Krdytkyu

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 what is usually called "solving a filtration 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.