A vector bundle associated to a codimension $1$ submanifold of a symplectic manifold

We consider the standard symplectic structure $$\omega=\sum dx_i\wedge dy_i$$ on $$\mathbb{R}^{2n}$$. To every codimension $$1$$ submanifold $$M\subset \mathbb{R}^{2n}$$ we associate a vector bundle $$E$$ over $$M$$ as follows:

$$E=\{(x,v)\in M\times \mathbb{R}^{2n}\mid \omega(v,N_x)=0,\;\;N_x\perp T_x M\}$$

Does the structure of this bundle $$E$$, or at least its characteristic classes, depend on the way $$M$$ is embedded in $$\mathbb{R}^{2n}$$? What can be said about orientability of this bundle? What is the structure of this bundle for the standard odd dimensional spheres?Is it trivial?is it isomorphic to the tangent bundle of sphere?

Note: One can extend this question by replacing $$\mathbb{R}^{2n}$$ with an arbitrary symplectic Riemannian manifold. Moreover the intersection of this bundle $$E$$ with $$TM$$ gives us a codimension $$1$$ distribution of $$M$$. So it would be interesting to study the dependency (and integrability and the structure of the foliation generated by) of this distribution on the way of embedding of $$M$$ in the ambient symplectic manifold. As an example: What is this distribution for $$S^3$$?

• Stupid question: what is $N_x$? Is it the Euclidean normal bundle, or the symplectic one? In the first case, you should change the title of your question. In the second, $E$ is just the tangent bundle. – Sebastian Goette May 5 at 11:17
• @SebastianGoette Your comment is helpful. May be I should be more precise in the question. Yes it is Euclidean normal bundle, as I wrote $N_x \perp T_x M$. So what should be the title in this case? – Ali Taghavi May 5 at 11:24
• I am not quite sure if there is anything better than "symplectic manifold with compatible metric". You probably don't want something strong like Kähler, do you? – Sebastian Goette May 5 at 11:28
• @SebastianGoette Very good suggestion: However I did not pay attention to this situation and I did not include in the title, but the main example I considered in the body of the question is avtually Kahler. But can the extra assumption Kahlerian help us to say somethong? – Ali Taghavi May 5 at 11:37
• The isomorphism in my answer below would be parallel, and hence even the Chern-Weil forms would agree. But then, for the question if $E\cap TM$ becomes a contact structure, the Kähler condition is irrelevant as far as I know. – Sebastian Goette May 5 at 11:43

We have to choose a compatible Riemannian metric $$g$$ to define the normal bundle $$N$$. Then the almost complex structure $$J$$ defined by $$\omega$$ and $$g$$ induces an isomorphism between $$E$$ and $$TM$$.
As for your second question, assume that $$M=f^{-1}(0)$$ for some function $$f$$ that has $$0$$ as a regular value. Then $$E\cap TM$$ is perpendicular to the Hamiltonian vector field of $$f$$ acting on $$M$$. Let $$\alpha\in\Omega^1(M)$$ be a (local) defining one-form, so $$\ker\alpha=E\cap TM$$. If $$E\cap TM$$ is integrable, we can find $$\alpha$$ such that $$d\alpha=0$$. Equivalently, we demand $$\alpha\wedge d\alpha=0$$, which is independent of the choice of $$\alpha$$.
On the other hand, if $$\alpha\wedge(d\alpha)^{n-1}$$ is a volume form (nonzero everywhere), then $$E\cap TM$$ is called a contact structure. Again, this condition does not depend on the choice of $$\alpha$$. This includes the case of the standard sphere $$S^{2n-1}\subset\mathbb R^{2n}$$.
• Thanks for your very interesting answer. $J$ was very key point which I dod mot pay attention to. – Ali Taghavi May 5 at 23:59
• BTW what can be said precisely about the distribution of the standard $S^3$?If it is integrable of course it can not have a leaf with non trivial holonomy sonce all datas are analytic. – Ali Taghavi May 6 at 10:03