Let's consider $M$ to be a smooth manifold with local coordinates $x_1,\ldots,x_n$ on a coordinate chart $U$. Denote by $\partial / \partial x_i$ the dual basis, $dx_j(\partial / \partial x_i)=\delta_{ij}$ and let $\xi_1,\ldots, \xi_n$ be such that $\xi_i: T^*M|_U\to \mathbb{R}$ defined by $\xi_i(x,\mu)=\mu(\partial / \partial x_i)(x), (x,\mu) \in T^*_x M.$
In many references they say that the form
$$\omega = \sum dx_i \wedge d\xi_i$$ is clearly non-degenerate, but I don't know how to show that !
Any help would be very appreciated!
Best Answer
Pick $X=\sum_{i=1}^n a_i\partial_{x_i}+b_i\partial_{\xi_i}$ a vector field of $T^*M$, then \begin{align*} \omega(X,Y)=0,\ \forall Y&\Longleftrightarrow 0=\omega(X,\partial_{x_j})=-b_j\ \text{and}\ 0=\omega(X,\partial_{\xi_j})=a_j,\ \forall i=1,...,n\\ & \Longleftrightarrow X=0 \end{align*} $$$$