Enveloping algebra and the bimodule–module dictionary

Let $k$ be a field and let $A$ be a unital associative $k$-algebra (Algebras over a commutative ring, central structure maps, and algebra homomorphisms). Its enveloping algebra is

$A^e:=A{\otimes }_k{A}^{\mathrm{o}\mathrm{p}}\mathrm{.}$

The multiplication from the tensor-product algebra structure (The tensor product of $R$-algebras has multiplication $(a\otimes b)(a^{\mathrm{\prime }}\otimes b^{\mathrm{\prime }})=a{a}^{\mathrm{\prime }}\otimes b{b}^{\mathrm{\prime }}$) and opposite-ring multiplication (The opposite ring $R^{\mathrm{o}\mathrm{p}}$) is

$(a\otimes b^{\mathrm{o}\mathrm{p}})(c\otimes d^{\mathrm{o}\mathrm{p}})=ac\otimes (db{)}^{\mathrm{o}\mathrm{p}},$

with unit $1\otimes 1^{\mathrm{o}\mathrm{p}}$. A $k$-central $A$-bimodule is a bimodule with commuting actions ($(S,R)$-bimodules and commuting left and right scalar actions) whose induced scalar actions agree as required by Associative graded algebras, bimodules, and internal shifts.

For such a bimodule $M$, the formula

$(a\otimes b^{\mathrm{o}\mathrm{p}})m:=amb$

defines a unital left $A^e$-module. Indeed,

$((a\otimes b^{\mathrm{o}\mathrm{p}})(c\otimes d^{\mathrm{o}\mathrm{p}}))m=(ac)m(db)=a(cmd)b=(a\otimes b^{\mathrm{o}\mathrm{p}})((c\otimes d^{\mathrm{o}\mathrm{p}})m),$

and the unit acts as $1_{A}m{1}_A=m$. Conversely, if $M$ is a left $A^e$-module, set $am:=(a\otimes 1)m$ and $mb:=(1\otimes b^{\mathrm{o}\mathrm{p}})m$. The two actions are unital, associative, commute because the two tensor factors commute in $A^e$, and are $k$-central because the two copies of a scalar $\lambda \in k$ define the same element of $A^e$. These constructions are inverse: $(a\otimes b^{\mathrm{o}\mathrm{p}})m=a(mb)=amb$.

The same bimodule is a unital right $A^e$-module by

$m(a\otimes b^{\mathrm{o}\mathrm{p}}):=bma\mathrm{.}$

Indeed, for $x=a\otimes b^{\mathrm{o}\mathrm{p}}$ and $y=c\otimes d^{\mathrm{o}\mathrm{p}}$,

$(mx)y=d(bma)c=(db)m(ac)=m((ac)\otimes (db{)}^{\mathrm{o}\mathrm{p}})=m(xy),$

and $m(1\otimes 1^{\mathrm{o}\mathrm{p}})=m$.

For the right-module converse, define $am:=m(1\otimes a^{\mathrm{o}\mathrm{p}})$ and $mb:=m(b\otimes 1)$. Right associativity gives the left and right $A$-module laws, and the two actions commute because $(1\otimes a^{\mathrm{o}\mathrm{p}})$ commutes with $(b\otimes 1)$. They are $k$-central for the same scalar-balancing reason. The constructions are inverse since

$m((a\otimes 1)(1\otimes b^{\mathrm{o}\mathrm{p}}))=(ma)(1\otimes b^{\mathrm{o}\mathrm{p}})=b(ma)=bma\mathrm{.}$

Thus $k$-central $A$-bimodules, left $A^e$-modules, and right $A^e$-modules have the same objects and morphisms under these dictionaries. In particular, the regular bimodule $A$ corresponds to $A$ with left action $(a\otimes b^{\mathrm{o}\mathrm{p}})c=acb$ and right action $c(a\otimes b^{\mathrm{o}\mathrm{p}})=bca$.