# Example of a projective bimodule with isomorphic left and right duals

What is an example of a non-free finitely generated $$R$$-bimodule $$M$$ satisfying

i) $$M$$ is projective as both a left and right $$R$$-module

ii) the right dual $$\mathrm{Hom}_R(M,R)$$ and the left dual $$_R\mathrm{Hom}(M,R)$$ are isomorphic as bimodules,

where $$R$$ is a noncommutative unital algebra defined over a field $$k$$ with non-zero characteristic.

• Why does $M=R$ not work? Feb 26 '20 at 23:17
• I guess I want a non-trivial example. Feb 26 '20 at 23:33
• You mean that $M$ is not free? Feb 26 '20 at 23:40
• yes, I have put this in the question Feb 26 '20 at 23:42
• Just a minor rant about terminology. $M=R$ is not free as a bimodule. Also, a bimodule that is projective on the left and on the right is not necessarily a projective bimodule. Feb 27 '20 at 9:49

Take $$G$$ a nonabelian finite group of size coprime to $$p$$, and $$k$$ a field of characteristic $$p>0$$. Then $$k[G]$$ is semisimple (Maschke's theorem), so every module on left or right is projective.
In particular, take $$\underline{k}$$ to be the trivial representation of $$G$$. So this is a bimodule projective on both sides, and $$\mathrm{Hom}_G(\underline{k}, k[G])={_G\mathrm{Hom}}(\underline{k}, k[G])=\underline{k}$$, as bimodules (both of these being naturally the same subspace of $$k[G]$$, i.e. $$\langle\sum_{g \in G}g\rangle$$).
• In fact, for any group algebra $k[G]$ of a finite group, semisimple or not, and for any bimodule $M$, the left dual and right dual are isomorphic. And there are plenty of natural bimodules that are projective on both sides: e.g., $k[G]\otimes_{k[H]}k[G]$ for a subgroup $H\leq G$. Feb 27 '20 at 9:49