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!



Cotangent bundle lift theorem
[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.


