Is there a variant of Hochschild homology?

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.

• 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. Aug 12 '20 at 17:12
• 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"? Aug 12 '20 at 17:16
• 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. Aug 12 '20 at 17:24

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 $$d$$.
$$f(x_1,m_3(x_2,x_3,x_4),x_5)$$
in the differential where $$m_3$$ is the component of $$d$$ corresponding to the map $$(sA)^{\otimes 3}\to sA$$.