Triviality of Sp(TM) Let M be a symplectic manifold of dimension $2n$ and $TM$ denote its tangent bundle. Let Sp(TM) denote the bundle over M whose fibers are linear maps preserving symplectic structure on M. Is Sp(TM) trivial (i.e can it be written as $M \times Sp(2n)$) because Id gives a cross section and existence of a global cross section imply the fiber bundle is trivial?
Edit: I will include more context. I want to understand the paper https://arxiv.org/pdf/1305.6810.pdf , what I am asking is defined at page 5 around the middle of the page starting with Letting Sp(TM).. Now I am thinking it is perhaps not a principal G bundle it is maybe rather just a fiber bundle with a fiber equal to Lie Group and since there is no equivariant G action, I cannot use the existence of Id section to trivialize the bundle. The concepts are rather new to me, so it would be great if someone can present their perspective.
 A: Adding this in response to Jason DeVito's comment: $\operatorname{Sp}(TM)$ as defined is not a principal fibre bundle. In fact, it is an associated bundle to the symplectic frame bundle (which is principal).
To see this: let me use the notation $\mathcal{F}_{\operatorname{Sp}}(M)$ for the symplectic frame bundle of $M$. One way of defining $\mathcal{F}_{\operatorname{Sp}}(M)$ is that its fibre $\mathcal{F}_{\operatorname{Sp}}(M)_p$ over $p\in M$ consists of all symplectic linear isomorphisms from $\mathbb{R}^{2n}$ with its standard symplectic form $\Omega$ to $(T_pM,\omega_p)$
$$
\mathcal{F}_{\operatorname{Sp}}(M)_p = \{ b_p:\mathbb{R}^{2n}\to T_pM \mid b_p^*\omega_p = \Omega\}
$$
(see for example Metaplectic-c Quantomorphisms by Jennifer Vaughan, Section 3.2, where the symplectic frame bundle is denoted $\operatorname{Sp}(M,\omega)$). This is a principal $\operatorname{Sp}(2n,\mathbb{R})$-bundle, with action given by composition
$$
(b_p, T)\in \mathcal{F}_{\operatorname{Sp}}(M) \times \operatorname{Sp}(2n,\mathbb{R}) \mapsto b_p\circ T \in \mathcal{F}_{\operatorname{Sp}}(M).
$$
I claim that (your) $\operatorname{Sp}(TM)$ is isomorphic to the associated bundle
$$
\mathcal{F}_{\operatorname{Sp}}(M)\times_{\operatorname{Sp}(2n,\mathbb{R})} \operatorname{Sp}(2n,\mathbb{R})
$$
where the action of $\operatorname{Sp}(2n,\mathbb{R})$ on itself is via conjugation. That is, for $b_p\in \mathcal{F}_{\operatorname{Sp}}(M), S\in\operatorname{Sp}(2n,\mathbb{R})$, the associated bundle is the quotient under the equivalence relation
$$
(b_p, S) \sim (b_p\circ T, T^{-1}\circ S \circ T)
$$
with $T\in \operatorname{Sp}(2n,\mathbb{R})$. The isomorphism between this associated bundle and $\operatorname{Sp}(TM)$ is given by
$$
[b_p,S] \mapsto b_p\circ S\circ b_p^{-1}
$$
where $[b_p,S]$ denotes the equivalence class of $(b_p,S)$. It is not too difficult to check that this is well-defined, and a bijection.
This hopefully clarifies that $\operatorname{Sp}(TM)$ is not a principal bundle, but rather an associated fibre bundle.
