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Stream: learning: questions

Topic: Isbell duality of schemes and rings

John Onstead (Jan 10 2025 at 12:22):

The classic example of Isbell duality is that between commutative rings and affine schemes. Yet, I think I'm still missing a few details on the specifics of how to get from the general statement of Isbell duality to the equivalence we expect.

The general statement of Isbell duality is an adjunction $[C^{op},\mathrm{Set}] \to [C,\mathrm{Set}]^{op}$. Now, let's replace $C$ with the Lawvere theory of commutative rings, which we will call $T$. With this, we now have the adjunction $[T^{op},\mathrm{Set}] \to [T,\mathrm{Set}]^{op}$. We know commutative rings are the product preserving functors from $T$, so we have an embedding $\mathrm{Mod}(T) \cong \mathrm{CRing} \to [T,\mathrm{Set}]$. We can dualize this to get an embedding $\mathrm{CRing}^{op} \to [T,\mathrm{Set}]^{op}$. I'm cheating a little bit here, but I will now invoke nlab Proposition 5.3 to say we now have an adjunction $[T^{op},\mathrm{Set}] \to \mathrm{CRing}^{op}$.

So far so good. But we still don't have our desired end-result: the equivalence $\mathrm{AffScheme} \cong \mathrm{CRing}^{op}$. What is the next step and how do I do it? Is this a case of finding the "fixed points" of the adjunction $[T^{op},\mathrm{Set}] \to \mathrm{CRing}^{op}$ in which case the restriction of this adjunction to an equivalence will give us what we want? And if so, then how do I "know" that affine schemes are the objects of the resulting category? Thanks for your help!

John Baez (Jan 10 2025 at 17:15):

Btw, I read an nLab article that seemed to hint that "the classic example of Isbell duality is that between commutative rings and affine schemes", but I never understood what that meant and I didn't believe it.

As far as I can tell, the duality between commutative rings and affine schemes is just that we define the category of affine schemes to be $\mathsf{CommRing}^{\text{op}}$. It's not a theorem. Isbell duality is a theorem, namely that for any category $\mathsf{C}$ we have an adjunction between $[\mathsf{C}^{\text{op}}, \mathsf{Set}]$ and $[\mathsf{C}, \mathsf{Set}]^{\text{op}}$.

If I wanted to relate the two I'd take $\mathsf{C} = \mathsf{CommRing}$. Then I'd get an adjunction between $[\mathsf{CommRing}^{\text{op}}, \mathsf{Set}]$ and $[\mathsf{CommRing}, \mathsf{Set}]^{\text{op}}$.

So, I get a contravariant adjunction between presheaves on the category of commutative rings and set-valued functors on the category of affine schemes. :thinking:

Or equivalently, but maybe better: a contravariant adjunction between presheaves on the category of affine schemes and set-valued functors on the category of commutative rings.

Is this saying something interesting about algebraic geometry? At least presheaves on the category of affine schemes include among them the schemes. And any scheme $X$ has a 'functor of points', which is a set-valued functor on the category of commutative rings, assigning to each commutative ring $k$ the 'set of $k$-points of $X$'.

So maybe I'm finally getting the point (pardon the pun). For any scheme we can take its functor of points, but there's also a way to turn any set-valued functor on $\mathsf{CommRing}$ and get from it, if not a scheme, at least something close: a presheaf on the category of commutative rings. (Then we can sheafify that, but I'm just talking about what we get out of Isbell duality.)

Mike Shulman (Jan 10 2025 at 17:38):

I think maybe you inserted an extra op. It seems to me that Isbell duality is a contravariant adjunction between set-valued functors on the category of affine schemes and set-valued functors on the category of commutative rings, or equivalently between presheaves on both of those categories.

Mike Shulman (Jan 10 2025 at 17:39):

The two categories you mentioned are just equivalent.

Mike Shulman (Jan 10 2025 at 17:45):

Someone else was also confused about the nLab's claim that algebra-geometry dualities are instances of Isbell duality. I suppose probably https://ncatlab.org/nlab/show/Isbell+duality#FunctionAlgebrasOnPresheaves is the attempt to make it precise, but I don't quite follow it yet.

John Baez (Jan 10 2025 at 18:02):

Blecch, an extra op - let me try again:

Isbell gives us an adjunction between $[\mathsf{CommRing}^{\text{op}}, \mathsf{Set}]$ and $[\mathsf{CommRing}, \mathsf{Set}]^{\text{op}}$.

In other words, an adjunction between $\mathsf{Psh}(\mathsf{CommRing})$ and $\mathsf{Psh}(\mathsf{AffSch})^{\text{op}}$.

Or in other words, a contravariant adjunction between $\mathsf{Psh}(\mathsf{CommRing})$ and $\mathsf{Psh}(\mathsf{AffSch})$.

Is this some fact about algebraic geometry that I wanted to know?

John Onstead (Jan 11 2025 at 01:01):

I admit I've never been a fan of saying that affine schemes are "by definition" the objects of the opposite category to commutative rings. It doesn't explain the why- that is, why the opposite categories of so many categories of algebraic structures seem to be categories of space-like structures. That's precisely why I wanted to study this from the perspective of Isbell duality.

John Baez said:

In other words, an adjunction between $\mathsf{Psh}(\mathsf{CommRing})$ and $\mathsf{Psh}(\mathsf{AffSch})^{\text{op}}$.

This does make sense, at least from a straightforward application of Isbell duality (at least, if the size issues associated with trying to define presheaves on a large category somehow work out). But again it seems to rely on affine schemes being the opposite of commutative rings "by definition". That is, you have to "already know" that affine schemes are equivalent to the opposite of commutative rings for this to work. I could put any category in the place of commutative rings and get a similar result, so it still doesn't answer the question of why, specifically, I should feel motivated to put a category of algebraic objects into this duality.

John Onstead (Jan 11 2025 at 01:10):

Mike Shulman said:

I suppose probably https://ncatlab.org/nlab/show/Isbell+duality#FunctionAlgebrasOnPresheaves is the attempt to make it precise, but I don't quite follow it yet.

This was the article I was using to base my question off of! If I'm understanding it correctly, I think it makes almost perfect sense. If it works, then I think it perfectly explains why so many opposites of categories of algebraic objects specifically behave like spaces. According to the article, there's an adjunction between the opposite of the category of models of any Lawvere theory (and thus many good notions of algebraic object) and presheaves on the Lawvere theory. By the general logic of "space and quantity" proposed and employed in Isbell duality, we can think of objects of a presheaf or sheaf category as "generalized spaces" and objects of a copresheaf category as "generalized quantities". Thus, models of a Lawvere theory can be seen as "generalized quantities", which makes sense since they are algebraic structures. On the other hand, presheaves of the Lawvere theory are "generalized spaces" modeled on the objects of the Lawvere theory. The Isbell duality then asserts the duality between these two.

If my assumption is correct that the adjunction between presheaves and the opposite of models of a Lawvere theory restricts to an equivalence involving the whole of the latter, then it means the latter embeds into the former. Since we can view the former as a category of generalized spaces, this implies the latter can also be viewed as generalized spaces. Since this logic works for any Lawvere theory, and specifically for Lawvere theories, it's basically stating the opposite of any category of algebraic objects embeds into a category of spaces and are thus space-like. Which then is the explanation for why opposite categories of algebraic objects are space-like I was looking for!

John Baez (Jan 11 2025 at 05:11):

I don't see how this explains that the category of affine schemes is "space-like", because I don't see how the opposite of the category of algebras of any Lawvere theory is space-like. In what sense is the opposite of the category of groups space-like, for example?

I only think of a category as resembling a category of spaces if products distribute over coproducts, and more so if it's extensive. I don't claim to have a precise definition of when a category is space-like, but I take these as pieces of evidence that the category of affine schemes, defined as the opposite of $\mathsf{CommRing}$, is space-like.

Mike Shulman (Jan 11 2025 at 05:28):

Maybe I'm old-fashioned, but I find it more explanatory to first define the category of locally ringed spaces, then construct the spectrum of a commutative ring as a locally ringed space, and observe that the spectrum functor is a contravariant embedding and define the category of affine schemes to be its image. Defining "affine scheme" to mean "object of $\mathsf{CommRing}^{\rm op}$" is a cute trick, but I get more intuition about what exactly an affine scheme is, and why they are the opposite of $\mathsf{CommRing}$, by thinking about locally ringed spaces.

Mike Shulman (Jan 11 2025 at 05:29):

Kind of like how defining "$\infty$-groupoid" to mean "Kan complex" is a cute trick, and useful, but doesn't really explain the homotopy hypothesis.

Mike Shulman (Jan 11 2025 at 05:37):

That said, I would guess that one way to think of the opposite of the category of algebras of a Lawvere theory as space-like is to think of a $T$-algebra $R$ as some algebra of "continuous functions" $X\to A$ for some "continuous $T$-algebra" $A$ and some "space" $X$, with pointwise $T$-structure.

For some Lawvere theories we can make literal sense of this for some notion of space, e.g. commutative $C^*$-algebras are continuous $\mathbb{C}$-valued functions on locally compact Hausdorff spaces. For others you have to futz with it a bit, or it's almost true but not quite, e.g. arbitrary commutative rings are kind of like function algebras on certain locally ringed spaces (affine schemes), but the codomain of these "functions" varies by point; and some frames (the spatial ones) are $2$-valued functions on (sober) topological spaces.

So we look at that and say, hey, maybe for any $T$ we can pretend there's a notion of "space" for which the $T$-algebras are precisely the "functions" on those spaces. Noncommutative rings? Sure, we can imagine "noncommutative spaces". I dunno exactly about bare groups, but probably one could develop some intuition in that direction if one tried hard enough.

Mike Shulman (Jan 11 2025 at 05:39):

(Is there a general theory of for which Lawvere theories the opposite of the category of algebras is distributive or extensive?)

John Baez (Jan 11 2025 at 07:19):

Mike Shulman said:

Maybe I'm old-fashioned, but I find it more explanatory to first define the category of locally ringed spaces, then construct the spectrum of a commutative ring as a locally ringed space, and observe that the spectrum functor is a contravariant embedding and define the category of affine schemes to be its image. Defining "affine scheme" to mean "object of $\mathsf{CommRing}^{\rm op}$" is a cute trick, but I get more intuition about what exactly an affine scheme is, and why they are the opposite of $\mathsf{CommRing}$, by thinking about locally ringed spaces.

Personally I'm happiest to first define an affine scheme to mean an object of $\mathsf{CommRing}^{\text{op}}$ and then discuss the various ways of defining points of affine schemes (as homomorphisms to other rings, or maximal/prime ideals), discuss how to put a topology on the set of prime ideals to get a locally ringed space, develop various geometrical structures such as Kahler differentials, and so on.

John Baez (Jan 11 2025 at 07:21):

Mike Shulman said:

(Is there a general theory of for which Lawvere theories the opposite of the category of algebras is distributive or extensive?)

Hmm, there's someone who asked a question like this on Math Overflow:

• When is the opposite of the category of algebras of a Lawvere theory extensive?

John Onstead (Jan 11 2025 at 14:44):

Mike Shulman said:

Maybe I'm old-fashioned, but I find it more explanatory to first define the category of locally ringed spaces, then construct the spectrum of a commutative ring as a locally ringed space, and observe that the spectrum functor is a contravariant embedding and define the category of affine schemes to be its image.

That's an interesting approach! But this is quite the interesting coincidence. A ringed space is a space with a sheaf into commutative rings, and in turn, the spectra of a commutative ring is a ringed space (a locally ringed one). So a notion of "ring" appears on both sides of the metaphorical equation. What if I, say, instead defined a "grouped space" to be a space with a sheaf into groups- would there be some sort of notion of "spectrum of a group" that would allow me to contravariantly embed groups into the "grouped spaces"? Or more generally, a "$T$-ed space" for $T$ some Lawvere theory, a space with a sheaf into the category of models of $T$? Are these even sensible questions to ask?

For instance, given some $T$, maybe there's some consistent way to define subsets of the powerset of the underlying sets of models of $T$ such that one could put a topology on those sets of subsets analogous to the Zariski topology. So that would be a generalization of the concept of "prime ideal" from commutative ring theory. Then the corresponding $T$-ed space for some model $A$ of $T$ would be $A$'s topological space equipped with a sheaf into the category of models of $T$ such that, via the stalk construction, every point in the space (each representing a subset of the underlying set of $A$) would be sent to $A$ localized at that subset of $A$ (I'm pretty sure that can be done for a generic algebraic category). Anyways I'm a complete newcomer to algebraic geometry so I probably messed up somewhere, but hopefully you get the gist of what I'm trying to say!

John Onstead (Jan 11 2025 at 14:49):

John Baez said:

Hmm, there's someone who asked a question like this on Math Overflow:

Ah, how funny! I had already finished going through some of the answer when I noticed who asked it.

John Baez said:

I only think of a category as resembling a category of spaces if products distribute over coproducts, and more so if it's extensive. I don't claim to have a precise definition of when a category is space-like,

This reminds me of the discussion a few weeks ago in the thread on finding a notion of "category of space". I wonder why you never brought this up there, that would have been a valid answer to my question! It also feels a lot more concrete than "a category with a site structure" or "a category with a class of weak equivalences" or even "any infinity category". Would you happen to know if all topological concrete categories over Set are extensive? That's the other more "concrete" notion of "category of space" I had in mind, I'd like to see if there's a direct inclusion there!

Also, I see some of your justification for why you think extensive categories are a good notion of "categories of spaces" from the Math Overflow question (distribution of products over coproducts, objects with a morphism to the initial object are initial themselves, and objects over X + Y are the same as an object over X with one over Y). But what do you think makes these good qualities for "spaces"? Is there some inherent notion of "locality" in these properties? Or is it just as simple as "a lot of categories that happen to have spaces also happen to exhibit these properties"?

John Baez (Jan 11 2025 at 18:26):

I wonder why you never brought this up there, that would have been a valid answer to my question!

I guess I don't believe your question, something like "what is the most general concept of space?", has a good answer - at least not at the current stage of mathematics. So it didn't engage my own personal thoughts about "space". If you'd asked "what are some characteristic properties of categories of spaces?" my reaction would have been very different.

Also, I see some of your justification for why you think extensive categories are a good notion of "categories of spaces" from the Math Overflow question (distribution of products over coproducts, objects with a morphism to the initial object are initial themselves, and objects over X + Y are the same as an object over X with one over Y). But what do you think makes these good qualities for "spaces"?

To me a "space" is something I can reason about visually, in analogy with the regions of space in the physical world. If I close my eyes and visualize "spaces" as blobs, I can see a picture proof that products distribute over coproducts, and also, with a bit more work, the extensivity law.

I haven't tried to extract a complete list of laws that follow from this (highly nonrigorous and somewhat questionable) strategy of trying to "see" what spaces are like.

Mike Shulman (Jan 11 2025 at 22:08):

John Onstead said:

Are these even sensible questions to ask?

I think they are, but I don't know the answers. You can try to work them out yourself!

Mike Shulman (Jan 11 2025 at 22:09):

One thing that's a bit special about rings, topologically speaking, is that they have two operations "multiplication" and "addition" with a distributive law, analogous to "intersection" and "union" in an open-set lattice. That may make it trickier to do the same sort of spectrum construction with groups or another Lawvere theory.

John Onstead (Jan 12 2025 at 04:08):

Mike Shulman said:

One thing that's a bit special about rings, topologically speaking, is that they have two operations "multiplication" and "addition" with a distributive law, analogous to "intersection" and "union" in an open-set lattice. That may make it trickier to do the same sort of spectrum construction with groups or another Lawvere theory.

Ah, I see!

John Baez said:

I guess I don't believe your question, something like "what is the most general concept of space?", has a good answer - at least not at the current stage of mathematics. So it didn't engage my own personal thoughts about "space". If you'd asked "what are some characteristic properties of categories of spaces?" my reaction would have been very different.

Many times as an icebreaker I get asked what I would do if I won the lotto. Currently, among other things, I would decide to plan and host an international conference on categorical topology and homotopy. I would invite the best and the brightest mathematicians across all these different fields of math from all over the world. And what would be the topic of the conference? The whole purpose of the entire event would be, by the very end, to finally, once and for all, come up with a unifying definition for what a "space" is mathematically, that at least most of the attendees can agree on, and that is a purely formal abstract logical definition rather than any vague imprecise definition. That is, some specific "a space is a X equipped with the structure of Y such that..." kind of definition. But I don't have any money so unfortunately this is one problem that might remain unsolved! Maybe they should make it a Millennium prize problem... (that's a joke)

Mike Shulman (Jan 12 2025 at 04:34):

I don't know how many mathematicians you'd get to come to something like that. I think most of us have come to the understanding that there are many different kinds of space and don't expect there to be any single unifying definition.

John Baez (Jan 12 2025 at 06:56):

If you look at history, you see that the concept of "space" keeps expanding beyond our attempts to confine it. If we had a conference that arrived at a unified definition of "space", the very next day mathematicians would start trying to study more general "spaces" that didn't quite fit that definition.

Many times as an icebreaker I get asked what I would do if I won the lotto.

Really? Nobody has ever asked me that! But now that you mention it, it's fun to think about.

Mike Shulman (Jan 12 2025 at 10:03):

(Was [[New Spaces for Mathematics and Physics]] mentioned in the "what is a space" discussion already?)

Damiano Mazza (Jan 12 2025 at 11:06):

John Onstead said:

If my assumption is correct that the adjunction between presheaves and the opposite of models of a Lawvere theory restricts to an equivalence involving the whole of the latter, then it means the latter embeds into the former. Since we can view the former as a category of generalized spaces, this implies the latter can also be viewed as generalized spaces. [...] Which then is the explanation for why opposite categories of algebraic objects are space-like I was looking for!

Maybe I am misinterpreting your reasoning, but to me it sounds like you are taking an embedding $A^{\mathrm{op}}\hookrightarrow\mathbf{PSh}(C)$ as evidence that $A^{\mathrm{op}}$ is a "category of spaces", but then any locally small category would be a category of spaces by letting $C:=A^{\mathrm{op}}$. I'm sure this is not what you are looking for...

In the specific case of Proposition 5.3 of the nLab page on Isbell duality, we have, for any Lawvere theory $T$ and small category $C$ with fully faithful functors $T\hookrightarrow C\hookrightarrow T\text{-}\mathbf{Alg}^{\mathrm{op}}$ whose composition is the Yoneda embedding, a functor

$\mathrm{Spec}:T\text{-}\mathbf{Alg}^{\mathrm{op}}\to\mathrm{PSh}(C)$

which has a left adjoint. There are indeed cases in which $\mathrm{Spec}$ is fully faithful (for example, if I am not mistaken, when $C=T\text{-}\mathbf{Alg}_{\mathrm{fp}}^{\mathrm{op}}$, the opposite of the category of $T$-algebras of finite presentation). I don't know, maybe you find this embedding more satisfactory because it is less "tautological" than the Yoneda embedding? In particular, because the category of "spaces" $T\text{-}\mathbf{Alg}^{\mathrm{op}}$ is large but the "controlling category" $C$ is small?

To me, this is still a bit tautological, because:

1. if you look at the definition on the nLab, $\mathrm{Spec}$ is really still the Yoneda embedding;
2. usually the intuition that presheaves on some $C$ are "generalized spaces" works best when $C$ is already a category of spaces (even better, a site, which allows us to speak of sheaves). In this case, presheaves on $T\text{-}\mathbf{Alg}_{\mathrm{fp}}^{\mathrm{op}}$ would be generalized spaces because we are already considering a category as a "category of spaces" because it is the opposite of a category of algebras...

I don't know but my personal conclusion is that Isbell duality doesn't have much to say on why, in certain cases, the opposite of a category of algebras of a Lawvere theory is a category of space-like objects.

John Onstead (Jan 12 2025 at 14:54):

Mike Shulman said:

I don't know how many mathematicians you'd get to come to something like that. I think most of us have come to the understanding that there are many different kinds of space and don't expect there to be any single unifying definition.

John Baez said:

If you look at history, you see that the concept of "space" keeps expanding beyond our attempts to confine it. If we had a conference that arrived at a unified definition of "space", the very next day mathematicians would start trying to study more general "spaces" that didn't quite fit that definition.

Right, I understand. My post was more of a joke, but I intended it to reveal that, in a way, the nebulous notion of "space" in mathematics doesn't seem "satisfying" to me. That is, when I think of this lack of unification of a concept of "space", it just seems "off" to me. I can't explain why this is- maybe it's some instinctual gut-feeling reaction!

In any case, as I promised at the end of last year, I wanted to work on a solution to this "problem" (which again, might only be a problem for me, where other mathematicians don't have a problem with it). I had a lot of time over the break to think of something, and I actually came up with something that might work. But that'll be for a future discussion!

Mike Shulman (Jan 12 2025 at 17:50):

Probably as you get more used to the way different kinds of spaces are used in mathematics, that gut-feeling reaction will dissipate.

John Baez (Jan 12 2025 at 18:41):

I've said this before, but I'll say it again: it's important for mathematics to have some terms whose definitions aren't precisely pinned down, so we can think about things vaguely - a crucial but sometimes neglected prerequisite for having new ideas. These terms include "space", "algebra", "structure", etc.

If someone gave a precise definition of "space" and everyone foolishly adopted it as the official definition, we'd need to make up a new word for the vaguer, broader concept surrounding the precise definition.

John Onstead (Jan 12 2025 at 22:11):

Damiano Mazza said:

But then any locally small category would be a category of spaces by letting $C := A^{op}$. I'm sure this is not what you are looking for...

Thanks for your response! Yes, you bring up a good point here. I guess this adds evidence to the whole idea that "space" is a nebulous term since really anything can be a space from a certain point of view.

Damiano Mazza said:

In the specific case of Proposition 5.3 of the nLab page on Isbell duality, we have,

Thanks for the clarification on this section, since I was wondering what this intermediate category $C$ was doing. But now I see it gives a whole spectrum of "Spec" adjunctions (pun intended). This would still include the construction I was trying to do, since I could set $C$ to be $T$ itself.

Damiano Mazza said:

To me, this is still tautological, because:

1. if you look at the definition on the nLab, Spec is really still the Yoneda embedding:

I suppose, although this seems to follow from the notion of density and how $C$ would be a dense subcategory. I don't think the concept of density is tautological!

Damiano Mazza said:

1. So even T itself is already the opposite of a category of algebraic objects and so, again, tautologically a category of "spaces"...

I don't know but my personal conclusion is that Isbell duality doesn't have much to say on why, in certain cases, the opposite of the category of algebras of a Lawvere theory is a category of space-like objects.

This reasoning makes sense. I suppose this will be something I have to think about a bit more!

John Onstead (Jan 13 2025 at 03:37):

Giving it some more thought, I want to try a different approach to understanding the duality between geometry and algebra- which is to start with some concrete information and try to work backwards into something like Isbell duality, rather than trying to define it directly in terms of Isbell duality. To this extend, I returned to the MO answer above, as well as this nlab page, and I noticed a common theme. Whenever there was a duality between some algebraic and geometric thing, there was a "special object" that mediated it. In the case of commutative rings and affine schemes, this "special object" was $\mathbb{A}^1$, the "affine line". I now want to understand more about why exactly the affine line is so important.

On the MO page, the $\mathrm{Spec}$ functor is literally given as a sort of hom functor of the form $\mathrm{Hom}(-,\mathbb{A}^1): \mathrm{CRing}^{op} \to \mathrm{AffSch}$. That is, the "spectrum" of a ring is meant to be interpreted as a space structure on the set of homomorphisms from a ring into this special affine line ring. However, we know the underlying set of the spectrum of a ring is precisely its set of prime ideals! Thus, if we take the MO answer at full face value, this means composing the $\mathrm{Spec}$ functor into affine schemes with the underlying set functor should give the functor $\mathrm{Prime}: \mathrm{CRing}^{op} \to \mathrm{Set}$. And if this is the case, it means that the object $\mathbb{A}^1$ represents this functor- that is, every ring homomorphism from a commutative ring into the affine line corresponds to a prime ideal in that ring. But I have no way of knowing if any of this is true, so that's my question: is it true, and if is, is it exactly this representability that makes the affine line object so "special" to this duality?

John Onstead (Jan 13 2025 at 03:37):

Also, checking above, maybe this is what Mike Shulman was getting at with the following (which I did not understand at the time):
Mike Shulman said:

That said, I would guess that one way to think of the opposite of the category of algebras of a Lawvere theory as space-like is to think of a $T$-algebra $R$ as some algebra of "continuous functions" $X\to A$ for some "continuous $T$-algebra" $A$ and some "space" $X$, with pointwise $T$-structure.

David Egolf (Jan 13 2025 at 03:49):

If $k$ is a field and $R$ is a commutative ring, then a ring homomorphism $f:R \to k$ I suspect picks out a prime ideal of $R$. Specifically, I think the kernel of $f$ is a prime ideal. (Note that the kernel of $f$ can't be all of $R$, as $f$ needs to map $1$ to $1$).

Indeed if $ab \in \ker f$, then $f(ab)=0$. Since $f$ is a ring homomorphism, we have $f(ab)=f(a)f(b)=0$. Since $k$ is a field, for this to be true either $f(a)=0$ or $f(b)=0$ (or both). Thus, either $a \in \ker f$ or $b \in \ker f$ (or both).

David Egolf (Jan 13 2025 at 03:51):

These notes on algebraic geometry lectures by Peter Scholze discuss this further, on pages 10-11.

John Onstead (Jan 13 2025 at 04:46):

David Egolf said:

These notes on algebraic geometry lectures by Peter Scholze discuss this further, on pages 10-11.

Ah, thanks for the resource! I'll certainly be trying to read through it when I have the time. On page 10 it does seem to try and define prime ideals and spectra as maps out of a ring as I was doing. However, it defines them in terms of maps from a ring into any field, but there's a huge number of different fields you can choose. I want to confirm if there's a single object that can play that role.

Peva Blanchard (Jan 13 2025 at 06:13):

There is a notion of dualizing object involved in, e.g., Stone duality or Gelfand-Naimark duality.

In the case of Isbell duality, it looks like the dualizing object is $\text{Set}$. But it's not quite right: the dualizing object is supposed to live in two different categories, whereas in the case of Isbell duality, both sides live in $\text{Cat}$.

Mike Shulman (Jan 13 2025 at 06:33):

There is not a single field that works to map into for all rings. You can see this because any field has a characteristic, and there are no maps at all between two fields of different characteristic. Indeed the category of fields is a disjoint union of the categories of "fields of characteristic $p$" for all primes $p$ and 0 (and each of those subcategories has an initial object).

The "affine line" is a scheme, namely $\mathrm{Spec}(\mathbb{Z}[x])$. So while you map out of a scheme into the affine line, the corresponding thing to do with rings is map out of $\mathbb{Z}[x]$ into a ring, which is the identity functor since $\mathbb{Z}[x]$ is the ring freely generated by one element.

John Onstead (Jan 13 2025 at 07:45):

Peva Blanchard said:

There is a notion of dualizing object involved in, e.g., Stone duality or Gelfand-Naimark duality.

Ah thanks, that seems to be exactly the term I was looking for!

Mike Shulman said:

The "affine line" is a scheme, namely $\mathrm{Spec}(\mathbb{Z}[x])$. So while you map out of a scheme into the affine line, the corresponding thing to do with rings is map out of $\mathbb{Z}[x]$ into a ring, which is the identity functor since $\mathbb{Z}[x]$ is the ring freely generated by one element.

I think I see. Let me try to talk through this. By the adjunction between rings and sets, every morphism from the free ring on the singleton set to any other ring is in bijection with morphisms from the singleton to the underlying set of that ring. And the free ring on the singleton happens to be $\mathbb{Z}[x]$, so every morphism out of it in the category of rings represents an element of the target ring. When we go to the opposite category, now the elements of a ring are in bijection with maps from its spectrum into the affine line. So at no point are we discussing the elements of the spectrum of rings, we are always discussing the elements of the rings themselves. This also explains why the adjoint to the spectrum functor acts like a hom functor (to the affine line), since we know that morphisms into the affine line correspond with elements of some ring.

David Corfield (Jan 13 2025 at 09:29):

John Onstead said:

There is a notion of dualizing object involved in, e.g., Stone duality or Gelfand-Naimark duality.

Ah thanks, that seems to be exactly the term I was looking for!

A great part of Peter Johnstone's book on 'Stone Spaces' is about equipping $\mathbf{2}$ with an array of different structures to provide a slew of dualities. $Set$ is a natural next target up the ladder, and allows for, e.g., [[Gabriel–Ulmer duality]].

Lawvere and Rosebrugh in 'Sets for Mathematics' have a nice treatment of the general idea:

image.png

Peva Blanchard (Jan 13 2025 at 10:44):

When you say

$Set$ is a natural next target up the ladder

do you know if there is a mathematical formalization of this "ladder"?

For instance, is there a way to relate dualities built arount $\textbf{2}$ and those built around $Set$?

John Onstead (Jan 13 2025 at 11:07):

David Corfield said:

Lawvere and Rosebrugh in 'Sets for Mathematics' have a nice treatment of the general idea:

Quite interesting! I saw that quote on the nlab page for Isbell duality. The nlab seems to assert that Isbell duality is the driving force behind the duality between space and quantity. But it's really interesting to see that was never the intention- the duality between space and quantity was originally, by its original authors, meant to refer to these "concrete dualities" mediated by a dualizing object. Nlab has been wrong before as I've seen many times on this server, so it really wouldn't surprise me if nlab is once again wrong- this time asserting the connection between Isbell duality and space-quantity duality when it should instead be attributing the connection to be between dualizing objects and space-quantity duality.

Peva Blanchard said:

do you know if there is a mathematical formalization of this "ladder"?

I think it's a case of "categorification". The object 2 can be viewed as the 0-category of (-1) categories, and Set can be viewed as the 1-category of 0-categories. So the next level up would be Cat as the 2-category of 1-categories, and so on.

David Corfield (Jan 13 2025 at 11:16):

I meant the $n$-category ladder.

image.png

If $\mathbf{2}$ forms a poset of truth values, $(0, -1)$-categories, $Set$ forms the category of sets, $(0, 0)$-categories.

If you prefer HoTT, think of hlevels:
image.png.

So the last stop up this ladder is $\infty Grpd$. Here's a question I pose about using that to dualize.

John Onstead (Jan 13 2025 at 11:27):

With "dualizing objects instead of Isbell duality" in mind, how about this for a definition of "category of space" that's hopefully at least a bit less tautological than the one from Isbell duality. Here it is: A "category of spaces" $C$ is a category in a pair of categories and functors of form $F,G: C,D \to \mathrm{Set}$ with a dualizing object $d$ between them, in which case it's the domain of the left adjoint in the resulting adjunction. The left adjoint can then be interpreted to be taking a space to the algebra of $d$-valued morphisms on it. Is this a good definition for a "category of spaces"?

Also, here's a second question. When it is a sufficient condition for an object $d$ in a concrete category $C$ that represents the point (IE, morphisms from it are in bijection with the elements of the underlying sets of objects) to be a dualizing object for some such duality? It certainly seems that $\mathbb{Z}[x]$ is sufficient, since if I take the category of rings and find its opposite category, it satisfies my definition for being a "category of spaces" above with $\mathrm{Spec}(\mathbb{Z}[x])$ as the dualizing object. Will this be true for any category of algebraic structures (since all categories monadic over Set should have an object that "picks out points" in this way)?

John Onstead (Jan 13 2025 at 11:29):

And I guess a third question: do you think nlab made a mistake by making the association space/quantity <-> Isbell duality instead of space/quantity <-> concrete (dualizing object) duality?

David Corfield (Jan 13 2025 at 11:34):

John Onstead said:

do you think nlab made a mistake by making the association space/quantity <-> Isbell duality instead of space/quantity <-> concrete (dualizing object) duality?

Good question about the relationship between Isbell duality and dualizing objects, one that seems to be in the air on MO periodically, e.g., here and here, without, as I can see, a clear answer.

David Corfield (Jan 13 2025 at 12:12):

Hmm, the nLab is rather lacking in connections. Neither Isbell duality nor dualizing object mention each other. Then there's a page duality between algebra and geometry which at least links to 'Isbell duality' but not to 'dualizing object'.

David Corfield (Jan 13 2025 at 12:23):

To add to the complication, 'Isbell duality' is also used to denote a specific concrete duality:

Many years ago, one of the authors heard Bill Lawvere say that a potential duality arises when a single object lives in two different categories. When asked, Lawvere credited the statement to John Isbell. From the literature, it seems that what is called “Isbell duality” is the one between frames with enough points and sober spaces.

• Michael Barr, John Kennison, Robert Raphael, Isbell Duality Theory and Applications of Categories 20 15 (2008) 504-542 [tac:20-15]

Damiano Mazza (Jan 13 2025 at 13:47):

John Onstead said:

Damiano Mazza said:

To me, this is still tautological, because:

1. if you look at the definition on the nLab, Spec is really still the Yoneda embedding:

I suppose, although this seems to follow from the notion of density and how $C$ would be a dense subcategory. I don't think the concept of density is tautological!

I understand that you are now taking a different angle, so this is no longer relevant for the ongoing discussion, but just for the sake of clarifying: I don't think that this has anything to do with density (unless I am misunderstanding what you mean by "density").

The nLab defines $\mathrm{Spec}: T\text{-}\mathbf{Alg}^{\mathrm{op}}\to\mathbf{PSh}(C)$ as the functor sending a $T$-algebra $A$ to $T\text{-}\mathbf{Alg}(A,-)$. Let us spell this out, and see how this is a presheaf on $C$ (rather than a copresheaf on $T\text{-}\mathbf{Alg}$, as the notation seems to suggest):

• given an algebra $B$ in $C$, $\mathrm{Spec}(A)(B)$ is the set $T\text{-}\mathbf{Alg}(A,B)$ (I treat objects of $C$ as $T$-algebras because of the embedding $C\hookrightarrow T\text{-}\mathbf{Alg}^{\mathrm{op}}$);
• given a morphism $f:B'\to B$ in $C$, since $C\hookrightarrow T\text{-}\mathbf{Alg}^{\mathrm{op}}$, we may see $f$ as an arrow $B\to B'$ of $T\text{-}\mathbf{Alg}$, and so $\mathrm{Spec}(A)(f)$ is the function sending an arrow $h:A\to B$ of $\mathrm{Spec}(A)(B)$ to the arrow $f\circ h$ of $\mathrm{Spec}(A)(B')$.

So, for any $T$-algebra $A$, $\mathrm{Spec}(A)$ is a presheaf on on $C$, as desired. Notice that, contrarily to what the notation $T\text{-}\mathbf{Alg}(A,-)$ seems to suggest, such a presheaf is not necessarily representable.

However, if $A$ happens to be in $C$, which is the case when we are in the image of the embedding $C\hookrightarrow T\text{-}\mathbf{Alg}^{\mathrm{op}}$, then this presheaf is the one represented by $A$ itself! In fact, in that case, thanks to the embedding being contravariant and fully faithful, we have $T\text{-}\mathbf{Alg}(A,-)=C(-,A)$. So the composition of the inclusion $C\hookrightarrow T\text{-}\mathbf{Alg}^{\mathrm{op}}$ followed by $\mathrm{Spec}$ sends $A$ to $C(-,A)$, which is just the Yoneda embedding $C\hookrightarrow\mathbf{PSh}(C)$.

Damiano Mazza (Jan 13 2025 at 14:18):

The thing is, Isbell duality works because of extremely general manipulations which, in the end, boil down to Yoneda, co-Yoneda and the like, so it's not surprising that this is what you end up with.

A possible high-level justification of where the $\mathrm{Spec}$ functor from algebraic geometry comes from is the following (maybe you already know this, I'm saying it just in case). There is a category $\mathbf{LRS}$ of locally ringed spaces (topological spaces equipped with a sheaf of commutative rings such that the stalk at every point is a local ring). There is little doubt that this is a "category of spaces", albeit it is already somehow "loaded up" with algebra. Then we have the category $\mathbf{CRing}$ of commutative rings, undoubtedly a category of algebraic objects. There is an obvious functor

$\Gamma:\mathbf{LRS}\to\mathbf{CRing}^{\mathrm{op}}$

which sends a locally ringed space $(X,\mathcal O_X)$ to the commutative ring $\mathcal O_X(X)$, the "ring of global sections" (that is, $\mathcal O_X$ associates a commutative ring with every open set of $X$, and $X$ itself in particular is open, so we take that ring. Morphisms of ringed spaces happen to go "the other way around" with respect to the underlying rings, so the functor is contravariant). It turns out that $\Gamma$ has a left adjoint

$\mathrm{Spec}:\mathbf{CRing}^{\mathrm{op}}\to\mathbf{LRS}$

which is precisely the functor sending a commutative ring to its spectrum. This is definitely a space/algebra duality, although I do not see at all how one may see it as an instance of Isbell duality.

This tells us, in particular, that the affine scheme $\mathrm{Spec}(R)$ is the "free" locally ringed space associated with the commutative ring $R$ ("free" because of the left adjoint, although the terminology maybe is a bit abusive here, hence the scare quotes).

Damiano Mazza (Jan 13 2025 at 14:28):

There is an MO answer which discusses this and motivates the use of locally ringed spaces on the "space" side. In particular, it warns us that if you take spaces equipped with sheaves of $T$-algebras for a generic Lawvere theory $T$ (which is something you asked about earlier), the duality can be extremely degenerate. For example, in the case of commutative rings, it is crucial that one looks at locally ringed spaces, as opposed to just ringed spaces, because otherwise the left adjoint (which still exists) is trivial.

John Baez (Jan 13 2025 at 17:46):

David Egolf said:

If $k$ is a field and $R$ is a commutative ring, then a ring homomorphism $f:R \to k$ I suspect picks out a prime ideal of $R$. Specifically, I think the kernel of $f$ is a prime ideal. (Note that the kernel of $f$ can't be all of $R$, as $f$ needs to map $1$ to $1$).

Indeed if $ab \in \ker f$, then $f(ab)=0$. Since $f$ is a ring homomorphism, we have $f(ab)=f(a)f(b)=0$. Since $k$ is a field, for this to be true either $f(a)=0$ or $f(b)=0$ (or both). Thus, either $a \in \ker f$ or $b \in \ker f$ (or both).

That's true. But you don't need $k$ to be a field to conclude that $\ker f$ is a prime ideal, and if $k$ is a field you get more: $\ker f$ is a 'maximal' ideal. This turns out to be important in algebraic geometry.

An integral domain is a commutative ring $k$ with no zero divisors: if $a, b \in k$ have $ab = 0$ then $a = 0$ or $b = 0$. Examining your argument, we see that if $f: R \to k$ is a homomorphism with $k$ an integral domain, then $\ker f$ is a prime ideal.

This a significant strengthening of what you said, because there are lots of integral domains that aren't fields: e.g. the integers, or the ring $k[x]$ of polynomials in one variable with coefficients in some integral domain $k$. By using the second one inductively, we see that rings of polynomials like $\mathbb{R}[x,y,z]$ or $\mathbb{Z}[x,y]$ are integral domains.

John Baez (Jan 13 2025 at 18:00):

This means that when Grothendieck (or whoever it was) defined the spectrum $\mathrm{Spec}(R)$ of a commutative ring to be the set of prime ideals, and treated elements of $\mathrm{Spec}(R)$ as 'points', they were significantly generalizing the earlier concept of point.

For example take $R = \mathbb{R}[x,y]$, the ring of real polynomials in two variables. These are polynomial functions on the plane $\mathbb{R}^2$ so you might guess that $\mathrm{Spec}(R)$ should be the plane. And indeed the set of maximal ideals of $R$ is just the plane! But the set of prime ideals is bigger. For example the kernel of the homomorphism

$f: \mathbb{R}[x,y] \to \mathbb{R}[x]$

which sends $y$ to zero is a prime ideal, since $\mathbb{R}[x]$ is an integral domain. But geometrically this kernel is the set of functions that vanishes on the line $y = 0$. So now $\mathrm{Spec}(R)$ is not merely including points of the plane: it's also including lines in the plane (and other curves).

John Baez (Jan 13 2025 at 18:03):

This was one of the more shocking things I ran into when I started learning algebraic geometry: the spectrum of a commutative ring is some sort of space, but the 'points' of this space are more general than I would have guessed (or wanted). It takes longer to explain why this is a good idea.

If we don't want these more general points, we can define a spectrum that consists only of maximal ideals.

David Egolf (Jan 13 2025 at 18:42):

That's interesting @John Baez ! I was recently browsing "An Introduction to Invariants and Moduli" by Mukai, which somewhat surprisingly (at least to me) chooses to just use the spectrum of maximal ideals. So I assume there's a quite a bit that can be done with only the maximal ideals!

It would be interesting to know what examples historically motivated people to use the spectrum of prime ideals.

Damiano Mazza (Jan 13 2025 at 20:10):

John Baez said:

[...] you don't need $k$ to be a field to conclude that $\ker f$ is a prime ideal, and if $k$ is a field you get more: $\ker f$ is a 'maximal' ideal.

Actually this is not necessarily the case. Take $f:R\to D$ with $D$ an integral domain, and suppose that $\mathrm{ker}\,f$ is prime but not maximal. Take now an injective homomorphism $i:D\to k$ with $k$ a field, which always exists (for example, $k$ can be the field of fractions of $D$). Then, $\mathrm{ker}(i\circ f)=\mathrm{ker}\,f$, so it is not maximal.

(A concrete example: the homomorphism $f:\mathbb Z[X]\to\mathbb Q$ sending $X$ to $0$ is such that $\mathbb Z[X]/\mathrm{ker}\,f=\mathbb Z$, which is not a field, showing that $\mathrm{ker}\,f$ is not maximal).

In fact, considering points as morphisms into fields gets you exactly all prime ideals. More precisely, the prime ideals of a commutative ring $R$ are in bijection with equivalence classes of homomorphisms $R\to k$ with $k$ a field, where $f:R\to k$ is equivalent to $f':R\to k'$ if there exists a common field extension $i:k\to K$, $i':k'\to K$ such that $i\circ f=i'\circ f'$. Proving that this is actually an equivalence relation is not hard but not completely trivial either (some time ago I asked a question about this on this Zulip, if I have time I'll try and find it...).

John Baez (Jan 13 2025 at 20:37):

Damiano Mazza said:

John Baez said:

[...] you don't need $k$ to be a field to conclude that $\ker f$ is a prime ideal, and if $k$ is a field you get more: $\ker f$ is a 'maximal' ideal.

Actually this is not necessarily the case.

I meant to write onto a field.

Damiano Mazza (Jan 13 2025 at 21:12):

David Egolf said:

It would be interesting to know what examples historically motivated people to use the spectrum of prime ideals.

I am in no way an expert in algebraic geometry and even less in its history, but I can give you an explanation that I found helpful. (Apologies if you already know it and are looking for something else!)

Around the time when the notion of scheme came around, algebraic geometers were looking for solutions of systems of polynomial equations in all sorts of fields, and it was interesting to be able to switch from one field to another. Now, a solution may be seen as a homomorphism, as follows.

Suppose that we have a system of polynomial equations $S:=p_1,\ldots,p_n$ in $m$ variables, and suppose for simplicity that these are polynomials with integer coefficients, so that we may look for solutions in any field whatsoever (because the integers uniquely map into every ring, hence into every field). Then, if we let $R:=\mathbb Z[X_1,\ldots,X_m]/\langle p_1,\ldots,p_n\rangle$, a solution of $S$ in the field $k$ is precisely a ring homomorphism $R\to k$. This gives us the idea that, if we manage to see all maps $R\to k$ as the points of a space, then we have a way of studying at the same time all possible solutions of $S$ we will ever wish to consider, which sounds very convenient! But the class of all maps $R\to k$ is not a set, we have too many "points"...

Luckily, there is a way around this problem: after all, if $i:k\to K$ is a morphism between fields (that is, we map the field $k$ into a bigger field $K$), and if $f:R\to k$ is a solution, it does not make sense to say that $i\circ f$ is a different solution. For example, the rational point $(1,0)$ of the unit circle, given by the map $\mathbb Z[X,Y]/\langle X^2+Y^2-1\rangle\to\mathbb Q$ sending $X\mapsto 1$ and $Y\mapsto 0$, should be "the same" point when we look at it in $\mathbb R$, or $\mathbb C$, etc. This motivates the equivalence relation I talked about above. Since the equivalence classes of this relation are in bijection with the prime ideals of $R$, of which there is just a set worth, we have found the underlying set of our space of all possible solutions!

Of course, we still have to equip this set with a topology, and then the resulting space with a local sheaf of rings, in order to get the affine scheme associated with $R$ and the $\mathrm{Spec}$ functor discussed above. But at least it motivates how we found the points!

Damiano Mazza (Jan 13 2025 at 21:27):

The above, and much more, is explained in countless places (for example on MO, in the book by Eisenbud and Harris, etc.), and in fact even by Grothendieck himself in the introduction to the second edition of EGA!

Unfortunately, that book seems to be very rare, I never saw a copy myself, the only accessible source I know of is this English translation on Daniel Murfet's web site. It is a rudimentary translation (it seems to have been made by an automatic translation tool in the early 2000s at best :big_smile:), but it is still quite readable and very informative. In particular, it introduces the "functorial" viewpoint of algebraic geometry which, if I understand correctly, Grothendieck advocated in the final years of his "normal" career, and which is extremely interesting for us category-minded people.

John Baez (Jan 13 2025 at 21:45):

David Egolf said:

It would be interesting to know what examples historically motivated people to use the spectrum of prime ideals.

This is a really interesting question. To add a little to Damiano's excellent answer, I think people started by mainly doing algebraic geometry 'over the complex numbers', working with algebras over $\mathbb{C}$, i.e. commutative rings equipped with a homomorphism from $\mathbb{C}$. Then they generalized to algebras over other fields, and found that algebraically closed fields work better. But they didn't want to work separately over every field. When they tried to work over 'all fields at once' they were led to think about maximal ideals... but eventually they decided to drop the focus on fields (to some extent) and develop everything in a way that works nicely for arbitrary commutative rings. This pushed them to use prime ideals.

As a result, 20th-century algebraic geometry has several historical 'strata', and it helps to know a tiny bit about the old approaches to understand the point of the new approaches.

I don't know all this stuff as well as I'd like, but some key players include Weil, Grothendieck and various members of Bourbaki. An important part of the story is the growing importance of sheaves.

John Onstead (Jan 13 2025 at 22:01):

Damiano Mazza said:

I understand that you are now taking a different angle, so this is no longer relevant for the ongoing discussion, but just for the sake of clarifying: I don't think that this has nothing to do with density (unless I am misunderstanding what you mean by "density").

Thanks for the clarification! For the record, by "density", I meant how a subcategory $C \to D$ is dense if the "restricted Yoneda embedding" automatically gives rise to a full and faithful embedding of $D$ in a the presheaf category of $C$, that is $D \to [C^{op}, \mathrm{Set}]$.

Damiano Mazza said:

This motivates the equivalence relation I talked about above. Since the equivalence classes of this relation are in bijection with the prime ideals of R, of which there is just a set worth, we have found the underlying set of our space of all possible solutions!

Very insightful! I didn't even stop to ask the motivation for all this, but I'm glad it was brought up!

John Onstead (Jan 13 2025 at 22:06):

David Corfield said:

Hmm, the nLab is rather lacking in connections. Neither Isbell duality nor dualizing object mention each other. Then there's a page duality between algebra and geometry which at least links to 'Isbell duality' but not to 'dualizing object'.

In an interesting twist, the page on the "duality between algebra and geometry" only talks about dualizing objects, and not Isbell duality! It just does so by going to seemingly great lengths to avoid actually saying "dualizing object" aloud (the page calls it a "seemingly simple algebra" $k$). Maybe if that nlab page gave a name to what it was talking about, a mysterious math curse might be unleashed?

Jokes aside, I do feel the nlab is in much need of an update of some form in this domain. But I think what that will come in the form of will require further discussion.

John Onstead (Jan 13 2025 at 22:08):

David Corfield said:

Good question about the relationship between Isbell duality and dualizing objects, one that seems to be in the air on MO periodically, e.g., here and here, without, as I can see, a clear answer.

Maybe a good place to start would be in finding the dualizing object for Isbell duality? The nlab page here hints such a thing might exist: "That this (Isbell duality) is an adjunction can be understood as a special case of abstract Stone duality induced by a dualizing object" But of course it doesn't elaborate.

David Corfield (Jan 14 2025 at 11:57):

John Onstead said:

Maybe a good place to start would be in finding the dualizing object for Isbell duality?

Here's something from some notes by Urs Schreiber:
image.png.

(Back in the day we were experimenting with terms such as 'ambimorphic' and 'Janusian', but settled on 'dualizing'.)

Schreiber's account tallies with an interesting generalization of Isbell duality/conjugacy in

• @Paul-André Melliès, @Noam Zeilberger , An Isbell Duality Theorem for Type Refinement Systems

image.png

where $R$ acts as the dualizing entity.

Mike Shulman (Jan 14 2025 at 16:04):

I prefer 'ambimorphic', personally. "Dualizing" refers to how we tend to use it, not about the fundamental property of "living in two categories".

Amar Hadzihasanovic (Jan 14 2025 at 19:28):

Ambimorphic is ugly because it's a Latin-Greek hybrid; you would want "amphimorphic" or "ambiform" instead.

John Onstead (Jan 14 2025 at 19:50):

Mike Shulman said:

I prefer 'ambimorphic', personally. "Dualizing" refers to how we tend to use it, not about the fundamental property of "living in two categories".

My favorite variation is "an object with summer and winter homes". It's a little too lengthy for proper use but it sure is humorous!

David Corfield said:

Here's something from some notes by Urs Schreiber:

Thanks for the resource it seems very helpful in settling some of these questions. But maybe all the "ops" are confusing me as I'm not fully understanding. This functor is meant to be a "co-presheaf valued presheaf"- that makes it sound like the functor part of the Isbell adjunction $[C^{op}, \mathrm{Set}] \to [C, \mathrm{Set}]^{op}$. In any case the dualizing object can't be $\mathrm{Hom}(-,-)$ itself (which it almost seems to be implied by the word choice) since that doesn't even live in either category. But I'm still not seeing what the dualizing object actually is. What object in $[C^{op}, \mathrm{Set}]$ is such that mapping a representable $A$ into it gives the copresheaf $\mathrm{Hom}(A, -)$? I know if you map into another representable $B$, then this gives you part of the information (the set of maps out of $A$ targeted in $B$), but not all of it (and also you would have to do this for every representable, when we want a singular object).

Todd Trimble (Jan 14 2025 at 20:09):

Amar Hadzihasanovic said:

Ambimorphic is ugly because it's a Latin-Greek hybrid; you would want "amphimorphic" or "ambiform" instead.

Do you also dislike "television"? Or "sociology"?

Peva Blanchard (Jan 14 2025 at 20:18):

About vision, there is an interesting etymology.

It comes from Latin “video” (to see), with “v” probably pronounced like the English “w”.
In Greek, there is also “eidos” (idea).
But some etymologists argued that the Greek word has ancient root "$\digamma\text{eidos}$" (see here) where $\digamma$ is digamma is close, phonetically, to the english "w".

All that to say that video and eidos, hence vision and idea, have a common ancestor which is about “seeing”.

With this reading, television becomes “remote ideas”. Nonsensical but funny.

ps: sorry I’m in linguistics mode.

Kevin Carlson (Jan 14 2025 at 21:25):

"v" probably pronounced like "w"? I was under the impression that we know quite definitely how classical Latin was pronounced. (Unlike Attic Greek.)

Peva Blanchard (Jan 14 2025 at 21:28):

Oh thanks for the clarification. The "probably" was me not being sure about the pronunciation (my memories of latin are getting old).

John Onstead (Jan 14 2025 at 22:17):

John Onstead said:

But I'm still not seeing what the dualizing object actually is.

I'm still not sure what the answer to this is overall, but I made some progress. A dualizing object exists relative to some functor $D \to \mathrm{Set}$ so that contravariantly composing some $C \to D^{op}$ with it is representable for some object in $C$. Now, let's choose the evaluation functor for some object $A$, that is, $ev_A: [C, \mathrm{Set}] \to \mathrm{Set}$, and choose a functor $[C^{op}, \mathrm{Set}] \to [C, \mathrm{Set}]^{op}$ to send a representable $B$ to $\mathrm{Hom}(B, -)$. Now let's see what happens. For any $B$, mapping into $A$ gives the set $\mathrm{Hom}(B, A)$, all entirely within $[C^{op}, \mathrm{Set}]$. Now, let's contravariantly compose the other two functors we define: the first sends any $B$ to $\mathrm{Hom}(B, -)$ as covered, and then the evaluation would send that to $\mathrm{Hom}(B, A)$- the same set as before! Therefore, mapping into $A$ represents the composition of these two operations, and thus $A$ is a dualizing object with respect to its own evaluation functor. Again, this doesn't solve the overall problem since there's as many evaluation functors as there are objects in the original category, but it is something that I noticed!

Todd Trimble (Jan 15 2025 at 00:42):

The dualizing object is just $\hom_C: C^{op} \times C \to \mathsf{Set}$.

The whole idea is that the desired contravariant adjunction works like this: as a contravariant functor, define $\Phi: \mathsf{Set}^C \to \mathsf{Set}^{C^{op}}$ as the functor that takes $G \in \mathsf{Set}^C$ to the functor

$(X \in C^{op}) \mapsto \mathrm{Nat}(G, \hom_C(X, -))$

and as a contravariant functor, define $\Psi: \mathsf{Set}^{C^{op}} \to \mathsf{Set}^C$ as the functor that takes $F \in \mathsf{Set}^{C^{op}}$ to

$(Y \in C) \mapsto \mathrm{Nat}(F, \hom_C(-, Y)).$

There is a natural bijection between natural transformations

$\frac{F \to \Phi(G)}{G \to \Psi(F)}$

because there are natural bijections between natural families consisting of

1. Transformations $FX \to \Phi(G)(X)$ natural in $X$,

2. Transformations $FX \times GY \to \hom_C(X, Y)$ natural in $X, Y$,

3. Transformations $GY \to \Psi(F)(Y)$ natural in $Y$.

and going back and forth between family 1 and family 3 is precisely the Isbell adjunction.

John Onstead (Jan 15 2025 at 02:41):

Todd Trimble said:

The whole idea is that the desired contravariant adjunction works like this: as a contravariant functor, define $\Phi: \mathsf{Set}^C \to \mathsf{Set}^{C^{op}}$ as the functor that takes $G \in \mathsf{Set}^C$ to the functor

$(X \in C^{op}) \mapsto \mathrm{Nat}(G, \hom_C(X, -))$

and as a contravariant functor, define $\Psi: \mathsf{Set}^{C^{op}} \to \mathsf{Set}^C$ as the functor that takes $F \in \mathsf{Set}^{C^{op}}$ to

$(Y \in C) \mapsto \mathrm{Nat}(F, \hom_C(-, Y)).$

After looking over this a few times I think that, at least according to Yoneda, I believe this is equivalent to what I was doing above (It's just that you expressed it in a more efficient and concise way than I currently have the ability to do). For instance it seems the second line is stating that given some presheaf $F$, $\Psi$ sends it to the copresheaf that itself sends an object $Y$ of $C$ to the set of morphisms from $F$ to $Y$ in the presheaf category. If $F$ is itself a representable $X$, then indeed this would send it to $\mathrm{Hom}(X, -)$ by Yoneda, which is what I got.

Todd Trimble said:

The dualizing object is just $\hom_C: C^{op} \times C \to \mathsf{Set}$.

I think this is what was confusing me about the original Urs Schrieber article. $\hom_C: C^{op} \times C \to \mathsf{Set}$ cannot be the dualizing object since the dualizing object must be thought of as an object of both categories (IE, it has a "summer and winter home" in two different categories). However, $\mathrm{Hom}(-,-)$, while an object of the category $[C \times C^{op}, \mathrm{Set}]$, is not an object of either of the two categories of focus, $[C, \mathrm{Set}]^{op}$ or $[C^{op}, \mathrm{Set}]$. But unless I'm mistaken, I think I understand what this is trying to say. From your write-out above it seems no matter the representable $Y$ you choose, the resulting adjunction formed by selecting it as a dualizing object is the same. So maybe there is no problem after all, and the lesson is that any representable (even one picked at random!) can be a dualizing object for Isbell duality (interestingly, this would mean any representable can be seen as "living in two categories at the same time"- quite cool!)

Todd Trimble (Jan 15 2025 at 04:40):

$\hom_C: C^{op} \times C \to \mathsf{Set}$ cannot be the dualizing object since the dualizing object must be thought of as an object of both categories (IE, it has a "summer and winter home" in two different categories).

My, my -- at times you really do express yourself with marvelous and provocative self-assurance!

Let's step back a moment and talk about dualizing objects. A really famous example is the classical Stone duality, which starts off by setting up a contravariant adjunction between Boolean algebras and topological spaces. In this case the dualizing object is often abbreviated to $\mathbf{2}$. Yes, on the one hand it can be regarded as carrying a Boolean algebra structure, and on the other as carrying topological space structure (as a discrete space). More importantly, though, it carries both: it is a Boolean algebra object internal to topological spaces, or in other words a topological space $B$ equipped with continuous maps $\wedge: B \times B \to B$, $\vee: B \times B \to B$, $\neg: B \to B$, etc. Or, if you prefer, a topological Boolean algebra. If $\mathrm{Bool(Top)}$ denotes the category of topological Boolean algebras, then we have forgetful functors

$\mathrm{Bool(Top)} \to \mathrm{Top}, \qquad \mathrm{Bool(Top)} \to \mathrm{Bool}$

where the first forgets the Boolean algebra structure and the second forgets the topological structure. Here in the fanciful terminology of "summer and winter homes", you could term say $\mathrm{Bool}$ is the summer home and $\mathrm{Top}$ is the winter home, but that's not really the point -- the point is more like that it lives in both summer and winter homes simultaneously, and the summer and winter home structures are suitably compatible with each other. The slogan "summer and winter homes" slogan, while cute and picturesque, sort of misses that point altogether. It's not even so much that "the dualizing object must be thought of as an object of both categories"; it's the compatibility between the two disparate structures borne by the object that counts.

Continuing the Stone duality story: if for example you temporarily ignore the Boolean algebra structure, and think of $\mathbf{2}$ as a topological space, then you get can speak of the contravariant representable

$\mathrm{Top}(-, \mathbf{2}): \mathrm{Top}^{op} \to \mathrm{Set}$

but then if you re-remember that $\mathbf{2}$ is not just a topological space but an internal Boolean algebra in $\mathrm{Top}$, well, that very fact tells you that the representable functor lifts through the forgetful functor $\mathrm{Bool} \to \mathrm{Set}$, to give a functor $\mathrm{Top}^{op} \to \mathrm{Bool}$, and bam -- right there is half of the contravariant adjunction induced by this dualizing object. .

A similar story pertains to $\hom_C$ in its role as a dualizing object for Isbell duality. It lives in the category $\mathrm{Set}^{C^{op} \times C}$. It's interesting to contemplate how there are forgetful functors

$\mathrm{Set}^{C^{op} \times C} \to \mathrm{Set}^C, \qquad \mathrm{Set}^{C^{op} \times C} \to \mathrm{Set}^{C^{op}}$

and there are various ways to think about this. One way is this: to get the first forgetful functor, think of the domain category as equivalent (even isomorphic) to $(\mathrm{Set}^{C^{op}})^C$, and then raise the forgetful functor $\mathrm{Set}^{C^{op}} \to \mathrm{Set}$, which assigns to a functor $F: C^{op} \to \mathrm{Set}$ its set of elements $\sum_{c \in C^{op}} Fc$, to the power $C$. The second can be described similarly.

(This gets into a whole thing, where $\mathrm{Set}^{C^{op}}$ can be regarded as comonadic over $\mathrm{Set}$: it is the category of coalgebras for a polynomial comonad associated with the category $C$. But I digress.)

I'll leave it at that for now.

John Onstead (Jan 15 2025 at 10:02):

Todd Trimble said:

My, my -- at times you really do express yourself with marvelous and provocative self-assurance!

Sorry about that! It's not my intention to come across this way. I've just found that sometimes this phrasing is the most useful, at least for me, for clearly designating where my problem / thought-roadblock is so it can best be directly addressed. Regardless, I'd like to thank you for your help and for your very in depth answers! Also, apologies for the long post, I really wanted to spend some time thinking about this!

Todd Trimble said:

the point is more like that it lives in both summer and winter homes simultaneously, and the summer and winter home structures are suitably compatible with each other. The slogan "summer and winter homes" slogan, while cute and picturesque, sort of misses that point altogether. It's not even so much that "the dualizing object must be thought of as an object of both categories"; it's the compatibility between the two disparate structures borne by the object that counts.

This has certainly changed by outlook on dualizing objects. I thought that they were something that you'd be lucky to find even once in a category or family of categories. That's partially why I thought they might make for a good notion of "category of space". But if you have any "$X$-object internal to $Y$-objects" then there's a chance it might be dualizing? That certainly opens up a lot of other questions... Which I should probably only address once I've finished this current line of thinking!

John Onstead (Jan 15 2025 at 10:03):

Todd Trimble said:

A similar story pertains to $\hom_C$ in its role as a dualizing object for Isbell duality. It lives in the category $\mathrm{Set}^{C^{op} \times C}$. It's interesting to contemplate how there are forgetful functors

$\mathrm{Set}^{C^{op} \times C} \to \mathrm{Set}^C, \qquad \mathrm{Set}^{C^{op} \times C} \to \mathrm{Set}^{C^{op}}$

and there are various ways to think about this. One way is this: to get the first forgetful functor, think of the domain category as equivalent (even isomorphic) to $(\mathrm{Set}^{C^{op}})^C$, and then raise the forgetful functor $\mathrm{Set}^{C^{op}} \to \mathrm{Set}$, which assigns to a functor $F: C^{op} \to \mathrm{Set}$ its set of elements $\sum_{c \in C^{op}} Fc$, to the power $C$. The second can be described similarly.

Ok, back to the main problem at hand. Let me try and see if I can work through this. I guess I can think of a "functor into $\mathrm{Set}$" as a sort of structured set- for instance, you can think of functors from the walking graph as being "graph objects" in $\mathrm{Set}$. So a functor from $C$ can be thought of as a "$C$-set". So what we want to do is consider sets with both a $C$ and a $C^{op}$ structure. That would amount to considering the category $[C, [C^{op}, \mathrm{Set}]]$. I think I can rearrange that via currying into $[C \times C^{op}, \mathrm{Set}]$, so indeed we can think of that as the category of sets with both $C$ and $C^{op}$ structure.

John Onstead (Jan 15 2025 at 10:03):

Since $\mathrm{Hom}(-,-)$ is an object of $[C \times C^{op}, \mathrm{Set}]$, it might be a dualizing object. I just have to find what its underlying object is in each category via the projections, and if the desired adjunction is induced. Given your hint, if you raise the "forgetful functor" $[C, \mathrm{Set}] \to \mathrm{Set}$ by $C^{op}$ you get a functor $[C \times C^{op}, \mathrm{Set}] \to [C^{op}, \mathrm{Set}]$. Generally taking a power of a morphism gives you some sort of composition, in this case it would send a functor $C^{op} \to [C, \mathrm{Set}]$ to its composition with this "forgetful functor". So first I need to view $\mathrm{Hom}(-,-)$ as a functor $C^{op} \to [C, \mathrm{Set}]$. This functor is the one that sends the object $A$ to the representable $\mathrm{Hom}(A, -)$ (the opposite Yoneda embedding). Composing with the "forgetful functor" would then send $A$ to the set of all morphisms out of $A$ in $C$. So then the answer is, the "underlying" presheaf of $\mathrm{Hom}(-,-)$ is the functor that sends an object to the set of all morphisms out of it. And so I guess that would be the object to "map into" during the dualization construction.

I should at least check to make sure this makes sense. If I pick a representable $\mathrm{Hom}(-, A)$, then via Yoneda the hom into this special presheaf will give the set of all morphisms out of $A$. Or, I can apply the adjunction we covered above to arrive at $\mathrm{Hom}(A, -)$, then take its underlying set. Indeed that's the same thing, the set of all morphisms out of $A$. So this "set of morphisms out of" presheaf passes the "sanity check" for being a good dualizing object candidate. I'm guessing that the "other half" is the copresheaf that sends an object to the set of all morphisms into it. Thus constituting the adjunction. Phew! I hope I did everything correctly. I tried showing my work so hopefully my thought process came through...

Amar Hadzihasanovic (Jan 15 2025 at 10:52):

Todd Trimble said:

Do you also dislike "television"? Or "sociology"?

My dislike is not that serious :) but I think that in both your examples there had been a long drift from their etymological origins in both the prefix "tele-" and the suffix "-logy", which have both come to be attached to all sorts of words, e.g. "telemarketing" or "ufology". A similar recent example is the drift of "cyber-" which, as known, does not even contain its complete Greek root...

Amar Hadzihasanovic (Jan 15 2025 at 10:55):

It seems a bit different from the creation of a new technical word which is modelled on scientific terminology "of old", which used to be more consistently fully Greek or fully Latin if anything because it came from texts written in one of the two.

Amar Hadzihasanovic (Jan 15 2025 at 10:58):

The main bits of mathematical terminology which use "morphic" or "morphism" are also consistent in using Greek... homo-, homeo-, iso-, endo-, auto-, even crypto- in matroid theory are all Greek prefixes.

Amar Hadzihasanovic (Jan 15 2025 at 11:03):

If you feel like there's still too little choice, perhaps I'd add "bilocated object" as a suggestion. :)

Todd Trimble (Jan 15 2025 at 13:24):

Not Latin + Greek, but there's also eigenvalue. :-)

Back in the early days of the n-Category Cafe, Tom Leinster argued that the older categorical term for the concept we're discussing, "schizophrenic object", is objectionable because it helps perpetuate wrong ideas about a serious mental illness. A few people then began to brainstorm. If you read further down the thread, you can witness the birth of "ambimorphic". :-)

Anyhow, thanks for pointing out the hybrid; I actually hadn't noticed it before. Although I've had my prescriptivist moments when it comes to language use, this one doesn't seem to bother me even after you point it out. But I'll remember it now.

Mike Shulman (Jan 15 2025 at 15:09):

I also tend towards the prescriptivist in many ways, but I try never to be prescriptive for its own sake; the innovations in language that I object to are those that destroy useful distinctions or reduce expressivity, like the misuse of "literally" to mean "figuratively", or the misuse of "exponentially" to mean "growing very fast". But I don't see any practical downside to combining roots from different languages; in fact rather the reverse, as it widens the field of possible neologisms. English is a marvelous mismash of etymology and I'm happy to embrace it.

Madeleine Birchfield (Jan 15 2025 at 16:12):

I've always had a problem with 'sheafification' because it is a combination of a Germanic root "sheaf" with a Latinate suffix "-ification".

David Egolf (Jan 15 2025 at 17:32):

Thanks for your comment, @Damiano Mazza !
Damiano Mazza said:

Around the time when the notion of scheme came around, algebraic geometers were looking for solutions of systems of polynomial equations in all sorts of fields, and it was interesting to be able to switch from one field to another.

It's interesting to me that people care about solving polynomial equations in a variety of fields. Just thinking of my small amount of experience of solving polynomial equations in high school, I think we only ever considered solutions in one field - the real numbers! More on this below.

Now, a solution may be seen as a homomorphism, as follows.... Then, if we let $R:=\mathbb Z[X_1,\ldots,X_m]/\langle p_1,\ldots,p_n\rangle$, a solution of $S$ in the field $k$ is precisely a ring homomorphism $R\to k$.

I think this works because:

1. $\Z[X_1, \dots, X_m]$ is (I think) the free commutative ring generated by a set with $m$ elements, so a ring homomorphism to $k$ from it amounts to a choice of $m$ elements of $k$. Basically, we "set a value" in $k$ for each $X_i$ with such a ring homomorphism. And we care about such ring homomorphisms that send each $p_i$ to zero, and hence all of $\langle p_1,\dots, p_n \rangle$.
2. The universal property of quotients ensures that a map from $\Z[X_1, \dots, X_m]$ that includes $\langle p_1,\dots, p_n \rangle$ in its kernel factors uniquely through $R$. And any map $:R \to k$ induces (by precomposition with the projection map) a map $:\Z[X_1, \dots, X_m] \to k$ that includes $\langle p_1, \dots, p_n \rangle$ in its kernel.

It is very cool that we can view a solution of $S$ in $k$ as a ring homomorphism $:R \to k$. If we move to the opposite category, a solution of $S$ in $k$ is then a $k$-shaped generalized element of $R$. This perhaps gives some intuition as to why we wish to study solutions over different fields: these corresponds to $k$-shaped generalized elements of $R$ (in this opposite category) as $k$ varies. So we can build up a deeper understanding of $R$ by studying solutions of $S$ over different fields. (This situation also reminds me a little bit of the notion of classifying topos, which I was reading about yesterday! Except instead of looking for solutions for polynomial equations in different fields, in that setting we look for models of a theory in different topoi.)

Luckily, there is a way around this problem: after all, if $i:k\to K$ is a morphism between fields (that is, we map the field $k$ into a bigger field $K$), and if $f:R\to k$ is a solution, it does not make sense to say that $i\circ f$ is a different solution. ... This motivates the equivalence relation I talked about above. Since the equivalence classes of this relation are in bijection with the prime ideals of $R$, of which there is just a set worth, we have found the underlying set of our space of all possible solutions!

That is very cool! Thanks for explaining that! To my understanding the punchline is roughly:

• ring morphisms from the quotient ring $R$ above to various fields correspond to solutions of $S$ over those various fields
• but a lot of solutions are related, because fields can be "nested" inside one another
• this induces an equivalence relationship on ring morphisms from our quotient ring $R$ of interest to various fields... we can roughly think of each equivalence class as a solution across potentially multiple related fields at once
• and these equivalence classes (hence solutions) are in bijection with the prime ideals of $R$!

Mike Shulman (Jan 15 2025 at 17:47):

David Egolf said:

This perhaps gives some intuition as to why we wish to study solutions over different fields: these corresponds to k-shaped generalized elements of R (in this opposite category) as k varies.

That's pretty backwards, speaking historically and probably for most algebraic geometers!

Mike Shulman (Jan 15 2025 at 17:48):

One intrinsic reason to consider solutions over different fields is Galois theory: if you know precisely in which fields the solutions to certain equations lie, the automorphisms of those fields can tell you useful things, like the unsolvability of the general quintic.

Mike Shulman (Jan 15 2025 at 17:50):

Another reason is number theory: knowing that an equation has solutions in fields like $\mathbb{Q}$ or $\mathbb{F}_p$ or $\mathbb{Q}_p$ can tell you things about whether it has solutions in rings like $\mathbb{Z}$.

Kevin Carlson (Jan 15 2025 at 18:33):

Madeleine Birchfield said:

I've always had a problem with 'sheafification' because it is a combination of a Germanic root "sheaf" with a Latinate suffix "-ification".

So the fully Germanic version would be something like "sheafmaking", I guess? I do quite enjoy the vibe of that one.

Mike Shulman (Jan 15 2025 at 18:38):

Unfortunately, in English the ending "-making" sounds like a verb form, not something that could describe the result of the process (e.g. "aX is the sheafification of X").

Mike Shulman (Jan 15 2025 at 18:40):

I really don't understand the objection to combining roots from different languages. When we import a word from another language, we often start pluralizing it using English grammar rules rather than keeping the plural from its original language. Why is it different to combine it with some other suffix that exists in English, regardless of where that suffix came from originally?

Kevin Carlson (Jan 15 2025 at 18:48):

Uh-oh, we're getting perilously close to the old saw about topoi vs toposes.

Kevin Carlson (Jan 15 2025 at 18:48):

No, "sheafmaking" is clearly not a plausible coinage, I just think it sounds charming. I also have trouble seeing how mixing etymologies is anything more than amusing trivia.

Damiano Mazza (Jan 15 2025 at 19:15):

David Egolf said:

Thanks for your comment, Damiano Mazza !

I'm glad you found it useful!

I think this works because:

Yes, that is exactly it!

John Onstead (Jan 15 2025 at 19:24):

Todd Trimble said:

I'll leave it at that for now.

Hi Todd Trimble, sorry to bug you, but I want to know: did I end up getting the correct answer in my work above? I spent a lot of time thinking about this and I want to make sure I understand everything correctly before moving on.

If it helps, here's the tl;dr of my work: I came to the conclusion that $\mathrm{Hom}(-,-)$ is a dualizing object for the categories of presheaves and copresheaves since you can think of $[C \times C^{op}, \mathrm{Set}]$ as the category of objects with both a "$C$-set" and "$C^{op}$-set" structure, as per your example with Boolean algebras internal to topological spaces. I found the underlying object of $\mathrm{Hom}(-,-)$ in both of those categories as the presheaf sending an object to the set of all morphisms out of it, and the copresheaf sending an object to the set of all morphisms into it respectively. Lastly, I then checked to make sure this was a sensible choice of pair of dualizing objects.

Thanks again for your help!

Damiano Mazza (Jan 15 2025 at 19:42):

David Egolf said:

It is very cool that we can view a solution of $S$ in $k$ as a ring homomorphism $:R \to k$. If we move to the opposite category, a solution of $S$ in $k$ is then a $k$-shaped generalized element of $R$.

This is actually something that was mentioned in a recent thread on the "field with in element". Every field $k$ has exactly one prime ideal (the zero ideal), so the underlying set of the space $\mathrm{Spec}\,k$ is going to be a singleton, no matter what the field $k$ is. The topology is of course also unique, so the only thing that changes is the sheaf on the point, which (unsurprisingly) associates $k$ with the unique non-empty open set.

In usual topology, if $X$ is a topological space and $\ast$ the one-point space, a (necessarily continuous) function $\ast\to X$ is exactly a point of the underlying set of $X$. In algebraic geometry, there are lots of "one-point spaces", one for each field, so one can consider "$k$-points" as maps $\mathrm{Spec}\,k\to X$. The collection of these "points" is, modulo the equivalence we described, the set underling $X$. However, none of the "one-point spaces" $\mathrm{Spec}\, k$ really behaves like the one-point space $\ast$, in the sense that none of them is the terminal object.

What algebraic geometers do then is to "relativize" with respect to $k$: they take the slice of the category of schemes over $\mathrm{Spec}\,k$. It is in this category, where $\mathrm{Spec}\,k$ really becomes like the one-point space we are used to (it is the terminal object), that algebraic geometers do "geometry over the field $k$". And if you have a field extension $K$ of $k$, "change of base" (i.e., pullback) along $\mathrm{Spec}\,K\to\mathrm{Spec}\,k$ will let you switch to the world of "geometry over $K$".

Like @Mike Shulman said, I'm not sure how much this "lots-of-one-point-spaces" perspective is useful to do algebraic geometry in practice (which I have never done!), but I certainly find it beautiful conceptually. Also, if I understand correctly, algebraic geometers before Grothendieck didn't have such a smoothly working machinery for switching between fields, and I find it really cool that a good theory can be found by saying that different points of view correspond to different one-point spaces.

Mike Shulman (Jan 15 2025 at 19:44):

I didn't mean that it isn't useful! I believe it is quite useful, and as you say it was a significant innovation of Grothendieck. I meant that it's backwards in terms of motivation -- we start out wanting to deal with solutions over different fields, and then discover that this is a good way to do it.

Amar Hadzihasanovic (Jan 15 2025 at 20:33):

Kevin Carlson said:

Madeleine Birchfield said:

I've always had a problem with 'sheafification' because it is a combination of a Germanic root "sheaf" with a Latinate suffix "-ification".

So the fully Germanic version would be something like "sheafmaking", I guess? I do quite enjoy the vibe of that one.

Sheafcraft.

Amar Hadzihasanovic (Jan 15 2025 at 20:44):

Mike Shulman said:

I really don't understand the objection to combining roots from different languages. When we import a word from another language, we often start pluralizing it using English grammar rules rather than keeping the plural from its original language. Why is it different to combine it with some other suffix that exists in English, regardless of where that suffix came from originally?

To me it seems that, in choosing the sort of style that "ambimorphic" has, one is inserting themselves into a long tradition, as if writing in imitation of the classical and medieval authors, and in this context it feels like a literary faux pas to get one's Greek and Latin mixed up.

Amar Hadzihasanovic (Jan 15 2025 at 20:45):

I am not against combining different languages in general, nor of course to the free use of prefixes of suffixes that have been incorporated into English or any other language.

Amar Hadzihasanovic (Jan 15 2025 at 20:47):

I wouldn't have the same objection to "ambidoodle" or "pastamorphic" because they are clearly not in imitation of the classics.

Mike Shulman (Jan 15 2025 at 20:54):

Huh.

Well, I guess we should stop hijacking this thread to argue about etymology. (-:O

John Baez (Jan 15 2025 at 22:02):

David Egolf said:

It's interesting to me that people care about solving polynomial equations in a variety of fields. Just thinking of my small amount of experience of solving polynomial equations in high school, I think we only ever considered solutions in one field - the real numbers!

That may have been true in my high school too, but in college they started caring about solutions in the complex numbers: for example in physics we'd take the equation of a damped harmonic oscillator, which is of the form

$a y'' + by' + cy = 0$

and by making the guess

$y = \exp(rt)$

we'd get

$ar^2 + br + c=0$

which gives exponential decay or growth when it has 2 distinct real roots, but oscillation when it has 2 distinct non-real roots.

Likewise when starting with a finite field like the integers mod p for some prime p, we are forced to think about a larger field, its 'algebraic closure' when we want to solve polynomial equations, or know which such equations have solutions in the original field. And we discover new and interesting things when we start thinking about the algebraic closure.

So for a while, algebraic geometers focused on algebraically closed fields. But that provides only limited help when trying to study integer solutions to polynomial equations, like in Fermat's Last Theorem. So they had to go further....

John Onstead (Jan 16 2025 at 07:08):

Mike Shulman said:

Huh.

Well, I guess we should stop hijacking this thread to argue about etymology. (-:O

It's no problem! I think- or, at least, I seem to think- that I understand the big picture with dualizing objects (at least, assuming I completed the above exercise correctly, as I haven't been told otherwise!) There's a lot more to get into with them of course (for instance, I've found a really interesting concept called the Chu construction I'm currently learning about), but I want to get back into sheaves which I'll probably do with a few more days of preparation!

Also it's always fun to learn about how mathematical concepts get their name, since in some cases that's when mathematicians show their humor! I remember reading somewhere about a list of funny named mathematical objects, but I'll have to look for it again.

David Egolf (Jan 17 2025 at 17:10):

Damiano Mazza said:

What algebraic geometers do then is to "relativize" with respect to $k$: they take the slice of the category of schemes over $\mathrm{Spec}\,k$. It is in this category, where $\mathrm{Spec}\,k$ really becomes like the one-point space we are used to (it is the terminal object), that algebraic geometers do "geometry over the field $k$". And if you have a field extension $K$ of $k$, "change of base" (i.e., pullback) along $\mathrm{Spec}\,K\to\mathrm{Spec}\,k$ will let you switch to the world of "geometry over $K$".

Thanks for your comment! I don't follow all the details, but it's interesting stuff! It was particularly fun to see the change of base functor show up again. That seems like a really powerful tool for switching between different but related settings.

David Egolf (Jan 17 2025 at 17:13):

John Baez said:

That may have been true in my high school too, but in college they started caring about solutions in the complex numbers: for example in physics we'd take the equation of a damped harmonic oscillator, which is of the form

$a y'' + by' + cy = 0$

and by making the guess

$y = \exp(rt)$

we'd get

$ar^2 + br + c=0$

which gives exponential decay or growth when it has 2 distinct real roots, but oscillation when it has 2 distinct non-real roots.

Oh yes! I remember doing this now that you point it out. I guess I never really thought of this as actually searching for a solution in the complex numbers. Somehow I always viewed all the uses of complex numbers in engineering contexts as a helpful shorthand - but fundamentally secondary to the "actual" solution we get by taking the real part at the end of some process.

David Egolf (Jan 17 2025 at 17:17):

That reminds a little bit of an example I recently worked through, where we can find solutions to a certain polynomial equation (a version of the Pell equation) in the integers, by considering the units in a ring $\mathbb{Z}[\sqrt{3}]$, the smallest subring of $\mathbb{C}$ that contains $\mathbb{Z}$ and $\sqrt{3}$. All the $\sqrt{3}$ stuff goes away when we get our final answer, but the setting with $\sqrt{3}$ still provides help for solving our equation!

John Baez (Jan 17 2025 at 18:20):

Somehow I always viewed all the uses of complex numbers in engineering contexts as a helpful shorthand - but fundamentally secondary to the "actual" solution we get by taking the real part at the end of some process.

You sound like a hard-nosed pragmatist focused on the "real world" - that brutal place people keep warning me about. :upside_down:

That reminds a little bit of an example I recently worked through, where we can find solutions to a certain polynomial equation (a version of the Pell equation) in the integers, by considering the units in a ring $\mathbb{Z}[\sqrt{3}]$, the smallest subring of $\mathbb{C}$ that contains $\mathbb{Z}$ and $\sqrt{3}$. All the $\sqrt{3}$ stuff goes away when we get our final answer, but the setting with $\sqrt{3}$ still provides help for solving our equation!

This reminds me of how Cardano invented (discovered?) complex numbers when solving cubic equations: some intermediate steps in the cubic formula can give complex numbers even when the final answers are real, and he had the guts to proceed nonetheless.

We can also take this attitude in quantum mechanics, and say that complex amplitudes are a helpful trick that's secondary to the actual probabilities we compute from them.... but most physicists start taking complex numbers as "real" by this point.

By now, in many branches of mathematics, like algebraic geometry or the theory of Lie algebras, the real numbers tend to be regarded as an annoyingly tricky subfield of the much better behaved $\mathbb{C}$. For example, studying the real solutions of a system of polynomial equations is something that you'd only want to mess with after you understand the complex solutions. "Real algebraic geometry" is practically a niche specialty compared to the complex case.

Madeleine Birchfield (Jan 17 2025 at 19:45):

David Egolf said:

Somehow I always viewed all the uses of complex numbers in engineering contexts as a helpful shorthand - but fundamentally secondary to the "actual" solution we get by taking the real part at the end of some process.

I view most uses of the complex imaginary unit $i \in \mathbb{C}$ in physics and engineering as representing the unit bivector of the Clifford algebra $\mathrm{Cl}_{2, 0}(\mathbb{R})$ representing a two-dimensional plane in 3D space, since the even grade subalgebra of $\mathrm{Cl}_{2, 0}(\mathbb{R})$ is isomorphic to the complex numbers.

John Baez (Jan 17 2025 at 22:09):

I'm not sure that's the clearest explanation of the power of Laplace and Fourier transforms in engineering, where we use a lot of facts about analytic functions on the complex plane, and use the algebraic completeness of $\mathbb{C}$ to solve linear ODE with constant coefficients (as in my example earlier).

Clifford algebra enthusiasts have tried to generalize the theory of complex analytic functions to other Clifford algebras - the approach seems to be Fueter's - but I've never figured out how well things like the theory of poles and residues works.

Morgan Rogers (he/him) (Jan 18 2025 at 19:30):

Amar Hadzihasanovic said:

Ambimorphic is ugly because it's a Latin-Greek hybrid; you would want "amphimorphic" or "ambiform" instead.

Ambimorphic is the adjective representing the duality between Latin and Greek, obviously :stuck_out_tongue_wink: