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I'm working through Hatcher's unfinished book Spectral Sequences in Algebraic Topology and have found myself stuck on Exercise 2 on page 23:

Compute the Serre spectral sequence for homology with $\mathbb{Z}$ coefficients for the fibration $K(\mathbb{Z}_2,1) \rightarrow K(\mathbb{Z}_8,1) \rightarrow K(\mathbb{Z}_4,1).$

The question asks to compare with Example 1.6 in the text (a similar computation), but when I try to write out the first few pages, I can't see anything close to a nice pattern. Is there something I'm missing?

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1 Answer 1

It would be nice to see what you've got for the $E_2$-page. Firstly, note that $K(\mathbb{Z}/m,1)$ is just an infinite dimensional lens space. We know (pp. 146 of Hatcher - Algebraic Topology) that these have homology $$H_i(K(\mathbb{Z}/m,1)) = \begin{cases} \mathbb{Z} &\text{ for } i=0 \\ \mathbb{Z}/m &\text{ for } i \text{ odd} \\ 0 &\text{ for } i \text{ even.} \\ \end{cases} $$

Thus we can calculate $E_2^{p,q} = H_p(K(\mathbb{Z}/4,1),H_q(K(\mathbb{Z}/2,1)))$. Note that because everything disappears in even dimensions our $E_2$-term looks like similar to the spectral sequence given in Example 1-6, except now we have $\mathbb{Z}/4$'s on the row $q=0$ (and $n \ne 0$).

Once we have calculated the $E_2$-term we now need to work out the differentials. Again a similar methodology to the example in Hatcher now works. For example there is a $\mathbb{Z}/2$ in the $n=2$ diagonal, which must be killed. There is only one possible way to kill this, and that is for there to be a map $d_2:E_2^{3,0} = \mathbb{Z}/4 \to E_2^{1,1} = \mathbb{Z}/2$. This leaves a $\mathbb{Z}/2$ in the $(3,0)$ position - but this is nice, because then this leaves us with three $\mathbb{Z}/2$'s in the $n=3$ diagonal, which is where we get the $\mathbb{Z}/8$ we need in the homology.

Proceed from here!

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