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Is there a notion of algebraic geometry for these objects? If we take the dual category of the category of cocommutative corings with counit, is there geometry in it in a sense dual to affine schemes? Can we look at the set of coideals of a coring, put a space structure on it and sheaves (maybe cosheaves) of sections?

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Don't you mean coalgebraic cogeometry? –  Qiaochu Yuan Dec 16 '09 at 17:35
    
I suggest accepting Leonid's answer. –  Greg Kuperberg Dec 23 '09 at 7:02
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up vote 13 down vote accepted

Let's consider coalgebras over a field rather than corings. There is a theorem that every (coassociative) coalgebra over a field is the union of its finite-dimensional subcoalgebras. So the category of coalgebras over a field k is the category of ind-objects in the category of finite-dimensional coalgebras, while the latter is the opposite category to the category of finite-dimensional algebras. Now if we restrict ourselves to the commutative case, then the category of finite-dimensional commutative algebras with unit over k is the opposite category to the category of 0-dimensional schemes of finite type over k, or just schemes finite over Spec k. Combining it all, the category of cocommutative coalgebras with counit over k is equivalent to the category of ind-0-dimensional ind-schemes of ind-finite type over k, or just ind-finite ind-schemes over Spec k. This explains, in particular, that the "underlying topological space" functor maps cocommutative coalgebras to discrete sets rather than anything else, and the coalgebra itself is simply the infinite direct sum of the coalgebras sitting at the points of this set.

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It seems that Soibelman did algebraic geometry in category of coalgebra. He called noncommutative thin scheme. –  Shizhuo Zhang Dec 16 '09 at 22:37
    
Very interesting, so in that case the spaces that arise are not that interesting. But still I think it is nice that the points of the set are subcoalgebras, does look a bit like a spectrum. So maybe it could work for corings as well. –  jef Dec 16 '09 at 22:49
    
combining the two answers, maybe one can say that coring, and k-coalg already behave 'geometrically', maybe it just doesn't make sense to take the opposite category to get more geometry. Indeed in the paper of Kontsevich and Soibelman (arxiv.org/abs/math/0606241) they define noncommutative thin schemes as covariant functors from finite dimensional k algebras (not necessarily commutative) to sets commuting with finite projective limits, they go on to prove that this category is equivalent to the category of all k-coalgebras. –  jef Dec 16 '09 at 22:58
    
Yes, there is this term "thin schemes" being used in connection with coalgebras. Yes, points of the spectrum set correspond to cosimple subcoalgebras, a cosimple coalgebra being defined as a coalgebra having no nonzero proper subcoalgebras. Yes, I do think it is better to take the category of coalgebras rather than its opposite category as a category of geometric objects. –  Leonid Positselski Dec 16 '09 at 23:36
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Kontsevich-Rosenberg also worked a little bit on category of coalgebras as an example of their framework Q-category for generalized sheaf condition.The reference is their paper "Noncommutative spaces and flat descent" –  Shizhuo Zhang Dec 17 '09 at 9:53
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I like to think of algebraic geometry as being born out of the fact that Ring behaves a lot like Setop. For instance, in Ring, A ∐ (B × C) = (A ∐ B) × (A ∐ C), where ∐ is the coproduct in Ring, which is just the tensor product. This formula is also true in Set after we swap ∐ and ×. This suggests that we can use our intuition about Set to think about Ring if we replace Ring by its opposite, the category of affine schemes.

Cocommutative corings with counit are monoid objects in the opposite category of (Ab, ⊗). However, to get the morphisms to point in the right direction, we need to take the opposite category again: so Coring is (CAlg((Ab, ⊗)op))op. It follows that products in Coring are computed by tensor products in Ab, and colimits are formed by taking colimits of underlying abelian groups; and in fact Coring is a closed cartesian category, even more like Set than Ringop is. In particular, we don't want to take the opposite category of Coring. Maybe this isn't surprising, since every set is already a cocommutative comonoid (w.r.t. ×) in a unique way, and we have a functor Set → Coring taking a set to the free abelian group on it.

These are purely formal observations, and I don't know whether anyone has built a more concrete geometric theory, with say a functor from Coring to some kind of topological spaces with extra structure.

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Are you sure that the coproduct in Ring is the tensor product? –  Qiaochu Yuan Dec 16 '09 at 19:41
    
Yes (where by Ring I meant commutative rings, otherwise it is false). –  Reid Barton Dec 16 '09 at 19:52
    
Right, sorry. I got a little confused. –  Qiaochu Yuan Dec 16 '09 at 20:19
    
coproduct in Rings should be free product –  Shizhuo Zhang Dec 16 '09 at 22:40
    
The tensor product over Z is the coproduct, since Z is the initial object, so it's all very concrete. In fact, the tensor product is also the pushout of rings. –  Harry Gindi Dec 17 '09 at 16:25
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One can do (relative) algebraic geometry with respect to any symmetric monoidal category, with affine schemes corresponding to the opposite category of (commutative) monoid objects. This viewpoint is developed in the paper Au-dessous de Spec Z (in French) by Toen and Vaquie. By taking the category of $\mathbb{Z}$--modules with tensor products over $\mathbb{Z}$ one recovers usual algebraic geometry.

For (cocommutative) corings, you just have to do the same in the opposite category $(\mathbb{Z}-Mod)^{op}$. The big question is: what do you take as your monoidal structure? If you just dualize diagrams, your new coproduct would be the old product, i.e. direct product of the underlying abelian groups. Or instead of this you take usual tensor products. Each of these approaches will produce a different geometry.

I remember a course given by Lieven Le Bruyn on Kontsevich-Soibelman (the noncommutative coalgebra thing) and he mentioning that by the "discrete" nature of coalgebras one could never expect them to describe anything but the etale topology of a scheme, which I think is precisely what Leonid observed in his answer.

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Leonid speaks with great authority. I'd like to point out that you get a different answer in the category of graded coalgebras. If $A$ is a graded vector space, it has a graded dual $A'$ which is smaller than its full vector space dual $A^*$. Indeed, if the grading is locally finite, then this $A'$ has the same dimension sequence as $A$ and $A'' = A$. If $A$ is a locally finite, graded coalgebra, then its ring structure is given by an infinite sequence of finite tensors, and $A'$ is equivalently a graded algebra. Graded coalgebra homomorphisms also transpose to graded algebra homomorphisms. Recall that if $A'$ is also finitely generated (and scalar in degree $0$), then it corresponds to a projective variety with a choice of an ample line bundle. The morphisms between these are a perfectly good non-full subcategory of the projective schemes.

The full dual $A^*$ is also the graded completion of $A'$. So what Leonid is saying in this case is that if you take the graded completion of a finitely generated, graded algera, it localizes the projective variety to the apex of its affine cone. This is similar to what Leonid describes for $\text{Spec}(A^*)$ in general.

I said in a previous version of this answer that $\text{Spec}(A^*)$ atomizes the variety $\text{Proj}(A')$ to a "Cantor-set-like" structure. As Leonid points out in a comment, this is totally wrong in context. However, it is true that if $A$ is a finitely generated algebra, it has a completion $\hat{A}$ related to coalgebras such that $\text{Spec}(\hat{A})$ is an atomized form of $\text{Spec}(A)$. Namely, let $A'$ be the vector space of those dual vectors on $A$ that factor through finite-dimensional algebra quotients of $A$. Then $A'$ is a coalgebra, and $\hat{A} = (A')^*$ is an atomization in the sense that $\text{Spec}(\hat{A})$ is a 0-dimensional scheme whose points are the closed points of $\text{Spec}(A)$. (It isn't a Cantor set though.)

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This is a good example. I am not sure how to atomize a projective variety into a Cantor-set-like structure, though. If C is a nonnegatively graded coalgebra over k with C_0=k, then the coalgebra C is conilpotent and its spectrum, as defined in my answer, consists of a single point. When the dual graded algebra to C is finitely generated, the ind-scheme corresponding to C can be also thought of as a formal scheme. This formal scheme is basically the affine cone over the corresponding projective variety, formally completed at the origin. So it has a single closed point. –  Leonid Positselski Dec 17 '09 at 18:22
    
Once again, you have quickly caught me in an absurd error. Maybe I should have you pre-referee my papers. Anyway, thank you; I will fix it. –  Greg Kuperberg Dec 17 '09 at 19:03
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