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A cool construction of Hochschild homology (that I saw on B. Antieau's website here ) is the following:

Let $k$ be a commutative ring, then denote by $\mathfrak{a}\text{CAlg}_k$ the category of animated commutative rings. For any $x\in BS^1$, we have a natural forgetful functor $$x^*:\mathfrak{a}\text{CAlg}_k^{BS^1}\to \mathfrak{a}\text{CAlg}_k.$$ Then Hochschild homology is the left adjoint $x_!:\mathfrak{a}\text{CAlg}_k\to \mathfrak{a}\text{CAlg}_k^{BS^1}$. I've not found a proof that this does agree with the "usual" Hochschild homology of commutive rings, is there a good reference for this?

My main question is in the non-commutative setting. Hochschild homology has a natural definition for non-commutative DG-algebras (see for instance here). DG-algebras are, at least in characteristic zero, equivalent to simplicial commutative algebras, so we can (probably) use $x_!$ as a definition for Hochschild homology of commutative DG-algebras (even if I'm not sure if this is backwards-compatible). My question is if we can use $x_!$ to define Hochschild homology for non-commutative DG-algebras. I think non-commutative DG-algebras are equivalent to simplicial non-commutative rings, and so we can consider the category of animated non-commutative rings $\mathfrak{a}\text{Alg}_k$ and the forgetful functor $$x^*:\mathfrak{a}\text{Alg}_k^{BS^1}\to \mathfrak{a}\text{Alg}_k$$ and the left adjoint $x_!$ thereof.

Edit: Maybe animated non-commutative rings is not the right way of going about it, but what I'm probably most interested in is having a universal property for non-commutative Hochschild homology, which is maybe a more amenable quesion.

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    $\begingroup$ As a reference for the commutative statement, see Toën-Vezzosi "Algèbres simpliciales $S^1$-équivariantes ..." and references therein. In the non-commutative setting, how would you propose to define an $S^1$-action? The relevant complex isn't $A\otimes S^1$ in the non-commutative category. $\endgroup$
    – Jon Pridham
    Commented Apr 2, 2022 at 18:47
  • $\begingroup$ @JonPridham I certainly don't have a proposition for that, but I was thinking that $\mathfrak{a}\text{Alg}_k$ and every other object involded purely formally, and get some object $x_! R$. I don't know how this consturction relates to anything on the non-commutative side to be honest. $\endgroup$
    – curious math guy
    Commented Apr 2, 2022 at 18:54
  • $\begingroup$ Hopefully I’m interpreting the question correctly, but one of the key reasons HH is the left adjoint in the commutative case is that the coproduct in CAlg agrees with the symmetric monoidal product. This fact does not hold for associative algebras. $\endgroup$
    – Liam Keenan
    Commented Apr 3, 2022 at 2:26
  • $\begingroup$ Ah sorry I basically repeated what Jon said above. $\endgroup$
    – Liam Keenan
    Commented Apr 3, 2022 at 2:31
  • $\begingroup$ @LiamKeenan What about the coproduct of $E_n$-algebras for $n\ge2$? $\endgroup$
    – Z. M
    Commented Apr 3, 2022 at 9:45

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No, this does not work in the non-commutative case. In general we have $HH(A)=A\otimes_{A\otimes A^{\mathrm{op}}} A$, and this is only a $k$-module, not an algebra. If $A$ is commutative, the tensor product happens to compute coproducts/pushouts of commutative $k$-algebras, and we have $HH(A)=\operatorname{colim}_{S^1}A=x_!(A)$ since $S^1=*\coprod_{*\sqcup *}*$, but in the non-commutative case the tensor product does not have such an interpretation.

To see that the group $S^1$ acts on $HH(A)$ in general, one can use the formalism of factorization homology, $HH(A)=\int_{S^1} A$, which makes the functoriality on $BS^1$ apparent. A more classical approach is to use the fact that the usual "Hochschild complex" extends to a cyclic $k$-module (a functor on Connes' cyclic category $\Lambda$), whose geometric realization acquires an action of $S^1$ due to the $\infty$-groupoid completion of $\Lambda$ being $BS^1$. A reference for the latter approach is Appendix B of the article by Nikolaus and Scholze: https://arxiv.org/pdf/1707.01799.pdf

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    $\begingroup$ What happens if the input is $E_n$ for $n\ge2$? $\endgroup$
    – Z. M
    Commented Apr 3, 2022 at 9:42
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    $\begingroup$ @z.m. It shifts n by -1. $\endgroup$
    – Fernando Muro
    Commented Apr 3, 2022 at 9:49
  • $\begingroup$ @Z.M You can see this by doing factorization homology in $E_{n-1}$-algebras. An $E_n$-algebra is an $E_1$-algebra in $E_{n-1}$-algebras, so you can integrate it over the circle to get an $E_{n-1}$-algebra (with $S^1$-action). $\endgroup$
    – Marc Hoyois
    Commented Apr 3, 2022 at 11:10

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