# Group cohomology revisited: Quaternionic unit group

Published: Tue 20 June 2006

In a previous installment, we calculated $$H^*(D_8)$$ with some amount of success. For that post, I said that I was going to calculate the cohomologies of $$D_8$$ and of $$U(\mathbb H)$$ by hand - and I've been at it for the latter group since then. With some help from my advisor - mainly with executing the obvious algorithms far enough that I get decent material to work with - I know have it.

## The group

So, for starters, we need a presentation of $$\mathbb H$$ such that we can work well with it. We all know that $$\mathbb H=\langle i,j,k|ij=-k, jk=-i, ki=-j, ijk=i^2=j^2=k^2=-1\rangle$$. So due to ij=-k and $$i^2=-1$$, we can just pick any two of the i,j,k and call them x and y. Then $$x^4=e$$, $$y^2=x^2$$ and iji=ik=j so xyx=y. This gives us the presentation $$\mathbb H=\langle x,y|x^4,y^2=x^2,xyx=y\rangle$$

## The algebra

With a presentation of the group, we need a good presentation of the group algebra. We set X=x+1, Y=y+1. Then the relations turn into $$X^4=0$$, $$X^2+Y^2=0$$ and if we want to express xyx+y with X and Y, we can start by calculating XYX=xyx+xy+x^2+yx+y+1. Adding to this XY, YX and $$X^2$$ we get $$XYX+XY+YX+X^2=xyx+y$$. With a term ordering that prefers Y to X and works lexicographic within equal degree, we thus get the basis $$X^4$$, $$Y^2+X^2$$, $$YX+XY+X^2+XYX$$. It can be verified, as well, that any word of length 5 vanishes under these relations - and by reducing the XYX term in the last relation repeatedly, we end up with the presentation
$$R=\mathbb F_2\mathbb H=\mathbb F_2\langle X,Y\rangle/\langle X^4,Y^2+X^2,YX+XY+X^2+X^2Y+X^3+X^3Y\rangle$$
with a basis, as vector space, consisting of $$1, X, Y, XY, X^2, X^2Y, X^3, X^3Y$$.

## The resolution

Calculating a resolution is, as always, a tedious matter of Gröbner basis eliminations in order to find minimal kernel generators. The first step is easy - the augmentation map has a kernel generated by X and Y. So, the next step is to calculate the kernel of the map
$$\begin{diagram}R^2 &\rTo^{\begin{pmatrix}X & Y\end{pmatrix}}& R\end{diagram}$$
This is, formulated as a Gröbner basis problem of modules, the same as eliminating $$e_0$$from
$$e_0X+e_1$$
$$e_0Y+e_2$$
$$e_0X^4$$
$$e_0Y^2+e_0X^2$$
$$e_0(YX+XY+X^2+XYX)$$
The key ingredient here is discovering that of the three obvious resulting relations
$$e_1X^3$$
$$e_1X+e_2Y$$
$$e_1(Y+X+YX)+e_2X$$
the first can be expressed in terms of the other two. So the actual map of this step of the resolution ends up being
$$\begin{diagram}R^2 &\rTo^{\begin{pmatrix}X & Y+X+YX\\Y&X\end{pmatrix}}& R^2\end{diagram}$$
For the next step, we calculate an eliminating Gröbner basis of
$$e_1X+e_2Y+e_3$$
$$e_1(Y+X+YX)+e_2X+e_4$$
with the ring relations in mind.
The additional relation $$e_2X^3Y+e_3X^3$$ is easy enough to find, and then the a long and tedious calculation starting off with multiplying the $$e_4$$-relation with X and then eliminating leading terms repetitively with the $$e_3$$-relation , we end up with
$$e_2X^2Y+e_3(Y+X^2+X^2Y)+e_4X$$
This relation, multiplied with X and eliminated with the $$e_2X^3Y+e_3X^3$$ gives us a first kernel element
$$e_3(XY+X^2+X^2Y+X^3)+e_4X^2$$
If we just keep on going; multiplying instead the $$e_4$$ relation with Y, and then eliminating everything in sight, we can in the end find another kernel element:
$$e_3(X+XY)+e_4(X+Y)$$
and by eliminating between these two kernel elements, we discover that this second one is a Gröbner basis for the kernel on its own.

(note, for those non-familiar with this particular algorithm - the surviving $$e_3$$ and $$e_4$$ terms end up being a kind of log over which things in what order you may want to do to get rid of everything with $$e_1$$ or $$e_2$$ in them...)

So we have our next map
$$\begin{diagram}R&\rTo^{\begin{pmatrix}X+XY\\X+Y\end{pmatrix}}&R^2\end{diagram}$$
Now, if we allow us a tiny bit of cheating, we can from Betti number considerations (the resolution we're hunting has the Betti numbers 1,2,2,1,1,2,2,1,1,2,2,...) and dimensions conclude that our next kernel should be one-dimensional. Thus, as soon as we have some element of the kernel, we're done. A quick calculation reveals that $$X^3Y$$ kills everything in sight, and thus more particularily our map. So we get
$$\begin{diagram}R&\rTo^{\begin{pmatrix}X^3Y\end{pmatrix}}&R\end{diagram}$$

And finally, this kernel, of course, is generated by X and Y, so we're back where we started in degree 5. This shows immediately that the resolution has period 4; which will be of some assistance later on.

This, being boring and automatable and everything, is a good point to moving forward.

## The chain maps and cocycles

Since the resolution is minimal (trust me, it is), we can just throw out cocycles at will. So we recognize that since the first map goes from an R^2, we have two cocycles in degree 1. Let's call them $$\xi$$ and $$\eta$$, and let $$\xi=\epsilon^1_1$$ and $$\eta=\epsilon^1_2$$, where $$\epsilon^i_j$$ means the augmentation map on the j:th component of the i:th degree module.

So, to calculate all possible products of degree 1 cocycles, we lift $$\xi$$ and $$\eta$$ to chain maps. Or at least the beginning of one.

So, take the cocycle $$a\xi+b\eta$$, with $$a,b\in\{0,1\}$$. Then the chain map lifting starts out
$$\begin{diagram} \rTo & R^2 & \rTo^{\begin{pmatrix}X & Y+X+YX\\Y&X\end{pmatrix}}& R^2 & \rTo \\ & \dTo^\psi && \dTo^{\begin{pmatrix}a & b\end{pmatrix}} & \rdTo \\ \rTo & R^2 & \rTo_{\begin{pmatrix}X&Y\end{pmatrix}} & R & \rTo & \mathbb F_2 &\rTo& 0 \end{diagram}$$
For the next function, we calculate the composition map in two ways. Set $$\psi=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}$$. Then
$$\partial\psi=\begin{pmatrix}X\alpha+Y\gamma & X\beta + Y\delta\end{pmatrix}$$
and$$\begin{pmatrix}a & b\end{pmatrix}\partial=\begin{pmatrix}aX+bY & aY+aX+aYX+bX\end{pmatrix}$$
Equality of these matrices gives us
$$\psi=\begin{pmatrix}a & a+b\\b & a+aX\end{pmatrix}$$
Thus we have the maps
$$\xi^2:(\alpha\quad\beta)\overset{\psi_\xi}{\mapsto}(\alpha+\beta\quad\beta+\betaX)\overset{\xi}{\mapsto}=\epsilon(\alpha+\beta)$$
and thus
$$\xi^2=\epsilon_1^2+\epsilon_2^2$$
$$\xi\eta=\eta\xi=(\alpha\quad\beta)\mapsto(\alpha+\beta\quad\beta+\betaX)\overset{\eta}{\mapsto}\epsilon(\beta)$$
and thus
$$\xi\eta=\eta\xi=\epsilon_2^2$$
$$\eta^2=(\alpha\quad\beta)\mapsto(\beta\quad\alpha)\overset{\eta}{\mapsto}\alpha$$
and thus
$$\eta^2=\epsilon_1^2$$
From this, we can read off a relation: $$\xi^2=\xi\eta+\eta^2$$ or differently $$\xi^2+\xi\eta+\eta^2$$. Letting $$\xi$$ be more significant than $$\eta$$, we now know that the only interesting cocycles in higher degrees will be $$\xi\eta^n$$ and $$\eta^n$$.

Thus, in degree 3, we know we have only one cocycle, and we can expect to find $$\eta^3$$ and $$\xi\eta^2$$ there. So, let's lift $$\eta^2$$ and $$\xi\eta+\eta^2$$ to chain maps, and then see what happens when we compose this with $$\xi$$ and with $$\eta$$.

$$\begin{diagram} \rTo & R & \rTo^{\begin{pmatrix}X+Y+XY\\X+Y\end{pmatrix}} & R^2 & \rTo \\ & \dTo^\psi && \dTo^{\begin{pmatrix}a&b\end{pmatrix}} & \rdTo\\ \rTo & R^2 & \rTo_{\begin{pmatrix}X&Y\end{pmatrix}}& R&\rTo&\mathbb F_2&\rTo&0 \end{diagram}$$

We set $$\psi=\begin{pmatrix}\alpha\\\beta\end{pmatrix}$$

Now, the two compositions end up being
$$\begin{pmatrix}a&b\end{pmatrix}\partial=\begin{pmatrix}a(X+Y+XY)+b(X+Y)\end{pmatrix}$$
$$\partial\psi=\begin{pmatrix}X\alpha+Y\beta\end{pmatrix}$$
and from this we can read off
$$\alpha=a(1+Y)+b$$ and $$\beta=a+b$$
Thus, we get the relation ($$\eta^2+\xi\eta: (a\quad b)=(1\quad 1)$$)
$$\xi^2\eta+\xi\eta^2: a\mapsto\begin{pmatrix}a+aY+a\\a+a\end{pmatrix}\mapsto 0$$
So $$\xi\eta^2=\xi^2\eta$$.
But then $$\xi^2\eta=\xi\eta^2+\eta^3$$, since $$\xi^2=\xi\eta+\eta^2$$; so $$\xi\eta^2=\xi^2\eta=\xi\eta^2+\eta^3$$. Thus $$\eta^3=0$$. Now, the only relevant basis elements in degree 4 are, due to the relation $$\xi^2=\xi\eta+\eta^2$$, $$\xi\eta^3$$ and $$\eta^4$$. Since $$\eta^3=0$$, both of these vanish, and so the single basis element in degree 4 has to be a new algebra generator: $$\zeta=\epsilon^4_1$$.
Now, due to the periodicity and the degrees of the generators, this is all we need. The homogenous components repeat with a period of four, and the generators for each slice will be
$$\zeta^n$$
$$\xi\zeta^n, \eta\zeta^n$$
$$\xi\eta\zeta^n, \eta^2\zeta^n$$
$$\xi\eta^2\zeta^n$$
$$\zeta^{n+1}$$
...
Thus, we have the group cohomology
$$H^*(U(\mathbb H),\mathbb F_2)=k[\xi^1,\eta^1,\zeta^4]/\langle \xi^2+\xi\eta+\eta^2,\xi\eta^2+\xi^2\eta\rangle$$