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Consider a $k$-algebra $A$ and a bimodule $M$. One can construct two complexes, the Hochschild complex $C_n(A,M)$ and the Chevalley-Eilenberg complex $C'_n(A,M)=M\otimes \Lambda^n(A)$. Given an element $m\otimes a_1\otimes \cdots\otimes a_n$ and a permutation $\sigma\in S_n$, there is an action $$\sigma(m\otimes a_1\otimes\cdots\otimes a_n)=m\otimes a_{\sigma^{-1}1}\otimes\cdots\otimes a_{\sigma^{-1}n}$$ of $k[S_n]$ on $C_n(A,M)$. The element $\varepsilon_n = \sum_{\sigma\in S_n} (-1)^\sigma \sigma$ induces well defined maps $C_n'(A,M)\to C_n(A,M)$ which turn out to give a chain map $C'(A,M)\to C(A,M)$. When $A$ is commutative and $M$ is symmetric the differential of $C'$ is trivial and taking homology there is a map $\varepsilon : M\otimes \Lambda^n(A)\to H_n(A,M)$.

If $\Omega^1(A)$ is the $A$-module of Kähler differentials of $k\to A$, one defines $\Omega^n(A) = \Lambda_A^n(\Omega^1(A))$ (where the exterior product is taken over $A$), so it is spanned by the elements $a_0da_1\cdots da_n$. There is a canonical $k$-map $d:A\to \Omega^1(A)$ that sends $a\mapsto da$. In his book Cyclic Homology, Loday claims that $\varepsilon$ factors through $M\otimes_A \Omega^n(A)$, and to check this is suffices to check that any element of the form $$\varepsilon(mx,y,a_3,a_4,\cdots,a_{n+1})+\varepsilon(my,x,a_3,\cdots,a_{n+1})-\varepsilon(m,xy,a_3,\cdots,a_{n+1})$$ is a boundary.

I have two problems:

  1. First, I assume the map $\varepsilon$ wants to factor through is the exterior power of $d:A\to \Omega^1(A)$ over $k$ followed by the projection to the exterior power over $A$.

  2. Assuming this, I really cannot see how checking the above suffices. Perhaps he wants to use that $d$ is a universal derivation? For example, one needs to check that one can replace $\otimes=\otimes_k$ with $\otimes_A$, and for this one needs to show an element of the form

$$\varepsilon(my,x,a_3,\cdots,a_{n+1})-\varepsilon(m,xy,a_3,\cdots,a_{n+1})$$

also goes to a boundary. I am not sure how this is included in Loday's verification.

Pedro
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  • Can you tell me why this is boundary precisely what Loday mentioned in his text? Because I couldn’t able to prove it. – Sunny Nov 06 '19 at 05:47

1 Answers1

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The $n$-th exterior power $\Omega^nA=\wedge_A^n(\Omega^1_A)$ represents the functor $$N\mapsto\left\{ d : A^n\to N\ |\ d\text{ antisymmetric, }k\text{-multilinear, derivation in all arguments}\right\}\\\ \ \ \ \ \ \ \ =\left\{ d : A^n\to N\ |\ d\text{ antisymmetric, }k\text{-multilinear, derivation in first argument}\right\}.$$ In particular, to construct the desired map $$\tilde{\varepsilon}: M\otimes_A \Omega^n A\to H_n(A,M)$$ it suffices to construct a map $$\hat{\varepsilon}: A^n\to\text{Hom}_A(M,H_n(A,M))$$ which is antisymmetric, $k$-multilinear and a derivation in the first component; then $\tilde{\varepsilon}$ is the unique morphism with $$\tilde{\varepsilon}(m\otimes\text{d}a_1\wedge\ldots\wedge\text{d}a_n)=\hat{\varepsilon}(a_1,\ldots,a_n)(m).$$ Here, for $\hat{\varepsilon}$ you take the adjoint of $\varepsilon$, which we already know to be $k$-multilinear and antisymmetric, and the Leibniz rule for the first argument is what Loday claims needs to be checked.

Hanno
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