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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!

Mira
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    Verify that the matrix representation of $\omega$ relative to the basis $\frac{\partial}{\partial x^i},\frac{\partial}{\partial \xi^i}$ is $\pm\begin{pmatrix}0&I_n\-I_n&0\end{pmatrix}$ ($\pm$ because I can't for the life of me remember the conventions). – peek-a-boo Jul 02 '21 at 11:19
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    Thank you very much to both of you peek-a-boo and Amitesh Data for your helpful comments, Now I get the idea! – Mira Jul 02 '21 at 11:30

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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*} $$$$

Anonymous
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