Homology of Covering Spaces - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T21:21:05Z http://mathoverflow.net/feeds/question/94465 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94465/homology-of-covering-spaces Homology of Covering Spaces Zuriel 2012-04-18T21:18:23Z 2012-04-18T22:24:55Z <p>Let $A$ be a subgroup of a group $G$. Then since $A$ is a subgroup of the fundamental group $\pi_1(K(G,1))=G$, there is a covering space $p\colon Y\to K(G,1)$ with $p_*(\pi_1(Y))=A$. So the homology of $Y$ should be completely determined by $A$ and $G$. Suppose that $A$ and $G$ is known, how can one compute $H_*(Y)$, the homology groups of $Y$? </p> http://mathoverflow.net/questions/94465/homology-of-covering-spaces/94470#94470 Answer by Igor Rivin for Homology of Covering Spaces Igor Rivin 2012-04-18T21:40:17Z 2012-04-18T21:40:17Z <p>Short answer: it is not so easy (especially for infinite index $A$) Long answer: Read Ken Brown's "Cohomology of groups". </p> http://mathoverflow.net/questions/94465/homology-of-covering-spaces/94475#94475 Answer by Jim Conant for Homology of Covering Spaces Jim Conant 2012-04-18T22:24:55Z 2012-04-18T22:24:55Z <p>This is not a trivial problem. A favorite example of mine is the case of a knot complement, $S^3\setminus K$. (It is known that these are Eilenberg-Maclane spaces.) If you pick $A$ to be the commutator subgroup of $\pi_1(S^3\setminus K)=G$, then $Y$ is called the universal Abelian cover, and $H_1(Y;\mathbb Q)$ turns out to be a torsion $\mathbb Q[t,t^{-1}]$-module, called the Alexander module of the knot. The order of the Alexander module is called the Alexander polynomial. One can calculate the Alexander module from the fundamental group using Fox calculus.</p>