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Stream: deprecated: algebra & CT

Topic: Category of polynomial functions

Evan Patterson (Feb 05 2023 at 20:40):

In their paper "Differential categories" (p. 1062), Blute et al describe a neat way to construct the category of vector spaces and polynomial maps between them. Let $S: \mathsf{Vect}_k \to \mathsf{Vect}_k$ be the free commutative $k$-algebra monad, which sends a vector space $V$ to the (underlying vector space of) the symmetric algebra $S(V)$. Then the opposite of the Kleisli category of $S$, or equivalently the coKleisli category of the comonad $S^{\mathrm{op}}$, is a category whose morphisms $V \to W$ are polynomial functions from $V$ to $W$. Call this category $\mathsf{Poly}_k$.

The authors go on to say that "it is well known that" $\mathsf{Poly}_k$ is a distributive category. Is this true? In fact, I am confused on a more basic point. Does $\mathsf{Poly}_k$ even have coproducts? If so, what are they?

John Baez (Feb 05 2023 at 21:07):

Hmm. I'm not sure.

I would often think of the category of vector spaces and polynomial maps between them in a somewhat different way. I would think of the Eilenberg-Moore category of your monad $S$ as the category of commutative $k$-algebras. Then its opposite is called the category of affine schemes. This is the usual way algebraic geometer's way of thinking of any commutative $k$-algebra as the "algebra of functions" on some sort of space, called an affine scheme.

Sitting inside the Eilenberg-Moore category is the Kleisli category. Its opposite is thus a full subcategory of the category of affine schemes, where the only objects are those coming from vector spaces. These special objects are called the "affine spaces" $\mathbb{A}^n$, at least in the finite-dimensional case. They are the affine schemes that look like vector spaces! (See the third definition here).

John Baez (Feb 05 2023 at 21:08):

So... so far, I'm just telling you that $\mathsf{Poly}_k$ is familiar in algebraic geometry as the category of affine spaces and arbitrary maps of affine schemes between these. This is suppose to help us visualize whether this category should have colimits.

John Baez (Feb 05 2023 at 21:09):

We should think of the objects as spaces that look like vector spaces, and the maps as functions that are defined by polynomials.

John Baez (Feb 05 2023 at 21:10):

That doesn't sound like it should have coproducts: the coproduct $\mathbb{A}^1 + \mathbb{A}^1$ should be an affine scheme that looks like the disjoint union of two lines!

John Baez (Feb 05 2023 at 21:12):

Yes, that seems right. So unwinding a bit, I'm saying the opposite of $\mathsf{Poly}_k$ doesn't have binary coproducts, or in other words, the Kleisli category of $S$ doesn't have binary products.

John Baez (Feb 05 2023 at 21:14):

And I'm suggesting that if you take $V$ to be a 1-dimensional vector space, then $S(V)$ is an object in the Kleisli category that doesn't have a product with itself.

John Baez (Feb 05 2023 at 21:15):

However all I've really shown is that if you take its product with itself in the Eilenberg-Moore category, you get an object that's no longer in the Kleisli category!

John Baez (Feb 05 2023 at 21:16):

Still, I'm offering a possible counterexample to the claim that $\mathsf{Poly}_k$ has coproducts, and it should be pretty easy to check this, by seeing whether $S(V)$ fails to have a product with itself in the Kleisli category of $S$.

Evan Patterson (Feb 05 2023 at 21:45):

Thanks John, this is helpful! I'll have to think about that last part more, but it seems plausible that $\mathsf{Poly}_k$ does not have coproducts. What most interests me is not necessarily this particular category but having a "good" category whose morphisms are polynomial functions, where "good" should include having limits and colimits. Perhaps the lesson from your comments is that I should be looking at a bigger category, such as the category of affine schemes. Algebraic geometry has always seemed scary to me, but I might need to bite the bullet and learn some basic things.

John Baez (Feb 05 2023 at 22:25):

Algebraic geometry is a very big subject, but affine schemes are not!

The category of affine schemes is just the opposite of the category of commutative rings, where we take the opposite because a map $X \to Y$ between spaces induces a map sending functions on $Y$ to functions on $X$. This way of talking lets us think of commutative algebras as spaces.

But right now you are working over a field, so you are dealing with commutative algebras over a field $k$... and the opposite of the category of those is called the category of affine schemes over $k$.

Affine schemes over $\mathbb{R}$ were basically invented by Descartes when he realized that algebra could be seen as geometry. E.g., the commutative algebra $\mathbb{R}[x,y]/\langle x^2 + y^2 = 1\rangle$ is the algebra of functions on an affine scheme over $\mathbb{R}$ called the 'circle'.

Jean-Baptiste Vienney (Feb 06 2023 at 05:43):

Evan Patterson said:

In their paper "Differential categories" (p. 1062), Blute et al describe a neat way to construct the category of vector spaces and polynomial maps between them. Let $S: \mathsf{Vect}_k \to \mathsf{Vect}_k$ be the free commutative $k$-algebra monad, which sends a vector space $V$ to the (underlying vector space of) the symmetric algebra $S(V)$. Then the opposite of the Kleisli category of $S$, or equivalently the coKleisli category of the comonad $S^{\mathrm{op}}$, is a category whose morphisms $V \to W$ are polynomial functions from $V$ to $W$. Call this category $\mathsf{Poly}_k$.

The authors go on to say that "it is well known that" $\mathsf{Poly}_k$ is a distributive category. Is this true? In fact, I am confused on a more basic point. Does $\mathsf{Poly}_k$ even have coproducts? If so, what are they?

I think that the coKleisli category of the monad $S:\mathsf{Vect}_{k} \rightarrow \mathsf{Vect}_{k}$ in fact has finite coproducts.

It comes from what is a model of linear logic.

Usually to interpret the connectors $\otimes,\mathbb{K},!,\times,\top$ of linear logic, we require a symmetric monoidal category with products and a comonad with the Seely isomorphisms $!(A \times B) \cong !A \otimes !B$ and $!\top \cong I$, that is $!$ is a strong monoidal functor between the cartesian monoidal structure and the symmetric monoidal structure . This is definition 3.1 on the nLab page $!$-modality.

If we have a Seely comonad, then the coKlesli category of $!$ have products which are still $\times$.

Now, $S:\mathsf{Vect}_{k} \rightarrow \mathsf{Vect}_{k}$ is a monad, we have coproducts in $\mathsf{Vect}_{k}$ and the Seely isomorphism (which are now for coproducts but in fact look exactly the same since $\mathsf{Vect}_{k}$ has biproducts) $S(A \oplus B) \cong S(A) \otimes S(B)$ and $S(0) \cong \mathbb{K}$ are well-known (actually, there are isomorphisms like this everywhere in mathematics). We thus have a Seely monad and the Kleisli category $\mathsf{Poly}_{k}$ of $S$ has coproducts.

These coproducts are given on the objects like in $\mathsf{Vect}_{k}$ by $A \oplus B$ and on morphisms like this:

If you take
$f:A \rightarrow C$
$g:B \rightarrow C$
in the Kleisli category of $S$.

You want a pairing $(f,g): A \oplus B \rightarrow C$

ie. if you take
$f:A \rightarrow S(C)$
$g:B \rightarrow S(C)$
in $\mathsf{Vec}_{k}$.

You want $(f,g):A \oplus B \rightarrow S(C)$ in $\mathsf{Vec}_{k}$.

This is simply given by the pairing in $\mathsf{Vec}_{k}$:
$(f,g):A \oplus B \rightarrow S(C)$.

Now the Seely isomorphism and the monad are useful to build the injections and verify that everything work in order that it really gives you coproducts in $\mathsf{Poly}_{k}$. This is here that the whole mechanism of linear logic with proofs made equivalent by cut elimination and the translation in the category theory world under the form of this categorical semantics plays its role.

Jean-Baptiste Vienney (Feb 06 2023 at 07:14):

Oups I did a mistake

Jean-Baptiste Vienney (Feb 06 2023 at 07:16):

You are looking for coproducts in the opposite category ahah

Jean-Baptiste Vienney (Feb 06 2023 at 07:23):

It looks more complicated indeed.

Jean-Baptiste Vienney (Feb 06 2023 at 07:34):

I'm interested by an answer to your question.

Jean-Baptiste Vienney (Feb 06 2023 at 07:50):

I look dumb but I learned something, I'm very used to that.

Evan Patterson (Feb 06 2023 at 20:59):

Thanks for the comment and no worries, it's easy to get mixed up by the op-ing going on here.

In order to have coproducts in $\mathsf{Poly}_k$, it seems that we would need to be able to find, for any two vector spaces $V_1$ and $V_2$, another vector space $V$ such that $S(V) \cong S(V_1) \oplus S(V_2)$. I don't see how to do that and I'm doubtful that it's possible.

John Baez (Feb 06 2023 at 22:22):

Okay, good: my remarks on affine schemes give a geometrical intuition for why this is not possible, and maybe also point the way to a quick proof:

In a polynomial algebra $S(V)$ we just have two elements $p$ that are idempotent (have $p^2 = p$), namely 0 and 1. But in $S(V_1) \oplus S(V_2)$ we have four, namely $(0,0), (0,1), (1,0)$ and $(1,1)$.

John Baez (Feb 06 2023 at 22:24):

This is the algebraic way of saying "the affine scheme $V$ is connected, but the coproduct of two copies of this affine scheme is not".

Evan Patterson (Feb 06 2023 at 22:49):

OK, nice! I think that settles it. Thanks John!