# A symplectic version of critical points

According to the interesting comment of Mohammad F Tehrani, I revise the question as follows:

Assume $n>2$. For what type of compact n dimensional manifolds $M$ we can say:

For every smooth embedding $f:M \to \mathbb{R}^{2n}$, there exist a point $p \in M$such that $Df_{p}(T_{p} M)$ is a lagrangian subspace of $R^{2n}$?

Is there an example of a manifold with this property? For example: does $S^{3}$ satisfy this property?

• just $dim M=1$; otherwise, you can rotate any embedding to make its tangent space include a symplectic two-plane at some arbitrary point. – Mohammad F. Tehrani Jan 15 '14 at 19:07
• @MohammadF.Tehrani "Some arbitray point" or "all points"? if you mean the first, obviously your comment is not an answer to my question. could you please read my question again? – Ali Taghavi Jan 15 '14 at 20:09
• to avoid the triviality, lets assume $n>1$ – Ali Taghavi Jan 15 '14 at 20:15
• @ Taghavi: You are right, I did not read it correctly. Lets put n=2, and $M$ a Riemann surface. Suppose $M$ does not have this property, then it mean for some embedding $f$, $T_pM$ is symplectic. But then $\int_M \omega\neq 0$ which is contradiction since $\omega$ is exact. So for every Riemann surface this is true as well. For n>2, thing get more complicated, since you can have small dimensional symplectic spaces inside $T_pM$, but I cant give you an example. Still, $\omega|M$ would be a non-zero but may be degenerate two-form and I am not sure to use it to get what you want. – Mohammad F. Tehrani Jan 15 '14 at 20:39
• I think it follows pretty quickly from the jet transversality theorem (see e.g. Hirsch's differential topology book) that if $n>3$ generic embeddings $f:M\to R^{2n}$ have no points at which they are Lagrangian. Indeed the Lagrangian Grassmannian has, by my count, codimension $n(n-1)/2$ in the whole Grassmannian $Gr(2n,n)$, and so if $n(n-1)/2>n$ then the "expected dimension" of the Lagrangian locus is negative, and jet transversality implies that generic perturbations of a given map will realize this expected dimension. So that leaves only the case $n=3$ unresolved. – Mike Usher Jan 17 '14 at 0:58