Model for symplectic geometry
$begingroup$
An almost symplectic structure on a smooth even dimension manifold $M$ can be viewed as a reduction of structure group $Sp(2n,mathbb{R}) hookrightarrow GL(2n,mathbb{R})$ for the principal frame bundle $mathcal{F}(M)$. If this structure group reduction corresponds to a closed section of $mathcal{F}(M)/Sp(2n,mathbb{R})$ then it is a symplectic structure.
It is claimed in here that, as an integrable $G$-structure, a symplectic manifold is a Cartan geometry.
So, if we want to describe a symplectic manifold as a Cartan geometry of type $(G,H)$ what are the possible choices of Lie groups $G$ and $H$?
Different choices of $G$ and $H$ will give rise to homogeneous spaces $G/H
$ that are related by model mutation. I am primarily concerned with the groups that give rise to a compact model $G/H$ for symplectic manifolds.
differential-geometry symplectic-geometry cartan-geometry
$endgroup$
add a comment |
$begingroup$
An almost symplectic structure on a smooth even dimension manifold $M$ can be viewed as a reduction of structure group $Sp(2n,mathbb{R}) hookrightarrow GL(2n,mathbb{R})$ for the principal frame bundle $mathcal{F}(M)$. If this structure group reduction corresponds to a closed section of $mathcal{F}(M)/Sp(2n,mathbb{R})$ then it is a symplectic structure.
It is claimed in here that, as an integrable $G$-structure, a symplectic manifold is a Cartan geometry.
So, if we want to describe a symplectic manifold as a Cartan geometry of type $(G,H)$ what are the possible choices of Lie groups $G$ and $H$?
Different choices of $G$ and $H$ will give rise to homogeneous spaces $G/H
$ that are related by model mutation. I am primarily concerned with the groups that give rise to a compact model $G/H$ for symplectic manifolds.
differential-geometry symplectic-geometry cartan-geometry
$endgroup$
add a comment |
$begingroup$
An almost symplectic structure on a smooth even dimension manifold $M$ can be viewed as a reduction of structure group $Sp(2n,mathbb{R}) hookrightarrow GL(2n,mathbb{R})$ for the principal frame bundle $mathcal{F}(M)$. If this structure group reduction corresponds to a closed section of $mathcal{F}(M)/Sp(2n,mathbb{R})$ then it is a symplectic structure.
It is claimed in here that, as an integrable $G$-structure, a symplectic manifold is a Cartan geometry.
So, if we want to describe a symplectic manifold as a Cartan geometry of type $(G,H)$ what are the possible choices of Lie groups $G$ and $H$?
Different choices of $G$ and $H$ will give rise to homogeneous spaces $G/H
$ that are related by model mutation. I am primarily concerned with the groups that give rise to a compact model $G/H$ for symplectic manifolds.
differential-geometry symplectic-geometry cartan-geometry
$endgroup$
An almost symplectic structure on a smooth even dimension manifold $M$ can be viewed as a reduction of structure group $Sp(2n,mathbb{R}) hookrightarrow GL(2n,mathbb{R})$ for the principal frame bundle $mathcal{F}(M)$. If this structure group reduction corresponds to a closed section of $mathcal{F}(M)/Sp(2n,mathbb{R})$ then it is a symplectic structure.
It is claimed in here that, as an integrable $G$-structure, a symplectic manifold is a Cartan geometry.
So, if we want to describe a symplectic manifold as a Cartan geometry of type $(G,H)$ what are the possible choices of Lie groups $G$ and $H$?
Different choices of $G$ and $H$ will give rise to homogeneous spaces $G/H
$ that are related by model mutation. I am primarily concerned with the groups that give rise to a compact model $G/H$ for symplectic manifolds.
differential-geometry symplectic-geometry cartan-geometry
differential-geometry symplectic-geometry cartan-geometry
edited Sep 11 '18 at 5:54
asked Sep 11 '18 at 5:43
user564024
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1 Answer
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$begingroup$
It is not true in general that an integrable $G$-structure is a Cartan geometry. Indeed, a symplectic structure cannot be described by a finite dimensional Cartan geometry, since it has infinite dimensional automorphism group (the group of symplectomorphisms). It is also not clear, why such a description should be of interest, since symplectic structures do not have local invariants.
In the standard picture of $G$-structures you would be led to consider Cartan geometries of type $(Sp(2n,mathbb R)ltimesmathbb R^{2n},Sp(2n,mathbb R))$. (This is like passing from the orthogonal group to the group of rigid motions.) Cartan geometries of this type have an underlying almost symplectic structure which is symplectic if and only if the Cartan geometry is torsion-free. But such a Cartan geometry also comes with a connection on the tangent bundle with preserves the almost symplectic structure. In the torsion-free case, these are so-called Fedosov connections (torsion-free symplectic connections), so you get Fedosov structures rather than just symplectic structures.
answered Dec 11 '18 at 8:35
Andreas CapAndreas Cap
11.2k923
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
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votes
active
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votes
$begingroup$
It is not true in general that an integrable $G$-structure is a Cartan geometry. Indeed, a symplectic structure cannot be described by a finite dimensional Cartan geometry, since it has infinite dimensional automorphism group (the group of symplectomorphisms). It is also not clear, why such a description should be of interest, since symplectic structures do not have local invariants.
In the standard picture of $G$-structures you would be led to consider Cartan geometries of type $(Sp(2n,mathbb R)ltimesmathbb R^{2n},Sp(2n,mathbb R))$. (This is like passing from the orthogonal group to the group of rigid motions.) Cartan geometries of this type have an underlying almost symplectic structure which is symplectic if and only if the Cartan geometry is torsion-free. But such a Cartan geometry also comes with a connection on the tangent bundle with preserves the almost symplectic structure. In the torsion-free case, these are so-called Fedosov connections (torsion-free symplectic connections), so you get Fedosov structures rather than just symplectic structures.
answered Dec 11 '18 at 8:35
Andreas CapAndreas Cap
11.2k923
$endgroup$
add a comment |
$begingroup$
It is not true in general that an integrable $G$-structure is a Cartan geometry. Indeed, a symplectic structure cannot be described by a finite dimensional Cartan geometry, since it has infinite dimensional automorphism group (the group of symplectomorphisms). It is also not clear, why such a description should be of interest, since symplectic structures do not have local invariants.
In the standard picture of $G$-structures you would be led to consider Cartan geometries of type $(Sp(2n,mathbb R)ltimesmathbb R^{2n},Sp(2n,mathbb R))$. (This is like passing from the orthogonal group to the group of rigid motions.) Cartan geometries of this type have an underlying almost symplectic structure which is symplectic if and only if the Cartan geometry is torsion-free. But such a Cartan geometry also comes with a connection on the tangent bundle with preserves the almost symplectic structure. In the torsion-free case, these are so-called Fedosov connections (torsion-free symplectic connections), so you get Fedosov structures rather than just symplectic structures.
answered Dec 11 '18 at 8:35
Andreas CapAndreas Cap
11.2k923
$endgroup$
add a comment |
$begingroup$
It is not true in general that an integrable $G$-structure is a Cartan geometry. Indeed, a symplectic structure cannot be described by a finite dimensional Cartan geometry, since it has infinite dimensional automorphism group (the group of symplectomorphisms). It is also not clear, why such a description should be of interest, since symplectic structures do not have local invariants.
In the standard picture of $G$-structures you would be led to consider Cartan geometries of type $(Sp(2n,mathbb R)ltimesmathbb R^{2n},Sp(2n,mathbb R))$. (This is like passing from the orthogonal group to the group of rigid motions.) Cartan geometries of this type have an underlying almost symplectic structure which is symplectic if and only if the Cartan geometry is torsion-free. But such a Cartan geometry also comes with a connection on the tangent bundle with preserves the almost symplectic structure. In the torsion-free case, these are so-called Fedosov connections (torsion-free symplectic connections), so you get Fedosov structures rather than just symplectic structures.
answered Dec 11 '18 at 8:35
Andreas CapAndreas Cap
11.2k923
$endgroup$
It is not true in general that an integrable $G$-structure is a Cartan geometry. Indeed, a symplectic structure cannot be described by a finite dimensional Cartan geometry, since it has infinite dimensional automorphism group (the group of symplectomorphisms). It is also not clear, why such a description should be of interest, since symplectic structures do not have local invariants.
In the standard picture of $G$-structures you would be led to consider Cartan geometries of type $(Sp(2n,mathbb R)ltimesmathbb R^{2n},Sp(2n,mathbb R))$. (This is like passing from the orthogonal group to the group of rigid motions.) Cartan geometries of this type have an underlying almost symplectic structure which is symplectic if and only if the Cartan geometry is torsion-free. But such a Cartan geometry also comes with a connection on the tangent bundle with preserves the almost symplectic structure. In the torsion-free case, these are so-called Fedosov connections (torsion-free symplectic connections), so you get Fedosov structures rather than just symplectic structures.
answered Dec 11 '18 at 8:35
Andreas CapAndreas Cap
11.2k923
answered Dec 11 '18 at 8:35
Andreas CapAndreas Cap
11.2k923
answered Dec 11 '18 at 8:35
Andreas CapAndreas Cap
11.2k923
answered Dec 11 '18 at 8:35
Andreas CapAndreas Cap
11.2k923
11.2k923
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