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This is mostly a question to make sure I have some signs correct. Let a group $G$ act on a space $X$. As notation, for $g \in G$ let $\phi_g : X \rightarrow X$ be the map induced by the action of $g$. It is then clear that we get an action on the homology of $X$ as follows: $$g \cdot \vec{v} = (\phi_g)_{\ast}(\vec{v}) \quad \quad (g \in G, \vec{v} \in H_{\ast}(X)).$$ Here $(\phi_g)_{\ast} : H_{\ast}(X) \rightarrow H_{\ast}(X)$ is the induced map.

In the situation I'm in, I want an action on the cohomology of the space. The problem then is that the naive recipe $$g \cdot \vec{v} = (\phi_g)^{\ast}(\vec{v}) \quad \quad (g \in G, \vec{v} \in H^{\ast}(X))$$ does not yield and action since we have $$(\phi_{g g'})^{\ast} = (\phi_{g'})^{\ast} (\phi_g)^{\ast}.$$ My guess is that the right thing to do would be to use $$g \cdot \vec{v} = (\phi_{g^{-1}})^{\ast}(\vec{v}) \quad \quad (g \in G, \vec{v} \in H^{\ast}(X)).$$ Am I right that this is the reasonable thing to do? Thanks!

Mike
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1 Answers1

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Yes, this is correct.

To convince yourself that this is the right thing to do, you can use the example of how a group acts on a representation $V$ and its linear dual (one has to use inverse as you did). The standard way to calculate cohomology of a space at the level of chains is to take the simplicial chain complex of the space and dualize it.

Steven Sam
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