Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question is, in some sense, a variant of this, but for certain cases.

The opposite category of an abelian category is abelian. In particular, if $R-mod$ is the category of $R$-modules over a ring $R$ (say left modules), its opposite category is abelian. The Freyd-Mitchell embedding theorem states that this opposite category can be embedded in a category of modules over a ring $S$. This embedding is usually very noncanonical though.

Question: Is there any way to choose $S$ based on $R$?

My guess is probably not, since these notes cite the opposite category of $R-mod$ as an example of an abelian category which is not a category of modules. I can't exactly tell if they mean "it is (for most $R$) (provably) not equivalent to a category of modules over any ring" or "there is no immediate structure as a module category." If it is the former, how would one prove it?

I also have a variant of this question when there is additional structure on the category.

The module category $H-mod$ of a Hopf algebra $H$ is a tensor category. A finite-dimensional Hopf algebra can be reconstructed from the tensor category of finite-dimensional modules with a fiber functor via Tannakian reconstruction. Now $(H-mod)^{opp}$ is a tensor category as well satisfying these conditions (namely, the hom-spaces in this category are finite-dimensional). The dual of the initial fiber functor makes sense and becomes a fiber functor from $(H-mod)^{opp} \to \mathrm{Vect}$ (since duality is a contravariant tensor functor on the category of vector spaces). In this case, $(H-mod)^{opp}$ is the representation category of a canonical Hopf algebra $H'$.

Question$\prime$ What is $H'$ in terms of $H$?

share|improve this question
2  
The obvious guess for the second question is the dual, at least when H is finite-dimensional. Have you checked this e.g. for H = C[G], G a finite group? –  Qiaochu Yuan Jun 25 '10 at 0:23
    
That sounds reasonable (I had forgotten that Hopf algebras had dualizable axioms) but I'll have to convince myself. –  Akhil Mathew Jun 25 '10 at 0:53
1  
Hopf algebra axioms are only dualizable (in the finite-dimesional case) when the group object is commutative, unless one wants to give up commutativity on both sides (and then it doesn't correspond to a finite group scheme, so is Tannaka nonsense applicable?) –  Boyarsky Jun 25 '10 at 2:15
    
Well, I had restricted myself to the finite-dimensional hom-space case. I wasn't really clear about it in the question, so I'll add it. –  Akhil Mathew Jun 25 '10 at 13:03
    
@Boyarsky: Yes, Tannaka works much more generally. See the paper by Joyal and Street (they have many together: look for ones with "Tannaka" in the title). –  Theo Johnson-Freyd Jun 29 '10 at 6:58
add comment

4 Answers

up vote 17 down vote accepted

One can prove that for any non-zero ring $R$ the category $R$-Mod$^{op}$ is not a category of modules. Indeed any category of modules is Grothendieck abelian i.e., has exact filtered colimits and a generator. So for $R$-Mod$^{op}$ to be a module category $R$-Mod would also need exact (co)filtered limits and a cogenerator. It turns out that any such category consists of just a single object.

I believe this is stated somewhere in Freyd's book Abelian Categories but I am not sure exactly where off the top of my head. Edit it is page 116.

Further edit: For categories of finitely generated modules here is something else, although it is more in the direction of the title of your question than what is in the actual body of the question. Suppose we let $R$ be a commutative noetherian regular ring with unit and let $R$-mod be the category of finitely generated $R$-modules. Then we can get a description of $R$-mod$^{op}$ using duality in the derived category.

Since $R$ is regular every object of $D^b(R) \colon= D^b(R-mod)$ is compact in the full derived category. The point is that $$RHom(-,R)\colon D^b(R)^{op} \to D^b(R) $$ is an equivalence (usually this is only true for perfect complexes, but here by assumption everything is perfect). So one can look at the image of the standard t-structure (which basically just "filters" complexes by cohomology) under this duality. The heart of the standard t-structure is $R$-mod sitting inside $D^b(R)$ so taking duals gives an equivalence of $R$-mod$^{op}$ with the heart of the t-structure obtained by applying $RHom(-,R)$ to the standard t-structure.

In the case of $R =k$ a field then this just pointwise dualizes complexes so we see that it restricts to the equivalence $k$-mod$^{op}\to k$-mod given by the usual duality on finite dimensional vector spaces.

As another example consider $\mathbb{Z} $-mod sitting inside of $D^b(\mathbb{Z})$ as the heart of the standard t-structure given by the pair of subcategories of $D^b(\mathbb{Z})$ $$\tau^{\leq 0} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i>0\}$$ $$\tau^{\geq 1} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i\leq 0\} $$ It is pretty easy to check that $RHom(\Sigma^i \mathbb{Z}^n, \mathbb{Z}) \cong \Sigma^{-i} \mathbb{Z}^n$ and $RHom(\Sigma^i \mathbb{Z}/p^n\mathbb{Z}, \mathbb{Z}) \cong \Sigma^{-i-1}\mathbb{Z}/p^n\mathbb{Z}$ so that this t-structure gets sent to $$\sigma^{\leq 0} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i> 0, \; H^{0}(X) \; \text{torsion}\}$$ $$\sigma^{\geq 1} = \{X \; \vert \; H^i(X) = 0 \; \text{for} \; i< 0, H^0(X) \; \text{torsion free}\} $$ using the fact that objects of $D^b(\mathbb{Z})$ are isomorphic to the sums of their cohomology groups appropriately shifted.

So taking the heart we see that $$\mathbb{Z}-mod^{op} \cong \sigma^{\leq 0} \cap \Sigma\sigma^{\geq 1} = \{X \; \vert \; H^{-1}(X) \; \text{torsion free}, H^0(X) \; \text{torsion}, H^i(X) = 0 \; \text{otherwise}\} $$ with the abelian category structure on the right coming from viewing it as a full subcategory of $D^b(\mathbb{Z})$ with short exact sequences coming from triangles. It is the tilt of $\mathbb{Z}$-mod by the standard torsion theory which expresses every finitely generated abelian group as a torsion and torsion free part.

share|improve this answer
    
Great, thanks! $ $ –  Akhil Mathew Jun 25 '10 at 0:37
2  
The proof is not so hard I don't think - consider the canonical map from the countable coproduct of copies of $M$ to the countable product in $R$-Mod and its opposite. Then deduce that it is an isomorphism and use an Eilenberg swindle to show $M = 0$. –  Greg Stevenson Jun 25 '10 at 0:44
    
Greg, a naive question here (just betraying my misunderstanding I suspect) - in the category of vector spaces over a field $k$, why isn't the one-dimensional space both a generator and cogenerator? And isn't this category the same as $k$-mod? –  Yemon Choi Jun 25 '10 at 1:37
16  
More simply: If a category has products and coproducts, and if the same object is both initial and final, then you get a map from the coproduct of any set of objects to their product. In R-Mod this map is always a monomorphism. Therefore in R-Mod^{op} this map is always an epimorphism. But in S-Mod it is never an epimorphism in the case of infinitely many nontrivial objects –  Tom Goodwillie Jun 25 '10 at 1:53
1  
@Yemon: I just included the requirement for generators/cogenerators for completeness - you are right that a one dimensional vector space is both a generator and a cogenerator in k-Mod and that is not a problem. The problem arises in trying to require that both filtered limits and colimits be simultaneously exact. –  Greg Stevenson Jun 25 '10 at 2:16
show 2 more comments

I am three months late for the party, but I'll still add an answer (to the first question):

There is the notion of "locally finitely presentable category" which can be described equivalently as

(i) a category such that there exists a set of finitely presentable objects such that every object is a directed colimit of these.

(Explanation: $X$ being finitely presentable means that $Hom(X,-)$ commutes with directed colimits, for modules it is equivalent to the usual notion of finitely presented, intuitively because for a given morphism into the colimit each of the finitely many generators has to go into some finite stage of the colimit diagram, then one can form a cone over these finitely many objects, as the diagram is directed, and the morphism factors through there)

(ii) a category of all models in SET of a finite limit sketch

(Explanation: A finite limit sketch is a small diagram, or small category, with distinguished cones, a model in SET is a functor from this small category to sets which maps the distinguished cones to limit cones - example: the limit sketch of groups is a category $C$ with an object $G$, a map $G \times G \rightarrow G$ a commutative diagram stating associativity etc and the cones which e.g. say that the object $G \times G$ is the product of $G$ with itself. A group is a functor from this to SET which takes $G \times G$ to the product, i.e. maps the distinguished cones to limit cones. Likewise for a ring $R$ the limit sketch for $R$-modules is given by an abelian group object $G$ as above plus one group endomorphism of $G$ for each element of $R$, behaving as dictated by the multiplication in $R$)

By either description you see that the category of $R$-modules is a locally finitely presentable category and now there is a theorem (e.g. Adamek/Rosicky Thm 1.64) saying that the opposite of a locally finitely presentable category is never itself locally presentable unless we are in a trivial case where our category is a poset.

The cool thing about this result is that it not only answers your question about modules, but also applies to all other algebraic structures (the theory of locally presentable cats gives a characterization of categories of algebraic structures) and I like to read it as is a provable mathematical statement which reflects the duality between algebra and geometry: The dual of an algebraic category is not itself algebraic, but geometric...

share|improve this answer
    
Very interesting, thanks. –  Akhil Mathew Sep 26 '10 at 14:16
    
Just to clarify: the limit sketch of groups needs to have more than just one object, right? It should have $G$, $G\times G$, $G\times G\times G$, etc. Otherwise I don't understand how you have a map in that category from $G\times G\rightarrow G$. –  David White Jul 27 '12 at 14:38
    
How does this theorem of Adamek and Rosicky relate to Ryan Reich's answer? If $H$ is a Hopf algebra, is $H-mod$ a locally finitely presentable category? Seems like it should be, since I can write down all its defining properties via commutative diagrams. But $(H-mod)^{opp} \cong (H-mod)$ so that seems to contradict the theorem. What am I missing? –  David White Jul 27 '12 at 14:40
    
@DavidWhite: 1. Yes, you need more than one object, G is just one generating object. 2. My answer was only about the first part, not about the Hopf algebra part. I don't think H-mod is locally presentable; it seems that you need more than just finite limits to describe that structure. –  Peter Arndt Aug 11 '12 at 21:55
add comment

I won't outdo Greg Sevensen's excellent answer, but I wanted to make an extended comment in the case when $R$ is a $\mathbb K$-algebra for a field $\mathbb K$ and you are interested only in the category of finite-dimensional (over $\mathbb K$) modules. This is the standard situation in particular when $R$ is a Hopf algebra.

One version of the Tannaka theorem says the following (c.f. Joyal and Street, An introduction to Tannaka duality and quantum groups): Let $\mathcal C$ by a category and $F: \mathcal C \to \text{f.d.Vect}$ any functor (I'll write $\text{f.d.Vect}$ for the category of finite-dimensional vector spaces). Then there is a coalgebra $\operatorname{End}^\vee(F)$ so that $F$ factors through $\operatorname{forget}: \operatorname{End}^\vee(F)\text{-comods} \to \text{f.d.Vect}$, and it is universal with this property. When $\mathcal C$ is $\mathbb K$-linear abelian and $F$ is $\mathbb K$-linear exact faithful, then $\mathcal C \to \operatorname{End}^\vee(F)\text{-f.d.comods}$ is an equivalence of categories. (Further structure on $\mathcal C,F$, e.g. monoidal structure, gives further structure on $\operatorname{End}^\vee(F)$, e.g. bialgebra.) When $\mathcal C$ is the category of finite-dimensional $R$-modules for a $\mathbb K$-algebra $R$ and $F$ is the forgetful functor, then $\operatorname{End}^\vee(F)$ is (usually) a "predual" to $R$. Without categories, it is the coalgebra of linear functions on $R$ that factor through some finite-dimensional representation, and if $R$ is nice enough (unital, complete in the ways it should be, etc), it is the dual $R = \operatorname{End}^\vee(F)^*$.

In any case, if you reverse all the arrows in $\mathcal C$, I think that this exactly reverses the comultiplication in $\operatorname{End}^\vee(F)$ — I think that this "coopposite coalgebra" satisfies the correct universal property —, and hence it reverses the multiplication in $\operatorname{End}^\vee(F)^*$. So if $R$ is reconstructible from its category of finite-dimensional representations, then the opposite of this category is the category of finite-dimensional representations of $R^{\rm op}$, the algebra with the reversed multiplication.

share|improve this answer
add comment

To answer your question about Hopf algebras: suppose $H$ is a Hopf algebra and $\mathbf{H-mod}$ is its category of finite-dimensional representations. Then this category has duals and dualization expresses $\mathbf{H-mod} \cong \mathbf{H-mod}^\mathrm{op}$. So actually, you get the same Hopf algebra from the opposite category. Rather, I should say that you get the "opposite" Hopf algebra whose multiplication and comultiplication are reversed, and that the antipode map from $H$ to itself, which is an antiisomorphism and an anticoisomorphism, expresses the opposite Hopf algebra as the same as the original.

share|improve this answer
    
But then we have to give up commutativity of multiplication in Hopf algebras. Does the Tannakian stuff work in that generality, and is it useful? (I though the Tannakian stuff always spits out affine algebraic groups, albeit not necessarily of finite type, so the coordinate ring is a commutative ring. Maybe I am ignorant.) –  Boyarsky Jun 26 '10 at 4:17
    
The opposite of a commutative ring is still commutative, no? It's just that I've chosen an unusual isomorphism between the two. –  Ryan Reich Jun 26 '10 at 4:25
1  
Maybe I should clarify: I am not swapping the multiplication with the comultiplication, but replacing each operation individually with its opposite operation with the arguments (or co-arguments) reversed. –  Ryan Reich Jun 26 '10 at 4:26
    
Aha, so your point is that "dual representation" involves both linear duality and group inversion, by separating these two ingredients it's all trivial in the end (and so of course is sure to be useless, sad to say). Thanks for the clarification. –  Boyarsky Jun 26 '10 at 5:18
2  
There is actually a lot of separation in the Tannakian duaity theorem. You can prove any number of pieces almost independently: that a fiber functor is ind-representable; that the representing object is a coalgebra; that when the source is abelian then the fiber functor gives an equivalence with comodules over this coalgebra; that the tensor structure makes it an algebra; that the rigidity makes it a Hopf algebra. Then there are the recognition theorems for properties of the corresponding group.... –  Ryan Reich Jun 26 '10 at 5:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.