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Michi’s blog » D8 revisited

 D8 revisited

  • February 7th, 2007
  • 6:36 pm

I have previously calculated the A-structure for the cohomology ring 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, m4, with m4(x,y,x,y)=m4(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.

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