# Cotangent bundle lift theorem

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!

• This is exercise 3.2F in Abraham and Marsden where you will find a sketch of the proof. Dec 7, 2013 at 7:30
• @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.
– Danu
Apr 2, 2019 at 13:07
• There's also a detailed proof in the book by Marsden & Ratiu, Introduction to Mechanics and Symmetry. (See the section on cotangent lifts.) Sep 30 at 15:08
• @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. Oct 5 at 19:52

[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.

• 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! Dec 8, 2013 at 0:47
• 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 $\alpha_p\in T_p^*M$.
– glS
Jan 16, 2022 at 18:49
• @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.
– 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 $d\theta$.