# 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

• Aug 27, 2019 at 11:18
• There is a theory of dualizing complexes over (certain) noncommutative rings. These dualizing complexes are actually complexes of bimodules, so the dual of a bimodule is a complex of bimodules. Aug 27, 2019 at 11:27
• So the situation for bimodules is significantly more difficult than for left/right-modules? Aug 27, 2019 at 11:48
• In general, an $A$-$B$ bimodule $M$ is the same thing -in the sense that there is an equivalence of categories- as a (right) $B\otimes A^{op}$ module. So, proceeding in a definition of a dual bimodule, depends on the notion of duality you have in mind for a right module. Aug 27, 2019 at 11:53
• 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 $B$ --- $(r\cdot f)(b)=f(br)$ --- and right $R$-module structure coming from the right $R$-module structure on $R$ --- $(f\cdot r)(b) = f(b)r$. Aug 27, 2019 at 12:33

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 $$R$$ --- $$(f⋅r)(b)=f(b)r$$.

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).

• These duals, I should mention, really exhibit duality in the sense that the left dual of the right dual is the original, and same with the right dual of the left dual. Aug 28, 2019 at 5:19

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.