Let $A$ be a commutative (unital, complex) algebra and let $\varphi$ be a $n+1$-linear functional on $A$ (we will call it cochain). Define $$(b\varphi)(a_0,a_1,...,a_{n+1}):=\sum_{j=0}^{n}(-1)^j\varphi(a_0,...,a_ja_{j+1},...,a_{n+1})+(-1)^{n+1}\varphi(a_{n+1}a_0,a_1,...,a_n).$$ It can be shown that this map satisfies $b^2=0$ (I've checked it carefully). Let $A_n\varphi(a_0,...,a_n):=\frac{1}{n!}\sum_{\sigma \in S_n}(-1)^{|\sigma|}\varphi(a_0,a_{\sigma(1)},...,a_{\sigma(n)})$ where $S_n$ is a set of all permutations of $\{1,...,n\}$ and $(-1)^{|\sigma|}$ is the sign of a permutation. The claim is the following: if $b\varphi=0$ then also $b(A_n\varphi)=0$. I tried first with the simplest (nontrivbial) case $n=2$: we have then only two permutations and the corresponding sums are of the form $b\varphi$ evaluated on some permutation of variables $a_0,...,a_{n+1}$ (commutativity is used). However when I gather in the general case, all terms corresponding to a given permutation $\sigma \in S_n$ it won't produce the term of the form $b\varphi(a_{\tau(0)},...,a_{\tau(n+1)})$ where $\tau$ is some permutation of $\{0,1,...,n+1\}$. So my question is the following:

How all arising terms finally cancel out to give again a cocycle?

I would appreciate any suggestion: I'm pretty sure that someone has computed it at least once in his life and remeber the trick. EDIT: I corrected the sign in the final term defining $b$ (as pointed out in comments).

Cyclic homology, 2nd edition, Proposition 1.3.5 (last sentence), you should always have $b\left(A_n\varphi\right) = 0$ for $n > 0$. The $b\varphi=0$ condition is not required! $\endgroup$ – darij grinberg Apr 29 '15 at 1:31