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D8 revisited

  • February 7th, 2007

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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Michi is a recent PhD working in homological algebra. This blog is his outlet for texts with some manner of thought put into them. Over at his LiveJournal intimate details and streams of consciousness might be found.
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