# What does equivalence of categories really tell us?

Apology for the vague question, but I hope I can make it clearer by considering some examples. Also, I added algebraic geometry to the tags because I'll borrow my examples from there, but feel free to change the tags.

From a formal point of view, an equivalence between two categories tells us that the two categories are 'essentially the same'. Nevertheless it may still happen that two equivalent categories come from different fields of mathematics, both with their own sets of definitions or theorems, which do not seem to translate to each other quite easily. Here are three examples which exhibit the just-mentioned phenomenon in different degrees.

• The category of prevarieties over an algebraically closed field $k$ is equivalent to the category of integral schemes of finite type over $k$. (The definitions are as in Mumford's Red Book.) In both categories, objects are represented by topological spaces with a sheaf. These spaces are constructed very similarly (the former from maximal ideals, and the latter from prime ideals) and look very much alike (the latter being like the former but with some hidden points, some reducedness). Many definitions, constructions, and proofs in the former category just carry over to the latter.

• The category of smooth projective curves over $\mathbb{C}$ is equivalent to the category of compact Riemann surfaces. There are strong similarities between the two categories. Both have meromorphic functions and divisors, both have differential forms, both have a Riemann--Roch theorem, and so on. It seems to me (but I may well be wrong here) that these similarities are more of a heuristic observation than something formal. For instance you can't apply the analytification functor to transform Riemann--Roch for curves into Riemann--Roch for Riemann surfaces, let alone transform a proof of one into a proof of the other.

• For many suitable ringed spaces $(X,\mathcal{O}_X)$, there's an equivalence between the category of finite-rank locally free sheaves on $X$, and the category of finitely generated projective $\mathcal{O}_X(X)$-modules, including for affine schemes and smooth manifolds. The objects representing both categories are very different in nature, and it seems to me that concepts from one category don't have a nice translation to the other category. (Stupid examples: How do I say that a smooth vector bundle $E$ on $M$ has vanishing curvature in terms of $C^\infty(E)$? How does saying a projective $A$-module is the natural module over a matrix ring translate to coherent sheaves?)

• There is an equivalence between the category of commutative $C^*$-algebras and the opposite category of compact Hausdorff topological spaces. Yet the objects representing the two categories are entirely different. There are concepts in one category that do not translate well to the other. (Stupid examples: To which $C^*$-algebras do manifolds correspond? What is the Spectral Theorem in the context of compact Hausdorff topological spaces?)

It is often said that the distinction between classical and modern algebraic geometry is artificial, and the equivalence between classical varieties and finite-type integral schemes should back up this statement. But no-one would say that the distinction between functional analysis and topology is artificial because of Gelfand--Naimark. So how much should I still think of the equivalence between varieties and integral schemes? Exactly how happy should I be when I prove that two seemingly different categories are equivalent? In short:

What does an equivalence of categories really tell us?

• Concepts in one category carry over to concepts in an equivalent one if they can be expressed in categorical terms. For instance I don't know whether "being a manifold" can be expressed in terms of isomorphisms and morphisms (it's a "local" property, which is hardly expressible -at least at first sight- in categorical terms), but if it can't then this explains why there's no corresponding concept in $C^*$-algebras; and similarly for the spectral theorem. However, many concepts are actually categorical ones, and they can be transported along an equivalence – Maxime Ramzi Mar 30 '18 at 17:11
• Relevant nLab article: ncatlab.org/nlab/show/evil – Clive Newstead Mar 30 '18 at 17:16
• Dear Max, thanks for the explanation, that makes sense. But why then are there sometimes very strong analogies, like between curves and Riemann surfaces, despite the concepts involved in that example not being categorical? – user542740 Mar 30 '18 at 17:20
• – Arnaud D. Mar 30 '18 at 18:22
• This is not something I have "hard evidence" to support, but I'd say that having equivalent categories that have properties that don't translate naturally between them is precisely why the equivalence is useful. Say you are working on a problem set in some category and you exhaust known results and seem to get nowhere. Then you switch to equivalent category that has some other natural notions inside of it that the original category doesn't. You can now use this new machinery to help solve your problem. If you could do it in the first category, then what would be the point of equivalence? – Ennar Mar 30 '18 at 20:48

Coming back to this question a couple of months on: how could I have neglected to mention Morita theory in my answer above? I'm posting again to include it, rather than editing my answer above, because it's of a rather different nature than the examples either of us has mentioned so far, but I think it's a very worthwhile example to see.

Let $R$ and $S$ be two rings, and consider the abelian categories of left modules $R\text{-Mod}$ and $S\text{-Mod}$. Suppose we have an equivalence of abstract categories $F: R\text{-Mod}\to S\text{-Mod}$. (Note: $F(R)$ isn't necessarily $S$, or even a free $S$-module. When I first learnt that fact, I thought "well, what good is the existence of such an $F$, if it doesn't even preserve such a basic property?".)

Firstly: if an $F$ like this exists, it's automatically an additive functor. (That is, $\mathrm{Hom}(A,B) \cong \mathrm{Hom}(FA,FB)$ is not just a bijection, it's actually a group isomorphism.)

Secondly: if the left-module categories are equivalent, then the right-module categories are equivalent too, and there are alternative characterisations of this property that make it easy(ish) to check.

Thirdly: such an $F$ actually preserves many properties of individual modules, as well as of the rings $R$ and $S$. Wikipedia has two nice lists of them, which I quote/paraphrase here:

the R module M has any of the following properties if and only if the S module F(M) does: injective, projective, flat, faithful, simple, semisimple, finitely generated, finitely presented, Artinian, and Noetherian

[but not: free, cyclic]

and

[R has any of the following properties if and only if S does:] simple, semisimple, von Neumann regular, right (or left) Noetherian, right (or left) Artinian, right (or left) self-injective, quasi-Frobenius, prime, right (or left) primitive, semiprime, semiprimitive, right (or left) (semi-)hereditary, right (or left) nonsingular, right (or left) coherent, semiprimary, right (or left) perfect, semiperfect, semilocal

[but not: commutative, local, reduced, domain, right (or left) Goldie, Frobenius, invariant basis number, and Dedekind finite]

Much like my example of "3-dimensionality" being a categorical property of vector spaces in my older answer, all of these properties are preserved precisely because they are categorical.

• Thank you. I accepted this answer because the list of preserved properties made it click what it means for a property to be 'expressible in category-theoretic language'. (In hindsight, that is exactly the point that Max made in the comments, and that you made in the previous answer, so I guess I'm just slow.) – user542740 Jun 6 '18 at 15:12

I think this is a great question, and I don't pretend to have a full answer for it, but here are some of my thoughts:

Exactly how happy should I be when I prove that two seemingly different categories are equivalent?

This depends on exactly how "seemingly different" they are - or why you're trying to prove this in the first place. I'll illustrate this with respect to your examples:

the equivalence between classical varieties and finite-type integral schemes

You shouldn't be surprised that these the same thing: the latter was practically designed to be a new way of encoding the former! But from the perspective of a researcher who only knows the former, and is trying to invent a new formalism to encode it, you should certainly be happy about it, because it tells you that the formalism is in some sense "correct" - it doesn't gain or lose too much information categorically.

commutative $C^∗$-algebras and the opposite category of compact Hausdorff topological spaces ... To which $C^∗$-algebras do manifolds correspond?

This is really a question about manifolds inside the category of compact Hausdorff topological spaces, and not about the equivalence of categories. Is the property of being a manifold a categorical property? That is, suppose I hand you the abstract category of compact Hausdorff topological spaces, and I don't allow you to actually use any topological properties (e.g. I just give you a bunch of vertices and arrows with composition data, and I don't tell you which spaces the vertices represent or which morphisms the arrows represent). Can you pick out the manifolds from this?

A stupid example of my own: let C be the category of finite-dimensional vector spaces. It's easy to pick out the vector spaces of dimension 3, as follows. The spaces of dimension 0 will be initial objects, so you know which those are; then, inductively, the spaces of dimension n+1 will be the objects x that aren't spaces of dimension < n+1, and such that the only morphisms into x are those coming from spaces of dimension < n+1 or those coming from isomorphic objects.

Now, maybe you can pick out the manifolds categorically. I don't actually know. But my suspicion is, if you can, it's not in any nice way, and so when you translate it into the language of $C^*$-algebras, you probably don't get anything usable.

Here's an example closer to my own heart: Lazard's famous equivalence of categories between certain nice $\mathbb{Z}_p$-Lie algebras and certain nice pro-$p$ groups. The devil's in the detail: I've hidden, behind both instances of the word "certain", a bunch of technical conditions that are more or less designed to make this true. But it's still remarkable that it's true at all.

Moreover, the real-life usefulness of this fact is guaranteed by the fact that all closed subgroups of $GL_n(\mathbb{Z}_p)$ do actually contain one of these certain nice pro-$p$ groups as a closed subgroup of finite index. In other words, no matter how horrible your closed subgroup of $GL_n$ is, viewed from the right angle it's more or less just (a finite amount of complexity away from) a Lie algebra. Whereas you previously only had the tools of group theory to apply to this object, you now also have the tools of Lie theory. That's definitely a win: even just knowing that finite-rank $\mathbb{Z}_p$-modules have a basis tells you a lot about the group and its completed group algebra, e.g. a PBW-type theorem.