# 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. – alvarezpaiva Dec 7 '13 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 at 13:07

We just need to relax and remember definitions.

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 $d(f^*) \circ d\pi=d\pi \circ df$, where $df$ is the usual differential and $d(f^*):TT^*M \to TT^*M$. From this commutativity it's easy to see that indeed $(f^*)^* \theta=\theta$, 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 $v$ lies in the tangent space of the `point' $(p,\alpha) \in T^*M$ in the cotangent bundle.

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! – Pengfei Dec 8 '13 at 0:47

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 $d\theta$.