# D8 revisited

Published: Wed 07 February 2007

I have previously calculated the A:sub:∞-structure for the cohomology ring of D8 <http://blog.mikael.johanssons.org/archive/2006/11/an-a-structure-on-the-cohomology-of-d8/>__. Now, while trying to figure out how to make my work continue from here, I tried working out what algebra this would have come from, assuming that I can adapt Keller's higher multiplication theorem to group algebras.

A success here would be very good news indeed, since for one it would indicate that such an adaptation should be possible, and for another it would possibly give me a way to lend strength both to the previous calculation and to a conjecture I have in the calculation of group cohomology with A means.

So, we start. We recover, from the previous post, the structure of the cohomology ring as k[x,y,z]/(xy), with x,y in degree 1, and z in degree 2. Furthermore, we have a higher operation, m:sub:4, with m:sub:4(x,y,x,y)=m:sub:4(y,x,y,x)=z.

Thus, with the theorem, stating that the maps
$$m_n\colon(\operator{Ext}^1(S,S))^{\otimes n}\to\operator{Ext}^2$$
are actually the duals of the maps
$$i_n\colon R_n\to A_1^{\otimes n}$$
embedding the relators of the original algebra into the tensor algebra over the generators.
So, for our case, we have the maps
$$m_2(x,x)=x^2$$
$$m_2(y,y)=y^2$$
$$m_2(x,y,x,y)=z$$
$$m_2(y,x,y,x)=z$$
and we look to dualizing them. Considering for a while what this all means, and fixing notation, with a,b the generators, dual to x,y, we end up with the maps
$$i_2((x^2)^*)=a^2$$
$$i_2((y^2)^*)=b^2$$
$$i_4(z^*)=abab+baba$$
from which we may recover a presentation of the group algebra as
$$k\langle a,b\rangle/\langle a^2,b^2,abab+baba\rangle$$

As such, this is an unqualified success. We recover our original group algebra presentation from the A-structure, and thus should be able to do similarily as a test for completion of future calculations as well. This, of course, needs to be proven before relied upon, but it lends credence to my hopes.