9
$\begingroup$

Let $M$ be a smooth manifold and $T^\ast M$ be its cotangent bundle. Consider the tautological 1-form $\theta$ on $T^\ast M$ ($\theta=\sum y_i dx^i$ in local canonical coordinate systems).

A diffeomorphism $f:M\to M$ induces a pull-back lift $F=f^\ast:T^\ast M\to T^\ast M$.
It seems that we always have $F^\ast\theta=\theta$. But I don't know how to verify this.

I just heard the following cotangent bundle lift theorem:

If a diffeomorphism $F:T^\ast M\to T^\ast M$ preserves $\theta$, then $F=f^\ast$ for some diffeomorphism $f:M\to M$.

This looks too strong to me. Do you know how to prove this?

Thank you!

$\endgroup$
4
  • 5
    $\begingroup$ This is exercise 3.2F in Abraham and Marsden where you will find a sketch of the proof. $\endgroup$ Dec 7, 2013 at 7:30
  • 1
    $\begingroup$ @alvarezpaiva They refer to it as a result by Robbin-Weinstein, but do not provide the precise reference. I am unable to find any collaborative papers by the two (on MathSciNet), and was wondering if you know the paper that is being referred to. $\endgroup$
    – Danu
    Apr 2, 2019 at 13:07
  • 1
    $\begingroup$ 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 15:08
  • $\begingroup$ @Danu It might be Robbin-Weinstein in the same sense as Newton-Raphson or Cauchy-Riemann - you'd be hard pushed to find any collaborative papers in either of those pairs. $\endgroup$
    – JamesM
    Oct 5, 2023 at 19:52

2 Answers 2

15
$\begingroup$

[Edited typo 01/16/2022]

Let $\pi:T^*M \to M$ be the canonical projection. Given a diffeomorphism of the base $f:M\to M$, the pullback mapping $f^*:T^*M \to T^*M$ is again a diffeomorphism, and one has commutativity $d\pi\circ d(f^*) = df \circ d\pi$, where $df$ is the usual differential and $d(f^*):TT^*M \to TT^*M$ is the differential of the pullback diffeomorphism $f^*$.

From the commutativity of the pullback diagram, we find that $(f^*)^* \theta=\theta$, where we recall that the $1$-form $\theta$ on $T^*M$ is defined as $\theta_{(p,\alpha)}(v)=\alpha_p(d\pi_{(p,\alpha)}(v))$, where $v \in T_{(p,\alpha)} T^*M$.

Suppose now we are given a diffeomorphism $F:T^*M \to T^*M$ of the cotangent bundle. If $F$ does not preserve the fibres of $\pi$, then there is no chance of $F$ being induced by a diffeomorphism of the base (this even if $F^*\theta=\theta$). Supposing that $F$ $does$ preserve fibres, then we obtain a diffeomorphism $f:M\to M$ of the base. Now our task is to show that if $F^* \theta=\theta$, then $F=f^*$. So let's say we're given a diffeomorphism $G:T^*M \to T^*M$ with $G^*\theta=\theta$ and which induces the identity map on the base $M$. Taking differentials gives us the handy relation $d\pi \circ dG =d\pi$. Of course, I'm thinking of $G=F^{-1}\circ f^*$ and we want to find if $G$ is the identity mapping.

As $G$ induces the identity mapping on the base, the diffeomorphism $G$ induces at every point $p\in M$ a fibre-diffeomorphism $G_p:T^*_p M \to T^*_p M$. That is, $G$ acts like $(p,\alpha)\mapsto (p, G_p\alpha)$. We want to show that $G_p$ is the identity mapping. For this, let $(p,\alpha) \in T^*M$ and $v\in T_{(p,\alpha)}T^*M$. Now we compute

$G^*\theta_{(p,\alpha)}(v)=\theta_{(p, G_p\alpha)}(dG(v)) =G_p\alpha(d\pi(dG(v))).$

Our handy relation making this last part equal to $G_p\alpha(d\pi(v))$. If $G^*$ preserves $\theta$, then we must have equality $G_p\alpha(d\pi(v))=\alpha(d\pi(v))$ for all $v$, i.e. $G_p$ is the identity.

$\endgroup$
3
  • 3
    $\begingroup$ I checked the reference provided by alvarezpaiva. By considering the symplectic vector field induced by $\theta$, such $F$ must preserve the fibers. The pullback relation gives $\pi\circ F=f^{-1}\circ \pi$, which is used to prove $f^\ast\theta=\theta$. There is another typo at the end of the second paragraph: $\pi\circ G=\pi$. Now I understand your answer. Thank you! $\endgroup$
    – Pengfei
    Dec 8, 2013 at 0:47
  • $\begingroup$ isn't $d(f^*)\circ d\pi$ ill-defined here? We have $\pi:T^* M\to M$, thus $d\pi:TT^* M\to TM$, but $d(f^*): TT^*M\to TT^* M$. So $d(f^*)$ takes as input elements of $TT^*M$ while $d\pi$ outputs elements of $TM$. Same for $d\pi\circ df$: $df$ gives things in $TM$ wihle $d\pi$ wants things in $TT^* M$. Am I misunderstanding something? The relation makes more sense if written as $d\pi\circ d(f^*)=df^{-1}\circ d\pi$, I think, which should follow from $f^{-1}\circ \pi=\pi\circ f^*$, due to $f^*(\alpha_p)\equiv\alpha_p\circ df|_{f^{-1}(p)}\in T^*_{f^{-1}(p)}M$ if $\alpha_p\in T_p^*M$. $\endgroup$
    – glS
    Jan 16, 2022 at 18:49
  • $\begingroup$ @glS You are correct I wrote composition in the wrong order, thank you for clarifying! I was trying to say $d\pi \circ d(f^*)=df \circ d\pi$, as per the commutative diagram $f \circ \pi = \pi \circ f^*$. But rereading my answer, I dont think I fully understood the question when I wrote it. $\endgroup$
    – JHM
    Jan 16, 2022 at 19:08
5
$\begingroup$

Just as a side-note. The proposition does not imply that every symplectomorphism of $T^*M$ is a cotangent lift of a diffeomorphism on $M$. As a counter example, let $\sigma$ be a non-zero closed one-form on $M$ and consider the map $F(x,p) = (x,p + \sigma(x))$. Notes that $F$ is not the cotangent lift of any diffeomorphism on $M$. Also $F^*d \theta = d(F^*\theta) = d( \sigma^H + \theta) = d\sigma^H + d\theta = d\theta$ where $\sigma^H$ is the one-form on $T^*M$ obtained by horizontal lift. Needless to say, if $\sigma$ is closed, then so is $\sigma^H$. Thus $F$ preserves the canonical symplectic form $d\theta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.