In a previous installment, we calculated with some amount of success. For that post, I said that I was going to calculate the cohomologies of and of 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.
So, for starters, we need a presentation of such that we can work well with it. We all know that . So due to ij=-k and , we can just pick any two of the i,j,k and call them x and y. Then , and iji=ik=j so xyx=y. This gives us the presentation
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 , 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 we get . With a term ordering that prefers Y to X and works lexicographic within equal degree, we thus get the basis , , . 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
with a basis, as vector space, consisting of .
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
This is, formulated as a Gröbner basis problem of modules, the same as eliminating from
The key ingredient here is discovering that of the three obvious resulting relations
the first can be expressed in terms of the other two. So the actual map of this step of the resolution ends up being
For the next step, we calculate an eliminating Gröbner basis of
with the ring relations in mind.
The additional relation is easy enough to find, and then the a long and tedious calculation starting off with multiplying the -relation with X and then eliminating leading terms repetitively with the -relation , we end up with
This relation, multiplied with X and eliminated with the gives us a first kernel element
If we just keep on going; multiplying instead the relation with Y, and then eliminating everything in sight, we can in the end find another kernel element:
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 and 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 or in them…)
So we have our next map
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 kills everything in sight, and thus more particularily our map. So we get
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 and , and let and , where 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 and to chain maps. Or at least the beginning of one.
So, take the cocycle , with . Then the chain map lifting starts out
For the next function, we calculate the composition map in two ways. Set . Then
Equality of these matrices gives us
Thus we have the maps
From this, we can read off a relation: or differently . Letting be more significant than , we now know that the only interesting cocycles in higher degrees will be and .
Thus, in degree 3, we know we have only one cocycle, and we can expect to find and there. So, let’s lift and to chain maps, and then see what happens when we compose this with and with .
So we start with the diagram
Now, the two compositions end up being
and from this we can read off
Thus, we get the relation ()
But then , since ; so . Thus . Now, the only relevant basis elements in degree 4 are, due to the relation , and . Since , both of these vanish, and so the single basis element in degree 4 has to be a new algebra generator: .
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
Thus, we have the group cohomology