# Definition of equivariant chain complex.

What do we mean by equivariant chain complex? Is it a chain complex with some property ? I looked in many references and i did not find a definition of the expression "equivariant chain complex", I met this expression when reading about homology with local coefficients and it seems that they use it as a name for the cellular chain complex of the universal covering of a finite connected CW-complex $$X$$. It seems like it has a link with the action of $$\pi_1(X)$$ on the universal covering $$\tilde X$$ by Deck transformations but in what sense is this action equivariant ? thank you for your help !

• Usually this means it's kosher with respect to some group acting on it. – Randall Aug 14 '19 at 17:44

It means the complex of chains $$c$$ on $$\widetilde X$$ which are "equivariant with respect to deck transformations", meaning that for each deck transformation $$f : \widetilde X \to \widetilde X$$ the formula $$f_\#(c)=c$$ holds.
• Do you mean that the induced homomorphism on the chain level $f_\#:C_p(\widetilde X)\longrightarrow C_p(\widetilde X)$ is the identity map ? where $C_p(\widetilde X)$ is the free abelian group on the $p-$cells of $\widetilde X$. – palio Aug 15 '19 at 14:00
• Yes, that's what it means. The one thing you have to be careful about is that this actually only makes sense when $\pi_1(X)$ is a finite group, because the free abelian group on the $p$-cells only allows finitely many cells to have nonzero coefficients (that problem, by the way, does not arise for equivariant cohomology, which is not a free abelian group over the cells, but instead a direct product over the cells). – Lee Mosher Aug 15 '19 at 14:42
• There is also one other thing to say. Although coefficients have not been mentioned, what makes things more interesting is that one is often (usually?) interested in "twisted coefficients" which means that the coefficients are some kind of $\pi_1(X)$-module; see this question. – Lee Mosher Aug 15 '19 at 14:44