# Why is there a duality between spaces and commutative algebras?

1) The category of affine varieties over $\mathbb{C}$ is equivalent to the opposite category of finitely generated reduced algebras over $\mathbb{C}$. The equivalence associates to an affine variety its algebra of regular functions to $\mathbb{A}^1$, and to each finitely generated reduced algebra its (ringed) space of algebra homomorphisms to $\mathbb{C}$. This is almost a tautology, but in some sense a deep one.

2) The category of compact Hausdorff spaces is equivalent to the opposite category of $\mathrm{C}^*$-algebras. The equivalence associates to a compact Hausdorff space its algebra of continuous functions to $\mathbb{C}$ (equipped with the sup norm), and to each $\mathrm{C}^*$-algebra its space of $\mathrm{C}^*$-algebra homomorphisms to $\mathbb{C}$. This is due to Gelfand.

3) The category of totally-disconnected compact Hausdorff spaces is equivalent to the opposite category of Boolean algebras. The equivalence associates to a totally-disconnected compact Hausdorf space its algebra of continuous functions to $\{0,1\}$, and to each Boolean algebra its space of Boolean algebra homomorphisms to $\{0,1\}$. This is due to Stone.

4) The $\infty$-category of simply-connected rational spaces of finite type is equivalent to the opposite $\infty$-category of simply-coconnected coconncetive $\mathbb{E}_\infty$-algebras of finite type over $\mathbb{Q}$. The equivalence associates to a rational space $X$ its cochain complex $C^*(X,\mathbb{Q})$, and to each $\mathbb{E}_\infty$-algebra over $\mathbb{Q}$ its space of $\mathbb{Q}$-linear $\mathbb{E}_\infty$-algebra maps to $\mathbb{Q}$. This is an $\infty$-categorical reformulation due to Lurie of a classical theorem of Sullivan.

5) The $\infty$-category of pro-$p$-finite spaces is equivalent to the opposite category of solvable $\mathbb{E}_\infty$-algebras over $\overline{\mathbb{F}}_p$. The equivalence associates to a pro-$p$-finite space $\{X_i\}_{i \in I}$ the corresponding colimit of cochain complexes $colim_i C^*(X_i,\overline{\mathbb{F}}_p)$. If $X$ is a $p$-finite space then $X$ can be reconstructed as the space of $\overline{\mathbb{F}}_p$-linear $\mathbb{E}_\infty$-algebra maps from $C^*(X,\overline{\mathbb{F}}_p)$ to $\overline{\mathbb{F}}_p$. Here $\overline{\mathbb{F}}_p$ is the algebraic closure of $\mathbb{F}_p$. The theorem in this form is due to Lurie, but a the original version of the theorem (working with $p$-complete spaces) is due to Mandell.

Indeed, if $R$ is a commutative-ring type object in some category of geometric objects, and $X$ is a geometric object carrying a sufficient supply of maps to $R$, then it is worthwhile to look at the ring of functions from $X$ to $R$, as this object will carry a lot of information on $X$. In the other direction, it is often useful to study commutative-ring objects by trying to realize them as rings of functions on some geometric object, their "spectrum". However, even with this in mind, it is still quite striking that this approach yields so many anti-equivalences between full subcategories of geometric objects and full subcategories of commutative-algebraic objects. It is like the geometer/topologist/homotopy-theorist is walking on one continent, and the commutative-algebraist is walking on another, and yet they keep stumbling upon the exact same species of categories (only in reverse).

What is going on here? Is there any conceptual explanation to this phenomenon? Is there any general argument which suggests, even heuristically, that the above equivalences are to be expected?

• My opinion is that, for each of these equivalences, half of the construction follow from "abstract categorical non-sense", and the rest is obtain by carefully choosing the definition on each side. To be a bit more precise, by this general idea of looking at morphisms into a structured object you obtain a functor which as soon as your category of geometric object has enough projective limits will have an adjoint functor (the spectrum). And as soon as you have an adjunction it can be turn into an equivalence by carefully choosing the categories involved on each side. – Simon Henry Sep 10 '15 at 11:48
• BTW, the fact that a topos is monadic over its opposite category can also be seen as an example of this : The geometric objects are the objects of the topos, and the "algebraic structure" are the $P^2$ algebra in the topos. – Simon Henry Sep 10 '15 at 11:51
• A sort of naive philosophical statement which captures at least some items on your list is "A ring is just a structured collection of functions on a space, and a space is just a structured collection of reasons why a ring has non-invertible elements." – Paul Siegel Sep 10 '15 at 12:11
• Something that surprised me in the case of compact hausdorff topological spaces is the fact the field $\mathbb{C}$ show up in topology and it is the only field realizing the anti equivalence. – Ilias A. Sep 10 '15 at 13:18
• I thought adjunctions do always restrict to equivalences between full subcategories: just take the full subcategories of objects for which the unit (or counit on the other side) is an isomorphism. – Omar Antolín-Camarena Sep 10 '15 at 14:41

## 4 Answers

I don't claim to have a complete answer but here are some miscellaneous comments.

1. Note that topological spaces are already very nearly defined to be dual to certain commutative algebra-like structures, namely their frames of open subsets. The simplest interesting case of this duality is a duality between finite sets and finite Boolean rings / algebras. One way to think about this conceptually is that the open subsets of a topological space axiomatize verifiable or semidecidable properties of a point in that space: that is, properties such that if they hold you can check that that's true, but such that if they don't hold you can't necessarily check that that's true. See this math.SE question for more discussion on this point. For example, the open sets in the product topology on $\{ 0, 1 \}^{\mathbb{N}}$ correspond precisely to those properties of an infinite sequence of zeroes and ones that you can verify by looking at finitely many terms of the sequence. (Once we agree that verifiable properties are a cool thing to look at, we should also agree that we care about logical operations on them, like AND and OR, and this is where the commutative algebra-like structure comes from.)

2. One way to define the category of commutative $k$-algebras is as follows. Let $\text{Poly}(k)$ be the category whose objects can be thought of as the affine spaces $\mathbb{A}^n$ over $k$ and whose morphisms $\mathbb{A}^n \to \mathbb{A}^m$ are $m$-tuples of polynomials in $n$ variables over $k$, with composition given by composition of polynomials. On the one hand, this is a Lawvere theory, and the category of commutative algebras over $k$ can be defined as the category of models of it, or more explicitly as the category of product-preserving functors $\text{Poly}(k) \to \text{Set}$. On the other hand, this is a full subcategory of the category of varieties over $k$, and one can try to probe varieties by mapping affine spaces into them; this turns a variety into a presheaf $\text{Poly}(k)^{op} \to \text{Set}$. There is a general relationship between functors and presheaves on the same category called Isbell duality, and the nLab suggests that this is the general setting for adjunctions between things that look like spaces and things that look like commutative algebras, although I haven't really internalized this.

3. The two points made above can be related as follows. One way to think about the category of sets is that it is the category of ind-objects of the category of finite sets; equivalently, it's the category of presheaves $\text{FinSet}^{op} \to \text{Set}$ sending finite colimits to finite limits. Now, $\text{FinSet}^{op}$ is the category of finite Boolean rings, and the category of Boolean rings can be thought of as the category of ind-objects in this; equivalently, it's the category of functors $\text{FinSet} \to \text{Set}$ preserving finite limits. These two descriptions should be related by Isbell duality.

4. Why is $\text{FinSet}$ so important, anyway? Well, one categorical property that categories of spaces often share that isn't true in general is that binary products tend to distribute over binary coproducts. This is true in particular in any cartesian closed category, such as $\text{Set}$, and any category with this property behaves in some sense like a categorified commutative ring (!). It turns out that $\text{FinSet}$ is the free distibutive category on a point.

• Thanks! I really like this answer. One can even expend a bit on point 4 as follows. As said, in geometric categories products tend to distribute over coproducts. On the other hand, in commutative-algebraic categories, coproducts tend to distribute over products. Since $\mathrm{FinSet}$ is free in that sense, we get that $\mathrm{FinSet}$ should map to any reasonable geometric category $C$, and $\mathrm{FinSet}^{op}$ should map to any reasonable commutative-algebraic category $D$. One may then try to extend these maps into a map $C\to D^{op}$, either by left or right Kan extension. -> – Yonatan Harpaz Sep 11 '15 at 9:57
• Right Kan extension is what's going on in the example of Boolean algebras. Left Kan extension (in the $\infty$-categorical sense), is a bit what's going on in the example of rational spaces. I'm still not completely sure how to accommodate for the case of $\mathrm{C}^*$-algebras and the case of pro-$p$-finite spaces. – Yonatan Harpaz Sep 11 '15 at 10:00
• The case of pro-$p$-finite spaces is a bit illusive. It looks like a right Kan extension case, but for this to hold the "base" category must include all the $p$-finite Eilenberg-MacLane spaces. In order for the opposite of this category to map into a ring of commutative-algebra objects one needs to realize various Steenrod powers. Curiously enough, Steenrod powers can actually be defined intrinsically for every $\mathbb{E}_\infty$-algebra over $\mathbb{F}_p$ using cohomology of the symmetric group. – Yonatan Harpaz Sep 11 '15 at 10:31
• It seems I should also mention the notion of an extensive category (ncatlab.org/nlab/show/extensive+category) somewhere in this discussion; this is a more conceptual condition satisfied by categories of spaces which in particular implies distributivity, but e.g. unlike the condition of being cartesian closed, it holds for $\text{Top}$. In any case, one might ask whether / why opposites of categories of commutative algebra-like objects tend to be extensive. – Qiaochu Yuan Sep 11 '15 at 19:41
• Here's why a category of commutative algebras is co-distributive: Suppose $(C,\otimes)$ is a symmetric monoidal category with biproducts $\oplus$ such that $\otimes$ distributes over $\oplus$ (guaranteed by closedness or by $\otimes$ respecting the unique enrichment of $C$ in commutative monoids). Then in the category $\mathrm{Comm}(C)$ of commutative monoid objects in $C$, $\otimes$ becomes the binary coproduct while $\oplus$ becomes the binary product, so binary coproduct distributes over binary product. Co-extensivity seems a bit more complicated – Tim Campion Sep 20 '15 at 23:13

An addendum to my comment to Qiaochu's answer, which amounts to expanding on Qiaochu's answer more generally: If we just add splitting of idempotents to our requirements on $(\mathcal{C}, \otimes)$, then $\mathrm{Comm}(\mathcal{C})$ is co-extensive! That is, the following equivalent conditions on a symmetric monoidal category $(\mathcal{C},\otimes)$ guarantee that $\mathrm{Comm}(\mathcal{C})$ is co-extensive:

• $\mathcal{C}$ has finite biproducts and split idempotents, and $\otimes$ preserves finite biproducts in each variable separately.

• $\mathcal{C}$ is enriched in commutative monoids and is Cauchy-complete and symmetric monoidal category in the enriched sense.

I find these conditions to be surprisingly mild and conceptual, compared to the definition of an extensive category, which I find a bit "fussy". I think these are pretty natural and minimal requirements of a category $\mathcal{C}$ to think of it as "a category of commutative modules", and being of the form $\mathrm{Comm}(\mathcal{C})$ for such a $\mathcal{C}$ is a pretty natural requirement for a category to be considered "a category of commutative algebras" -- although admittedly this is probably not the most natural way to think about commutative $C^\ast$-algebras. So the upshot is that any category of commutative algebras is dual to an extensive category, which is a good start toward being "a category of spaces", whatever that means.

But I don't know about the reverse direction -- given a "category of spaces", how likely is it to be dual to a category of algebras? This seems tricky, because there are certainly categories of spaces, such as projective varieties, which don't seem to be dual to categories of algebras -- we need to formulate some kind of "affineness criterion" in order to have a shot.

Anyway, it is not hard to show that $\mathrm{Comm}(\mathcal{C})$ is co-extensive when $\mathcal{C}$ is as above (it hardly can be, given how "clean" the hypotheses are!):

• The 0-ary case of co-extensivity says that we have a strict terminal object (it admits no maps out except isomorphisms), which is true because the terminal object in $\mathrm{Comm}(\mathcal{C})$ is $0$, and if $0 \to A$ is a ring homomorphism, then the unit of $A$ is $0$ and the identity on $A$ factors through $0$.
• The binary case says that the pushouts of the legs of a product decomposition again form a product decomposition. The product in $\mathrm{Comm}(\mathcal{C})$ is the biproduct in $\mathcal{C}$, so a map $A \times B \to C$ is induced by maps $f: A \to C$ and $g: B \to C$ in $\mathcal{C}$. Then composing $f$ and $g$ with the units on $A$ and $B$ we get mutually annihilating idempotents on $C$ which span $C$, and splitting the idempotents, we get a biproduct decomposition of $C$ in $\mathcal{C}$, which is a product decomposition in $\mathrm{Comm}(\mathcal{C})$.
• Yeah, it seems like idempotents are pretty key here. To show that $\text{CRing}$ is coextensive the key is that if $R \times S$ is a product ring then there are two idempotents $(1, 0)$ and $(0, 1)$ you can play with to show that the category of morphisms $R \times S \to (-)$ is equivalent to the product of the categories of morphisms $R \to (-)$ and $S \to (-)$. (Hence affine schemes over $\text{Spec } R \sqcup \text{Spec } S$ correspond to pairs of an affine scheme over $\text{Spec } R$ and an affine scheme over $\text{Spec } S$.) – Qiaochu Yuan Sep 23 '15 at 23:01
• Projective varieties are 2-affine, though... – David Roberts Sep 24 '15 at 3:14
• What does 2-affine mean? – Tim Campion Sep 24 '15 at 18:15
• The argument for distributivity no longer works if you work with the category $\mathrm{Mon}(\mathcal{C})$ of all monoids in $\mathcal{C}$ (because coproduct is no longer given by $\otimes$), but the argument for extensivity does still work. And any extensive category where finite products exist is automatically distributive. – Tim Campion Aug 25 '16 at 13:05
• @Tim my best guess is that I was referring to work of Martin Brandenburg, eg arxiv.org/abs/1202.5147 – David Roberts Sep 3 '17 at 0:07

All these five equivalences $\mathcal{C}^{op} \simeq \mathcal{D}$ follow the following pattern: There is an object $P$ which, in some sense, belongs to both $\mathcal{C}$ and $\mathcal{D}$. More precisely, there are objects $P_{\mathcal{C}} \in \mathcal{C}$ and $P_{\mathcal{D}} \in \mathcal{D}$ which are "similar" in the sense that for example they share the same underlying set, maybe more. Moreover, the hom-functor $\hom(-,P_{\mathcal{C}})$ lifts to a functor $$\underline{\hom}(-,P_{\mathcal{C}}) : \mathcal{C}^{op} \to \mathcal{D},$$ and likewise $\hom(-,P_{\mathcal{D}})$ to a functor $$\underline{\hom}(-,P_{\mathcal{D}}) : \mathcal{D} \to \mathcal{C}^{op}.$$ Then in most examples it is rather easy to construct an adjunction between these functors, since unit and counit may be defined as "evaluation" $$X \to \underline{\hom}(\underline{\hom}(X,P),P),\,x \mapsto (f \mapsto f(x)).$$ For more details, see the nlab article on dualizing objects. This is the purely formal part. The nontrivial part is to determine the fixed objects of this adjunction. If we are lucky, every object is a fixed object, so that $\mathcal{C}^{op} \simeq \mathcal{D}$. But often, we have to shrink $\mathcal{C}$ and $\mathcal{D}$ further.

In Gelfand duality (2), $P=\mathbb{C}$ "is" both an object of $\mathsf{Top}$ and of $*\mathsf{Alg}$ (unital $^*$-algebras), and the Gelfand spectrum $$\underline{\hom}_{*\mathsf{Alg}}(-,\mathbb{C}) : *\mathsf{Alg} \to \mathsf{Top}^{op}$$ is left adjoint to the function algebra $$\underline{\hom}_{\mathsf{Top}}(-,\mathbb{C}) : \mathsf{Top}^{op} \to *\mathsf{Alg}.$$ The fixed objects are (at least) the unital commutative $C^*$-algebras resp. the compact Hausdorff spaces, so that these categories are anti-equivalent. Similarly, in algebraic geometry (1), $\mathbb{C}$ "is" both a variety and an algebra, and in Stone duality (3), $\{0,1\}$ "is" both a space and a Boolean algebra. I am not familiar with the other examples (4), (5), but here the dualizing objects seem to $\mathbb{Q}$ and $\overline{\mathbb{F}_p}$.

• Sure, but why should these objects exist, and why should they relate things that look like spaces and things that look like commutative algebras? – Qiaochu Yuan Sep 12 '16 at 17:39
• Do you have an answer for this? To be honest, I didn't understand the other answers (and their relation to the question). – HeinrichD Sep 12 '16 at 18:18
• Tim Campion's suggestion is that under mild hypotheses, categories of commutative algebras are coextensive, so their opposites are extensive. Extensivity is a categorical property that basically ensures that coproducts behave the way coproducts of spaces ought to (that they are really "disjoint" in a particular sense), so it's a reasonable candidate for a necessary condition a category should satisfy to be considered a category of "spaces." – Qiaochu Yuan Sep 12 '16 at 18:20

I have not yet considered the new answer in full detail, so apologies for not addressing it. I'm coming back to this question after a while, so I thought I'd share some observations which came up during that time, and which might interest someone (at least, those someones who thought the original question was of any interest).

1) Example (3) above is actually a particular case of Example (5) with $p=2$, in the sense that it is obtained by restricting from 2-pro-finite spaces to pro-finite sets. Indeed, recall the notion of "solvable" which appears in Lurie's $p$-adic homotopy theory paper. In the discrete case, the data of a solvable $\overline{\mathbb{F}}_2$-algebra is essentially equivalent to the data of a Boolean algebra. In fact, example (3) admits a version for every prime $p$, replacing Boolean algebras with algebras which satisfy the relation $a^p=a$ for every $a$. These categories of "$p$-Boolian algebras" are in fact all equivalent to the opposide category of pro-finite sets (or totally disconnected compact Hausdorff spaces).

2) In every context where there is an interesting notion of geometric objects and commutative algebra objects which are connected by some duality, there is usually also a related category of "abelian group objects" underlying the situation. In particular, the commutative algebras are commutative algebra objects in this category, and there is typically an adjunction between geomtric objects and abelian group objects which associates to an abelian group its "underlying space" and to a geometric obejct the free abelian group generated from it. This later abelian group should be dual in some since to the underlying abelian group of the algebra of functions on the space. The adjunction between geometric objects and abelian group objects then induces a comonad on the category of abelian group objects. In that case one can ask two types of questions: (i) to what extent is this adjunction comonadic? in other words, how close is it to exhibiting spaces as coalgebras over this comonad? and (ii) how close is this comonad to being the cocommutative coalgebras comonad? When both answers are sufficiently positive we get that spaces look a lot like cocommutative coalgebras in abelian groups. Since abelian groups often have something that is close to self duality, this makes spaces close to the opposite of commutative algebras.

3) If one is working in the $\infty$-categorical setting, and $\mathcal{C}$ is some $\infty$-category of geometric objects, then one has a natural choice for the $\infty$-category $\mathcal{A}$ of abelian group objects, namely, the stabilization $Sp(\mathcal{C})$ of $\mathcal{C}$. Alternatively, one can tensor the stabilization with a field, such as $\mathbb{Q}$ or $\overline{\mathbb{F}}_p$, the choices underlying examples (4) and (5) above. If $\mathcal{C}$ is furthermore presentable then we will have an adjunction $\mathcal{C} \leftrightarrows \mathcal{A}$ where we can consider the left functor as an analogue of the free abelian group functor. If $\mathcal{C}$ carries a symmetric monoidal structure which preserves colimits in each variable separately then $\mathcal{A}$ will inherit the same type of structure and the left functor $\mathcal{C} \to \mathcal{A}$ will be monoidal.

4) Let $K: \mathcal{A} \to \mathcal{A}$ be the comonad corresponding to the adjunction $\mathcal{C} \leftrightarrows \mathcal{A}$. Since $\mathcal{A}$ is monoidal it makes since to ask if $K$ is the coalgebra comonad of some operad enriched in $\mathcal{A}$. This question can be tackled using Goodwillie calculus and obstruction theory, see work of Gijs Heuts (e.g., https://arxiv.org/abs/1510.03304). In particular, these obstructions vanish when working over a field of characteristic $0$, so if we take $\mathcal{A}$ to be the stabilization tensored with $\mathbb{Q}$, then the comonad $K$ will always come from some operad in $\mathcal{A}$. The spaces of $n$-ary operations of this operad (considered as $\Sigma_n$-objects in $\mathcal{A}$) are given by the Goodwillie derivatives of $K$ of $\mathcal{A}$. For this operad to be the commutative operad we need that the derivatives are all the unit object. This is likely to happen for completely abstract reasons when the monoidal structure on $\mathcal{C}$ is the Cartesian one, i.e., when $\mathcal{C}$ is Cartesian closed (a typical property of categories of geometric objects). Here is a cool heuristic explanation of this phenomenon that I learned from Tomer Schlank: the right adjoint $\mathcal{A} \to \mathcal{C}$ preserves Cartesian products and under our assumptions the left adjoint sends Cartesian products to tensor products in $\mathcal{A}$. Since $\mathcal{A}$ is stable products and coproducts in $\mathcal{A}$ coincide, and are often known simply as direct sums. One may then conclude that the comonad $K: \mathcal{A} \to \mathcal{A}$ sends direct sums to tensor products. In other words, in the analogy between functors and functions underlying the Goodwille calculus, $K$ is analogous to a function $\mathbb{R} \to \mathbb{R}$ which sends addition to multiplication. In other words, $K$ is analogous to an exponential function. As such, its derivatives at $0$ are the sequence of powers of some number $a$. There is a natural family of operads which have the same property. If $A \in \mathcal{A}$ is a cocommutative bialgebra object then the theory of commutaive $A$-algebras is controlled by an operad whose $\Sigma_n$-object of $n$-ary operations is $A^{\otimes n}$. It seems likely (though I don't have a proof of this), then for every Cartesian closed $\infty$-category $\mathcal{C}$ there is a cocommutative bialgebra object in $\mathcal{A} = Sp(\mathcal{C}) \otimes \mathbb{Q}$ such that the comonad associated to the adjunction $\mathcal{C} \leftrightarrows \mathcal{A}$ is the comonad of cocommutative $A$-coalgebras. For example, when $\mathcal{C}$ is the $\infty$-category of spaces this $A$ is the unit object of $\mathcal{A}$ and one gets the commutative operad.

Edit: A closer look at 1 shows that this $A$ is actually always the unit, see Lemma B.3. On the other hand, the situation is more complicated than I thought, so remark (4) should really be taken loosely.