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Edit: According to the comment of Michael Albanese we revise the question.

Assume that $n$ is an odd integer and $M\subset \mathbb{R}^{2n}$ is a compact orientable $n$ dimensional submanifold.

Does there exist a non vanishing smooth vector field $X$ on $M$ with the following property?

$$\forall p\in M\quad \omega(X(p), V_p)=0,\quad \forall V_p\in T_pM$$

where $\omega$ is the standard symplectic structure of $\mathbb{R}^{2n}$.

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    $\begingroup$ Do you want $M$ to be a hypersurface? As written, $M$ need not be odd-dimensional (which your title suggests you intend it to be). $\endgroup$ Commented May 7, 2019 at 12:10
  • $\begingroup$ @MichaelAlbanese Thanks for your comment. I forgot to mention that $M$ is $n$ dimensional. Now oddness assumption is meaning full by two reasons: first we are sure that $M$ admit a non vanishing section second fiberise there is a vector which lies in $\omega$- angilator of the fiber. Now the question seeks for a global anhilator section. Please see my last comment to the answer to this questio mathoverflow.net/questions/330686/… $\endgroup$ Commented May 7, 2019 at 18:34

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This is not necessarily true, even locally. Consider $M \subset \mathbb{R}^6$ given by the graph of the function $f : \mathbb{R}_x^3 \rightarrow \mathbb{R}_y^3$ with

$$f(x_1,x_2,x_3) = (x_2x_3, 0, -x_1x_2)$$

Then at a point $p = (\vec{x},f(\vec{x})) \in M \subset \mathbb{R}_x^3 \times \mathbb{R}_y^3$, we compute $T_pM$ is given by the span of the vectors

$$e_1 = \partial_{x_1} -x_2\partial_{y_3}$$ $$e_2 = \partial_{x_2} + x_3\partial_{y_1} - x_1\partial_{y_3}$$ $$e_3 = \partial_{x_3} + x_2\partial_{y_1}$$

Explicitly, we have that in this basis,

$$\omega = \begin{pmatrix} 0 & -x_3 & -x_2 \\ x_3 & 0 & -x_1 \\ x_2 & x_1 & 0 \end{pmatrix}$$

from which it is easy to compute that so long as $p \neq 0$, any $X$ such that $\omega(X,v) = 0$ for all $v \in T_pM$ (i.e. $X$ is in the nullspace of the above matrix) must satisfy that $$X(p) \in \mathrm{span}\left\langle\begin{pmatrix}x_1 \\ - x_2 \\ x_3\end{pmatrix}\right\rangle$$ By continuity, $X(0)$ would need to be a multiple of $e_1$ (by looking at $X(p)$ along the axis $x_2 = x_3 = 0$) and also a multiple of $e_3$ (similarly). So there can be no nonzero $X(p)$ in a neighborhood of $0$.

For $M$ in higher dimensions, one can repeat the argument by successively taking $M \times \mathbb{R}^2 \subset \mathbb{R}^{2n} \times \mathbb{R}^4$, where the $\mathbb{R}^2 \subset \mathbb{R}^4$ we choose is symplectic. For a compact $M$, simply choose one which matches this model near $0$.

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    $\begingroup$ I have answered this in the last sentence: yes. Simply take a compact $M$ which matches this model near $0$. $\endgroup$
    – KSackel
    Commented May 10, 2019 at 19:58
  • $\begingroup$ Thanks again for your interesting answer. i erroneously deleted my previous comment which was a result of not reading your last sentence. $\endgroup$ Commented May 10, 2019 at 20:27
  • $\begingroup$ what about if we modify the question as follows: can a compact odd($n$) dimensional manifold be embedded in $\mathbb{R}^{2n}$ which admit a non zero section which $\omega$-annihilates the tangent space? $\endgroup$ Commented May 10, 2019 at 20:39

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