for more notes about the polar decomposition. Actually if you follow the notes there, he uses pd to construct $sqrt{-A^2}$ just as you are doing, so I'm pretty sure this is the construction you are referring to.
$endgroup$
– Ben
Jan 24 at 15:52
1
$begingroup$
@klirk Yes that's right.
$endgroup$
– Ben
Jan 24 at 16:02
|
show 1 more comment
1
$begingroup$
So you are saying that if I take two different metrics and construct the associated almost complex structure, then the resulting bundles are isomorphic, because the almost complex structures are both compatible with $omega$?
$endgroup$
– klirk
Jan 24 at 15:26
$begingroup$
@klirk Sorry - I'm not carefully analyzing your description of how to produce the almost complex structure from the metric. However, if it's really canonical then it will be the same canonical assignment as mentioned in Auroux's note I linked to, which does produce a compatible (see the proposition there) ACS. I'll leave it to you to make sure that's really the construction you are talking about.
$endgroup$
– Ben
Jan 24 at 15:33
$begingroup$
thx for taking your time
$endgroup$
– klirk
Jan 24 at 15:44
$begingroup$
@klirk No problem. You might want to see the previous lecture in Auroux's course (ocw.mit.edu/courses/mathematics/…) for more notes about the polar decomposition. Actually if you follow the notes there, he uses pd to construct $sqrt{-A^2}$ just as you are doing, so I'm pretty sure this is the construction you are referring to.
$endgroup$
– Ben
Jan 24 at 15:52
1
$begingroup$
@klirk Yes that's right.
$endgroup$
– Ben
Jan 24 at 16:02
1
1
$begingroup$
So you are saying that if I take two different metrics and construct the associated almost complex structure, then the resulting bundles are isomorphic, because the almost complex structures are both compatible with $omega$?
$endgroup$
– klirk
Jan 24 at 15:26
$begingroup$
So you are saying that if I take two different metrics and construct the associated almost complex structure, then the resulting bundles are isomorphic, because the almost complex structures are both compatible with $omega$?
$endgroup$
– klirk
Jan 24 at 15:26
$begingroup$
@klirk Sorry - I'm not carefully analyzing your description of how to produce the almost complex structure from the metric. However, if it's really canonical then it will be the same canonical assignment as mentioned in Auroux's note I linked to, which does produce a compatible (see the proposition there) ACS. I'll leave it to you to make sure that's really the construction you are talking about.
$endgroup$
– Ben
Jan 24 at 15:33
$begingroup$
@klirk Sorry - I'm not carefully analyzing your description of how to produce the almost complex structure from the metric. However, if it's really canonical then it will be the same canonical assignment as mentioned in Auroux's note I linked to, which does produce a compatible (see the proposition there) ACS. I'll leave it to you to make sure that's really the construction you are talking about.
$endgroup$
– Ben
Jan 24 at 15:33
$begingroup$
thx for taking your time
$endgroup$
– klirk
Jan 24 at 15:44
$begingroup$
thx for taking your time
$endgroup$
– klirk
Jan 24 at 15:44
$begingroup$
@klirk No problem. You might want to see the previous lecture in Auroux's course (ocw.mit.edu/courses/mathematics/…) for more notes about the polar decomposition. Actually if you follow the notes there, he uses pd to construct $sqrt{-A^2}$ just as you are doing, so I'm pretty sure this is the construction you are referring to.
$endgroup$
– Ben
Jan 24 at 15:52
$begingroup$
@klirk No problem. You might want to see the previous lecture in Auroux's course (ocw.mit.edu/courses/mathematics/…) for more notes about the polar decomposition. Actually if you follow the notes there, he uses pd to construct $sqrt{-A^2}$ just as you are doing, so I'm pretty sure this is the construction you are referring to.
$endgroup$
– Ben
Jan 24 at 15:52
1
1
$begingroup$
@klirk Yes that's right.
$endgroup$
– Ben
Jan 24 at 16:02
$begingroup$
@klirk Yes that's right.
$endgroup$
– Ben
Jan 24 at 16:02
|
show 1 more comment
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