Let $M$ be a smooth n dimensional manifold. Is there an smooth embedding $f:M \to \mathbb{R}^{2n}$ whose image is a Lagrangian submanifold of $\mathbb{R}^{2n}$?
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asked Jan 11, 2014 at 10:47
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3$\begingroup$ No there has been a lot of work on this by Gromov, Luttinger, Fukaya ... for a survey of Fukaya's work you can look here math.uni-hamburg.de/home/latschev/forschung/fukaya.pdf $\endgroup$– Daniel PomerleanoCommented Jan 11, 2014 at 11:19
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$\begingroup$ Probably that's jumping to the end of the story and it might be worthwhile to look at Gromov's contribution. $\endgroup$– Daniel PomerleanoCommented Jan 11, 2014 at 11:24
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1 Answer
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No: being a Lagrangian submanifold of $\mathbb{R}^{2n}$ imposes strong conditions on $M$. If $M$ is embedded in $\mathbb{R}^{2n}$, the bundle $T_M\oplus N$ (normal bundle) is trivial; if $M$ is Lagrangian, the symplectic form induces an isomorphism $N\cong T_M^*$. Thus $T_M\oplus T_M$ is trivial; this implies for instance that the Pontryagin classes of $M$ are trivial in $H^*(M,\mathbb{Q})$. So e.g. $\mathbb{CP}^2$ cannot be embedded as a Lagrangian submanifold of $\mathbb{R}^{8}$.
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$\begingroup$ @abx thank you for the answer.Is triviality of $TM \oplus TM$ a sufficient condition for existence of an embedding with lagrangian image? $\endgroup$ Commented Jan 11, 2014 at 12:04
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$\begingroup$ @Ali Taghavi: Good question, I don't know. $\endgroup$– abxCommented Jan 11, 2014 at 12:56
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3$\begingroup$ No, see my remark above. I should have mentioned these elementary topology obstructions. What Gromov proved is that M must have non-trivial $H^1(M,\mathbb{R})$. His idea was precisely to go beyond these algebraic topology obstructions using pseudo-holomorphic curves. Then Fukaya proved some strong refinements of that, using more subtle properties of pseudo-holomorphic curves combined with some algebra. $\endgroup$ Commented Jan 11, 2014 at 13:09
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2$\begingroup$ If the manifold is connected and closed Gromov says no (as others have mentioned). If the manifold is connected and open then h-principle says that $T_M\otimes \mathbb{C}$ being a trivial complex vectorbundle over $M$ is enough for there to be a Lagrangian immersion, then perturbe to get only simple double points, then use the openness to move the intersections points of the image of $M$ - so yes. However, the interesting question then becomes: which open connected manifolds admits a PROPER Lagrangian embedding. $\endgroup$ Commented Jan 14, 2014 at 17:14
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1$\begingroup$ I am so used to working with exact Lagrangians that I was assuming this in the first part above. Obviously $S^1$ embeds into the plane. $\endgroup$ Commented Jan 15, 2014 at 10:10