# Rational Homology of a Covering Space

I have heard that the rational homology of a covering space is easy to compute, compared with the ordinary homology. However, I don't know any details about that. Can anyone help me? Any reference will be greatly appreciated.

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## 1 Answer

There are several reasons why this is true. Here's one: For a finite cover $p:\tilde X\to X$, there is a transfer map $t:H_i(X)\to H_i(\tilde X)$ which, on the chain level, takes a chain $\sum a_i \sigma_i$ to $\sum a_i \sum g\sigma_i$, where the inner sum is over all lifts of $\sigma_i$. This holds with any coefficients, but over the rationals, $p\circ t$ is multiplication by the index of the cover, an isomorphism. Hence the transfer is injective, and so the homology of $\tilde X$ contains a copy of the homology of $X$.

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In particular, if the rational homology of a finite cover vanishes, then so does the rational homology of the original space. Corollary: the rational homology of finite groups vanishes. –  Jim Conant Feb 27 '12 at 3:39