Let $A$ be a unital algebra, defined over the complex numbers. Any bimodule $M$ over $A$ must, by definition, be a left, and right, module satisfing $$ a.(m.b) = (a.m).b, ~~~~~~~ \textrm{ for all } a,b \in A, ~ m \in M $$ What is a "natural" or "well-motivated" example of a an object which is both a left and right module, but does not satisfy the above condition?
$\begingroup$
$\endgroup$
asked Dec 19, 2019 at 22:25
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Suppose $A$ is a non-commutative Hopf algebra. Then you can use $M=A$ with the left adjoint and right regular actions: $$ a\cdot m = \sum_{(a)}a_{(1)}mS(a_{(2)}), \ m\cdot b = mb . $$ In particular, you can use the group algebra of a a group $G$ so that $$ g\cdot m = gmg^{-1}, \ m\cdot h = mh, \ g,h,m\in G. $$
answered Jan 6, 2020 at 11:58
$\begingroup$
$\endgroup$
1
$\newcommand{\C}{\mathbb C}$ Let $G$ be a nonabelian group and $\rho\colon G\to\mathrm{Aut}(V)$ be a faithful representation of $G$ over $\C$. Then $V$ is naturally a left $\C[G]$-module via the action given by $g\cdot v := \rho(g)(v)$, and is naturally a right $\C[G]$-module via the action given by $v\cdot g := \rho(g^{-1})(v)$. (Here we use the fact that $\C[G]$ is generated as an algebra by the elements of $G$, so one can define a module action using the elements of $G$ and extend to a $\C[G]$-action.)
If $V$ is a bimodule, then for all $g,h\in G$, $g\cdot (v\cdot h^{-1}) = (g\cdot v)\cdot h^{-1}$, but the left-hand side is $\rho(gh)v$ and the right-hand side is $\rho(hg)v$. But $\rho$ is faithful and $G$ is nonabelian, so this cannot occur.
answered Dec 19, 2019 at 22:39
-
4$\begingroup$ More generally, if $M$ is some (faithful) left $A$-module and $\theta : A \to A^{op}$ is a homomorphism (or even isomorphism) of algebras, then, since $M$ has naturally a right $A^{op}$-module structure, we can use $\theta$ to get a right $A$-module structure. This gives a bimodule iff $a$ and $\theta(a)$ commute for all $a \in A$. $\endgroup$ Dec 25, 2019 at 11:57