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Let $M$ be a compact manifold and $\varphi:M\rightarrow M$ a diffeomorphism. The invariant differential forms $$ \Omega^{k}_{inv}(M)=\{\alpha\in\Omega^{k}(M):\varphi^{*}\alpha=\alpha\} $$ form a subcomplex $(\Omega_{inv}(M),d)$ of the de Rham complex. Denote by $H_{inv}(M)$ its cohomology.

Question: Is there an example where the map $H^{1}_{inv}(M)\rightarrow H^{1}(M)$ is not injective?

It seems like the map is injective when forms can be averaged with respect to $\varphi$ in some way. For instance if $\varphi$ has finite order, or more generally if $\varphi$ is an isometry for some Riemannian metric $g$, in which case one can average over the closure $\overline{\{\varphi^{n}\}}$ in the compact Lie group $Isom(M,g)$. But I think there should be an example where the map is not injective.

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    $\begingroup$ If you allow non-compact manifolds, it seems to me that the diffeomorphism $x \mapsto x+1$ of $\mathbb{R}$ should provide a counterexample. $\endgroup$
    – Francesco Polizzi
    Commented Oct 25, 2022 at 17:31
  • $\begingroup$ Is this the same question: mathoverflow.net/questions/294918/… $\endgroup$
    – Chris McDaniel
    Commented Oct 25, 2022 at 20:36
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    $\begingroup$ @ChrisMcDaniel Not really. My question is indeed about a specific group action $(G=\mathbb{Z})$, but the counterexample to injectivity in your link concerns a non-compact manifold (essentially what Francesco Polizzi says in his comment above) $\endgroup$
    – studiosus
    Commented Oct 25, 2022 at 20:47
  • $\begingroup$ @studiosus yes but he also says that the cohomology of invariant differential forms is isomorphic to the invariant cohomology classes, which does indeed inject into the cohomology ring. $\endgroup$
    – Chris McDaniel
    Commented Oct 26, 2022 at 0:35
  • $\begingroup$ @ChrisMcDaniel Only when the group acting is a compact Lie group, isn't it? This doesn't apply to the case $G=\mathbb{Z}$. $\endgroup$
    – studiosus
    Commented Oct 26, 2022 at 8:43

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There is no such thing: any $\phi$-invariant exact 1-form is a differential of $\phi$-invariant function.

Indeed, let $\alpha$ be an exact $\phi$-invariant form, $\alpha=df$, where $f$ is not $\phi$-invariant. Then $d(\phi^* f -f)=0$, hence $\phi^* f = f + C$, where $C$ is a constant. This gives $\sup f = \sup \phi^* f = \sup f + C$, which is a contradiction.

For higher cohomology I think there are some examples.

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