Let $f: M \to M$ define an automorphism on the smooth manifold M.
Given a differential form $\omega \in \Omega^k$ is it true that the de Rham cohomology class of $\omega$ and $f^*\omega$ are the same? That is, does $[\omega]=[f^*\omega]$.
Let $f: M \to M$ define an automorphism on the smooth manifold M.
Given a differential form $\omega \in \Omega^k$ is it true that the de Rham cohomology class of $\omega$ and $f^*\omega$ are the same? That is, does $[\omega]=[f^*\omega]$.
No. One example: take the torus $X = \mathbb{R}^2/\mathbb{Z}^2$. The flip-flop on the factors interchanges the closed forms $dx$ and $dy$ which are linearly independent in $H^1(X)$.