# What does Rng^{op} look like?

There are several well-known dualization results in category theory, i.e. that such-and-such a well-known category D is isomorphic to the opposite C^{op}. Does anyone know of such a result concerning what the opposite category to Rng, rings (*-monoid on +-group) with their homomorphisms, looks like?

I ask (naively I should add) because I'm curious if there's a natural "algebraic" structure on homology dual to that of the ring-structure of cohomology.

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When working with not-necessarily-commutative rings, there is a sense in which we fail to have an adequate supply of morphisms between them. If you consider the category of rings where morphisms $A\to B$ are $A-B$ bimodules and composition is given by the tensor product, we obtain a more useful category. Two rings are isomorphic in this category if and only if they are Morita equivalent. We can then think of rings in terms of their abelian categories of modules. This is the starting point for one of the approaches to noncommutative geometry. – Harry Gindi Feb 8 '11 at 14:05
The affine objects in noncommutative geometry are then defined to be an appropriate form of a noncommutative spectrum of an abelian category. (Also, it's worth noting that Morita equivalence classifies commutative rings up to isomorphism (the proof is a Yoneda-type argument).) – Harry Gindi Feb 8 '11 at 14:10
Anyway, the way this relates to your question is that, I think, spectra of abelian categories should be the "correct version of the opposite category" consisting of appropriate affine noncommutative geometric objects. – Harry Gindi Feb 8 '11 at 14:29
@Nick: A co-ring is not the same as an object in the opposite category of the category of rings! – Rasmus Bentmann Feb 8 '11 at 15:07
Nick, for me a rng is a ring without identity, and a *-monoid is a monoid (which includes an identity) equipped with an anti-involution. Could you please clarify what you mean? – Todd Trimble Feb 8 '11 at 16:07

The category you're interested in, as noted by others, is the category of coalgebras / corings, which is emphatically not the opposite category of rings --- but we're going to see exactly what's different between the two in the nicest case. To start things off, here's a definition of a coalgebra so that we're all on the same page: an $R$-coalgebra is an $R$-module $A$ together with morphisms $\Delta: A \to A \otimes_R A$ and $\epsilon: A \to R$ satisfying

1. Coassociativity: $(\Delta \otimes \operatorname{id}) \circ \Delta = (\operatorname{id} \otimes \Delta) \circ \Delta$,
2. Counitality: $(\epsilon \otimes \operatorname{id}) \circ \Delta = (\operatorname{id} \otimes \epsilon) \circ \Delta = \operatorname{id}$ after identifying $R \otimes_R A$ and $A \otimes_R R$ with $A$.

Coalgebras are familiar objects in algebraic topology, and you've already found the biggest source of them. Suppose you have a cohomology theory $E$ whose coefficient ring is a (graded) field. For any space $X$, the constant map $X \to \mathrm{pt}$ induces a map in homology $E_* X \to E_*$ which serves as the counit for $E_* X$. Toward a comultiplication, we have maps $E_* X \xrightarrow{\Delta} E_* (X \times X) \leftarrow E_* X \otimes_{E_*} E_* X$, but asking for the right-hand map to have an inverse is the same as asking for a Kunneth isomorphism. This only happens under restrictions on $X$ or restrictions on $E$ --- such as when $E_*$ is a field, which is one reason we requested that.

Now, let's lay our cards on the table and just announce some dualities we see in front of us:

1. There's your opposite category $\mathsf{Rings}^{op}$, which is dual to rings in the sense that $(\mathsf{Rings}^{op})^{op}$ is equal to $\mathsf{Rings}$.
2. In the language of the comments below, coalgebras are Eckmann-Hilton duals of algebras. This is really a statement about how we produced the definition above: we took the definition for an algebra, and we flipped all the arrows around.
3. There's also the notion of linear algebraic duals: $V^\vee = \operatorname{Hom}_k(V, k)$. To avoid some very serious technicalities, we'll want to work in the nicest, most familiar setting possible: modules of finite rank over a ground ring that's a field.

Now, we would like to compare these three ideas. There is another category of interest floating around: the category of $k$-algebras has an associated category of presheaves $\widehat{\mathsf{Algebras}_k}$ $=$ $\operatorname{Functors}(\mathsf{Algebras}_k, \mathsf{Sets})$ which receives a map $\mathsf{Algebras}_k^{op} \to \widehat{\mathsf{Algebras}_k}$ described by the left side of the $\operatorname{Hom}$-functor: $X \mapsto \operatorname{Hom}(X, -)$. This assignment, called the Yoneda embedding, is a functor into a cocomplete category which is an equivalence onto its image and whose image is codense --- these are consequences of the Yoneda lemma. That the target of the Yoneda embedding is cocomplete makes it a much nicer category to play around in, and so it's worth considering what this embedding's use is.

I claim there's a relation between the category of $k$-coalgebras and $\widehat{\mathsf{Algebras}_k}$. Again, to make linear algebra behave nicely, we need to encode finiteness restrictions into our setup, and to make that happen we'll turn to "$k$-formal schemes". The classical $\operatorname{Spec}$ construction in algebraic geometry also gives a contravariant functor off the category of $k$-algebras which is an equivalence onto its image. Rather than fussing with what a Zariski spectrum is, since we're just playing around with categories, I will instead take the Yoneda embedding to be my definition of $\operatorname{Spec}$ and the presheaf category to be something dimly, vaguely, sorta like the category of schemes. Representable presheaves (i.e., those in the image of $\operatorname{Spec}$) are called affine schemes. Plenty of constructions from algebraic geometry transfer almost without comment; for instance, defining $\mathbb{A}^1 = \operatorname{Spec} \mathbb{Z}[x]$, we recover the functor $\mathcal{O}(X) = \operatorname{Hom}_{\widehat{\mathsf{Algebras}_k}}(X, \mathbb{A}^1)$, which in the case of an affine $X = \operatorname{Spec}(A)$ gives $\mathcal{O}(\operatorname{Spec} A) \cong A$.

A scheme $X$ will be called finite if it is $\operatorname{Spec}$ of an algebra of finite dimension as a $k$-module. These, too, are in ample supply in algebraic topology. If $X$ is a compact pointed space, then the algebra $H^* X$ will be finite in the sense we need. Of course, algebraic topology gets done on more than compact spaces, so we need to broaden our perspective a little bit: we can ask instead that $X$ be compactly generated, so that if $X_\alpha$ denotes the collection of compact subsets of $X$ directed by inclusion, we then have $X = \operatorname{colim} X_\alpha$. In the case that $X$ is a CW-complex, it is sufficient to take $X_\alpha$ to be just the finite subcomplexes of $X$. We might then be interested in the scheme $X_E := \operatorname{colim} \operatorname{Spec} E^* X_\alpha$. Such a scheme which occurs as the colimit of a directed system of finite $k$-schemes is called a $k$-formal scheme.

So now we have a suitable notion of a 'finite' scheme that still captures all our interesting (and frequently large) cohomology rings. Here's the comparison I promised early on: the functor taking a formal scheme $\operatorname{colim} \operatorname{Spec} A_i$ to the $k$-coalgebra $\operatorname{colim} A_i^\vee$ is an equivalence of categories. There's almost nothing to say because the sea of definitions we've made take care of most everything, but there is one key lemma: given a $k$-coalgebra, you need to know that you can write it as the colimit of finite $k$-coalgebras. The exact lemma that gets used is: if $E$ is a finite dimensional subspace of a $k$-coalgebra $A$, then there exists a subcoalgebra $F \subseteq A$ which is finite as a $k$-vector space and which contains $E$ (i.e., $E$ can be finitely enlarged so that it becomes closed under comultiplication). If you push around elements a bit you'll see that that's the case (and that this requires working over a ground field). Then, straight after, here's the second big assertion: the assignments $X \mapsto E_* X$ and $X \mapsto X_E$ are equivalent under the above equivalence of categories.

Now, this construction is delicate, and the limitations are not meant to be taken lightly! You need a ring structure on the underlying (co)homology theory to even dream of having products, you need Kunneth isomorphisms to make sense of the coalgebra structure, you need a coefficient field to have good linear-algebraic duality, and there are still potential problems with supercommutativity that we haven't addressed. But, when all the stars align and God smiles on us, this is what the coalgebra structure on homology is supposed to mean: it's another presentation of the formal scheme associated to the (perhaps more familiar) ring structure on cohomology.

To counterbalance that caveat, that's not to say that this point of view is not immensely useful. Here are some sample applications:

• All of chromatic homotopy: $\mathbb{C}\mathrm{P}^\infty$ ($= B\mathrm{U}(1) = \mathrm{Gr}_1$) carries the structure of an $H$-group, and there are a whole sea of cohomology theories $E$, called complex-orientable, for which $\mathbb{C}\mathrm{P}^\infty_E$ is (noncanonically) isomorphic to $\hat{\mathbb{A}}^1 = \operatorname{colim} \operatorname{Spec} E^*[x] / \langle x^n \rangle$, an object which behaves a lot like an infinitesimal one-dimensional Lie group. This "formal Lie group" carries an immense amount of information about the cohomology theory $E$, and the "space" of available formal Lie groups carries an immense amount of information about stable homotopy theory as a whole.
• There are a variety of partial theorems in the following spirit: if $E$ and $F$ are complex oriented cohomology theories and $F$ is represented by the spaces $F_k$ in the sense that $F^k X = [X, F_k]$, then $\bigoplus_k E_* F_k$ behaves like $\operatorname{Hom}(\mathbb{C}\mathrm{P}^\infty_E, \mathbb{C}\mathrm{P}^\infty_F)$. Goerss has shown this when $E = H\mathbb{F}_p$ and $F$ satisfies a certain condition on $F_* \Omega^2 S^3$ (which is also satisfied for a class of complex-oriented spectra called Landweber exact). (Addendum: Goerss spends a lot of time setting up a "super" version of Dieudonne modules, which is meant to address in part issues with supercommutativity ignored/avoided here.)
• Armed with an ample supply of Kunneth isomorphisms, Ravenel and Wilson, the progenitors of the idea above, computed these coalgebras in the cases where $E$ and $F$ ranged in $K(n)$ (Morava $K$-theory), $E(n)$ (Johnson-Wilson theories), $BP$ (Brown-Peterson theory), and $H\mathbb{Z}/p^j$ (singular theories / Eilenberg-Mac Lane spaces). For instance, one can define the free (supercommutative) algebra on a group object in the category of $k$-formal schemes, and it turns out that $H_*(K(\mathbb{Z}/p^j, q); \mathbb{F}_p)$ is the free alternating algebra in this sense on the formal group scheme $B\mathbb{Z}/p^j_{H\mathbb{F}_p}$. A similar statement can be made for $K(n)_* K(\mathbb{Z}/p^j, q)$.
• The above ideas have stable versions too: the homology of spectra $E_* F$ is to be thought of as the scheme of isomorphisms $\operatorname{Iso}(\mathbb{C}\mathrm{P}^\infty_E, \mathbb{C}\mathrm{P}^\infty_F)$, and so, for instance, the dual of the Steenrod algebra, an object of classical interest, can be thought of as $\mathcal{O}$ of the automorphisms of a particular formal Lie group $\hat{\mathbb{G}}_a$. Relatedly, the statement that the dual Steenrod algebra coacts on the homology coalgebras $H_* X$ is straightened out in the category of formal schemes by saying that $\operatorname{Aut} \hat{\mathbb{G}}_a$ acts on the formal scheme $X_{H\mathbb{F}_2}$. This has also received classical interest, though not in this language: the final part of Thom's thesis on calculations in the real bordism ring amount to showing that this $\operatorname{Aut} \hat{\mathbb{G}}_a$ action on the homology of the real bordism spectrum is free.
• Finally, this is an example of a broader phenomenon: often the schemes associated to rings have enlightening interpretations in moduli theoretic terms. A great many classical objects in stable homotopy theory lead double lives in this framework as moduli spaces, and the geometry of the moduli space frequently informs us on how the original topological objects behave.

Anyway, this is all to say that you should definitely care deeply about the coalgebra structure on homology, as it's one way to get into formal schemes, where everything is interesting and magical and great.

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could you give a reference for the result of Goerss that you mention? Excellent answer by the way. – Sean Tilson Feb 9 '11 at 2:24
Paul Goerss' Hopf rings, Dieudonné modules, and E_∗ Ω^2 S^3. Homotopy invariant algebraic structures (Baltimore, MD, 1998), 115–174, Contemp. Math., 239, Amer. Math. Soc., Providence, RI, 1999. – Eric Peterson Feb 9 '11 at 3:05
Thank you very much. – Sean Tilson Feb 9 '11 at 5:33
+1: Your answer has inspired me to learn more. A few questions. So there's no problem with understanding Spec of a ring which is only graded-commutative? Also, isn't it slightly misleading to describe coalgebras as dual to algebras in the "linear-algebraic" sense? They are "Eckmann-Hilton dual" (of course this distinction only matters when things aren't of finite type). – Mark Grant Feb 9 '11 at 7:08
Beautiful answer. – B. Bischof Jun 10 '11 at 14:10

If you restrict to commutative rings with identity, the opposite category is equivalent to the category of affine schemes. See the book "The geometry of schemes" by Eisenbud and Harris. I'll leave it to others more qualified to tell you more...

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There is a natural algebraic structure on homology that is dual to the algebra structure on cohomology. It is gotten the same way you get the cup product: use the Kunneth theorem and the diagonal to give you a coalgebra. $H_\ast(X) \to H_\ast(X \times X) \to H_\ast(X) \otimes H_\ast(X)$, the downside is that coalgebras seem strange.

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It's worth noting that this has essentially nothing to do with $Rng^{op}$. – Harry Gindi Feb 8 '11 at 15:03
at least if $H_*(X)$ is torsion free, otherwise there's no Kunneth isomorphism and no coalgebra... – Tilman Feb 8 '11 at 15:34
I have no idea why this answer was accepted, and I also don't know why Sean considers coalgebras "strange". (Less familiar than algebras for most people, sure.) I have to second Harry's response here. – Todd Trimble Feb 8 '11 at 16:07
@Todd: I think the answer was accepted because it was, to some extent, what the OP was looking for. I should rephrase the comment about coalgebras being strange. They are not strange, but the coproduct on homology "seems" harder to work with than the product in cohomology, or products in general for that matter. This is because we are taught algebras first. There is a quote of Moore's that Kathryn Hess posted on the n-Category Cafe a bit ago (6 months at least) which was my intended meaning. Also, I agree completely with Harry's comment on my answer, I voted it up even. – Sean Tilson Feb 8 '11 at 19:42
@Sean: The non-naturality is the sticking point. Using the AW map you can actually see that $C_*(X)$ itself is always a differential graded coalgebra, but being a coalgebra isn't preserved under taking homology. – Tyler Lawson Feb 9 '11 at 2:18