Let $k$ be the base field and $A$ be a unital associative $k$-algebra. Let's review the Hochschild homology theory: we have the Hochschild chain comple $C_{\cdot}(A)$ where $$ C_n(A):=A^{\otimes n+1} $$ and the differntial $b: C_n(A)\rightarrow C_{n-1}(A)$ is defined to be $$ b(a_0\otimes\ldots\otimes a_n):=\sum_{i=0}^{n-1}(-1)^i a_0\otimes \ldots\otimes a_ia_{i+1}\otimes\ldots\otimes a_n. $$ The Hochschild homology $HH^{\cdot}(A)$ is defined to be the homology of the complex $(C_{\cdot}(A),b)$.
Moreover, $C_{\cdot}(A)$ is a simplicial module where $d_i: C_n(A)\rightarrow C_{n-1}(A)$ is defined to be $$ d_i(a_0\otimes\ldots\otimes a_n):=a_0\otimes \ldots\otimes a_ia_{i+1}\otimes\ldots\otimes a_n $$ and $b=\sum_{i=0}^{n-1}(-1)^id_i$ according to standard simplicial homology theory.
On the other hand, we have a simplicial model of $S^1$, denoted by $S^1_{\cdot}$ (see Loday's Free loop space and homology). Where $$ S^1_0=\{*\}\\ S^1_1=\{s_0(*),\tau\} $$ where $d_0\tau=*$ and $d_1\tau=*$. For $n>1$ $$ S^1_n=\{s_0^n(*),s_{n-1}\ldots\widehat{s_{i-1}}\ldots s_0\tau,i=1,\ldots,n\}. $$ We see that each $S^1_n$ consists of $n+1$ elements. Anyway, it is not difficult to check that its geometric realization $|S^1_{\cdot}|$ is homeomorphic to $S^1$.
$\textbf{If the algebra A is commutative}$, the Hochschild chain complex $C_{\cdot}(A)$ can be considered as the tensor product of $S^1_{\cdot}$ and $A$, and the $d_i$'s in $C_{\cdot}(A)$ are induced by the $d_i$'s in $S^1_{\cdot}$.
For example, both $d_0$ and $d_1$ have to map the two elements in $S^1_1$ to the same element $*$ in $S^1_0$ and both of them correspond to $$ a_0\otimes a_1\mapsto a_0a_1. $$ Here we don't need to distinguish $a_0a_1$ and $a_1a_0$ since $A$ is commutative.
$\textbf{My question}$ is: when $A$ is NOT commutative, could we also get the simplicial module $C_{\cdot}(A)$ from $S^1_{\cdot}$?
