# A special non vanishing vector field on odd dimensional compact manifolds

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 $$\mathbb{R}^{2n}$$.

• 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). – Michael Albanese May 7 '19 at 12:10
• @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/… – Ali Taghavi May 7 '19 at 18:34

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 $$0$$.

• I have answered this in the last sentence: yes. Simply take a compact $M$ which matches this model near $0$. – KSackel May 10 '19 at 19:58
• Thanks again for your interesting answer. i erroneously deleted my previous comment which was a result of not reading your last sentence. – Ali Taghavi May 10 '19 at 20:27
• 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? – Ali Taghavi May 10 '19 at 20:39