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If we have a simplical complex $K$, then we are able to define $C_i(K)$ as the free abelian group over $\mathbb Z_2$ with the basis of all $i$-dimensional simplices. By using the boundary map we are able to define a long sequence of $C_i(K)$ and as result we can compute the ith homology group $H_i(K,\mathbb Z_2)$.

Now let us go to the group homology for a given group $G$. $H_n(G,\mathbb Z_2)$ can be computed using a projective resolution of $\mathbb Z_2G$.

Now my question is How can I construct a simplicial complex $K$ associated to $G$ such that $H^i(K,\mathbb Z_2)$ is exactly the same as $H^i(G,\mathbb Z_2G)$?

I was thinking of the Cayley complex but unfortunately the maximum dimension of cayley complex is two and so it is not a good choice.

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  • $\begingroup$ You seem to be switching between homology and cohomology here. Do you mean the group cohomology (which is what is most commonly considered)? In which case you want a resolution of the trivial module, not of the group algebra (which is already projective). $\endgroup$ – Tobias Kildetoft Jul 4 '16 at 18:11
  • $\begingroup$ @TobiasKildetoft you are right $\endgroup$ – Jivid Jul 4 '16 at 18:12
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Yes, the candidate is $BG$ the classifying space of $G$. You can endow $G$ with the discrete topology, you have the universal bundle $EG\rightarrow BG$ which is a covering and $EG$ is contractible $BG$ is a $K(G,1)$, apply Serre 1.5 p.91

Serre, Jean Pierre. Cohomologie des groupes discrets.

http://www.maths.ed.ac.uk/~aar/papers/serrecoh2.pdf

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