Dual of a bimodule For a noncommutative ring $R$, and an $R$-$R$-bimodule $B$, is there a "correct/natural" notion of a dual bimodule? I am interested, really, when $B$ is projective as a left $R$-module.
Note: Switched from Stackexchange, since no answers
 A: Copied from comments as requested.
There isn't enough context in the question but if you know that $B$ is a projective left $R$-module and nothing else there is a fair chance that you want $\mathrm{Hom}(B,R)$, where $\mathrm{Hom}$ means left $R$-linear maps, with left $R$-module structure coming from the right R-module structure on $B$ --- $(r⋅f)(b)=f(br)$ --- and right $R$-module structure coming from the right $R$-module structure on $R$ --- $(f⋅r)(b)=f(b)r$.
A: As explained in more detail in this blog post linked by Jakob in the comments, every $(A, B)$-bimodule $M$ has two natural duals:


*

*If $M$ is finitely generated projective as a left $A$-module, it has a left dual given by the $(B, A)$-bimodule $\text{Hom}_A(M, A)$.

*If $M$ is finitely generated projective as a right $B$-module, it has a right dual given by the $(B, A)$-bimodule $\text{Hom}_B(M, B)$.


These duals come from thinking of an $(A, B)$-bimodule as a 1-morphism in the Morita 2-category whose 


*

*objects are rings

*1-morphisms are bimodules

*2-morphisms are bimodule homomorphisms


and applying the general equational definition of dual or adjoint 1-morphisms in a 2-category given by the zigzag identities (the one which, applied to the 2-category of categories, produces left and right adjoints). 
A: I do not know if there is a "standard/correct/natural" method of defining the notion of dual bimodule. A little search i made in the literature did not lead me to smt "standard". In any case, expanding the idea in my comment above, here is what i have thought of:  
Given an $A$-$B$-bimodule $M$ we can always view this as a right $B\otimes A^{op}$-module setting $m(b\otimes a^{op})=amb$. Conversely, if $M$ is a right  $B\otimes A^{op}$-module then we get an $A$-$B$-bimodule by setting $am=m(1\otimes a^{op})$ and $mb=m(b\otimes 1)$.
 (Here $a^{op}$ stands for the element $a$ of the algebra $A$ viewed as an element of the opposite algebra/ring $A^{op}$).
Similarly, an $A$-$B$-bimodule $M$ can be viewed as a left $B^{op}\otimes A$ module.  
So, start with your $A$-$B$-bimodule, view it as a right $B\otimes A^{op}$-module and then get the dual module $Hom_{B\otimes A^{op}}(M,B\otimes A^{op})$. This will be a left $B\otimes A^{op}$-module. Following a strategy similar to the one mentioned in the preceding paragraph this can be viewed as an $A^{op}$-$B^{op}$-bimodule, or equivalently a $B$-$A$-bimodule which may be the notion of the dual of your initial $A$-$B$-bimodule you are looking for.  
P.S.: This is a general method, which i think works for modules over rings or algebras. It is not tied especially to the projective case.
