Let's say we're looking at $d_n: A^{\otimes n} \rightarrow A^{\otimes n -1}$ in the Hochschild homology chain complex defined by $d_n(a_0 \otimes a_1 \otimes \cdots \otimes a_{n+1}) = \sum_{i=0}^n (-1)^i a_0 \otimes \cdots \otimes a_ia_{i+1} \otimes \cdots \otimes a_{n+1}$. Is there a similarly defined chain complex where the differential does something which combines $3$ of the tensor components for example? (I'm interested in constructions that work for combining any $n$). So the differential would yield some linear combination of elements that look roughly like this $a_0 \otimes a_ia_{i+1}a_{i+2} \otimes \cdots \otimes a_{n+1}$. Obviously this would also require a change in the chain groups as well.

  • 1
    $\begingroup$ Maybe there is a Hochschild homology for $A_{\infty}$-algebras? But before you can have a variant of HH that uses $n$-ary operations, you need to have some sort of $n$-ary operations. Any natural operation on associative algebras is built using a binary operation that is subject to a compound ternary operation vanishing, which if you're thinking homologically leads you to viewing multiplication as being like a cocycle. It might be more fruitful to look for things that can be interpenetrated similarly. $\endgroup$
    – Aaron
    Aug 12, 2020 at 17:12
  • $\begingroup$ Could you elaborate on what you mean by "any natural operation on associative algebras is built using a binary operation that is subject to a compound ternary operation vanishing"? $\endgroup$
    – EgoKilla
    Aug 12, 2020 at 17:16
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    $\begingroup$ An associative algebra over $k$ is a $k$-module $A$ with a bilinear operation $m:A\otimes A\to A$ such that $m(m(a,b),c)-m(a,m(b,c))$ vanishes for all $a,b,c$. This is just a way to rewrite associativity. The fact that you have a differential at the far end of the hochschild complex is exactly the statement that you have an associative algebra. The fact that it lifts to a differential in higher degrees is perhaps surprising, or perhaps not, but you only have one basic operation in an associative algebra (outside of the $k$-linear structure) and it is built from one relation. $\endgroup$
    – Aaron
    Aug 12, 2020 at 17:24

1 Answer 1


Yes, there are variants of this as mentioned in the comments. If $A$ is instead an $A_\infty$-algebra, then the Hochschild boundary on $\hom(BA,A)$ takes a more involved form.

To be brief, an $A_\infty$-algebra structure on $A$ is the datum of a degree $-1$ coderivation $BA\to BA$, which in fact corresponds to a map $d : BA\to sA$. Then the Hochschild boundary is obtained by taking the bracket of coderivations with $d$.

This means, for example, that you will have elements of the form

$$ f(x_1,m_3(x_2,x_3,x_4),x_5)$$

in the differential where $m_3$ is the component of $d$ corresponding to the map $(sA)^{\otimes 3}\to sA$.

  • $\begingroup$ (If you do cohomology of other types of algebras, more complicated formulas may appear. But that's kind of a longer story.) $\endgroup$
    – Pedro
    Aug 19, 2020 at 21:29

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