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.

  • $\begingroup$ 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. $\endgroup$ – user2035 Aug 10 '11 at 7:17
  • $\begingroup$ 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? $\endgroup$ – 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 $M$.

  • $\begingroup$ 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? $\endgroup$ – Kamran Reihani Aug 10 '11 at 18:58
  • $\begingroup$ 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? $\endgroup$ – Kamran Reihani Aug 10 '11 at 19:00
  • $\begingroup$ 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? $\endgroup$ – Kamran Reihani Aug 10 '11 at 19:12
  • $\begingroup$ 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 $FM$. $\endgroup$ – Ben McKay Aug 11 '11 at 7:59
  • $\begingroup$ 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. $\endgroup$ – Ben McKay Aug 11 '11 at 8:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.