Half-dimensional torus fibration vs Lagrangian torus fibration
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Assume we have a closed symplectic manifold $M$ which is the total space of a smooth fibration by half-dimensional tori. Can we infer that $M$ is the total space of a smooth fibration by Lagrangian tori?
sg.symplectic-geometry fibration
asked Dec 12 '18 at 8:44
geometergeometer
3465
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Assume we have a closed symplectic manifold $M$ which is the total space of a smooth fibration by half-dimensional tori. Can we infer that $M$ is the total space of a smooth fibration by Lagrangian tori?
sg.symplectic-geometry fibration
asked Dec 12 '18 at 8:44
geometergeometer
3465
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add a comment |
$begingroup$
Assume we have a closed symplectic manifold $M$ which is the total space of a smooth fibration by half-dimensional tori. Can we infer that $M$ is the total space of a smooth fibration by Lagrangian tori?
sg.symplectic-geometry fibration
asked Dec 12 '18 at 8:44
geometergeometer
3465
$endgroup$
Assume we have a closed symplectic manifold $M$ which is the total space of a smooth fibration by half-dimensional tori. Can we infer that $M$ is the total space of a smooth fibration by Lagrangian tori?
sg.symplectic-geometry fibration
sg.symplectic-geometry fibration
asked Dec 12 '18 at 8:44
geometergeometer
3465
asked Dec 12 '18 at 8:44
geometergeometer
3465
asked Dec 12 '18 at 8:44
geometergeometer
3465
asked Dec 12 '18 at 8:44
geometergeometer
3465
asked Dec 12 '18 at 8:44
geometergeometer
3465
3465
add a comment |
add a comment |
1 Answer
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This doesn't need to hold. For example, if one takes a $(T^4,omega)$ with a constant symplectic structure $omega$, in order for it to have a fibration by Lagrangian tori one should be able to find a homologically non-trivial $T^2subset T^4$ such that $int_{omega} T^2=0$ which is impossible for general $omega$.
One can also give counter-examples when a half-dimensional torus bundle exists on the symplectic manifold but no Lagrangian torus bundle exists for any symplectic structure on the manifold. To construct such an example consider $M^4$ that fibers over $B=T^2$ with a fiber $F=T^2$ such that the action of $mathbb Z^2=pi_1(B)$ on $H_1(F)=mathbb Z^2$ contains a hyperbolic element. Such a manifold is symplectic by Thurston, but it is not a total space of a Lagrangian torus fibration by the classification of such fibrations in dimension $4$.
Just to add, that the classification of Lagrangian $T^2$ fibrations was done in
https://ac.els-cdn.com/S0926224596000241/1-s2.0-S0926224596000241-main.pdf?_tid=824690ea-affb-4cd2-98a6-6d7ed7f08d9f&acdnat=1544694026_3c6783fa389437767cd9c8246616d4dd
And there is as well a very nice paper classifying Lagrangian $T^2$ fibrations over the Klein bottle. As you can see from Theorem 2.1 stated in the paper, the representation of $pi_1(B)$ in $H^1(F)=mathbb R^2$ (for Lagragian $T^2$ fibrations over $T^2$) is unipotent.
https://arxiv.org/pdf/0909.2982.pdf
answered Dec 12 '18 at 9:53
Dmitri PanovDmitri Panov
19.4k458119
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1 Answer
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1 Answer
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$begingroup$
This doesn't need to hold. For example, if one takes a $(T^4,omega)$ with a constant symplectic structure $omega$, in order for it to have a fibration by Lagrangian tori one should be able to find a homologically non-trivial $T^2subset T^4$ such that $int_{omega} T^2=0$ which is impossible for general $omega$.
One can also give counter-examples when a half-dimensional torus bundle exists on the symplectic manifold but no Lagrangian torus bundle exists for any symplectic structure on the manifold. To construct such an example consider $M^4$ that fibers over $B=T^2$ with a fiber $F=T^2$ such that the action of $mathbb Z^2=pi_1(B)$ on $H_1(F)=mathbb Z^2$ contains a hyperbolic element. Such a manifold is symplectic by Thurston, but it is not a total space of a Lagrangian torus fibration by the classification of such fibrations in dimension $4$.
Just to add, that the classification of Lagrangian $T^2$ fibrations was done in
https://ac.els-cdn.com/S0926224596000241/1-s2.0-S0926224596000241-main.pdf?_tid=824690ea-affb-4cd2-98a6-6d7ed7f08d9f&acdnat=1544694026_3c6783fa389437767cd9c8246616d4dd
And there is as well a very nice paper classifying Lagrangian $T^2$ fibrations over the Klein bottle. As you can see from Theorem 2.1 stated in the paper, the representation of $pi_1(B)$ in $H^1(F)=mathbb R^2$ (for Lagragian $T^2$ fibrations over $T^2$) is unipotent.
https://arxiv.org/pdf/0909.2982.pdf
answered Dec 12 '18 at 9:53
Dmitri PanovDmitri Panov
19.4k458119
$endgroup$
add a comment |
$begingroup$
This doesn't need to hold. For example, if one takes a $(T^4,omega)$ with a constant symplectic structure $omega$, in order for it to have a fibration by Lagrangian tori one should be able to find a homologically non-trivial $T^2subset T^4$ such that $int_{omega} T^2=0$ which is impossible for general $omega$.
One can also give counter-examples when a half-dimensional torus bundle exists on the symplectic manifold but no Lagrangian torus bundle exists for any symplectic structure on the manifold. To construct such an example consider $M^4$ that fibers over $B=T^2$ with a fiber $F=T^2$ such that the action of $mathbb Z^2=pi_1(B)$ on $H_1(F)=mathbb Z^2$ contains a hyperbolic element. Such a manifold is symplectic by Thurston, but it is not a total space of a Lagrangian torus fibration by the classification of such fibrations in dimension $4$.
Just to add, that the classification of Lagrangian $T^2$ fibrations was done in
https://ac.els-cdn.com/S0926224596000241/1-s2.0-S0926224596000241-main.pdf?_tid=824690ea-affb-4cd2-98a6-6d7ed7f08d9f&acdnat=1544694026_3c6783fa389437767cd9c8246616d4dd
And there is as well a very nice paper classifying Lagrangian $T^2$ fibrations over the Klein bottle. As you can see from Theorem 2.1 stated in the paper, the representation of $pi_1(B)$ in $H^1(F)=mathbb R^2$ (for Lagragian $T^2$ fibrations over $T^2$) is unipotent.
https://arxiv.org/pdf/0909.2982.pdf
answered Dec 12 '18 at 9:53
Dmitri PanovDmitri Panov
19.4k458119
$endgroup$
add a comment |
$begingroup$
This doesn't need to hold. For example, if one takes a $(T^4,omega)$ with a constant symplectic structure $omega$, in order for it to have a fibration by Lagrangian tori one should be able to find a homologically non-trivial $T^2subset T^4$ such that $int_{omega} T^2=0$ which is impossible for general $omega$.
One can also give counter-examples when a half-dimensional torus bundle exists on the symplectic manifold but no Lagrangian torus bundle exists for any symplectic structure on the manifold. To construct such an example consider $M^4$ that fibers over $B=T^2$ with a fiber $F=T^2$ such that the action of $mathbb Z^2=pi_1(B)$ on $H_1(F)=mathbb Z^2$ contains a hyperbolic element. Such a manifold is symplectic by Thurston, but it is not a total space of a Lagrangian torus fibration by the classification of such fibrations in dimension $4$.
Just to add, that the classification of Lagrangian $T^2$ fibrations was done in
https://ac.els-cdn.com/S0926224596000241/1-s2.0-S0926224596000241-main.pdf?_tid=824690ea-affb-4cd2-98a6-6d7ed7f08d9f&acdnat=1544694026_3c6783fa389437767cd9c8246616d4dd
And there is as well a very nice paper classifying Lagrangian $T^2$ fibrations over the Klein bottle. As you can see from Theorem 2.1 stated in the paper, the representation of $pi_1(B)$ in $H^1(F)=mathbb R^2$ (for Lagragian $T^2$ fibrations over $T^2$) is unipotent.
https://arxiv.org/pdf/0909.2982.pdf
answered Dec 12 '18 at 9:53
Dmitri PanovDmitri Panov
19.4k458119
$endgroup$
This doesn't need to hold. For example, if one takes a $(T^4,omega)$ with a constant symplectic structure $omega$, in order for it to have a fibration by Lagrangian tori one should be able to find a homologically non-trivial $T^2subset T^4$ such that $int_{omega} T^2=0$ which is impossible for general $omega$.
One can also give counter-examples when a half-dimensional torus bundle exists on the symplectic manifold but no Lagrangian torus bundle exists for any symplectic structure on the manifold. To construct such an example consider $M^4$ that fibers over $B=T^2$ with a fiber $F=T^2$ such that the action of $mathbb Z^2=pi_1(B)$ on $H_1(F)=mathbb Z^2$ contains a hyperbolic element. Such a manifold is symplectic by Thurston, but it is not a total space of a Lagrangian torus fibration by the classification of such fibrations in dimension $4$.
Just to add, that the classification of Lagrangian $T^2$ fibrations was done in
https://ac.els-cdn.com/S0926224596000241/1-s2.0-S0926224596000241-main.pdf?_tid=824690ea-affb-4cd2-98a6-6d7ed7f08d9f&acdnat=1544694026_3c6783fa389437767cd9c8246616d4dd
And there is as well a very nice paper classifying Lagrangian $T^2$ fibrations over the Klein bottle. As you can see from Theorem 2.1 stated in the paper, the representation of $pi_1(B)$ in $H^1(F)=mathbb R^2$ (for Lagragian $T^2$ fibrations over $T^2$) is unipotent.
https://arxiv.org/pdf/0909.2982.pdf
answered Dec 12 '18 at 9:53
Dmitri PanovDmitri Panov
19.4k458119
edited Dec 13 '18 at 10:13
answered Dec 12 '18 at 9:53
Dmitri PanovDmitri Panov
19.4k458119
answered Dec 12 '18 at 9:53
Dmitri PanovDmitri Panov
19.4k458119
answered Dec 12 '18 at 9:53
Dmitri PanovDmitri Panov
19.4k458119
19.4k458119
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