In a previous
installment,
we calculated [tex]H^*(D_8)[/tex] with some amount of success. For
that post, I said that I was going to calculate the cohomologies of
[tex]D_8[/tex] and of [tex]U(\mathbb H)[/tex] 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 [tex]\mathbb H[/tex] such
that we can work well with it. We all know that [tex]\mathbb H=\langle
i,j,k|ij=-k, jk=-i, ki=-j, ijk=i^2=j^2=k^2=-1\rangle[/tex]. So due to
ij=-k and [tex]i^2=-1[/tex], we can just pick any two of the i,j,k and
call them x and y. Then [tex]x^4=e[/tex], [tex]y^2=x^2[/tex] and
iji=ik=j so xyx=y. This gives us the presentation [tex]\mathbb
H=\langle x,y|x^4,y^2=x^2,xyx=y\rangle[/tex]

## 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
[tex]X^4=0[/tex], [tex]X^2+Y^2=0[/tex] 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 [tex]X^2[/tex] we get
[tex]XYX+XY+YX+X^2=xyx+y[/tex]. With a term ordering that prefers Y to
X and works lexicographic within equal degree, we thus get the basis
[tex]X^4[/tex], [tex]Y^2+X^2[/tex], [tex]YX+XY+X^2+XYX[/tex]. 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

[tex]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[/tex]

with a basis, as vector space, consisting of [tex]1, X, Y, XY, X^2,
X^2Y, X^3, X^3Y[/tex].

## 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

[tex]\begin{diagram}R^2 &\rTo^{\begin{pmatrix}X &
Y\end{pmatrix}}& R\end{diagram}[/tex]

This is, formulated as a Gröbner basis problem of modules, the same as
eliminating [tex]e_0 [/tex]from

[tex]e_0X+e_1[/tex]

[tex]e_0Y+e_2[/tex]

[tex]e_0X^4[/tex]

[tex]e_0Y^2+e_0X^2[/tex]

[tex]e_0(YX+XY+X^2+XYX)[/tex]

The key ingredient here is discovering that of the three obvious
resulting relations

[tex]e_1X^3[/tex]

[tex]e_1X+e_2Y[/tex]

[tex]e_1(Y+X+YX)+e_2X[/tex]

the first can be expressed in terms of the other two. So the actual
map of this step of the resolution ends up being

[tex]\begin{diagram}R^2 &\rTo^{\begin{pmatrix}X &
Y+X+YX\\Y&X\end{pmatrix}}& R^2\end{diagram}[/tex]

For the next step, we calculate an eliminating Gröbner basis of

[tex]e_1X+e_2Y+e_3[/tex]

[tex]e_1(Y+X+YX)+e_2X+e_4[/tex]

with the ring relations in mind.

The additional relation [tex]e_2X^3Y+e_3X^3[/tex] is easy enough to
find, and then the a long and tedious calculation starting off with
multiplying the [tex]e_4[/tex]-relation with X and then eliminating
leading terms repetitively with the [tex]e_3[/tex]-relation , we end
up with

[tex]e_2X^2Y+e_3(Y+X^2+X^2Y)+e_4X[/tex]

This relation, multiplied with X and eliminated with the
[tex]e_2X^3Y+e_3X^3[/tex] gives us a first kernel element

[tex]e_3(XY+X^2+X^2Y+X^3)+e_4X^2[/tex]

If we just keep on going; multiplying instead the [tex]e_4[/tex]
relation with Y, and then eliminating everything in sight, we can in
the end find another kernel element:

[tex]e_3(X+XY)+e_4(X+Y)[/tex]

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 [tex]e_3[/tex] and [tex]e_4[/tex] 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 [tex]e_1[/tex] or [tex]e_2[/tex] in them...)

So we have our next map

[tex]\begin{diagram}R&\rTo^{\begin{pmatrix}X+XY\\X+Y\end{pmatrix}}&R^2\end{diagram}[/tex]

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
[tex]X^3Y[/tex] kills everything in sight, and thus more particularily
our map. So we get

[tex]\begin{diagram}R&\rTo^{\begin{pmatrix}X^3Y\end{pmatrix}}&R\end{diagram}[/tex]

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 [tex]\xi[/tex]
and [tex]\eta[/tex], and let [tex]\xi=\epsilon^1_1[/tex] and
[tex]\eta=\epsilon^1_2[/tex], where [tex]\epsilon^i_j[/tex] 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
[tex]\xi[/tex] and [tex]\eta[/tex] to chain maps. Or at least the
beginning of one.

So, take the cocycle [tex]a\xi+b\eta[/tex], with
[tex]a,b\in\{0,1\}[/tex]. Then the chain map lifting starts out

[tex]\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}[/tex]

For the next function, we calculate the composition map in two ways.
Set
[tex]\psi=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}[/tex].
Then

[tex]\partial\psi=\begin{pmatrix}X\alpha+Y\gamma & X\beta +
Y\delta\end{pmatrix}[/tex]

and[tex]

\begin{pmatrix}a & b\end{pmatrix}\partial=\begin{pmatrix}aX+bY &
aY+aX+aYX+bX\end{pmatrix}[/tex]

Equality of these matrices gives us

[tex]\psi=\begin{pmatrix}a & a+b\\b & a+aX\end{pmatrix}[/tex]

Thus we have the maps

[tex]\xi^2:(\alpha\quad\beta)\overset{\psi_\xi}{\mapsto}(\alpha+\beta\quad\beta+\betaX)\overset{\xi}{\mapsto}=\epsilon(\alpha+\beta)[/tex]

and thus

[tex]\xi^2=\epsilon_1^2+\epsilon_2^2[/tex]

[tex]\xi\eta=\eta\xi=(\alpha\quad\beta)\mapsto(\alpha+\beta\quad\beta+\betaX)\overset{\eta}{\mapsto}\epsilon(\beta)[/tex]

and thus

[tex]\xi\eta=\eta\xi=\epsilon_2^2[/tex]

[tex]\eta^2=(\alpha\quad\beta)\mapsto(\beta\quad\alpha)\overset{\eta}{\mapsto}\alpha[/tex]

and thus

[tex]\eta^2=\epsilon_1^2[/tex]

From this, we can read off a relation:
[tex]\xi^2=\xi\eta+\eta^2[/tex] or differently
[tex]\xi^2+\xi\eta+\eta^2[/tex]. Letting [tex]\xi[/tex] be more
significant than [tex]\eta[/tex], we now know that the only
interesting cocycles in higher degrees will be [tex]\xi\eta^n[/tex]
and [tex]\eta^n[/tex].

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

So we start with the diagram

[tex]\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}[/tex]

We set [tex]\psi=\begin{pmatrix}\alpha\\\beta\end{pmatrix}[/tex]

Now, the two compositions end up being

[tex]\begin{pmatrix}a&b\end{pmatrix}\partial=\begin{pmatrix}a(X+Y+XY)+b(X+Y)\end{pmatrix}[/tex]

[tex]\partial\psi=\begin{pmatrix}X\alpha+Y\beta\end{pmatrix}[/tex]

and from this we can read off

[tex]\alpha=a(1+Y)+b[/tex] and [tex]\beta=a+b[/tex]

Thus, we get the relation ([tex]\eta^2+\xi\eta: (a\quad
b)=(1\quad 1)[/tex])

[tex]\xi^2\eta+\xi\eta^2:
a\mapsto\begin{pmatrix}a+aY+a\\a+a\end{pmatrix}\mapsto 0[/tex]

So [tex]\xi\eta^2=\xi^2\eta[/tex].

But then [tex]\xi^2\eta=\xi\eta^2+\eta^3[/tex], since
[tex]\xi^2=\xi\eta+\eta^2[/tex]; so
[tex]\xi\eta^2=\xi^2\eta=\xi\eta^2+\eta^3[/tex]. Thus
[tex]\eta^3=0[/tex]. Now, the only relevant basis elements in degree
4 are, due to the relation [tex]\xi^2=\xi\eta+\eta^2[/tex],
[tex]\xi\eta^3[/tex] and [tex]\eta^4[/tex]. Since
[tex]\eta^3=0[/tex], both of these vanish, and so the single basis
element in degree 4 has to be a new algebra generator:
[tex]\zeta=\epsilon^4_1[/tex].

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

[tex]\zeta^n[/tex]

[tex]\xi\zeta^n, \eta\zeta^n[/tex]

[tex]\xi\eta\zeta^n, \eta^2\zeta^n[/tex]

[tex]\xi\eta^2\zeta^n[/tex]

[tex]\zeta^{n+1}[/tex]

...

Thus, we have the group cohomology

[tex]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[/tex]