So I tried to compute the cohomology of $D_{2n}$, for n odd , $H^{k}(D_{2n}, \Bbb Z)$. using Lyndon SS. I have obtained a few obstacles:

My computation, using the fact that there is a $C_2$ action $H^q(C_m, \Bbb Z)$, shows that my $E_2^{pq}$ page looks like this:

$$ \Bbb Z/m \quad 0 \quad 0 \quad 0 \quad 0 \quad 0 \quad \cdots $$ $$ \cdots 0 \cdots $$ $$ \cdots 0 \cdots $$ $$ \cdots 0 \cdots $$ $$ \Bbb Z/m \quad 0 \quad 0 \quad 0 \quad 0 \quad 0 \quad \cdots $$ $$ \cdots 0 \cdots $$ $$ \cdots 0 \cdots $$ $$ \cdots 0 \cdots $$ $$ \Bbb Z \quad 0 \quad \Bbb Z/2 \quad 0 \quad \Bbb Z/2 \quad 0 \quad \cdots $$

Is this correct?

I suspect the correct $E_2$ page is more or less similar. But I am still unclear how this gives $H^k(D_{2n},\Bbb Z)$.

a) How is this $E_2$ page related to $H^k(D_{2n},\Bbb Z)$? I only know the case when page 2 collapses to an axis.

b) I suppose one has to show all differential are $0$ to compute $E_\infty^{p,q}$. In my case this is simple, since all are $0$.

c) Even if we know the $E_\infty$ page does this give $H_n$ - surely we cannot just take direct sum? **

** Supposing my calucations were correct - then we can go to c) directly: What we have here is the relation when $n \equiv 0 \pmod 4$ $$ 0 \rightarrow C_2 \rightarrow H^n \rightarrow C_m \rightarrow 0 $$


Your $E_2$ page is correct.

To continue, observe that everything nonzero in your $E_2$ page lies in some even bidegree $(2p,2q)$, and that the grading of the differential $d^i$ on the $E_i$ page is of bidegree $(1-i, i)$, and in particular, changes the parity of at least one of the degrees; in particular, $d^i$ must be identically zero for all $i$. So $E_2 = E_\infty$. This is always true for spectral sequences whose elements are concentrated in bidegrees $(p,q)$ with $p+q$ even.

Next, let us recall what it means that this spectral sequence converges: the calculational part of that means that the homology groups $H^k(D_{2n}, \Bbb Z)$ have a filtration $0 = F^0 \supset \cdots \supset F^i \supset F^{i+1} \supset \cdots$ and the associated graded $$\text{gr}^p H^{p+q}(D_{2n}, \Bbb Z) = E^{p,q}_\infty.$$

For us, on the lines with $p + q = 4n+2$, we see that there is only one nontrivial subquotient $F^p/F^{p+1}$, and in particular, $H^{4n+2} = E_\infty^{4n+2,0}$.

On the lines with $p + q = 4n >0$, we have two terms. There is $E_\infty^{4n,0} = F^0/F^1 H^{4n}$; because there is nothing until $q = 4n$, we see that $F^1 = F^2 = \cdots = F^{4n}$, but that $F^{4n+1} = 0$, and thus $F^{4n} = E_\infty^{0,4n}.$

In particular, we have a short exact sequence $0 \to E_\infty^{0,4n} \to H^{4n} \to E_\infty^{4n,0} \to 0$. Resolving this is the simplest case of what is usually called "solving an extension problem". If there are more than two nonzero terms on the $p+q = n$ line, then one has to resolve these iteratively, starting from $E_\infty^{0,4n}$ and walking down the line.

In this particular case, $E_\infty^{4n,0} = \Bbb Z/2$, and $E_\infty^{0,4n} = \Bbb Z/m$ where $m$ is odd. Extensions of abelian groups where the subgroup and quotient are of coprime order always split, and so in particular $H^{4n} = \Bbb Z/2 \oplus \Bbb Z/m \cong \Bbb Z/2m$ in this case.

If you are also interested in the multiplicative structure, one can write this ring as $\Bbb Z[c_1, c_2]/(2c_1, c_1^2 = mc_2)$, following the strategy of this answer for $S_3 = D_6$; there is no essential difference between that case and $D_{4n+2}$ in general. Here $|c_1| = 2$ and $|c_2| = 4$.

  • $\begingroup$ Thanks a lot. Just curious, is there a reason why your notation went from $E_\infty $ to $E^\infty$? $\endgroup$ – Bryan Shih Nov 16 '18 at 7:38
  • $\begingroup$ @CL. I normally think about homology spectral sequences and so I am more used to writing $E^\infty$. So I reverted by mistake halfway through. I will edit. $\endgroup$ – user98602 Nov 16 '18 at 16:25

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