Show that the form $\omega$ defined locally as $$\omega = \sum dx_i \wedge d\xi_i$$ is globally well-defined on $T^*M$ and restricted to the zero section of $T^*M$ vanishes.
Here we consider $M$ to be a smooth manifold. $M$ has coordinates $x_1,\ldots,x_n$ on a coordinate chart $U$, i.e. if $\psi:U\to B^n$ is a local chart then the coordinate function $x_i(x)$ is the $i$-th coordinate of $\psi(x)$. We have that $dx_1,\ldots,dx_n$ form a local frame for the cotangent bundle. 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 $\xi_i(x,\mu)=\mu(\partial / \partial x_i)(x), (x,\mu) \in T^*_x M.$
I cannot figure out what it means for a form to be 'globally' well-defined. I thought I can show that $\omega$ is a closed non-degenerate $2$-form. For $1$-form $\alpha$ as $\alpha=\xi_i dx_i$ , then $-d\alpha= \omega$ so $\omega$ is exact and so closed.
Next I showed that $\alpha$ is independent of the choice of coordinates. For two charts $(U, \varphi = x_1,\ldots, x_n)$ and $(V, \psi = x_1',\ldots, x_n')$ on $M$, for $x \in U \cap V$ we have that $dx_i'\frac{\partial }{\partial x_j} = \sum \frac{\partial (\psi \circ \varphi^{-1})_i}{\partial x_j}(\varphi(x))dx_j$ and so for $\xi \in T_x^*M$ we have $\xi=\sum \xi_i dx_i=\sum \xi_j'dx_j;$ where $\xi_i$ is defined as above. So we conclude that for two coordinate charts on $(T^*U, x_1,\ldots,x_n,\xi_1,\ldots,\xi_n)$ and $(T^*V, x_1',\ldots,x_n',\xi_1',\ldots,\xi_n')$, on $T^*U\cap T^*V$ we have
$$\alpha=\sum \xi_i dx_i = \sum \xi_j'dx_j' = \alpha'.$$
The zero section of $T^*M$ is the set $M_0=\{(x,\xi)\in T^*X : \xi=0 \in T_x^*M\}$, which is the injection $M\to T^*M$. Then $\alpha$ restricts to $0$ on $M_0$ so $\omega|_{M_0}\equiv 0$.
Is this correct? Did I get the notation right? It gets a bit confusing with the dual basis.
Best Answer
Yes, very good.
I only have three comments to make.
1) the coordinate-free definition of $\alpha \in \Omega^1(T^*M)$ is given by $\alpha_{(x,\mu)}(Z_{(x,\mu)}) = \mu({\rm d}\pi_{(x,\mu)}(Z_{(x,\mu)}))$, where $(x,\mu) \in T^*M$, $Z_{(x,\mu)} \in T_{(x,\mu)}(T^*M)$ and $\pi\colon T^*M \to M$ is the projection.
2) $\alpha$ is called the tautological $1$-form because it is the unique $1$-form in $T^*M$ such that for all $\sigma \in \Omega^1(M)$ we have $\sigma^*\alpha = \sigma$. To understand the pull-back, actually think of $\sigma$ as a section $M \to T^*M$.
3) if $\iota\colon M_0 \to T^*M$ denotes the inclusion of the zero section, we have that $\iota^*\alpha = 0$. But pull-backs commute with ${\rm d}$, so applying $-{\rm d}$ it follows that $\iota^*\omega = 0$, as you wanted.