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
[tex]m_n\colon(\operator{Ext}^1(S,S))^{\otimes
n}\to\operator{Ext}^2[/tex]
are actually the duals of the maps
[tex]i_n\colon R_n\to A_1^{\otimes n}[/tex]
embedding the relators of the original algebra into the tensor
algebra over the generators.
So, for our case, we have the maps
[tex]m_2(x,x)=x^2[/tex]
[tex]m_2(y,y)=y^2[/tex]
[tex]m_2(x,y,x,y)=z[/tex]
[tex]m_2(y,x,y,x)=z[/tex]
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
[tex]i_2((x^2)^*)=a^2[/tex]
[tex]i_2((y^2)^*)=b^2[/tex]
[tex]i_4(z^*)=abab+baba[/tex]
from which we may recover a presentation of the group algebra as
[tex]k\langle a,b\rangle/\langle a^2,b^2,abab+baba\rangle[/tex]
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.