When does the pullback of differential forms preserve cohomology classes: $[\alpha]=[F^* \alpha]$? Let $M^n$ be a smooth manifold, compact with no boundary. Consider a diffeomorphism $F: M \rightarrow M.$
The pullback $F^*$ defines a linear isomorphism on the de Rham cohomogy spaces $F^*: H^k_{dR}(M)\rightarrow H^k_{dR}(M)$ where $0\leq k \leq n.$ 
My question is: under which conditions this is the identity map on $H^k_{dR}(M)$? Also, if $M$ is a complex closed manifold, under which conditions a holomorphic diffeomorphism $F$ would give the identity on $H^{r,r}(M)$?
Any help is greatly appreciated!
 A: First, consider the case when your complex manifold is biholomorphic to ${\mathbb C}P^n$. The group of biholomorphic automorphism of ${\mathbb C}P^n$ equals $PGL(n+1, {\mathbb C})$ and the latter is connected. Hence, every biholomorphic automorphism is homotopic to the identity map, hence, induces the identity map on the cohomology ring. 
(I am cheating here: The usual way to classify biholomorphic automorphisms of ${\mathbb C}P^n$ is to show that they send linear subspaces to linear subspaces.) 
More generally, if $f$ is a diffeomorphism of $M$ then one can say the following:
The integer cohomology ring of $M$ is isomorphic to the truncated polynomial ring ${\mathbb Z}[z]/(z^{n+1})$ with the degree shift by the factor of 2. Therefore, if $f: M\to M$ is a diffeomorphism it sends the ring generator $H\in H^2(M, {\mathbb Z})$ to $\pm H$. Accordingly, the induced ring automorphism is $Id$ if and only if $f^*(H)=H$. 
More generally, if you are interested in $M$ which is a smooth complex-projective variety whose integer cohomology ring is isomorphic to that of ${\mathbb C}P^n$, then the Poincare dual of the generator $H$ of $H^*(M, {\mathbb Z})$  is represented by a smooth complex hypersurface $C$. If $f: M\to M$ is biholomorphic then $f(C)$ is a smooth complex hypersurface and, hence, represents  the Poincare dual to  $H$  (rather than $-H$). Hence, $f$ induces the identity map on the cohomology ring of $M$. 
I am not sure what happens if $M$ is just a complex manifold with cohomology of ${\mathbb C}P^n$.  If $M$ is Kahler then the conclusion is the same as before. 
