# Diffeomorphism-invariant differential forms

We know if a compact Lie group $G$ acts smoothly on a smooth manifold $M$, then for each $k$-form $\omega$ on $M$, we can simply construct a $G$-invariant $k$-form by "averaging over translations by $G$".

Now, suppose we have a (discrete) subgroup $\Gamma$ of ${\rm diff}(M)$, the full diffeomorphism group of $M$. For example, $\Gamma$ may be all of ${\rm diff}(M)$. Then there may not exist any $\Gamma$-invariant differential forms on $M$. But, is there any (principal) fiber bundle over $M$, with a natural lifting of the action of $\Gamma$, which carries $\Gamma$-invariant differential forms of a certain order $k$, or of any arbitrary order? I suspect the Jet bundle $J_k(M)$ of order $k$, or ultimately, the infinite jet bundle $J_\infty(M)$ may give the answer. But I am not sure.

The case $\Gamma$ equals integers, namely, generated by a single diffeomorphism is also quite interesting to me. Namely, given a single diffeomorphism $\phi$ on a manifold, can we construct a fiber bundle on $M$ that carries a $\bar\phi$-invariant differential form? Here, $\bar\phi$ is the natural" lifting of $\phi$ to the fiber bundle in question.

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If $\Gamma$ is discrete and countable, take $M\times\Gamma$ with $\rho_1(\gamma)(m,\delta)=(\gamma m,\gamma\delta)$. It is a principal $\Gamma$-bundle via $\rho_2(\gamma)(m,\delta)=(m,\delta\gamma^{-1})$. To get an invariant form, just choose any one on $M\times\{1\}$ and translate. – user2035 Aug 10 '11 at 7:17
This is interesting. But what if we restrict to the case of connected bundle, and what if we take the full diffeomorphism group which is uncountable? – Kamran Reihani Aug 10 '11 at 18:03

Let $FM$ be the frame bundle, i.e. the bundle of pairs $(m,u)$ where $m \in M$ and $u : T_m M \to \mathbb{R}^n$ is a linear isomorphism. Then define the map $\pi : (m,u) \in FM \mapsto m \in M$. Then define a 1-form $\omega$ on $FM$ by $\omega_{(m,u)} = u \circ \pi'$, i.e. on any vector $v \in T_{(m,u)} FM$, we let $\omega(v) = u(\pi'(v))$. Then $\omega$ is invariant under any diffeomorphism of $M$. In fact, the group of diffeomorphisms of $FM$ leaving $\omega$ invariant and commuting with the right $GL(n,\mathbb{R})$-action $(m,u)g=(m,g^{-1}u)$ is precisely the diffeomorphism group of $M$.
This is great! It is indeed very close to what I had in mind, especially that $J_1(M)=FM$, by considering I assume the action of $\Gamma$ on $M$ naturally lifts to $FM$ by $\bar\phi(m,u):=(\phi(m),\phi'(m)u)$. I have three questions: 1) what do you think for obtaining invariant $k$-forms? Do we need higher order frame bundles, or higher order jet bundles? 2) Can we think this construction as "the canonical one"? I mean, is there any way we can consider this as some sort of a universal construction? – Kamran Reihani Aug 10 '11 at 18:58
3) If we start by a smaller group, say $\Gamma$ is generated by a single diffeomormism, or finitely many of them, do we still need a bundle as large as the frame bundle? This may have something to do with the comment made by a-fortiori above. What do you think? – Kamran Reihani Aug 10 '11 at 19:00
Ok, my formula for $\bar\phi$ makes sense if $u:\Bbb{R}^n\rightarrow T_mM$. But with your direction, I should write $\bar\phi(m,u):=(\phi(m),u\circ (\phi'(m))^{-1})$, right? – Kamran Reihani Aug 10 '11 at 19:12
Since this $\omega$ is an invariant 1-form valued in $\mathbb{R}^n$, it consists of $n$ linearly independent 1-forms valued in $\mathbb{R}$, say $\omega^i$. Any constant coefficient wedge product of these gives an invariant form, so there are invariant forms of all degrees up to $n$ on $FM$. – Ben McKay Aug 11 '11 at 7:59
If you have a group acting on a manifold, smaller than the diffeomorphism group, then there might be a more complicated family of invariant differential forms (and even invariant functions), and a smaller bundle could do. For example, a symplectomorphism preserves a 2-form on the manifold itself, while every diffeomorphism preserves a 1-form n $T^*M$. Every diffeomorphism preserves a $k$-form on $\Lambda^k T^*M$, etc. – Ben McKay Aug 11 '11 at 8:01