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Both Abraham-Marsden and Da Silva seem to imply that given a symplectomorphism $g:T^\ast X\to T^\ast X$ which preserves the tautological $1$-form $\alpha$, it must be that $g$ is fibre preserving.

In fact, this is addressed in the question Cotangent bundle lift theorem.

The answer given to this question stresses very clearly that g must preserve fibres to have any chance of being a lift, even if it preserves the canonical 1-form.

I would really like to see a counterexample to this. That is, a symplectomorphism which preserves the canonical $1$-form but is not the lift of a diffeomorphism.

I have struggled with trying to prove both Abraham-Marsden and Da Silva's claim. Any help would be greatly appreciated.

I can show the following:

If $V$ is the symplectic dual of $\alpha$ (i.e. $\omega(V,\cdot)=\alpha$), then $g_\ast V=V$, or equivalently $g\circ\theta_t=\theta_t\circ g$ where $\theta_t$ is the flow of $V$. Also I can show that $\theta_t$ is fibre preserving (i.e. $\theta_t(T_p^\ast X)=T_p^\ast X)$. This last fact follows from actually computing $\theta_t$ which is $\theta_t(p,\xi_p)=(p,e^t\xi_p)$.

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    $\begingroup$ You can find a detailed proof in my book on Poisson geometry and deformation quantization (Satz 3.2.11, iii.) which, however, is in german ;) $\endgroup$ Jul 11, 2014 at 6:00
  • $\begingroup$ There's also a detailed proof in the book by Marsden & Ratiu, Introduction to Mechanics and Symmetry. (See the section on cotangent lifts.) $\endgroup$
    – JamesM
    Sep 30, 2023 at 15:07

1 Answer 1

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The vector field $V$ is the radial vector field in the fibers, and it vanishes along the zero locus of $\alpha$, which is the zero section $Z\subset T^*M$. Thus, $g$ must preserve $Z$ and hence be a diffeomorphism of $Z$, say $f:Z\to Z$. Since $Z\to M$ is a diffeomorphism, this can be regarded as a diffeomorphism $f:M\to M$. Since $g$ preserves $V$, it must, for each $z\in Z$, i.e., $z = 0_x$ for $x\in M$, carry the $\alpha$-limit set $S_z$ of $z$ with respect to $V$ into (and diffeomorphically onto) the $\alpha$-limit set of $g(z) = 0_{f(x)}$, but $S_{0_x}=T^*_xM$ and $S_{0_{f(x)}}=T^*_{f(x)}M$. Thus, $g(T^*_xM) = T^*_{f(x)}M$. The rest is now clear, I think.

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    $\begingroup$ Professor Bryant, thanks very much for your answer. I now understand. Also, I'm actually a current attendant of the Geometry/Topology via G2 seminar for which you just gave your last lecture today. I really enjoyed and appreciated your lectures and so wanted to thank you for that as well! $\endgroup$
    – JonHerman
    Jul 11, 2014 at 12:37
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    $\begingroup$ You're welcome, and thanks for the feedback! $\endgroup$ Jul 11, 2014 at 12:47

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