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The dual of an abelian category is again abelian, since the axioms are all preserved by the reversing of arrows. For example, the category of finite-dimensional vector spaces over a field is easily seen to be dual to itself, since we can just take linear duals of vector spaces. However, this is the only example where I know a 'concrete' description of both an abelian category and its dual, where by concrete I mean describing the category as a variant of, say, modules over a ring or sheaves on some space.

What are other examples of dual pairs of abelian categories which can both be described 'nicely'? For example, is there a concrete description of the dual of the category of abelian groups?

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up vote 6 down vote accepted

A few (mostly trivial) examples (may be related to references in Tim Porter's answer?):

First of all, the self-duality of the category of vector spaces can be enriched to other self-dualities:

  • The category of finite-dimensional representations of a group is self-dual.

  • The category of finite-dimensional representations of a Lie algebra is self-dual.

(Of course, now we can generalize this to Hopf algebras.)

Another simple class of examples comes from taking the dual of some object, and asking what additional structure is necessary to reconstruct the original object. For instance:

  • Let $V$ be a vector space. Consider $V^*$. It has a natural topology, in which open subspaces are orthogonal complements of finite-dimensional subspaces of $V$. (This is essentially weak topology if $V$ is viewed as a discrete space.) This gives a duality between vector spaces and certain class of topological vector spaces that can be easily described explicitly. (A fancier way to view this: vector spaces are 'ind-finite-dimensional', so their dual are 'pro-finite-dimensional'.)

This can be modified to duality on Tate vector spaces, if you happen to like this sort of things.

There are also some examples in (algebraic) geometry (that are probably not as straightforward as the others).

  • Category of perverse sheaves is self-dual.

  • Category of holonomic $D$-modules is self-dual (this is related to the previous comment by the Riemann-Hilbert).

Finally, there is a (somewhat cheap) trick to start with a duality on the derived category, then take the abelian category sitting inside it, and realize that it is dual to its image. For instance:

  • Look at the derived category of constructible sheaves (say, on a scheme); it has Verdier's duality. Then the abelian subcategories corresponding to dual perversities are dual to each other. For instance, the category of constructible sheaves is dual to perverse sheaves for a certain perversity.

  • Look at the derived category of coherent sheaves on a scheme; it has Serre's duality. Then the abelian subcategories corresponding to dual perversities are dual to each other. For instance, the category of coherent sheaves is dual to perverse sheaves for a certain perversity.

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The dual of the category of abelian group is the category of compact topological abelian groups, by Pontryagin duality.

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This just shows how little I understand compact abelian groups. One of the main things that prevents $(R\text{-mod})^\text{op}$ from being a category of representations is that the canonical map $\coprod^\infty M \to \prod^\infty M$ in $(R\text{-mod})^\text{op}$ is an epi that is not a mono. So if I take the circle, say, and look at its countable, say, coproduct in the category of compact abelian groups --- I don't actually know what I get! But you're telling me that it <strike>surjects</strike>epimorphs onto the countable product? –  Theo Johnson-Freyd Sep 23 '10 at 16:17
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The majority of the results that I have seen relate to various generalisations of linearly compact modules and some indication of the theory can be gleaned from the paper by U. Oberst

Duality theory for Grothendieck categories, Bull. Amer. Math. Soc. Volume 75, Number 6 (1969), 1401-1407.

Note that if one has a category of modules, as direct limits are exact there, inverse limits will be exact in the dual category and that sort of phenomenon is typical of categories of 'compact' modules over some compact ring. The meaning of compact will usually be 'algebraic' in character but vary according to what extra properties the input abelian category has. In the Oberst case these include being a Grothendieck category, being Noetherian and so on.

A lot of this is well treated in N. Popescu, Abelian Categories with Applications to Rings and Modules, Academic Press, 1973. MR 0340375 (49:5130)

That is thorough and quite long so I won't attempt to describe it more fully here.

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See my answer to the question about construction of opposite categories. There I descripe the dual category of $R \text{-Mod}$ geometrically. But to be honest, I don't think that this description is useful at all. Note that $(R \text{-Mod})^{op}$ is never equivalent to some $S \text{-Mod}$, since in $R \text{-Mod}$ we have $\cup_i (F_i \cap G) = (\cup_i F_i) \cap G$ for a filtered system $F_i$, but we do not have the dual statement $\cap_i (F_i + G) = (\cap F_i) + G$.

Here is another example: Let $R$ be a PID, which is not a field. Then $Hom(-,Q(R)/R)$ yields an equivalence between the abelian category of finitely generated $R$-modules and its dual. Perhaps this also works for hausdorff locally compact abelian groups (Pontryagin duality); but perhaps this category is not abelian: Kernels and Cokernels exist by abstract nonsense, but I don't think that every epimorphism is a cokernel.

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Since the category of locally compact Hausdorff abelian groups is self-dual, we may as well ask whether every monomorphism is a kernel. And indeed that's false: kernels are closed subgroups, so the irrational line on a torus would be a counterexample. –  Todd Trimble Sep 24 '10 at 14:11
    
@Todd: Hm, I don't understand this example yet. Which homomorphism are you considering? And also, is it continuous? –  Martin Brandenburg Sep 24 '10 at 15:03
    
The homomorphism R --> R/Z x R/Z which takes t to (t mod Z, st mod Z) where s is irrational. This is a continuous injective homomorphism, with dense image. –  Todd Trimble Sep 24 '10 at 16:42
    
Ah ok. This is an example of a morphism in the category of loc. comp. top. groups, which is mono and epi, but not iso. Thus this category is not abelian, as I expected. –  Martin Brandenburg Sep 24 '10 at 16:52
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An $\epsilon$ generalization of your example: if $A$ is a $k$-algebra for a field $k$, then opping the category of finite-dimensional $A$-modules gives the category of $A^{\rm op}$ modules (the same algebra with the opposite multiplication). I think you know already that this patently fails when you move to infinite dimensional things: the canonical map $\coprod^\infty M \to \prod^\infty M$ is a mono that is not an epi in $A$-mod, whereas it is an epi that is not a mono in $(A$-mod$)^{\rm op}$.

Letting $A = \mathbb Z$ and $k = \mathbb F_1$ (yeah, that sounds right), I think the analogy is that finite abelian groups are closed under Pontrjagin duality. Hm, now I wonder what other "$\mathbb F_1$ algebras" have interesting finite representation theory.

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I guess I should mention to that if you want to study an algebra, its finite-dimensional representation theory generally does not know enough. What you can reconstruct from the finite-dimensional representation theory of an algebra is its "hopf dual", a coalgebra. The full dual of the "hopf dual" is an algebra that is generally smaller than your original algebra, but has the same finite-dimensional representations. –  Theo Johnson-Freyd Sep 23 '10 at 16:13
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