The automorphism group of a symplectic symmetric space
$begingroup$
Why is the automorphism group of a sympelctic symmetric space a Lie group?
$\$
A symplectic symmetric space is a triple $(M, omega, s)$, where $(M, omega)$ is a symplectic manifold and $ s ; colon M times M to M $, $(x, y) mapsto s_x(y)$, is such that $s_x$ is an involutive symplectic diffeomorphism with an isolated fixed point at $x$ and $s_xs_ys_x = s_{s_x(y)} ; forall ; x, y in M; $ $big($this can be read as $s_x s = ss_x ; forall x in M$$big)$.
An interesting fact about a symplectic symmetric space is the one that once $x$ is an isolated fixed point of $s_x$ and $s_x^2 = Id$, $ds_{x (x)} = -Id_{T_xM}$. Furthermore, a symplectic symmetric space admits a unique affine connection $nabla$ on it such that $s_x$ is an affinity for every $x in M$ and such that $nabla omega = 0$. This connection has no torsion and is thus a symplectic connection.
The automorphism group $Aut(M, omega, s)$ of $(M, omega, s)$ is the group of symplectic automorphisms $varphi$ of $(M, omega)$ that satisfy $varphi circ s_x = s_{varphi(x)} circ varphi ; forall ; x in M$ $big($ or, in other words, $varphi circ s = scirc varphi$ $big)$. It's simple to check that this group is the intersection of the symplectic automorphism group of $(M, omega)$ and the affine group of $(M, nabla)$.
My question then is: why is $Aut(M, omega, s)$ a Lie group?
sg.symplectic-geometry connections symmetric-spaces
asked Dec 12 '18 at 13:04
ValentinoValentino
1425
$endgroup$
add a comment |
$begingroup$
Why is the automorphism group of a sympelctic symmetric space a Lie group?
$\$
A symplectic symmetric space is a triple $(M, omega, s)$, where $(M, omega)$ is a symplectic manifold and $ s ; colon M times M to M $, $(x, y) mapsto s_x(y)$, is such that $s_x$ is an involutive symplectic diffeomorphism with an isolated fixed point at $x$ and $s_xs_ys_x = s_{s_x(y)} ; forall ; x, y in M; $ $big($this can be read as $s_x s = ss_x ; forall x in M$$big)$.
An interesting fact about a symplectic symmetric space is the one that once $x$ is an isolated fixed point of $s_x$ and $s_x^2 = Id$, $ds_{x (x)} = -Id_{T_xM}$. Furthermore, a symplectic symmetric space admits a unique affine connection $nabla$ on it such that $s_x$ is an affinity for every $x in M$ and such that $nabla omega = 0$. This connection has no torsion and is thus a symplectic connection.
The automorphism group $Aut(M, omega, s)$ of $(M, omega, s)$ is the group of symplectic automorphisms $varphi$ of $(M, omega)$ that satisfy $varphi circ s_x = s_{varphi(x)} circ varphi ; forall ; x in M$ $big($ or, in other words, $varphi circ s = scirc varphi$ $big)$. It's simple to check that this group is the intersection of the symplectic automorphism group of $(M, omega)$ and the affine group of $(M, nabla)$.
My question then is: why is $Aut(M, omega, s)$ a Lie group?
sg.symplectic-geometry connections symmetric-spaces
asked Dec 12 '18 at 13:04
ValentinoValentino
1425
$endgroup$
add a comment |
$begingroup$
Why is the automorphism group of a sympelctic symmetric space a Lie group?
$\$
A symplectic symmetric space is a triple $(M, omega, s)$, where $(M, omega)$ is a symplectic manifold and $ s ; colon M times M to M $, $(x, y) mapsto s_x(y)$, is such that $s_x$ is an involutive symplectic diffeomorphism with an isolated fixed point at $x$ and $s_xs_ys_x = s_{s_x(y)} ; forall ; x, y in M; $ $big($this can be read as $s_x s = ss_x ; forall x in M$$big)$.
An interesting fact about a symplectic symmetric space is the one that once $x$ is an isolated fixed point of $s_x$ and $s_x^2 = Id$, $ds_{x (x)} = -Id_{T_xM}$. Furthermore, a symplectic symmetric space admits a unique affine connection $nabla$ on it such that $s_x$ is an affinity for every $x in M$ and such that $nabla omega = 0$. This connection has no torsion and is thus a symplectic connection.
The automorphism group $Aut(M, omega, s)$ of $(M, omega, s)$ is the group of symplectic automorphisms $varphi$ of $(M, omega)$ that satisfy $varphi circ s_x = s_{varphi(x)} circ varphi ; forall ; x in M$ $big($ or, in other words, $varphi circ s = scirc varphi$ $big)$. It's simple to check that this group is the intersection of the symplectic automorphism group of $(M, omega)$ and the affine group of $(M, nabla)$.
My question then is: why is $Aut(M, omega, s)$ a Lie group?
sg.symplectic-geometry connections symmetric-spaces
asked Dec 12 '18 at 13:04
ValentinoValentino
1425
$endgroup$
Why is the automorphism group of a sympelctic symmetric space a Lie group?
$\$
A symplectic symmetric space is a triple $(M, omega, s)$, where $(M, omega)$ is a symplectic manifold and $ s ; colon M times M to M $, $(x, y) mapsto s_x(y)$, is such that $s_x$ is an involutive symplectic diffeomorphism with an isolated fixed point at $x$ and $s_xs_ys_x = s_{s_x(y)} ; forall ; x, y in M; $ $big($this can be read as $s_x s = ss_x ; forall x in M$$big)$.
An interesting fact about a symplectic symmetric space is the one that once $x$ is an isolated fixed point of $s_x$ and $s_x^2 = Id$, $ds_{x (x)} = -Id_{T_xM}$. Furthermore, a symplectic symmetric space admits a unique affine connection $nabla$ on it such that $s_x$ is an affinity for every $x in M$ and such that $nabla omega = 0$. This connection has no torsion and is thus a symplectic connection.
The automorphism group $Aut(M, omega, s)$ of $(M, omega, s)$ is the group of symplectic automorphisms $varphi$ of $(M, omega)$ that satisfy $varphi circ s_x = s_{varphi(x)} circ varphi ; forall ; x in M$ $big($ or, in other words, $varphi circ s = scirc varphi$ $big)$. It's simple to check that this group is the intersection of the symplectic automorphism group of $(M, omega)$ and the affine group of $(M, nabla)$.
My question then is: why is $Aut(M, omega, s)$ a Lie group?
sg.symplectic-geometry connections symmetric-spaces
sg.symplectic-geometry connections symmetric-spaces
asked Dec 12 '18 at 13:04
ValentinoValentino
1425
asked Dec 12 '18 at 13:04
ValentinoValentino
1425
asked Dec 12 '18 at 13:04
ValentinoValentino
1425
asked Dec 12 '18 at 13:04
ValentinoValentino
1425
asked Dec 12 '18 at 13:04
ValentinoValentino
1425
1425
add a comment |
add a comment |
1 Answer
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$begingroup$
The affine group of $(M,nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Differential Geometry). The dimension of $G$ is at most $n+n^2$ (where $n=dim M$).
The subgroup $H$ of $G$ consisting of those elements of $G$ that preserve the symplectic structure on $M$ is clearly a closed subgroup, and since every closed subgroup of a Lie group is a Lie group (see, for example, Helgason's proof in his Differential Geometry, Lie Groups, and Symmetric Spaces), it follows that $H$ is a Lie group as well.
answered Dec 12 '18 at 13:19
Robert BryantRobert Bryant
73.7k6216318
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$begingroup$
The affine group of $(M,nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Differential Geometry). The dimension of $G$ is at most $n+n^2$ (where $n=dim M$).
The subgroup $H$ of $G$ consisting of those elements of $G$ that preserve the symplectic structure on $M$ is clearly a closed subgroup, and since every closed subgroup of a Lie group is a Lie group (see, for example, Helgason's proof in his Differential Geometry, Lie Groups, and Symmetric Spaces), it follows that $H$ is a Lie group as well.
answered Dec 12 '18 at 13:19
Robert BryantRobert Bryant
73.7k6216318
$endgroup$
add a comment |
$begingroup$
The affine group of $(M,nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Differential Geometry). The dimension of $G$ is at most $n+n^2$ (where $n=dim M$).
The subgroup $H$ of $G$ consisting of those elements of $G$ that preserve the symplectic structure on $M$ is clearly a closed subgroup, and since every closed subgroup of a Lie group is a Lie group (see, for example, Helgason's proof in his Differential Geometry, Lie Groups, and Symmetric Spaces), it follows that $H$ is a Lie group as well.
answered Dec 12 '18 at 13:19
Robert BryantRobert Bryant
73.7k6216318
$endgroup$
add a comment |
$begingroup$
The affine group of $(M,nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Differential Geometry). The dimension of $G$ is at most $n+n^2$ (where $n=dim M$).
The subgroup $H$ of $G$ consisting of those elements of $G$ that preserve the symplectic structure on $M$ is clearly a closed subgroup, and since every closed subgroup of a Lie group is a Lie group (see, for example, Helgason's proof in his Differential Geometry, Lie Groups, and Symmetric Spaces), it follows that $H$ is a Lie group as well.
answered Dec 12 '18 at 13:19
Robert BryantRobert Bryant
73.7k6216318
$endgroup$
The affine group of $(M,nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Differential Geometry). The dimension of $G$ is at most $n+n^2$ (where $n=dim M$).
The subgroup $H$ of $G$ consisting of those elements of $G$ that preserve the symplectic structure on $M$ is clearly a closed subgroup, and since every closed subgroup of a Lie group is a Lie group (see, for example, Helgason's proof in his Differential Geometry, Lie Groups, and Symmetric Spaces), it follows that $H$ is a Lie group as well.
answered Dec 12 '18 at 13:19
Robert BryantRobert Bryant
73.7k6216318
answered Dec 12 '18 at 13:19
Robert BryantRobert Bryant
73.7k6216318
answered Dec 12 '18 at 13:19
Robert BryantRobert Bryant
73.7k6216318
answered Dec 12 '18 at 13:19
Robert BryantRobert Bryant
73.7k6216318
73.7k6216318
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add a comment |
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