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how to Prove that the set of fixed points of a Hamiltonian action of a torus on a symplectic manifold is a...
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who can explain how to Prove that the set of fixed points of a Hamiltonian action of a torus on a symplectic manifold is a symplectic submanifold?
symplectic-geometry
asked Jan 23 at 18:16
user220607user220607
42
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add a comment |
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who can explain how to Prove that the set of fixed points of a Hamiltonian action of a torus on a symplectic manifold is a symplectic submanifold?
symplectic-geometry
asked Jan 23 at 18:16
user220607user220607
42
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What have you tried? Questions without context or visible effort tend to get downvoted and closed.
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– postmortes
Jan 23 at 18:29
1
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I think this is true in the case where a compact Lie group $G$ acts by symplectomorphisms on a symplectic manifold $M$. In this case, a connected component of the fixed point subspace $M^G$ is a symplectic submanifold of $M$
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– C. Zhihao
Jan 23 at 19:03
add a comment |
$begingroup$
who can explain how to Prove that the set of fixed points of a Hamiltonian action of a torus on a symplectic manifold is a symplectic submanifold?
symplectic-geometry
asked Jan 23 at 18:16
user220607user220607
42
$endgroup$
who can explain how to Prove that the set of fixed points of a Hamiltonian action of a torus on a symplectic manifold is a symplectic submanifold?
symplectic-geometry
symplectic-geometry
asked Jan 23 at 18:16
user220607user220607
42
asked Jan 23 at 18:16
user220607user220607
42
asked Jan 23 at 18:16
user220607user220607
42
asked Jan 23 at 18:16
user220607user220607
42
asked Jan 23 at 18:16
user220607user220607
42
42
$begingroup$
What have you tried? Questions without context or visible effort tend to get downvoted and closed.
$endgroup$
– postmortes
Jan 23 at 18:29
1
$begingroup$
I think this is true in the case where a compact Lie group $G$ acts by symplectomorphisms on a symplectic manifold $M$. In this case, a connected component of the fixed point subspace $M^G$ is a symplectic submanifold of $M$
$endgroup$
– C. Zhihao
Jan 23 at 19:03
add a comment |
$begingroup$
What have you tried? Questions without context or visible effort tend to get downvoted and closed.
$endgroup$
– postmortes
Jan 23 at 18:29
1
$begingroup$
I think this is true in the case where a compact Lie group $G$ acts by symplectomorphisms on a symplectic manifold $M$. In this case, a connected component of the fixed point subspace $M^G$ is a symplectic submanifold of $M$
$endgroup$
– C. Zhihao
Jan 23 at 19:03
$begingroup$
What have you tried? Questions without context or visible effort tend to get downvoted and closed.
$endgroup$
– postmortes
Jan 23 at 18:29
$begingroup$
What have you tried? Questions without context or visible effort tend to get downvoted and closed.
$endgroup$
– postmortes
Jan 23 at 18:29
1
1
$begingroup$
I think this is true in the case where a compact Lie group $G$ acts by symplectomorphisms on a symplectic manifold $M$. In this case, a connected component of the fixed point subspace $M^G$ is a symplectic submanifold of $M$
$endgroup$
– C. Zhihao
Jan 23 at 19:03
$begingroup$
I think this is true in the case where a compact Lie group $G$ acts by symplectomorphisms on a symplectic manifold $M$. In this case, a connected component of the fixed point subspace $M^G$ is a symplectic submanifold of $M$
$endgroup$
– C. Zhihao
Jan 23 at 19:03
add a comment |
1 Answer
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I am aware of one way to see this, which I saw in McDuff and Salamons book introduction to symplectic geometry.
We can find a compatible almost complex structure which is invariant by the torus action (This basically follows from Gromov's result that a symplectic manifold always has a compatible almost complex structure plus an averaging argument). Note that together with the symplectic structure we also get an invariant Riemannian metric.
Now consider a point $p$ which is fixed by the action, note that $T_{p}M$ inherits the structure of a complex representation of the torus $mathbb{T}^n$. The points close to $p$ which are fixed by the action are the image under the exponential map at $p$ of the fixed point set of this representation say $V = Fix(mathbb{T}^n ) subset T_p M$. Note that $$ V = cap_{g in mathbb{T}^n} Fix(g)$$ where $Fix(g)$ is the fixed point set of the linear transformation associated to $g$.
Now note that since the representation is complex linear, each $Fix(g)$ is a complex subspace (since it is the eigenspace for the eigenvalue 1), and hence $V$ is a complex subspace. This shows that that the fixed point set is locally an almost complex submanifold and hence symplectic.
answered Jan 23 at 22:07
Nick LNick L
1,294210
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$begingroup$
I am aware of one way to see this, which I saw in McDuff and Salamons book introduction to symplectic geometry.
We can find a compatible almost complex structure which is invariant by the torus action (This basically follows from Gromov's result that a symplectic manifold always has a compatible almost complex structure plus an averaging argument). Note that together with the symplectic structure we also get an invariant Riemannian metric.
Now consider a point $p$ which is fixed by the action, note that $T_{p}M$ inherits the structure of a complex representation of the torus $mathbb{T}^n$. The points close to $p$ which are fixed by the action are the image under the exponential map at $p$ of the fixed point set of this representation say $V = Fix(mathbb{T}^n ) subset T_p M$. Note that $$ V = cap_{g in mathbb{T}^n} Fix(g)$$ where $Fix(g)$ is the fixed point set of the linear transformation associated to $g$.
Now note that since the representation is complex linear, each $Fix(g)$ is a complex subspace (since it is the eigenspace for the eigenvalue 1), and hence $V$ is a complex subspace. This shows that that the fixed point set is locally an almost complex submanifold and hence symplectic.
answered Jan 23 at 22:07
Nick LNick L
1,294210
$endgroup$
add a comment |
$begingroup$
I am aware of one way to see this, which I saw in McDuff and Salamons book introduction to symplectic geometry.
We can find a compatible almost complex structure which is invariant by the torus action (This basically follows from Gromov's result that a symplectic manifold always has a compatible almost complex structure plus an averaging argument). Note that together with the symplectic structure we also get an invariant Riemannian metric.
Now consider a point $p$ which is fixed by the action, note that $T_{p}M$ inherits the structure of a complex representation of the torus $mathbb{T}^n$. The points close to $p$ which are fixed by the action are the image under the exponential map at $p$ of the fixed point set of this representation say $V = Fix(mathbb{T}^n ) subset T_p M$. Note that $$ V = cap_{g in mathbb{T}^n} Fix(g)$$ where $Fix(g)$ is the fixed point set of the linear transformation associated to $g$.
Now note that since the representation is complex linear, each $Fix(g)$ is a complex subspace (since it is the eigenspace for the eigenvalue 1), and hence $V$ is a complex subspace. This shows that that the fixed point set is locally an almost complex submanifold and hence symplectic.
answered Jan 23 at 22:07
Nick LNick L
1,294210
$endgroup$
add a comment |
$begingroup$
I am aware of one way to see this, which I saw in McDuff and Salamons book introduction to symplectic geometry.
We can find a compatible almost complex structure which is invariant by the torus action (This basically follows from Gromov's result that a symplectic manifold always has a compatible almost complex structure plus an averaging argument). Note that together with the symplectic structure we also get an invariant Riemannian metric.
Now consider a point $p$ which is fixed by the action, note that $T_{p}M$ inherits the structure of a complex representation of the torus $mathbb{T}^n$. The points close to $p$ which are fixed by the action are the image under the exponential map at $p$ of the fixed point set of this representation say $V = Fix(mathbb{T}^n ) subset T_p M$. Note that $$ V = cap_{g in mathbb{T}^n} Fix(g)$$ where $Fix(g)$ is the fixed point set of the linear transformation associated to $g$.
Now note that since the representation is complex linear, each $Fix(g)$ is a complex subspace (since it is the eigenspace for the eigenvalue 1), and hence $V$ is a complex subspace. This shows that that the fixed point set is locally an almost complex submanifold and hence symplectic.
answered Jan 23 at 22:07
Nick LNick L
1,294210
$endgroup$
I am aware of one way to see this, which I saw in McDuff and Salamons book introduction to symplectic geometry.
We can find a compatible almost complex structure which is invariant by the torus action (This basically follows from Gromov's result that a symplectic manifold always has a compatible almost complex structure plus an averaging argument). Note that together with the symplectic structure we also get an invariant Riemannian metric.
Now consider a point $p$ which is fixed by the action, note that $T_{p}M$ inherits the structure of a complex representation of the torus $mathbb{T}^n$. The points close to $p$ which are fixed by the action are the image under the exponential map at $p$ of the fixed point set of this representation say $V = Fix(mathbb{T}^n ) subset T_p M$. Note that $$ V = cap_{g in mathbb{T}^n} Fix(g)$$ where $Fix(g)$ is the fixed point set of the linear transformation associated to $g$.
Now note that since the representation is complex linear, each $Fix(g)$ is a complex subspace (since it is the eigenspace for the eigenvalue 1), and hence $V$ is a complex subspace. This shows that that the fixed point set is locally an almost complex submanifold and hence symplectic.
answered Jan 23 at 22:07
Nick LNick L
1,294210
answered Jan 23 at 22:07
Nick LNick L
1,294210
answered Jan 23 at 22:07
Nick LNick L
1,294210
answered Jan 23 at 22:07
Nick LNick L
1,294210
1,294210
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$begingroup$
What have you tried? Questions without context or visible effort tend to get downvoted and closed.
$endgroup$
– postmortes
Jan 23 at 18:29
1
$begingroup$
I think this is true in the case where a compact Lie group $G$ acts by symplectomorphisms on a symplectic manifold $M$. In this case, a connected component of the fixed point subspace $M^G$ is a symplectic submanifold of $M$
$endgroup$
– C. Zhihao
Jan 23 at 19:03