# Explicit calculation to show a complex is exact

I know that the image of the preceding map, when equaling the kernel of the next map, means the complex is exact. Furthermore, this operator, listed below, is linear, but I am having a hard time explicitly calculating it. $E_n(A)$ is the set of all maps $f$ and has an abelian structure where map addition is as $$f+g: G^n \to A :: x \mapsto f(x) + g(x) .$$ The differntial map used is:

$$d_n: E_n(A) \to E_{n+1}(A):: d_n(f)(g_0,\ldots,g_n) = g_0\cdot f(g_1,\ldots,g_n) + \sum_{i=1}^{n}\,(-1)^i f(g_0,\ldots,g_{i-2},(g_{i-1}\,g_i),g_{i+1},\ldots,g_n) + (-1)^{n+1}\,f(g_0,\ldots,g_{n-1})$$ where $$f: G^n \to A \in E_n(A)$$ and the elements $$g_0,\ldots,g_n\in G$$. If $n=0$, this is meant to reduce to the following: if $a \in E_0(A) = A$ then $d_0(a) : G \to A$ is the map $$d_0(f)(g) = g\,a - a.$$

So in the explicit calculation, I start by showing, $$d_{n+1}(f)(g_0,\ldots,g_{n+1}) = g_0\,f(g_1,\ldots,g_{n+1}) + \sum_{i=1}^{n+1}(-1)^i\,f(g_0,\ldots,g_{i-1}g_i,\ldots,g_{n+1}) + (-1)^{n+2}\,f(g_0,\ldots,g_n)$$ and then apply on each of these three terms, $d_n(f)$.

For instance, the first term,

$$(d_n \circ d_{n+1})(f)(g_0,\ldots,g_{n+1}) = d_n \circ \big( g_0\,f(g_1,\ldots,g_{n+1}) \big) = g_0\,d_n(f)(g_1,\ldots,g_{n+1}) = g_0\,g_1\,f(g_2,\ldots,g_{n+1}) + g_0\,\sum_{i=1}^n\,(-1)^i\,f(g_1,\ldots,(g_{i-1}g_i),\ldots,g_{n+1}) + g_0\,(-1)^{n+1}f(g_1,\ldots,g_n)$$

(with corrections)

On the last term, I get, $$(d_n \circ d_{n+1})(f)(g_0,\ldots,g_{n+1}) = d_n \circ \big( (-1)^{n+2}\,f(g_0,\ldots,g_n)\big) = (-1)^{n+2}\,d_n(f)(g_0,\ldots,g_n) = (-1)^{n+2}\,(-1)^{n+1}\,f(g_0,\ldots,g_{n-1}) = (-1)^{n+2}\,g_0\,f(g_1,\ldots,g_{n}) + (-1)^{n+2}\,\sum_{i=1}^n\,(-1)^i f(g_0,\ldots,(g_{i-1}g_i),\ldots,g_n) + (-1)^{n+2}\,(-1)^{n+1}\,f(g_0,\ldots,g_{n-1}) = (-1)\,f(g_0,\ldots,g_{n-1}) .$$

(with corrections)

For the middle term I got,

$$d_n(f) \circ \left( \sum_{i=1}^{n+1}\,(-1)^i\,f(g_0,\ldots,(g_{i-1}g_i),\ldots,g_{n+1}) \right) = \sum_{i=1}^{n+1}\,(-1)^i\,d_n(f)(g_0,\ldots,(g_{i-1}g_i),\ldots,g_{n+1}) = nightmare! .$$

This is 3 terms for each $i$ - how does one do this in general?

I'm pretty sure the first and last terms will cancel out in the composition of these neighboring differentials and that the summations will cancel out, but I cannot even get the terms not involing the summation to cancel. If one could show this explicitly I would greatly appreciate.

P.S. I forgot to add... For $f: G^2 = G \times G \to A$, how do I show:

Let $Z^2(A)$ be the set of factor systems, denote the set of maps $f$ s.t. $$g\cdot f(g',g'') - f(g\,g',g'') + f(g,g'g'') - f(g,g') = 0$$ for the $g$'s in $G$. Show that every element in $ker\;d_2$ gives rise to a factor system.

"gives rise"? Thank you for any clarification on that part too!

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I assume $G$ is a group, $A$ is an abelian group, $G$ operates o $A$ and that's it, or are there additional constraints? Also, do you really want to show tha tthe complex is exact, or just that it is a complex in the first place? – Hagen von Eitzen Sep 30 '12 at 22:01
There are no other constraints than what you said, and I left out - A is a G-module. Umm. I guess to show it is a complex to begin with, i.e., that $d_{n+1} \circ d_n = 0$. – nate Sep 30 '12 at 22:14

To see that this really is a complex, note that $d_n(f)$ has $n+1$ summands:

• Summand 0 is $g_0\cdot f(g_1,\ldots,g_n)$,
• for $1\le i\le n$ summand $i$ is $(-1)^if(g_0,\ldots, (g_{i-1}g_i),\ldots, g_n)$ and
• finally the $n+1$st summand is $(-1)^{n+1}f(g_0,\ldots, g_{n-1})$

This can be written more uniformly: Let $$\tilde f(g_0,\ldots, g_{n+1})=g_0\cdot f(g_1,\ldots,g_n)$$ Then the $i$th summand, $0\le i\le n+1$ is just $$(-1)^i\tilde f(g_{-1},\ldots, (g_{i-1}g_i),\ldots, g_{n+1})$$ with $g_{-1}=g_{n+1}=1$. That is, we now have only summands of the "middle type".

Now if we consider $d_{n+1}\circ d_n$, then the summands in the result are obtained by combining two pairs of consecutive arguments - first one pair, then the other. However, the same combinig of arguments can be obtained in the other order. The sign factors cause thse summands to cancel, i.e. one time we obtain $(-1)^i(-1)^j$, say, and with reverse order $(-1)^j(-1)^{i+1}$ or $(-1)^{j-1}(-1)^{i}$. In summary, $d_{n+1}\circ d_n=0$.

Regarding your gives rise question. The condition written there is precisely that $f\in\ker d_2$, i.e. the left hand side is simply $d_2(f)(g, g', g'')$.

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Oh I see that now - it makes sense that $d_{n+1} \circ d_n = d_n \circ d_{n+1}$ - thank you!! Also, I see what the second part means - probably much clearer if I just saw it as a $d_2$ operation. – nate Sep 30 '12 at 22:43