# What should you do if $d^n \circ d^{n-1} \circ \dots d^{2} = 0$ but not $d^n \circ d^{n-1}$ in general?

by Shine On You Crazy Diamond   Last Updated August 20, 2019 14:20 PM

Here I have a free abelian group $$A$$ on $$\Sigma$$ and a grouping operator $$()$$ on $$\Sigma^*$$ which turns it into a magma. The grouping could mean arguments to a function (later) in which case this is just regular group cohomology. I came up with these formulas. We do have $$d^3 \circ d^2 = 0$$ but not for the higher ones (they have to compose down to $$2$$ for that to happen, I think):

I've identified strings over $$\Sigma$$ with tuples, so that's why the weird string notations:

$$\partial^3(abcd) = a(bcd) + (abc)d \\ \partial^2(abc) = a(bc) - (ab)c \\ \partial^3\partial^2(abcd) = a\partial^2(bcd) + \partial^2(abc)d \\ = a(b(cd) - (bc)d) + (a(bc) - (ab)c)d = \\ (ab)(cd) - a(bc)d + a(bc)d - (ab)(cd) = 0 \\ \ \ \\ \partial^4(\partial^3)(abcde) = a\partial^3(bcde) - \partial^3(abcd)e = \\ a (b(cde) + (bcd)e) - (a(bcd) + (abc)d)e = \\ ab(cde) + a(bcd)e - a(bcd)e + (abc)de \neq 0$$

The sign middle sign is $$+$$ in $$\partial^n$$ when $$n$$ is odd, else $$-$$. So in this case is cohomology possible to do, or not?

Thanks.

I guess a more standard way that still would end up grouping strings would be:

$$\partial^4(abcde) = a(bcde) - ((ab)cde) + (a(bc)de) - (ab(cd)e) + (abc(de)) - (abcd)$$ but it's not a very symmetric idea with respect to strings.

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