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Let $\pi:E\to X$ be a (smooth) vector bundle, and let $q\in X$. First of all, I have to find a canonical isomorphism $T_{(q,0)}E\cong T_qX\times E_q$. I proceed as follows.

Since $\pi$ is a submersion, we have an exact sequence (to ease the notation I drop the subscript on $d\pi$, it is taken at $(q,0)$) \begin{equation} 0\longrightarrow\ker(d\pi)\overset{\iota}{\longrightarrow}T_{(q,0)}E\overset{d\pi}{\longrightarrow}T_qX\longrightarrow 0 \end{equation} which is split since the zero section $s:X\to E,q\mapsto(q,0)$ yields a map $ds_q$ such that $id_{T_qX}=d\pi\circ ds_q$. Hence the map $\tilde\iota:T_{(q,0)}E\longrightarrow,v\mapsto v-ds_q(d\pi(v))$ gives us a map \begin{equation} \psi:T_{(q,0)}E\longrightarrow T_qX\times\ker(d\pi),v\mapsto(d\pi(v),v-ds_q(d\pi(v))) \end{equation} defines an isomorphism.

However, I am not sure how to construct an isomorphism from $\ker(d\pi)$ to $E_q$. Intuitevely it is clear to me that there is an isomorphism, since the vectors tangent to $\pi^{-1}(q)$ are "vertical" and hence projected down to 0, but I don't know how to construct it.

I need it because next I need to show that under this isomorphism the canonical symplectic form (defined as $\omega^{can}=-d\lambda^{can}$, with $\lambda^{can}_{(q,p)}=p\circ d\pi_{(q,p)}$ for any $(q,p)\in T^\ast X$) agrees with the canonical symplectic form $\omega_{T_qX}$ on $T_qX\times (T^\ast_qX)^\ast$, which is given by $\omega_{T_qX}((a,\phi),(b,\psi))=\psi(a)-\phi(b)$.

This is so far not working for me, and I feel like I'm simply not quite unraveling the definitions correctly and really getting confused at the technicalities. Any help or clarification is much appreciated.

EDIT If you consider the point $(q,0)\in T^\ast X$, why is $\lambda^{can}$ and thus $-d\lambda^{can}$ not trivial at this point? Since you act after everything with the zero map.

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1 Answer 1

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The isomorphism $ker(d\pi)\to E_q$ is easier to construct in the other direction. Given an element $v\in E_q$, consider $t\mapsto t\cdot v$ for $t\in\mathbb R$. This defines a smooth curve $\mathbb R\to E$, which actually has values in the fiber $E_q$. Taking the derivative at $t=0$ of this curve, one thus obtains a tangent vector in $T_{(q,0)}E$, which by construction lies in $\ker(\pi)$. Now you easily see that this defines a linear injection $E_q\to \ker(\pi)$, which must be an isomorphism since both spaces have the same dimension. (This also works outside the zero section.)

Concerning your Edit: It is true that $\lambda^{can}$ vanishes along the zero section, but this says nothing about its exterior derivative. Compare with the standard coordinate expression $\lambda^{can}=\sum p_idq_i$ vs. $d\lambda^{can}=\sum dp_i\wedge dq_i$ in points where all $p_i$ are zero.

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