Diffeomorphism-invariant differential forms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:08:45Z http://mathoverflow.net/feeds/question/72557 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72557/diffeomorphism-invariant-differential-forms Diffeomorphism-invariant differential forms Kamran Reihani 2011-08-10T06:02:39Z 2011-08-10T08:09:51Z <p>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 $G$".</p> <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/72557/diffeomorphism-invariant-differential-forms/72563#72563 Answer by Ben McKay for Diffeomorphism-invariant differential forms Ben McKay 2011-08-10T08:09:51Z 2011-08-10T08:09:51Z <p>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 $M$.</p>