I stumbled upon this in Bredon and don't know how to understand this. What I have given is a smooth action from $G$ on $M$. I did not find any special definiton for the above phrase, so I expect an action $Aut(G) x H^p (M) \to H^p (M) $, but do not have any clue how that would look like.
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The action $G \times H^p(M,\mathbb{R}) \to H^p(M,\mathbb{R})$ is as follows: each element $g \in G$ acts as a diffeomorphism of $M$, say $\phi(g) \colon M \to M$. Then this map $\phi(g)$ pulls back differential forms $\xi$, say as $\phi(g)^* \xi$. To get an action, since pullbacks change direction of composition, we need to define $g \xi$ to mean $\phi(g^{-1})^* \xi$. This action of $G$ on $p$-forms commutes with exterior derivative, because pullback of forms commutes with exterior derivative. So this action preserves closedness and exactness of differential forms, and preserves the degree of each form. Hence the forms in the same cohomology class of a form $\xi$ are mapped by $\xi \mapsto g \xi=\phi(g^{-1})^*\xi$ to forms in the same cohomology class. Hence we descend to a map on cohomology defined by $g[\xi]=[g\xi]$. For example, if we let the group $G={\pm 1}$ act on the circle $M=S^1$ by $(x,y) \mapsto (-x,y)$, then the action on the cohomology generators $d\theta$ is $(-1)d\theta=-d\theta$, since $$d\theta=\frac{x \, dy - y \, dx}{x^2+y^2}$$ and the action flips the signs of $x$, so also of $dx$.
