Definition of bivector field on a manifold.

Question

I am reading this article: http://arxiv.org/abs/1112.5037v1 . In this, it defines a symplectic manifold as a manifold equiped with a nondegenerate bivector field $\pi$ that is Poisson. l want to understand what is a bivector field on a manifold. I couldn't get a definition on manifolds. Thanks for your help.

Answer 1

Let $M$ be a manifold. A vector field assigns for each point $m\in M$ an element of the vector space $T_mM$.

A bivector field should assign for each point $m\in M$ a bivector (or at least a pair of vectors) in the tangent space $T_mM$. As in the case of vector field, we want this choice of pair of tangent vectors to vary smoothly. To formalise this notion of smoothness, we see a bisector field from an alternate point of view.

Let $\pi$ be a bivector field.

Let $\omega\in \Omega^1(M)$ (a $1$-form on $M$).

Fix $m\in M$.

The $1$-form $\omega$ gives a map $\omega(m):T_mM\rightarrow \mathbb{R}$. The bivector field $\pi$ gives a pair of vectors $(v_1,v_2)$.

Evaluating $\omega(m)$ at $v_1$ gives $\omega(m)(v_1)\in \mathbb{R}$. This in turn gives an element $\omega(m)(v_1)v_2\in T_mM$. Note that $\omega(m)(v_1)$ is a real number and $v_2$ is an element in the vector space $T_mM$. Thus, for each $m\in M$, we have an element in the tangent space $T_mM$. We can ask if these tangent vectors vary smoothly to give a vector field. So, we get a map $\Omega^1(M)\rightarrow \mathfrak{X}(M)$. Starting at a decent map $\Omega^1(M)\rightarrow \mathfrak{X}(M)$, one can go backwards and produce a bivector field.

So, a bivector field can be seen as a nice map $\Omega^1(M)\rightarrow \mathfrak{X}(M)$.

Doing similar kind of manipulations one can see a bivector field as a nice map $\Omega^1(M)\times \Omega^1(M)\rightarrow C^\infty(M)$.