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More group cohomology

  • June 8th, 2006
  • 10:10 pm

My advisor told me to go hit D_8 and U(\mathbb H) as my next two cohomology calculation projects; try to do them with resolutions by hand so that I get a feeling for what’s going on. After failing spectacularily both at getting a resolution of \mathbb F_2 with \mathbb F_2D_8, he walked me through his Shiny! Gr�bner base method to get resolutions with free modules over finite p-group algebras. Armed with the minimal resolution, I sat down and started hunting products; and finally found the cohomology ring.

Or … to be exact, I found H^{\leq 2}(D_8,\mathbb F_2) and then peeked into Carlson, et.al. for the Big List of 2-group cohomologies to see that all interesting stuff happens in H^{\leq 2} anyway.

So for the benefit of any and all readers who want to see what it looks like, I’m going to walk through it again here. Nonono, you don’t need to flee all of you – just skip this entry if it’s that scary!

So. Before starting, we need to be clear about what D_8 looks like, and what \mathbb F_2D_8 looks like. So, I will choose a presentation with merits for this particular calculation: D_8=\langle a,b\mid a^2=b^2=(ab)^4\rangle. This is a presentation by two mirrorings; and the “usual” presentation of a mirror and a 90-degree rotation can be recovered by the map \langle a,b\mid a^2=b^2=(ab)^4\rangle\to\langle a,b\mid a^2=b^4=(ab)^2\rangle that sends a\mapsto a and b\mapsto ab.

Now, the algebra, we choose to be presented as R=\mathbb F_2D_8\langle A,B\rangle/\langle A^2,B^2,ABAB+BABA\rangle with A=a+1 and B=b+1. Hacking at these relations for a few short moments convinces you that they are equivalent to the group relations.

C_0=R, and C_1=Re_1^1\oplus Re_2^1 follows rather immediately with the augmentation map, and from sending e_1^1\overset{\partial_1}\mapsto A and e_2^1\overset{\partial_1}\mapsto B to the minimal generators of the augmentation ideal.

Now, the kernel of \partial_1 can be calculated by eliminating the variable e_0 from the ideal generated by the group algebra relations and the expressions x_i-f_i(e_0). This is a (rather) straightforward Gr�bner base calculation, and we get at the end the eliminated generators
x_1A
x_2B
x_1BAB+x_2ABA
so we get a kernel \langle e_1^1A,e_2^1B,e_1^1BAB+e_2^1ABA\rangle with three generators.

By a similar calculation, we get the next kernel \langle e_1^2A,e_2^2B, e_1^2BAB+e_3^2A, e_2^2ABA+e_3^2B\rangle. This gives us the first few bits of the minimal free resolution as
R^4\overset{\begin{pmatrix}A&0&BAB&0\\0&B&0&ABA\\0&0&A&B\end{pmatrix}}\to R^3\overset{\begin{pmatrix}A&0&BAB\\0&B&ABA\end{pmatrix}}\to R^2\overset{\begin{pmatrix}A&B\end{pmatrix}}\to R\to\mathbb F_2\to 0

So, this calculation gives us a minimal resolution, and thus the presentation as a graded module of our cohomology ring (or at least the portion in degree lower or equal to 3); it remains to be seen what happens with the cohomology product.

So, in degree 0, we have just the constants of the cohomology ring, and nothing very exciting will happen. In degree 1, we have two elements – let’s call them x,y. For their interactions, we want to lift the maps given by ax+by in
\begin{diagram} R^4 &\rTo^{\begin{pmatrix}A&0&BAB&0\\0&B&0&ABA\\0&0&A&B\end{pmatrix}} & R^3 &\rTo^{\begin{pmatrix}A&0&BAB\\0&B&ABA\end{pmatrix}} & R^2 &\rTo^{(A \quad B)} & R & \rTo\\ \dTo^{f_3} && \dTo^{f_2} && \dTo^{f_1} & \rdTo^{(a \quad b)\circ\epsilon} \\ R^3 &\rTo^{\begin{pmatrix}A&0&BAB\\0&B&ABA\end{pmatrix}} & R^2 &\rTo^{\begin{pmatrix}A&B\end{pmatrix}} & R & \rTo^\epsilon &\mathbb F_2 & \rTo & 0 \end{diagram}
Now, obviously f_1=(a\quad b) will make the beginning commute. Then, f_2=\begin{pmatrix}a&0&bBA\\0&b&aAB\end{pmatrix} makes the square commute, and finally f_3=\begin{pmatrix}a&0&0&0\\0&b&0&0\\0&0&a&b\end{pmatrix} makes the last square commute. Finding these matrices is just a tedium of multiplying corresponding matrices to get the compositions of maps, and then setting the coefficients just so. I won’t enter into more details than that.

Similarily, for degree 2, we get a similar set of matrices, parametrized by a map (a\quad b\quad c)\circ\epsilon\colon R^3\to\mathbb F_2 as g_1=(a\quad b\quad c) and g_2=\begin{pmatrix}a&0&c&bBA\\0&b&aAB&c\end{pmatrix}.

How does this help us find the products? Well, for \zeta and \eta in the group cohomology, lifting \eta to a chain map – as we have done here for all possible maps in degrees 1 and 2 – and then composing at the corresponding position with a representative for \zeta gives a representative for \zeta\eta (or possibly the other way around). Now the cohomology in each degree is spanned by just the augmentation on each component; call them \epsilon_i^j for the augmentation of component i in degree j. So, we can calculate, first, all products of things in degree 1:
\epsilon_1^1\cdot\epsilon_1^1=x\cdot x=\epsilon\circ \begin{pmatrix}1&0\end{pmatrix}\circ \begin{pmatrix}1&0&0\\0&0&AB\end{pmatrix}=\epsilon_1^2
\epsilon_2^1\cdot\epsilon_1^1=y\cdot x=\epsilon\circ \begin{pmatrix}0&1\end{pmatrix}\circ \begin{pmatrix}1&0&0\\0&0&AB\end{pmatrix}=AB\cdot\epsilon_1^2=(ab+b+a+1)\epsilon_1^2=4\epsilon_1^2=0
\epsilon_1^1\cdot\epsilon_2^1=x\cdot y=\epsilon\circ \begin{pmatrix}1&0\end{pmatrix}\circ \begin{pmatrix}0&0&BA\\0&1&0\end{pmatrix}=BA\cdot\epsilon_2^2=4\epsilon_2^2=0
\epsilon_2^1\cdot\epsilon_2^1=x\cdot x=\epsilon\circ \begin{pmatrix}0&1\end{pmatrix}\circ \begin{pmatrix}0&0&BA\\0&1&0\end{pmatrix}=\epsilon_2^2

so we see that x^2=\epsilon_1^2, that y^2=\epsilon_2^2 and that xy=yx=0. So we have a relation, and two generators, and in addition two of the three generators of degree 2 already. Let’s call that third generator z, and keep in mind that z lives in degree 2.

This gives us some more products that we can calculate.
\epsilon_3^2\cdot\epsilon_1^1=z\cdot x=\epsilon\circ \begin{pmatrix}0&0&1\end{pmatrix}\circ \begin{pmatrix}1&0&0&0\\0&0&0&0\\0&0&1&0\end{pmatrix}=\epsilon_3^3
\epsilon_3^2\cdot\epsilon_2^1=z\cdot y=\epsilon\circ \begin{pmatrix}0&0&1\end{pmatrix}\circ \begin{pmatrix}0&0&0&0\\0&1&0&0\\0&0&0&1\end{pmatrix}=\epsilon_4^3
\epsilon_1^2\cdot\epsilon_1^1=x^2\cdot x=\epsilon\circ \begin{pmatrix}1&0&0\end{pmatrix}\circ \begin{pmatrix}1&0&0&0\\0&0&0&0\\0&0&1&0\end{pmatrix}=\epsilon_1^3
\epsilon_2^2\cdot\epsilon_2^1=y^2\cdot y=\epsilon\circ \begin{pmatrix}0&1&0\end{pmatrix}\circ \begin{pmatrix}0&0&0&0\\0&1&0&0\\0&0&0&1\end{pmatrix}=\epsilon_2^3

We could lift the second degree elements, and with that second lift verify these results, but it’s scarcely necessary. With these calculations, we have the cohomology ring in degrees lower than or equal to 3. Going further means more Gr�bner bases, more liftings to chain maps, and I haven’t the energy right now. On the other hand, sneakpeeking in Carlson et.al. reveals that those who have done this -with- a condition to test whether the whole ring is in there – also find three generators, in degrees 1 and 2, and with a single relation.

Thus I have calculated H^*(D_8,\mathbb F_2)=\mathbb F_2[x^1,y^1,z^2]/\langle xy\rangle

3 People had this to say...

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Another great group cohomology! I can see some places where you can improve this, but over all it was good work!

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[...] 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. [...]

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[...] Furthermore, we pick a canonical nice resolution P, continuing the one I calculated previously. This has the i:th component Λi+1, and the differentials looking like for differentials starting in odd degree, and for differentials starting in even degree. The first few you can see on the previous calculation, or if you don’t want to bother, they are [...]

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