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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

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    $\begingroup$ Related: qchu.wordpress.com/2015/10/26/dualizable-objects-and-morphisms $\endgroup$ Aug 27, 2019 at 11:18
  • $\begingroup$ 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. $\endgroup$
    – the L
    Aug 27, 2019 at 11:27
  • $\begingroup$ So the situation for bimodules is significantly more difficult than for left/right-modules? $\endgroup$ Aug 27, 2019 at 11:48
  • $\begingroup$ 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. $\endgroup$ Aug 27, 2019 at 11:53
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    $\begingroup$ 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 $R$ --- $(f\cdot r)(b) = f(b)r$. $\endgroup$ Aug 27, 2019 at 12:33

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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 $R$ --- $(f⋅r)(b)=f(b)r$.

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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).

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    $\begingroup$ 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. $\endgroup$ Aug 28, 2019 at 5:19
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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.

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