$\DeclareMathOperator\End{End}\DeclareMathOperator\Mod{Mod}\DeclareMathOperator\Ker{Ker}\newcommand{\bi}{\mathrm{bi}}\newcommand{\op}{\mathrm{op}}$Let $K$ be a field (say of characteristic zero), and $A$ be an associative $K$-algebra. We let $A^{\op}$ be its opposite algebra, and define its enveloping algebra by $B=A\otimes_K A^{\op}$. Let $\Mod^{\bi}_A$ denote the category of $A$-bimodules. There is a canonical equivalence of categories $\Mod^{\bi}_A \rightarrow \Mod^l_B$ given as follows:

For an $A$-bimodule $M$, we define an action of $B$ on $M$ by setting: $(a\otimes b)m=amb$, for $a,b\in A$ and $m\in M$.

In particular, the canonical structure of $A$, as an $A$-bimodule yields an action of $B$ on $A$. In particular, we get a map of $K$-algebras: $$\varphi:B \rightarrow \End_K(A).$$ I would like to obtain a description of the elements in the kernel of $\varphi$. Recall that for any associative $K$-algebra $A$, we have the bar resolution: $$C^{\bullet}=\left[\cdots\rightarrow A\otimes_K A\otimes_K A \rightarrow A\otimes_K A\rightarrow A\rightarrow 0 \right], $$ which is an exact complex of projective $B$-modules. The differential on degree one is given by the multiplication map $a\otimes b\mapsto ab$, and on degree two by the formula $a\otimes b\otimes c\mapsto ab\otimes c -a\otimes bc$. Clearly, every $f\in \Ker(\varphi)$ is mapped to zero under the multiplication map. Hence, we have: $$ f=\sum_{i=1}^r (a_ib_i\otimes c_i-a_i\otimes b_ic_i),$$ where $a_i,b_i,c_i\in A$ for all $1\leq i\leq r$.

Question: Is there any way of further characterizing the elements in $\Ker(\varphi)?$

For instance, if $A$ is commutative $\Ker(\varphi)$ is the kernel of the multiplication map. This, however, is clearly not the case in general. Let $Z(A)$ be the center of $A$. If $A$ is an associative $K$-algebra without zero-divisors, any elements of the form $ab\otimes c -a\otimes bc \in \Ker(\varphi)$ must satisfy that $b\in Z(A)$. However, I do not see a clear way to generalize this.

Remark: I am mainly interested in the case where $A$ is an associative $K$-algebra without zero divisors and $Z(A)=K$.



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