# Computing cohomology of dihedral group in detail

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 $$|c_2| = 4$$.

• Thanks a lot. Just curious, is there a reason why your notation went from $E_\infty$ to $E^\infty$? – Bryan Shih Nov 16 '18 at 7:38
• @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. – user98602 Nov 16 '18 at 16:25