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!

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    $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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 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 at 19:52

2 Answers 2


[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.

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    $\begingroup$ 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

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 $d\theta$.


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