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May 24, 2012 at 12:28 comment added Tyler Lawson @Drew: Right. In your initial formula for $d(\overline{\xi}_2 \otimes \overline{\xi}_2)$, the two elements would cancel with each other if you could commute tensor products of things, so you can decide to use differentials of these two listed elements to commute them; however, there's a correction factor in higher MSS filtration that's left over.
May 24, 2012 at 4:27 vote accept Drew Heard
May 24, 2012 at 3:06 comment added Drew Heard @Tyler: Thanks for that. So are you saying, in the example I have in mind, $d[\xi_1 \vert \xi_1^2\xi_2 ]=[\xi_1 \vert \xi_1^2 \vert \xi_2] + [\xi_1 \vert \xi_2 \vert \xi_1^2]+\cdots$?
May 24, 2012 at 1:31 comment added Tyler Lawson @Drew: It's actually very standard to use: (a) $\psi(xy) = \psi(x) \psi(y)$; (b) $\psi(x) = x \otimes 1 + 1 \otimes x + d(x)$; hence (c) $d(xy) = x \otimes y + y \otimes x + \text{(correction)}$ to commute elements across each other in the cobar complex.
May 24, 2012 at 0:46 comment added Drew Heard Ok, I've verified the relations in 2 now - thank you again. So it is really 'grunt work' - trying to find boundaries who have the right summand to produce a relation?
May 23, 2012 at 23:42 comment added Drew Heard Thank you very much! I've finally managed to locate your thesis, so I'll sit down this morning and try and work through the calculations. (Hopefully I've fixed in the typo's in the first question now. I guess my problem in part 3 is - how does one do the coproduct $\psi(\xi_1^2 \overline{\xi}_2)$?)
May 23, 2012 at 15:14 history answered Peter May CC BY-SA 3.0