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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).

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  • $\begingroup$ The $+(-1)^n$ should be a $-(-1)^n$, right? (Also, I recommend splitting the long equation into two so that it doesn't get aligned badly.) $\endgroup$ – darij grinberg Apr 29 '15 at 1:28
  • $\begingroup$ Also, according to Loday, 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
  • $\begingroup$ Two remarks: one about the sign in the final term: I corrected it as you pointed out but please note, that in order to get $b^2=0$ the sign is irrelevant (the differential with the final term omitted is often denoted by $b'$). And second remark: just not to confuse cohomology with homology I will denote $b^*,A^*$ and $b_*,A_*$ the corresponding maps in cohomology and homology (resp.). It is true that $b_*A_*=0$: I would like to know whether $b^*A^*\varphi=0$ in other words whether $(A^*\varphi) \circ b_*=\varphi \circ A_* \circ b_*=0$. I don't see how this follows from $b_*A_*=0$. $\endgroup$ – truebaran Apr 29 '15 at 1:39
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    $\begingroup$ $b^2 = 0$ does not hold with the wrong signs; check the case $n = 0$. $\endgroup$ – darij grinberg Apr 29 '15 at 2:22
  • $\begingroup$ You are right about homology vs. cohomology, though -- these are different problems. Sorry! $\endgroup$ – darij grinberg Apr 30 '15 at 5:02
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The fact that for commutative unital algebras, the antisymmetric cochains form a subcomplex of the Hochschild complex is a special case of the "Hodge-type decomposition" of Gerstenhaber and Schack, JPAA 1987 (am in a bit of a rush right now so will add a link later). This is all connected to the "Eulerian idempotents"; if I recall correctly, Loday has some discussion of this in his Cyclic Homology book.

In my opinion it is safest to do the calculation for $HH^n(A,M)$ where $M$ is a symmetric $A$-module, and only at the end take $M=A'$ as in your question; this avoids possible errors arising from forgetting that "the module variable is special", as my supervisor used to keep reminding me.

There should of course be a direct way to verify this — it's very possible that I indeed do this at some point between 2002 and 2006 but right now I don't remember I'm afraid.

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