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I would like to calculate the $\mathbb{Z}$-Homology of $K(\mathbb{Z}/n,1)$ via the Fibration $$K(\mathbb{Z},1)\hookrightarrow K(\mathbb{Z},1)\twoheadrightarrow K(\mathbb{Z}/n,1)$$ and the Serre-Spectral-Sequence.

If we knew, that the local coeffitient system in $H_p(K(\mathbb{Z}/n,1);H_q(K(\mathbb{Z},1)))$ was trivial, then we could easily apply the Serre-Spectral sequence and derive $$H_p(K(\mathbb{Z}/n,1);\mathbb{Z})=\begin{cases}\mathbb{Z}&p=0\\ \mathbb{Z}/n &p=2k+1 \\ 0& else\end{cases}$$ The Problem here is obviously the non-trivial fundamental group of $K(\mathbb{Z}/n,1)$. This Problem is easily resolved for odd $n$ as there is no non-trivial morphism $$\mathbb{Z}/n\rightarrow Aut(\mathbb{Z})=\lbrace\pm 1\rbrace$$ for odd $n$. For even $n$ there is a non-trivial morphism, so we can not proceed using only formal arguments. Using Künneth one can assume $n=2^k$ if one wishes, but this also does not solve the real problem.

Is there some way to deduce the triviality of the local coeffitients (and are they trivial at all)?

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Let's use the definition of $BG$ from Hatcher, built out of simplicies $[g_0,\ldots, g_n]$ whose vertices are in $G$, glued together along faces $[g_0,\ldots, \hat g_i,\ldots, g_n]$, and taken modulo the group action $g[g_0,\ldots, g_n] = [gg_0,\ldots, gg_n]$.

In order to compute the monodromy action on the fibers, we should find explicit paths in $BG$ representing the classes of $\pi_1(G)$. For $g\in G$, there is the linear path $\gamma: I \to EG$ along the 1-simplex $[1,g]$, whose projection in $BG$ is the loop in $\pi_1BG$.

Now, if we have a surjection of groups $H \to G$ and we wish to understand the monodromy of the fibration $BH \to BG$, then we need to lift the loop $[1,g] \in \pi_1(BG)$ to a path in $BH$. But it is easy to make this lift, simply take $\tilde g \in H$ and use $[1,\tilde g]$ which is a loop in $BH$. Since it's a loop, the induced action on the fiber is trivial.

The action on the homology groups is induced by the monodromy action, so the corresponding local coefficient system is trivial too.

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