Two Miracles of Algebraic Geometry

August 10, 2016

Posted by John Baez

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In real analysis you get just what you pay for. If you want a function to be seven times differentiable you have to say so, and there’s no reason to think it’ll be eight times differentiable.

But in complex analysis, a function that’s differentiable is infinitely differentiable, and its Taylor series converges, at least locally. Often this lets you extrapolate the value of a function at some faraway location from its value in a tiny region! For example, if you know its value on some circle, you can figure out its value inside. It’s like a fantasy world.

Algebraic geometry has similar miraculous properties. I recently learned about two.

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Suppose if I told you:

Every group is abelian.
Every function between groups that preserves the identity is a homomorphism.

You’d rightly say I’m nuts. But all this is happening in the category of sets. Suppose we go to the category of connected projective algebraic varieties. Then a miracle occurs, and the analogous facts are true:

Every connected projective algebraic group is abelian. These are called abelian varieties.
If $A$ and $B$ are abelian varieties and $f : A \to B$ is a map of varieties with $f(1) = 1$ , then $f$ is a homomorphism.

The connectedness is crucial here. So, as Qiaochu Yuan pointed out in our discussion of these issues on MathOverflow, the magic is not all about algebraic geometry: you can see signs of it in topology. As a topological group, an abelian variety is just a torus. Every continuous basepoint-preserving map between tori is homotopic to a homomorphism. But the rigidity of algebraic geometry takes us further, letting us replace ‘homotopic’ by ‘equal’.

This gives some interesting things. From now on, when I say ‘variety’ I’ll mean ‘connected projective complex algebraic variety’. Let $Var_*$ be the category of varieties equipped with a basepoint, and basepoint-preserving maps. Let $AbVar$ be the category of abelian varieties, and maps that preserve the group operation. There’s a forgetful functor

$U: AbVar \to Var_*$

sending any abelian variety to its underlying pointed variety. $U$ is obviously faithful, but Miracle 2 says that it’s is a full functor.

Taken together, these mean that $U$ is only forgetting a property, not a structure. So, shockingly, being abelian is a mere property of a variety.

Less miraculously, the functor $U$ has a left adjoint! I’ll call this

$Alb: Var_* \to AbVar$

because it sends any variety $X$ with basepoint to something called its Albanese variety.

In case you don’t thrill to adjoint functors, let me say what this mean in ‘plain English’ — or at least what some mathematicians might consider plain English.

Given any variety $X$ with a chosen basepoint, there’s an abelian variety $Alb(X)$ that deserves to be called the ‘free abelian variety on $X$ ’. Why? Because it has the following universal property: there’s a basepoint-preserving map called the Albanese map

$i_X \colon X \to Alb(X)$

such that any basepoint-preserving map $f: X \to A$ where $A$ happens to be abelian factors uniquely as $i_X$ followed by a map

$\overline{f} \colon Alb(X) \to A$

that is also a group homomorphism. That is:

$f = \overline{f} \circ i_X$

Okay, enough ‘plain English’. Back to category theory.

As usual, the adjoint functors

$U: AbVar \to Var_* , \qquad Alb: Var_* \to AbVar$

define a monad

$T = U \circ Alb : Var_* \to Var_*$

The unit of this monad is the Albanese map. Moreover $U$ is monadic, meaning that abelian varieties are just algebras of the monad $T$ .

All this is very nice, because it means the category theorist in me now understands the point of Albanese varieties. At a formal level, the Albanese variety of a pointed variety is a lot like the free abelian group on a pointed set!

But then comes a fact connected to Miracle 2: a way in which the Albanese variety is not like the free abelian group! $T$ is an idempotent monad:

$T^2 \cong T$

Since the right adjoint $U$ is only forgetting a property, the left adjoint $Alb$ is only ‘forcing that property to hold’, and forcing it to hold again doesn’t do anything more for you!

In other words: the Albanese variety of the Albanese variety is just the Albanese variety.

(I am leaving some forgetful functors unspoken in this snappy statement: I really mean “the underlying pointed variety of the Albanese variety of the underlying pointed variety of $X$ is isomorphic to the Albanese variety of $X$ ”. But forgetful functors often go unspoken in ordinary mathematical English: they’re not just forgetful, they’re forgotten.)

Four puzzles:

Puzzle 1. Where does Miracle 1 fit into this story?

Puzzle 2. Where does the Picard variety fit into this story? (There’s a kind of duality for abelian varieties, whose categorical significance I haven’t figured out, and the dual of the Albanese variety of $X$ is called the Picard variety of $X$ .)

Puzzle 3. Back to complex analysis. Suppose that instead of working with connected projective algebraic varieties we used connected compact complex manifolds. Would we still get a version of Miracles 1 and 2?

Puzzle 4. How should we pronounce ‘Albanese’?

( I don’t think it rhymes with ‘Viennese’. I believe Giacomo Albanese was one of those ‘Italian algebraic geometers’ who always get scolded for their lack of rigor. If he’d just said it was a bloody monad…)

Posted at August 10, 2016 8:40 AM UTC

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Re: Two Miracles of Algebraic Geometry

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Puzzle 4 is easy to knock off: it is pronounced “alba-nay-say”.

For Puzzle 3: yes, both miracles carry over to connected compact complex manifolds. There are proofs in the first few pages of the book “Complex Abelian Varieties” by Birkenhake–Lange.

The other two seem to require some actual thought.

Posted by: Artie Prendergast-Smith on August 10, 2016 10:35 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Great!

By the way, you said some things by email that emphasize how the ‘albanesization’ of a projective variety is similar to the ‘abelianization’ of a group, and I feel like broadcasting them here, just to spread wisdom:

What I wanted to say is that the Albanese functor is reminiscent of a free functor for some kinds of varieties, but for others it is very different. Indeed, as Wikipedia says, the dimension of $Alb(X)$ is the Hodge number $h^{1,0}(X)$ . So for projective varieties with finite fundamental group (which includes all kinds of interesting examples like smooth rational varieties, K3 surfaces, Calabi–Yaus,…) the Albanese variety is just a point. A bit more generally, if you take $X$ to be a bundle over a variety $Y$ whose fibres are of one of the types I just listed, then then Albanese map of $X$ factors through that of $Y$ . So in these settings the Albanese map is more like a quotient (although it is not so clear what subobject we are quotienting by).

I guess this is in the same spirit as your comments about $Alb$ being an idempotent functor, and the mention of abelianization.

Of course the other side of the coin is that for some classes of varieties the Albanese morphism really is injective, for example curves of genus $\ge 2$ . More generally, if the image of $X$ inside $Alb(X)$ has the same dimension as $X$ , it is said to be of maximal Albanese dimension. (Googling shows that there is now a lot written about this class of varieties, but it is pretty heavy stuff.)

Posted by: John Baez on August 10, 2016 10:43 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Puzzle 1 has an easy categorical explanation: If every basepoint preserving map between Abelian varieties is a group homomorphism, then for any Abelian variety $A$ the inverse image morphism $\iota: A \to A$ is a group homomorphism, and this property uniquely characterises Abelian group objects in a category.

Here is a great link for future puzzle-4s: http://forvo.com/search/albanese/

Posted by: Adrian Clough on August 10, 2016 2:11 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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For Puzzle 1 you can even make an argument on the level of monoids, without mentioning inversion: the multiplication map is a basepoint-preserving map, hence must also be a homomorphism, and this characterizes commutative monoids by the Eckmann-Hilton argument.

Posted by: Qiaochu Yuan on August 10, 2016 9:22 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Nice!!

Posted by: Adrian Clough on August 10, 2016 10:24 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Nice stuff! Based on you guys’ work, here’s my answer to Puzzle 1:

By ‘variety’ I’ll mean ‘connected projective complex algebraic variety’. Let $Var$ be the category of varieties and whatever people call the nice maps between them — maybe ‘regular’ maps?

Let $Var_*$ be the category of pointed varieties and basepoint-preserving maps. Let $GpVar$ be the category of group objects in $Var$ and maps that preserve the group structure.

Then there are adjoint functors

$U : GpVar \to Var_*, \qquad F : Var_* \to GpVar$

$U$ is a faithful functor, obviously. Moreoever $U$ is monadic: i.e., $GpVar$ is equivalent to the category of algebras of the monad $T = U F$ .

This would all be true if we replaced $Var$ by $Set$ . But here’s the miracle that’s special to our situation: $U$ is full.

By either of the two different arguments you guys have given, this implies that every group object in $Var_*$ is an abelian variety:

$GpVar = AbVar$

Thus, the free group object on a pointed variety $X$ is the free abelian group object on $X$ , namely the Albanese variety of $X$ .

Also, by some categorical stuff, it implies that $T$ is an idempotent monad: the multiplication

$\mu: T^2 \Rightarrow T$

is an equivalence. So, the Albanese of the Albanese is the Albanese.

In practice, I believe people have to work up from a series of nontrivial results to this final claim that $U$ is full, not down from this final claim to these ‘corollaries’. But it’s nice to see how much is packed into this statement that $U$ is full.

Posted by: John Baez on August 11, 2016 7:14 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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OK, here’s a possible take on Puzzle 1. (I work with varieties over the complex numbers for convenience, and fail to mention basepoints where I ought to.) Let me know if I made an error.

Suppose $X$ is a connected projective variety, and we ask whether there is a “groupification” of $X$ in the category of varieties. That thing, if it exists, should be an algebraic group (i.e. a variety that is compatibly a group) $G X$ together with a morphism $g: X \rightarrow G X$ such that any morphism $X \rightarrow G$ to another algebraic group $G$ should factor uniquely through $g$ .

The first thing to note is that the image $g(X)$ has to generate $GX$ as a group, because if $g(X)$ generated a smaller group $H \subset G X$ then we would be able to factor the identity $G X \rightarrow G X$ through a morphism $GX \rightarrow H$ , which is impossible. (In particular, since $X$ is connected, this implies $G X$ is too.)

So that means every element of $G X$ can be written as a word of finite length in elements of $g(X)$ and their inverses.

Put differently, for each natural number $n$ , we have $2^n$ morphisms

$X^n \rightarrow G X$

sending $(x_1,\ldots,x_n)$ to $g(x_1)^{\pm 1} \cdots g(x_n)^{\pm 1}$ , and the union over all $n$ of the images of all these morphisms covers $G X$ .

Now here’s the catch: since $X$ , and hence $X^n$ is projective, the image of each morphism is a closed subset of $G X$ . The countable union of proper closed subsets cannot be the whole of $G X$ , so at least one of the images must be the whole of $G X$ . In particular, $G X$ is itself projective, hence by Miracle 1 abelian.

So the fact that images of projective varieties are projective, combined with Miracle 1, tells us that the groupification of any projective variety, if it exists, must be an abelian variety.

Posted by: Artie Prendergast-Smith on August 10, 2016 2:53 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Unfortunately I don’t think this works as it is currently written, because it implicitly assumes that $g(X)$ generates a Zariski-closed subgroup of $GX$ .

I think it can be fixed, but I don’t have a full argument yet.

Posted by: Artie Prendergast-Smith on August 11, 2016 8:15 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Does this interest in Albanese varieties arise from the Abel-Jacobi theory described at the end here, intermediate jacobians being in the air?

That path leads to the dizzying precipice of number theory meets QFT, such as in Urs’s table and Minhyong Kim’s Arithmetic Chern-Simons Theory I.

Posted by: David Corfield on August 10, 2016 9:25 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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So something regarding Puzzle 2 is that Albanese and Picard varieties have minimal and maximal degree as intermediate Jacobians. Whatever is found out about this duality could be added there.

Posted by: David Corfield on August 10, 2016 9:49 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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David wrote:

Does this interest in Albanese varieties arise from the Abel-Jacobi theory described at the end here, intermediate jacobians being in the air?

Yes. The story goes like this: Greg Egan and I spent a few weeks proving that the obvious map from the maximal abelian cover of a graph $X$ into the vector space $H_1(X,\mathbb{R})$ is injective. Then people started telling me that what we’d “really” done is prove that a certain tropical analogue of the Abel–Jacobi map for curves is injective. Apparently this is “well-known” in some community, but I still haven’t seen a proof. (Click on the first link above the precise details.)

This got me wondering about the real meaning of the Abel–Jacobi map. I started thinking it looked like the unit of a monad.

As you probably know, the Jacobian variety of a curve is a special case of two things. First, it’s a special case of what might be called the ‘Picard variety’ $Pic_0(X)$ : the identity component of the moduli space of line bundles on a variety $X$ . But to make $Pic_0$ into a monad, I’d need a unit, meaning a natural way to map $X$ into $Pic_0(X)$ , and this didn’t seem to work. I also didn’t see a multiplication for this would-be monad: I don’t know how to take a line bundle on the moduli space of line bundles on $X$ and extract a line bundle on $X$ .

But then I learned that the Jacobian of a curve is also a special case of the ‘Albanese variety’ $Alb(X)$ , which is the universal way of taking $X$ equipped with a specific basepoint and universally making it into an abelian variety! And this clearly sounded like a monad, and after checking on MathOverflow I became convinced this is right.

I’m not very interested in ‘intermediate Jacobians’ yet, but it’s easy to believe that while the Jacobian is closely connected to $H^1$ , which is very nice and ‘self-dual’ for complex curves (basically because 1 is half of 2), we’ll have a whole series of Jacobian-like things connected to $H^k$ of a higher-dimensional variety, with the Picard variety and Albanese variety being the extreme cases, and being dual to each other.

Various flavors of $H^1$ are good at classifying various kinds of line bundles (or ‘invertible sheaves’.) Similarly various flavors of $H^k$ tend to be good at classifying various kinds of ‘line $k$ -bundles’ (or ‘invertible $(k-1)$ -gerbes’). Do people make the connection between intermediate Jacobians and these categorified analogues of line bundles?

Posted by: John Baez on August 11, 2016 6:22 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Do people make the connection between intermediate Jacobians and these categorified analogues of line bundles?

Urs certainly does. From the nLab entry I linked to:

conceptually we may

think of $H^{2k+1}(\Sigma,\mathbb{R})$ as the space of those $n$ -form connections on $\Sigma$ which are both flat and have trivial underlying line $n$ -bundle;

think of $H^{2k + 1}(\Sigma,\mathbb{Z})$ as the group of “large” (not connected to the identity) higher gauge transformations acting on these gauge fields;

and hence understand $J^{k+1}(\Sigma)$ as the moduli space of flat $n$ -form connections on trivial underlying line $n$ -bundles.

Posted by: David Corfield on August 11, 2016 7:45 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Good! I see he’s worked out a lot of nice connections between higher gauge theory and the things algebraic geometers have traditionally done with these intermediate Jacobians. It would be nice if the higher gauge theory either helped the algebraic geometers solve problems they’re already interested in, or gave them interesting new problems to work on (in particular, problems they could easily understand). There should be a lot of stuff to do here.

Posted by: John Baez on August 11, 2016 7:56 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Of course, who would stick with mere spaces when you can have intermediate Jacobian higher group stacks for any differential generalized cohomology theory?

Posted by: David Corfield on August 11, 2016 10:04 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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The answer to your question is, of course, “most mathematicians”.

Posted by: John Baez on August 11, 2016 11:34 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Thoughts on Puzzle 2. The dual of an abelian variety can be interpreted as the moduli space of degree 0 line bundles on it. Degree 0 line bundles on the Albanese variety $A(X)$ of $X$ pull back to degree 0 line bundles on $X$ , and so the claim that the dual of the Albanese variety is naturally the Picard variety means that this pullback establishes an equivalence: in other words, degree 0 line bundles on $X$ extend uniquely to $A(X)$ . (Actually I guess it means this statement in families but let’s ignore that.)

(There’s something funny going on here: this means that duality for abelian varieties is like a categorification of Cartier duality, where the dualizing object is something like $B \mathbb{G}_m$ instead of $\mathbb{G}_m$ . I wish I understood this better.)

This statement can, I think, be reinterpreted as the statement that Miracle 2 extends to the moduli stack of degree 0 line bundles; in other words, it would follow straightforwardly if the moduli stack of degree 0 line bundles were itself an abelian variety in some sense. If this were true, then morphisms of abelian varieties from $A(X)$ to this moduli stack would correspond to degree 0 line bundles on $X$ , but then by a suitably strong version of Miracle 2 we can replace “morphism of abelian varieties” with “basepoint-preserving map”… although I’m not sure about this, since it seems that to get a basepoint-preserving map I need to consider line bundles equipped with a trivialization at the basepoint?

An analogous result in topology is the following. Topological line bundles on a space $X$ correspond to maps $X \to BGL_1(\mathbb{C}) \cong B^2 \mathbb{Z}$ from $X$ to the classifying space of line bundles. Now, this classifying space can be realized as a topological abelian group, and so maps from $X$ to it correspond to maps of abelian groups from the free topological abelian group on $X$ to it. Of course there is a more invariant way to say this where we replace “topological abelian group” with “infinite loop space” or “spectrum.”

Unfortunately, here Miracle 2 is false; it would say something like, $\pi_0$ of the space of maps of pointed spaces between two connected spectra is $\pi_0$ of the space of maps of spectra between them. For Eilenberg-MacLane spectra this would mean that all cohomology operations are stable, which is definitely false.

Posted by: Qiaochu Yuan on August 10, 2016 9:51 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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if the moduli stack of degree 0 line bundles were itself an abelian variety in some sense

isn’t it a symmetric algebraic group stack?

Posted by: David Roberts on August 12, 2016 12:52 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Qiaochu wrote:

(There’s something funny going on here: this means that duality for abelian varieties is like a categorification of Cartier duality, where the dualizing object is something like $B\mathbb{G}_m$ instead of $\mathbb{G}_m$ . I wish I understood this better.)

Me too. I don’t even know what ‘Cartier duality’ is, but it sounds like the algebraic geometry analogue of Pontryagin duality where the dualizing object is $\mathbb{G}_m$ rather than the circle group $\mathrm{U}(1)$ .

(Since I’m still annoyed at how nobody ever came out and told me what $\mathbb{G}_m$ was, I’ll spill the beans: it’s the multiplicative group $k^\times$ of a field $k$ , or if you want to be fancy, it’s the multiplicative group scheme, whose $k$ -points form $k^\times$ .)

In topology, has someone tried some funny ‘shifted’ kind of Pontryagin duality where the dualizing object is $B\mathrm{U}(1)$ rather than $\mathrm{U}(1)$ ? Perhaps fans of spectra wouldn’t even blink an eyelid at this.

Posted by: John Baez on August 12, 2016 3:13 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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It goes without saying, but here goes anyway, please do fill in what’s seriously missing at nLab entries such Cartier duality.

Posted by: David Corfield on August 12, 2016 5:59 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Sylvain Brochard has a nice paper about this categorified Cartier duality and the Albanese morphism.

Posted by: Justin Campbell on August 25, 2016 10:46 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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There’s another nice variant of these ideas that might appeal to (or possibly revolt?) the topologically-inclined:

Call a variety $X$ an algebraic H-space if there is a map $m: X \times X \rightarrow X$ and a point $e \in X$ such that $m(x,e)=m(e,x)=x$ for all $x \in X$ .

Then:

Theorem: A connected proper algebraic H-space is an abelian variety.

A proof is in “Appendix to §4” on pp 44-45 of Mumford’s Abelian Varieties.

Posted by: Artie Prendergast-Smith on August 12, 2016 10:06 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Nice! That’s an even stronger version of Miracle 1. I’ll have to look at Mumford’s book, to see how he proves this and get a sense of what technology is involved.

Posted by: John Baez on August 12, 2016 2:56 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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(I think the reply I wrote got eaten, so let me try again. I am still getting used to this interface.)

The proof really uses very little, really just:

— universal closedness of proper varieties, and

— the “rigidity lemma” saying that if $X$ is proper and $f: X \times Y \rightarrow Z$ contracts some subset $X \times \{y_0\}$ to a point, then $f$ factors through the projection $X \times Y \rightarrow Y$ . (And this is itself more or less a consequence of universal closedness.)

It is amazing how much one gets almost for free in the world of projective (or proper) varieites!

Posted by: Artie Prendergast-Smith on August 12, 2016 3:24 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Thanks! I got ahold of that book and saw that the ‘rigidity lemma’ plays a big role in proving that miraculous statement about $H$ -spaces, together with some clever-looking calculations.

I still haven’t figured out what the rigidity lemma is saying in category-theoretic terms, but I guess it’s saying products have a property they don’t have in the category of sets, giving rise to various ‘miracles’.

I have no idea what ‘universal closedness’ is, but it sounds like I’d better learn about it!

Posted by: John Baez on August 13, 2016 4:24 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Isn’t universal closedness just ‘closed, plus all pullbacks are closed’? This is just properness in the topological sense.

Posted by: David Roberts on August 13, 2016 4:54 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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That’s right. I didn’t mean to use confusing jargon; I always remember this property as “universal closedness” rather than “properness” because the former actually says what the property is!

This property gets used countless times in proving basic (and not-so-basic) results in algebraic geometry. The “classical” proof that $\mathbf P^n$ has this property (which you can read a version of in Shafarevich) is not very pleasant: it uses elimination theory and, if I remember correctly, a fair amount of non-transparent algebraic manipulation.

There is a nicer proof using Chevalley’s valuative criterion for properness, which says that $X$ is proper if every map from the “algebraic punctured disk” to $X$ can be filled in to a map from the whole disk. That’s written down in Hartshorne.

Posted by: Artie Prendergast-Smith on August 13, 2016 7:02 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Just to aid my own thinking: the rigidity lemma says that under certain circumstances “a 2-variable function $f(x,y)$ that’s independent of $x$ for one value of $y$ is independent of $x$ for all values of $y$ .”

In other words:

In a category with products, say a morphism $f: X \to Y$ is constant if it factors through the unique morphism $X \to 1$ . Say a morphism $f : X \times Y \to Z$ is independent of $X$ if it factors through the projection $X \times Y \to Y$ .

Say a point of $Y$ is a morphism $p: 1 \to Y$ . Say a morphism $f: X \times Y \to Z$ is independent of $X$ at some point $p$ of $Y$ if $f \circ (1_X \times p) : X \to Y$ is constant.

Say a category with products obeys the rigidity lemma if any morphism $f: X \times Y \to Z$ that is independent of $X$ at some point of $Y$ is in fact independent of $X$ .

I’ll conjecture that any category $C$ obeying the rigidity lemma obeys this property:

Property 1. Given group objects $G,H$ in $C$ , any morphism in $C$ from $G$ to $H$ that preserves the identity is a group homomorphism.

We’ve seen that Property 1 implies:

Property 2. Any group object $G$ in $C$ is abelian.

The easy proof is that the inverse map $inv: G \to G$ preserves the identity, so by Property 1 it’s a homomorphism. It’s pretty easy to see that this implies $G$ is abelian.

Puzzle: Show that any category with products obeying the rigidity lemma has Property 1.

It’s also fun to derive Property 2 directly from the rigidity lemma.

Posted by: John Baez on August 13, 2016 7:15 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Thanks for the intriguing conjecture! I think that one can quite easily prove it if:

1) we strengthen the definition of rigidity to satisfy a universal property, which says that if we have a pair of arrows $f, g : X \times Y \rightarrow Z$ such that $f \circ (id \times p) = g \circ (id \times p)$ , then the arrows $Y \rightarrow Z$ which factor $f$ and $g$ by virtue of rigidity must be equal;

2) the category has ‘functional extensionality’, namely we can check if two arrows are equal by checking if they are equal on points.

I would be rather surprised if it is possible to give a proof without 1), and I would be surprised too if something like 2) were not required in addition.

Posted by: Richard Williamson on August 15, 2016 8:59 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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When I stated it, I thought I roughly saw how to prove that conjecture. Now I don’t! Can you sketch your proof? An argument using elements of a group $G$ , as if it were a set, should be fine; I can then worry about what formal properties are needed to justify reasoning with elements in that way.

‘Functional extensionality’ should be enough, but I’m actually thinking it’s more than enough, since one can often interpret an element like $g \in G$ as a ‘generalized element’ $g: X \to G$ , and use the Yoneda trick to derive equations between morphisms from equations between those morphisms applied to generalized elements.

Posted by: John Baez on August 16, 2016 6:08 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Yes, I agree that functional extensionality should be more than enough.

Here is the proof. Let $G$ and $H$ be group objects, and let $f : G \rightarrow H$ be an arrow which preserves the identity. We need to show that the arrow $\cdot \circ (f \times f) : G \times G \rightarrow \rightarrow H \times H \rightarrow H$ , which I’ll denote by $f_{1}$ , is equal to the arrow $f \circ \cdot : G \times G \rightarrow G \rightarrow H$ , which I’ll denote by $f_{2}$ .

Let $g: 1 \rightarrow G$ be any arrow. Let $p : G \rightarrow 1$ be the canonical arrow. Consider the arrow $f_{1} \circ (g \times id) \circ (p \times id) : G \times G \rightarrow 1 \times G \rightarrow G \times G \rightarrow H$ , which I’ll denote by $g_{1}$ . Let $e : 1 \rightarrow G$ be the identity of $G$ . Then $g_{1}$ is independent of $G$ at $e$ . Indeed, if we use elements for readability, we have, for any arrow $g' : 1 \rightarrow G$ of $G$ , that $\big( g_{1} \circ (id \times e) \big)\big( g' \big) = f_{1}\big( g, e) = f(g) \cdot f(e) = f(g)$ , where in the final equality we appeal to the fact that $f$ preserves the identity.

Consider now the arrow $f_{2} \circ (g \times id) \circ (p \times id) : G \times G \rightarrow 1 \times G \rightarrow G \times G \rightarrow H$ , which I’ll denote by $g_{2}$ . Then $g_{2}$ is also independent of $G$ at $e$ . Indeed, if we use elements for readability again, we have, for any arrow $g' : 1 \rightarrow G$ of $G$ , that $\big( g_{2} \circ (id \times e) \big)\big( g' \big) = f_{2}\big(g, e \big) = f( g \cdot e) = f(g)$ .

Now, by the universal property of rigidity of 1) of the comment of mine which you replied to, we deduce that $g_{1}$ is equal to $g_{2}$ (both factor through the projection of $G \times G$ onto its second factor by means of the same arrow).

To use elements again, this says that, for any element $g'$ of $G$ , we have that $f(g) \cdot f(g') = f(g \cdot g')$ . Since $g$ was any element of $G$ , we have, by functional extensionality, that is to say 2) of the comment of mine which you replied to, proven that $f$ is a homomorphism, as required.

It is of course worth noting that nothing in this argument using the fact that the entire category satisfies rigidity; we need only that rigidity holds for $G$ . Thus this argument should cover the algebraic geometry case, if the stronger form of rigidity of 1) of my comment holds. This seems plausible to me, but I’d need to look at it further.

Posted by: Richard Williamson on August 16, 2016 7:32 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Actually, having written the proof out, I see that functional extensionality is not in fact needed at all. One can carry out the proof for the arrows $f_{1}$ and $f_{2}$ , without introducing $g_{1}$ and $g_{2}$ .

Posted by: Richard Williamson on August 16, 2016 8:35 PM | Permalink | Reply to this

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Very nice! Now I think that maybe the rigidity lemma is enough, without need for that extra universal property.

The idea is to take the expression

$f(g \cdot g')(f(g) \cdot f(g'))^{-1},$

and note that it’s independent of $g$ when $g' = 1$ , since $f(1) = 1$ . We conclude that this expression is independent of $g$ .

Then we note that it’s independent of $g'$ when $g = 1$ , for the same reason. We conclude that it’s independent of $g'$ . We thus conclude that

$f(g \cdot g')(f(g) \cdot f(g'))^{-1}$

is $1$ for all $g,g'$ , since this is what it equals when either $g$ or $g'$ equals $1$ .

So, $f$ preserves multiplication. I believe any morphism between group objects that preserves multiplication must also preserve inverses, though I should check this point. Thus, $f$ is a homomorphism.

Of course it looks like I’m using functional extensionality a lot in this argument, but I don’t think it’s required…

Does this sound right? There are various things to check here.

Posted by: John Baez on August 17, 2016 3:46 AM | Permalink | Reply to this

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That’s a very clever idea! When I first read through your argument, I thought that it held. On later readings, however, I don’t think I quite see how you conclude that $f(g \cdot g') \cdot \big( f(g) \cdot f(g') \big)^{-1}$ is $1$ for all $g$ and $g'$ from the fact that it is independent of both $g$ and $g'$ seperately. Could you elaborate?

I agree that your argument, if this point is fine, should not require functional extensionality anywhere.

Posted by: Richard Williamson on August 17, 2016 7:26 AM | Permalink | Reply to this

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It’s true I don’t see yet how to make that part rigorous. I want a lemma like this:

Lemma. In a category with products, if $f : X \times Y \to Z$ factors through the projection $p_1: X \times Y \to X$ and also factors through the projection $p_2: X \times Y \to Y$ , then it factors through the unique morphism $!: X \times Y \to 1$ .

I’ll try to prove it. Maybe it only holds under some special circumstances. If it’s not always true, I’d really like to see a counterexample!

Posted by: John Baez on August 17, 2016 9:26 AM | Permalink | Reply to this

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I haven’t been reading this whole thread, but I think here is a counterexample: in a lattice, if $X\wedge Y \le Z$ factors through $X\wedge Y\le X$ and $X\wedge Y\le Y$ , then all that means is that $X\vee Y \le Z$ , which certainly doesn’t imply $\top\le Z$ .

In the positive direction, I think one should be able to prove using internal logic that in a regular category, if $f:X\times Y\to Z$ factors through both projections, then it factors through the support of $X\times Y$ , i.e. the regular image of $X\times Y\to 1$ . It seems unlikely to me that there is any positive result for plain categories with products, since this seems like a sort of exactness property; it’s almost saying that the pullback square of $X \times_1 Y$ is also a pushout.

Posted by: Mike Shulman on August 17, 2016 4:58 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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The lemma holds for all elements (i.e. arrows $1 \rightarrow X \times Y$ ) of $X \times Y$ . To see this, take any pair $(x,y)$ and any other pair $(x', y')$ . Since $f$ factors through the projection onto $X$ , we have that $f(x,y) = f(x,y')$ . And since $f$ factors through the projection onto $Y$ , we have that $f(x,y') = f(x', y')$ . Thus we have that $f(x,y) = f(x', y')$ , as required.

Thus, if we have functional extensionality (arrows are determined by what they do on elements), then the lemma holds. In particular, we have this for algebraic varieties, so we should now have a proof for that case.

More generally, it will hold, by a similar kind of argument, whenever the arrows are determined by arrows out of certain objects. For instance, it will hold for the category of categories, where we can view the objects as arrows out of the walking object, and arrows as arrows out of the walking arrow. More generally, it will hold for all presheaf categories, and all algebraic structures upon them.

It doesn’t hold completely generally though. For an extreme example, take any category with products in which there is an object $X$ for which $X \times X$ can be taken to be $X$ , and the projection arrows to be identities (for instance, $\mathbb{N}$ in the category of sets). Then any arrow from $X \times X$ to any other object $Y$ factors trivially through both projections.

Even though your very nice argument recovers the algebraic geometry result, I must say that I have a slight aesthetic preference for the one I gave (assuming that the stronger rigidity lemma can be proven, which I would guess it can): it avoids elements, and I have a preference for properties in category theory being universal!

Posted by: Richard Williamson on August 17, 2016 8:46 PM | Permalink | Reply to this

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Sorry, the counterexample involving $\mathbb{N}$ was nonsense! The general counterexample is fine, though: it applies to any preorder with products, for instance.

Posted by: Richard Williamson on August 17, 2016 9:01 PM | Permalink | Reply to this

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I think what you’re calling “functional extensionality” is more commonly called “1 is a generator” or well-pointedness. However, it seems to me that it isn’t required at all for this argument.

The expression $f(g\cdot g')\cdot (f(g) \cdot f(g'))^{-1}$ compiles to a particular morphism $G\times G \xrightarrow{k} H$ . The fact that $f$ preserves the identity $e_G:1\to G$ implies that both composites $G \xrightarrow{(id,e_G)} G\times G \xrightarrow{k} H$ and $G \xrightarrow{(e_G,id)} G\times G \xrightarrow{k} H$ are constant at the identity $e_H:1\to H$ . (This is a straightforward calculation in the internal logic of a category with products, or alternatively a slightly tedious diagram chase.)

In particular, $k$ is independent of the first $G$ in its domain at the point $e_G : 1\to G$ of the second $G$ in its domain. So by the rigidity lemma, there exists a morphism $r : G\to H$ such that the composite $G\times G \xrightarrow{\pi_2} G \xrightarrow{r} H$ is equal to $k$ . Now precompose both of these with $G \xrightarrow{(e_G,id)} G\times G$ : the first gives $r \circ \pi_2 \circ (e_G,id) = r$ and the second gives $k \circ (e_G,id) = e_H \circ !$ . Thus, $r$ is constant at the identity of $H$ , and hence so is $k$ . This implies $f$ preserves multiplication, and thus also inverses (again, by a calculation in internal logic or a diagram chase).

Posted by: Mike Shulman on August 18, 2016 5:02 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Excellent work, Mike!

I think what you’re calling “function extensionality” is more commonly called “1 is a generator” or well-pointedness. However, it seems to me that it isn’t required at all for this argument.

Good, that was my impression.

Let me just state what you’ve proved, for people having trouble following the twists and turns of this conversation:

Theorem. Suppose $C$ is a category with finite products obeying the rigidity lemma: if a morphism $f: X \times Y \to Z$ has the property that $f \circ (1_X \times p)$ factors through the terminal object for some $p: 1 \to Y$ , then $f$ factors through the projection from $X \times Y$ to $X$ . Then if $G,H$ are group objects in $C$ and $f : G \to H$ is a morphism that preserves the identity, $f$ is a homomorphism.

Corollary. In a category with finite products obeying the rigidity lemma, every group object is abelian.

There, that statement squeezed all the joy out of what we were doing — it’s almost dry enough to publish.

I’ve put a more interesting summary on the nLab articles abelian variety and Albanese variety.

Posted by: John Baez on August 18, 2016 6:15 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Nice argument, Mike! Aesthetically, I might still prefer the argument I gave; there is a little less trickery, and morally (if one subscribes to the morality of category theory at least :-)) it seems to me that if the rigidity lemma can be proven to hold, then the stronger one needed for my argument should also hold. I’m quite interested to see if it can be shown in the algebraic geometry case; I’ll investigate that when I get the opportunity.

It would be interesting if the homotopical version of this result, that John referred to, could be proven in the same way, but working up to homotopy.

Posted by: Richard Williamson on August 18, 2016 7:57 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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It’s really John’s argument; I just wrote it out using arrows to make it clear that it actually works. (In fact, actually I first wrote it out using the internal logic, then manually compiled the proof out to one in terms of arrows, since I thought the audience here might appreciate that more.) There doesn’t really seem to be any trickery in it to me; all you have to do is remember that to show $g = h$ in a group you can show $g h^{-1}=1$ , and then apply rigidity and follow your nose.

I do agree that the rigidity lemma as currently stated is not very category-theoretic, however.

Posted by: Mike Shulman on August 19, 2016 4:17 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Yes, I was referring specifically to your conclusion of the argument.

The use of the word ‘trickery’ was not intended to be pejorative, except with regard to the category theoretic aesthetics. On the contrary, what I meant by it was something like: a step where one does not just entirely follow one’s nose, where one needs some little creative idea. There seem to me to be two occurrences of this in the argument of John and yourself: the fact you mentioned that John had the idea to consider; and the way in which you show that $r$ is constant by pre-composing it with something that gives back $r$ .

By contrast, I think that one really does just follow one’s nose in the argument I gave: given the stronger version of rigidity, I think any category theorist would immediately try to use rigidity in the way I did, and then one shortly sees that the identity is the only element that can possibly help.

To put it slightly differently, somehow it seems ‘wrong’ to me that a category theoretic argument to show that $g=h$ relies instead on demonstrating that $gh^{-1} =1$ ; I would expect a proof using the latter to translate straightforwardly to a proof that does not use it.

Posted by: Richard Williamson on August 19, 2016 8:02 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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the way in which you show that $r$ is constant by pre-composing it with something that gives back $r$

Perhaps that is more obvious in the internal logic, since it’s just substitution. We know $x:G \vdash k(x,e) = e$ and $x:G \vdash k(e,x)=e$ , rigidity gives us $r$ with $x:G,y:G \vdash r(y) = k(x,y)$ , hence plugging in $e$ for $x$ we get $y:G \vdash r(y) = k(e,y) = e$ .

Posted by: Mike Shulman on August 20, 2016 11:40 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Ah, which is essentially the same point John made below by writing it out in $Set$ (that I didn’t see before writing this).

Posted by: Mike Shulman on August 20, 2016 11:43 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Richard wrote:

There seem to me to be two occurrences of this in the argument of John and yourself: the fact you mentioned that John had the idea to consider; and the way in which you show that $r$ is constant by pre-composing it with something that gives back $r$ .

The second step, ‘Mike’s trick’, seemed tricky to me when I first read it, but now I think it’s not.

The rigidity lemma is a way to show a function is independent of one variable. I couldn’t see how to use it to show a function is independent of two variables. A function of two variables that’s independent of each variable separately is independent of both, but my attempt to generalize this idea to any category with finite products failed miserably. Mike really just figured out the right way to proceed here; his idea seems ‘tricky’ because he explained it in general before outlining how it works when morphisms are functions, which I’ve done in the $n$ Lab, thus:

Let $k : G \times G \to H$ be given by

$k(g,g') = f(g\cdot g')\cdot (f(g) \cdot f(g'))^{-1}$

Assume $f(1) = 1$ . Then $k(1,g') = 1$ for all $g' \in G$ , so by the rigidity lemma $k(g,g')$ is independent of $g'$ and we can write $k(g,g') = r(g)$ . Similarly $k(g,1) = 1$ for all $g \in G$ , so by the rigidity lemma $k(g,g')$ is independent of $g$ . This means that $r(g)$ is independent of $g$ . But $r(1) = k(1,1) = 1$ so $r(g) = 1$ for all $g$ . Thus $k(g,g') = 1$ for all $g,g' \in G$ . This says that $f$ preserves multiplication.

This argument does easily generalize to any category with finite products.

Posted by: John Baez on August 20, 2016 12:35 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Richard wrote:

To put it slightly differently, somehow it seems ‘wrong’ to me that a category theoretic argument to show that $g = h$ relies instead on demonstrating that $g h^{-1} =1$ ; I would expect a proof using the latter to translate straightforwardly to a proof that does not use it.

Here I agree. You could argue that we’re doing group theory here, in a category with finite products, and among group theorists it’s bog-standard to prove that $g = h$ by showing $g h^{-1} = 1$ . This is why those rascals invented ‘kernels’ instead of equalizers.

But you could also agree that a certain chunk of group theory might as well be monoid theory, and it’s our job as category theorists to figure out how much math we can do without sneaky tricks like subtraction and division.

And in fact there’s a good reason to do that right here. As Artie pointed out, Mumford actually shows that any connected proper algebraic $H$ -space is an abelian variety. Here he’s pulling not just commutativity but also associativity and inverses out of thin air!

If there’s a nice category-theoretic explanation of this fact, it can’t rely on my trick, because we don’t have inverses to start with. Your approach sounds promising for proving associativity… though it may take even further ideas to get the inverses.

Quite generally, I’d like to know what ‘automatic improvements’ occur when we consider models of an algebraic theory $T$ in a category like ‘connected proper algebraic spaces’ or ‘connected projective algebraic varieties’, where the rigidity lemma and perhaps various generalizations like yours hold. Any model of $T$ becomes a model of some improved theory $T'$ : what is $T'$ , exactly?

It is amusing to let $T$ be the theory of rings. This reminds me slightly of Tom Leinster’s post Holy crap, do you know what a compact ring is?, but it seems to call for an even more obscene exclamation.

Posted by: John Baez on August 20, 2016 12:54 AM | Permalink | Reply to this

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Nice post, just an observation is that there is an error in the definition of the monad T at the right hand side of the equality.

Posted by: Eduardo Ruiz Duarte on August 15, 2016 9:30 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Thanks! It took me a long time to find it, but now it’s fixed.

Posted by: John Baez on August 16, 2016 4:33 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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John wrote:

Suppose if I told you:

Every group is abelian.

Every function between groups that preserves the identity is a homomorphism. You’d rightly say I’m nuts.

No, I’d say you were working in a unital category, that is, a pointed category so that the pair of canonical morphisms $X\xrightarrow{l_X}X\times Y\xleftarrow{r_Y}Y$ is strongly epimorphic, i.e. does not jointly factor through any subobject of $X\times Y$ . The definitions and results are taken from Borceux and Bourn’s Mal’cev, Protomodular, Homological and Semi-Abelian Categories.

Theorem. In a unital category, the following are equivalent properties for an object $X$

The trivial pair $X\xrightarrow{\id_X}X\xleftarrow{\id_X}$ factors through $X\xrightarrow{l_X}X\times X\xleftarrow{r_X}X$ .
$X$ has a morphism $X\times X\to X$ that is the multiplication of an internal magma.
$X$ has a morphism $X\times X\to X$ that is the multiplication of an internal commutative monoid.

Perhaps we’re even working with a strongly unital category, which is a pointed category with the more stringent condition that the pair $X\xrightarrow{\Delta}X\times X\xleftarrow{r_X}X$ is strongly epimorphic. If that were the case you another equivalent condition is

$X$ has a morphism $X\times X\to X$ that is the multiplication of an internal abelian group.

Except we are not working with a pointed category at all, but maybe we working with a Mal’cev category, that is, a category in which every reflexive relation is an equivalence relation.

Being a Mal’cev category is equivalent to requiring that that for every object $X$ , the (pointed) category of pointed objects of the slice category over $X$ is unital, which is equivalent to requiring it be strongly unital (in particular, the category of pointed objects of the slice over the terminal object is strongly unital).

In fact, the existence of an Albanese variety has an analog for cocomplete regular unital categories: for any object $X$ , the coequalizer of $l_X,r_X\colon X\rightrightarrows X\times X$ has a commutative monoid structure and is the reflection of the forgetful functor from internal commutative monoids to the category. Furthermore, the inclusion of abelian groups in commutative monoids also has a left adjoint given by the the cokernels of $M\xrightarrow{\Delta}M\times M$ . When the unital category is strongly unital, the second step is of course unnecessary.

I’d be shocked to learn that the category of complete varieties is Mal’cev. What actually transpires in the theory of abelian varieties passes through a lot of cohomological machinery, but seems to have some affinity to the above notions:

The rigidity lemma says that in the category of pointed varieties, $X\xrightarrow{l_X} X\times Y$ is not only the kernel of $X\times Y\xrightarrow{\pi_Y}Y$ but that $X\times Y\xrightarrow{\pi_Y}Y$ is the cokernel of $X\xrightarrow{l_X}X\times Y$ .
The theorem of the cube says that if a line bundle on the pointed variety $X\times Y\times Z$ , for $X$ and $Y$ complete and $Z$ connected is trivial on the three inclusions $X,Y,Z\hookrightarrow X\times Y\times Z$ , then $L$ is trivial. In particular, when $Z$ is a point this looks to be saying that the pair $X\xrightarrow{l_X}X\times Y\xleftarrow{r_Y}Y$ is at least “epimorphic relative to line bundles”.

Posted by: Vladimir Sotirov on August 30, 2016 9:30 PM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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I was reading a post somewhere else that reminded me that Miracle 2 is shared by endomorphisms of very small finite groups: aside from the trivial group, the cyclic groups with 2 and 3 elements, and the Klein viergruppe, have this interesting property (and no other groups, I think).

Posted by: L Spice on June 21, 2019 11:52 PM | Permalink | Reply to this

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Interesting! I don’t think every function $f : \mathbb{Z}/3 \to \mathbb{Z}/3$ with $f(0) = 0$ is a homomorphism: consider $f$ with $f(0) = 0$ , $f(1) = 0$ and $f(2) = 1$ . I’m also doubtful about the Klein four-group. But every function $f : \mathbb{Z}/2 \to \mathbb{Z}/2$ with $f(0) = 0$ is a homomorphism, and the same is true of any function $f : \mathbb{Z}/2 \to G$ where every element of $G$ has order 2.

Posted by: John Baez on June 22, 2019 2:50 AM | Permalink | Reply to this

Re: Two Miracles of Algebraic Geometry

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Yes, you’re right; although you clearly said, and I clearly repeated, that we were just talking about arbitrary set maps, somehow I had bijections in mind; and this protects us from counterexamples like your one for the cyclic group of order 3, but perhaps it’s too strong a requirement to regard this as a Miracle. (A mere Providence?)

Posted by: L Spice on June 22, 2019 12:50 PM | Permalink | Reply to this

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August 10, 2016