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This is a follow up to this question.

In the answer by Qmechanic, they state that the symplectic group, $Sp(2n,\mathbb{R})$, is the group of linear, time-independent canonical transformations.

If we consider a canonical transformation as a symplectomorphism on phase space (as per V. I. Arnold here), how can we restrict this to linear transformations? Since linearity is only defined if the phase space has a vector space structure, which in general it doesn't. More generally, how can we arrive at the symplectic group, from the symplectomorphisms on phase space?

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    $\begingroup$ Well, it's the Jacobian which satisfies the condition to be a symplectic group transformation, right? $\endgroup$ Feb 7, 2021 at 3:01

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Let there be given a $2n$-dimensional symplectic manifold $(M,\omega)$.

  1. OP is right that geometrically speaking the symplectic group $Sp(2n,\mathbb{R})$ is only guaranteed to be canonically/manifestly imbedded into the groupoid of symplectomorphisms if the manifold $M$ is a vector space.

  2. Given a preferred/distinguished/fixed/fiducial Darboux/canonical coordinate system $(q^1,\ldots,q^n,p_1,\ldots,p_n)$ with an origin, one can use it to define a linear structure in a neighborhood $U\subseteq M$, and in this way locally define a notion of linear transformations. (This is the pragmatic attitude of my linked Phys.SE answer.)

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