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Stream: learning: questions

Topic: algebra, geometry, coalgebra... cogeometry?

Jules Hedges (Apr 11 2021 at 10:40):

I'm going to ask a nontechnical handwave-y question, and hoping for nontechnical handwave-y answers

If you have a category of algebras, and you take its opposite category, it tends to be equivalent to a category of spaces. For example, you can make a specific monad (among other presentations) whose category of algebras is the category of commutative unital rings. If you take its opposite category, you get something equivalent to the category of affine schemes (I think... the exact details aren't important), which has a more direct presentation in terms of topological spaces + structure.

My question is, what sorts of things do you get if you take the opposite of a category of coalgebras, say coalgebras of a comonad?

Algebra is to geometry, as coalgebra is to...?

Jules Hedges (Apr 11 2021 at 10:46):

The algebra/geometry and algebra/coalgebra dualities are taken in orthogonal dimensions. An $F$-algebra is a morphism $FX \to X$, and a homomorphism of $F$-algebras is a map $X \to Y$ making a square commute. Taking the opposite category of algebras (yielding a category hopefully equivalent to some kind of spaces) yields a category whose objects are morphisms $FX \to X$ and whose morphisms are $Y \to X$ making a square commute. On the other hand, a category of coalgebras has objects like $X \to FX$ and homomorphisms like $X \to Y$, so its opposite category will have objects like $X \to FX$ and morphisms like $Y \to X$

Fawzi Hreiki (Apr 11 2021 at 12:11):

It’s important to note that not all categories of algebras look like the duals of categories of spaces. There’s a reason why algebraic geometry uses rings rather than say monoids.

Fawzi Hreiki (Apr 11 2021 at 13:44):

Also, another thing that may be worth looking into is the following. The general geometry-algebra duality hinges on the fact that we can talk about spaces in two (opposite?) ways; by shapes in them (i.e. by maps into them, usually from 'small' and 'simple' spaces) and by functions on them (i.e. by maps out of them, usually into some canonical 'algebraic' space).

The second (contravariant) approach is the one which is prevalent in topology (via the Sierpiński space), traditional differential geometry (via the real line or complex plane), and algebraic geometry (via spectra of rings ?). Here, the notion of monad comes into play via co-density monads. For example, the co-density monad of the full subcategory on the finite powers of the Sierpiński space is the soberification monad whose category of algebras is the category of sober spaces - or equivalently the category of locales with enough points. The idea is that you're restricting to those spaces which can be described perfectly by Sierpiński-space-valued functions - i.e. by their open subsets.

On the other hand, if you take a small subcategory of basic shapes and take the density co-monad, you get as co-algebras those spaces which can be described perfectly by probing them by these basic shapes. For example, in the category of (Lawvere) metric spaces (i.e. categories enriched in the extended non-negative real line), the full subcategory on the intervals $[0, d]$ has as density co-monad the geodesic re-metrisation monad which re-metrises spaces based on minimum distance paths.

Fawzi Hreiki (Apr 11 2021 at 13:46):

So in a sense, co-algebras for co-monads are already geometric to begin with, but this is relative to the base category on which the co-monad lives.

Fawzi Hreiki (Apr 11 2021 at 13:47):

Although perhaps it's more appropriate to think of co-algebras as actions rather than spaces.

John Baez (Apr 11 2021 at 15:25):

Categories of commutative algebras tend to look like duals of categories of spaces.

Categories of cocommutative coalgebras tend to look like categories of spaces.

Fawzi Hreiki (Apr 11 2021 at 15:26):

What are some nice examples of commutative coalgebras that look like spaces?

John Baez (Apr 11 2021 at 15:28):

A classic example is that given any set you can take the free vector space on that set and equip it with the structure of a commutative coalgebra, giving a functor

$F: \mathsf{Set} \to \mathsf{CoCommCoAlg}$

and if you look at the full image of this functor it's equivalent to Set.

John Baez (Apr 11 2021 at 15:28):

I consider a set as a very primitive and fundamental sort of "space".

John Baez (Apr 11 2021 at 15:33):

The idea is that the free vector space on a set $S$ becomes a cocommutative coalgebra with this comultiplication:

$\Delta(x) = x \otimes x$

for any $x \in S$.

John Baez (Apr 11 2021 at 15:33):

It's just borrowed from the diagonal in $\mathsf{Set}$.

John Baez (Apr 11 2021 at 15:35):

You can think of it this way:

A category of spaces should be a cartesian category. Every object $X$ in a cartesian category is automatically a cocommutative comonoid, using the diagonal $\Delta: X \to X \times X$. So it's already like a cocommutative coalgebra!

John Baez (Apr 11 2021 at 15:36):

If you then map your cartesian category to, say, $(\mathsf{Vect}, \otimes)$ via some symmetric monoidal functor, the objects of your category get mapped to cocommutative comonoids in $\mathsf{Vect}$, which are exactly cocommutative coalgebras.

John Baez (Apr 11 2021 at 15:39):

This is what I was doing earlier: I was mapping the cartesian category $(\mathsf{Set},\times)$ to $(\mathsf{Vect},\otimes)$ by the "free vector space" functor, which is symmetric monoidal. So, the free vector space on a set becomes a cocommutative coalgebra.

John Baez (Apr 11 2021 at 15:40):

The "twist" in algebraic geometry is to then dualize and work with commutative algebras instead of cocommutative coalgebras.

Mike Shulman (Apr 11 2021 at 18:13):

Jules Hedges said:

Algebra is to geometry, as coalgebra is to...?

cogeometry, of course.

Mike Shulman (Apr 11 2021 at 18:13):

(-:O

John Baez (Apr 11 2021 at 19:10):

That's true. But since algebra is like the dual of geometry, it's perhaps even more enlightening to say "algebra is to cogeometry as coalgebra is to geometry".

John Baez (Apr 11 2021 at 19:24):

By the way, there's been work on generalizing algebraic geometry to use just commutative monoids instead of commutative rings - for example, defining schemes at this level of generality.

John Baez (Apr 11 2021 at 19:25):

And then people like Toen and Vacquie go further and use commutative monoid objects in symmetric monoidal categories!

John Baez (Apr 11 2021 at 19:25):

And then people like Lurie go further and do the same in symmetric monoidal $(\infty,1)$-categories!

John Baez (Apr 11 2021 at 19:28):

So we could show off and algebraic geometry is the study of homotopy-coherent commutative monoid objects in symmetric monoidal $(\infty,1)$-categories. But this is after the "twist" I mentioned - the op that turns geometry into algebra. The geometry here is the study of homotopy-coherent commutative comonoid objects in symmetric monoidal $(\infty,1)$-categories.

Jules Hedges (Apr 11 2021 at 19:45):

Jules Hedges said:

The algebra/geometry and algebra/coalgebra dualities are taken in orthogonal dimensions. An $F$-algebra is a morphism $FX \to X$, and a homomorphism of $F$-algebras is a map $X \to Y$ making a square commute. Taking the opposite category of algebras (yielding a category hopefully equivalent to some kind of spaces) yields a category whose objects are morphisms $FX \to X$ and whose morphisms are $Y \to X$ making a square commute. On the other hand, a category of coalgebras has objects like $X \to FX$ and homomorphisms like $X \to Y$, so its opposite category will have objects like $X \to FX$ and morphisms like $Y \to X$

I find this all a bit surprising given that a priori these are totally unrelated dualities. And also because I think of objects in categories of coalgebras as machines which perform actions and have internal state, and machines are totally different sorts of things to spaces

John Baez (Apr 11 2021 at 20:46):

Jules Hedges said:

I find this all a bit surprising given that a priori these are totally unrelated dualities.

Note that I'm saying geometry is dual to the study of cocommutative coalgebras. Here "coalgebra" is being used in the algebraist's sense, the thing you'd see if you looked up "coalgebra" in Wikipedia:

• Wikipedia, Coalgebra.

This is a lot more specialized than the kind of coalgebra you're talking about, @Jules Hedges.

John Baez (Apr 11 2021 at 20:47):

It's just like how in "algebraic geometry", "algebra" does not mean "algebra of any endofunctor". Algebra means a place where you can add and multiply.

Morgan Rogers (he/him) (Apr 11 2021 at 20:48):

To the extent that categories of algebras can be characterized in terms of their categorical properties, the two should coincide: anything I can say about a generic category of algebras, the dual will be true for both the opposite of such a category and a generic category of coalgebras. However, there's an important distinction in the "domain" of these categories. If we're talking about algebras over Set, for example, this is saying that categories of coalgebras over Set^op will behave like spaces over Set.

Jules Hedges (Apr 11 2021 at 21:05):

Fair enough... I guess this leads to a related question like "how to intuitively think about the opposite of the category of $F$-algebras/whatever"

Matteo Capucci (he/him) (Apr 12 2021 at 07:08):

Fawzi Hreiki said:

It’s important to note that not all categories of algebras look like the duals of categories of spaces. There’s a reason why algebraic geometry uses rings rather than say monoids.

What if I want to have some $\mathbb F_1$? :upside_down:

JS PL (he/him) (Apr 12 2021 at 10:44):

John Baez said:

Categories of commutative algebras tend to look like duals of categories of spaces.

Categories of cocommutative coalgebras tend to look like categories of spaces.

Shameless plus/differential category advertisement: this idea is also captured in differential category theory.

JS PL (he/him) (Apr 12 2021 at 10:44):

Briefly, a differential category is a symmetric monoidal category equipped with a comonad $!$ where in particular $!(A)$ has a natural cocommutative comonoid structure on it. It follows that every $!$-coalgebra also comes equipped with a canonical cocommutative comonoid structure. So the coEilenberg-Moore category of $!$ can be interpreted as a category of cocommtuative comonoids with extra structure relating to $!$.

On the other end of the spectrum, you have tangent categories, which naively generalize the category of smooth manifolds. So a tangent category can be interpreted as a sort of category of smooth spaces, and there many examples linking tangent categories to differential geometry, algebraic geometry, synthetic differential geometry, etc.

In this paper of mine, with Robin Cockett and Rory Lucyshyn-Wright
https://drops.dagstuhl.de/opus/volltexte/2020/11660/pdf/LIPIcs-CSL-2020-17.pdf
we explained how the coEilenberg-Moore category is in fact a (representable) tangent category. So in this manner, many categories of certain kinds of cocommutative comonoids are properly interpreted as smooth spaces over some differential theory.

JS PL (he/him) (Apr 12 2021 at 10:46):

In particular, when $!$ is the cofree cocommutative coalgebra comonad on $VEC$, then the coEilenberg-Moore category in this case is of course the category of cocommutative coalgebras. Then the tangent category structure is the one associated to the differential theory studied by Clift and Murfet in:
https://arxiv.org/abs/1701.01285

JS PL (he/him) (Apr 12 2021 at 10:48):

Does this help with general intuition: probably not. But thought I'd share this story and hopefully some of you will find it interesting!

Reid Barton (Apr 12 2021 at 13:46):

Can you give an example?

Reid Barton (Apr 12 2021 at 13:47):

i.e. some category of cocommutive coalgebras, that looks like some category of spaces?

John Baez (Apr 12 2021 at 15:04):

@_Matteo Capucci (he/him)|275932 said:

Fawzi Hreiki said:

It’s important to note that not all categories of algebras look like the duals of categories of spaces. There’s a reason why algebraic geometry uses rings rather than say monoids.

What if I want to have some $\mathbb F_1$? :upside_down:

Yes, I was going to say (and later did) that algebraic geometry does use commutative monoids as a substitute for commutative rings, these days.

John Baez (Apr 12 2021 at 15:10):

Reid Barton said:

Can you give an example? i.e. some category of cocommutative coalgebras, that looks like some category of spaces?

I mentioned one earlier: $\mathsf{Set}$. This is a full subcategory of the category of cocommutative coalgebras, using the trick I explained: the free vector space functor $F: (\mathsf{Set},\times) \to (\mathsf{Vect},\otimes)$ embeds $\mathsf{Set}$ in $\mathsf{Vect}$, faithfully but not fully, but since every set $S$ is a cocommutative comonoid in a unique way, each vector space $F(S)$ gets the structure of a cocommutative coalgebra, and we get a full and faithful embedding of $\mathsf{Set}$ in $\mathsf{CocomCoalg}$.

John Baez (Apr 12 2021 at 15:19):

To make this a bit more explicit we can work out what sort of cocommutative coalgebras we get. To make this easy on myself I'll do the case of finite sets and work over field of complex numbers. Then the above trick turns each finite set into what we might call a finite-dimensional "semisimple" cocommutative coalgebra over $\mathbb{C}$: just the dualized version of a finite-dimensional semisimple commutative algebra.

There's a theorem that every finite-dimensional semisimple commutative algebra over $\mathbb{C}$ is a finite product of copies of $\mathbb{C}$. So, we can always think of it as the algebra of complex-valued functions on a finite set. We are just dualizing that idea here.

John Baez (Apr 12 2021 at 15:20):

The upshot is that $\mathsf{FinSet}$ is equivalent to the category of finite-dimensional semisimple cococommutative coalgebras over $\mathbb{C}$.

Reid Barton (Apr 12 2021 at 15:41):

Right, I meant that as a question to @JS Pacaud Lemay, sorry.

John Baez (Apr 12 2021 at 15:48):

I figured that might be true, but I had fun working out the category of finite sets as a category of coalgebras!

JS PL (he/him) (Apr 12 2021 at 16:09):

Reid Barton said:

Can you give an example?

If you take the coEilenberg-Moore category of the comonad on the category of convenient vector spaces from https://arxiv.org/pdf/1006.3140.pdf you get the category of convenient manifolds (https://ncatlab.org/nlab/show/convenient+manifold)

JS PL (he/him) (Apr 12 2021 at 16:15):

I would assume another good example is taking the coEilenberg-Moore category of https://arxiv.org/abs/1507.03262 -- but not 100% certain what the resulting spaces would be (maybe Mackey-complete bornological manifolds? uncertain...)

JS PL (he/him) (Apr 12 2021 at 16:20):

Then there are examples when you take $VEC^{op}$ as your starting category, where cocommutative coalgebra ares of course commutative algebras in $VEC$. Taking the $!(V) = Sym(V)$, then the coEilenberg-Moore category is the category of affine schemes. Taking instead $!(V)$ as the free $C^\infty$-ring over $V$, then the coEilenberg-Moore category is the category the dual category of $C^\infty$-rings which includes the category of smooth loci (which includes a full and faithful embedding of the category of smooth manifolds)

JS PL (he/him) (Apr 12 2021 at 16:23):

Reid Barton said:

i.e. some category of cocommutive coalgebras, that looks like some category of spaces?

So in most of these examples, there's a notion of "infinite dimensional" spaces. So these examples are much larger then what people are usually interested.

John Baez (Apr 12 2021 at 17:53):

Nice! Some of us love infinite-dimensional manifolds. Finite-dimensional manifolds are just a trivial special case. :upside_down: