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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.

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    $\begingroup$ $Sp(TM)$ is a principal $Sp(2n)$-bundle so $Sp(2n)$ acts simply transitively on the fibers, but the fibers aren't canonically identified with $Sp(2n)$ so it doesn't make sense to say which element corresponds to the identity. If $Sp(TM)$ admits a section, then the corresponding point in each fiber gives a choice of identity. $\endgroup$ Commented Mar 16, 2022 at 3:02
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    $\begingroup$ I see. I thought $Sp(TM)$ was the symplectic frame bundle (for which my comments stand), but you're describing a different bundle. $\endgroup$ Commented Mar 16, 2022 at 3:46
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    $\begingroup$ I also would have thought $\operatorname{Sp}(TM)$ stands for the symplectic frame bundle (which as Michael says, is not trivial in general). From the tags on your question, you seem to indicate that $\operatorname{Sp}(TM)$ is a principal bundle. If so, how does $\operatorname{Sp}(2n,\mathbb{R})$ act on it, i.e. if $L\in\operatorname{Sp}(TM)_p$ (so $L:T_pM\to T_pM$), and $g\in\operatorname{Sp}(2n,\mathbb{R})$, what is $L\cdot g$? $\endgroup$
    – user17945
    Commented Mar 16, 2022 at 5:11
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    $\begingroup$ @Lucky, I believe your amended assessment is correct - $\operatorname{Sp}(TM)$ is not a principal bundle, and so the existence of a global section does not guarantee it is trivial. $\endgroup$
    – user17945
    Commented Mar 16, 2022 at 16:24
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    $\begingroup$ Then it's not clear to me whether or not the bundle is principal. Locally, it's easy to define the action of $Sp(2n,\mathbb{R})$: using Darboux coordinates, we can locally write $\omega = \sum d p^i \wedge dq^i$. Then $Sp(2n,\mathbb{R})$ acts on $Sp(TM)_p$ by composition. But I haven't been able to work out whether these patch together nicely. $\endgroup$ Commented Mar 16, 2022 at 16:27

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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.

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  • $\begingroup$ This does show that it is an associated fibre bundle however I want to make sure I understand so I will ask: strictly speaking this does not prove that this fiber bundle (Sp(TM) does not admit some Sp(2n) principal bundle structure, right? $\endgroup$
    – user2222
    Commented Mar 16, 2022 at 17:57
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    $\begingroup$ There would certainly be cases where it could admit a principal $\operatorname{Sp}(2n,\mathbb{R})$ structure. For example, suppose $\mathcal{F}_{\operatorname{Sp}}(M)$ were trivial, i.e. isomorphic to $M\times\operatorname{Sp}(2n,\mathbb{R})$. Then we would have $\operatorname{Sp}(TM) \simeq M\times \operatorname{Sp}(2n,\mathbb{R})$, and this has an obvious right $\operatorname{Sp}(2n,\mathbb{R})$-action. But this is not canonical (since the trivialisation of $\mathcal{F}_{\operatorname{Sp}}(M)$ is not canonical). I think the point is there's no obvious general construction. $\endgroup$
    – user17945
    Commented Mar 16, 2022 at 18:09
  • $\begingroup$ I agree that this answer shows that it's unlikely that $Sp(TM)$ will be a principal bundle in general. Still, it would be nice to see an explicit counterexample. I'll think about this for a bit and see if I can come up with one on my own. Thanks! $\endgroup$ Commented Mar 16, 2022 at 19:19

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