# On the structure of an algebra as a bimodule

$$\DeclareMathOperator\End{End}\DeclareMathOperator\Mod{Mod}\DeclareMathOperator\Ker{Ker}\newcommand{\bi}{\mathrm{bi}}\newcommand{\op}{\mathrm{op}}$$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 $$Z(A)=K$$.