Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question

1 Answer 1

up vote 10 down vote accepted

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 $X$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.