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Stream: learning: questions

Topic: Monoidal category question

John Baez (Jul 24 2024 at 22:52):

@Jean-Baptiste Vienney raised a fun question which I posted to MathOverflow. Check it out:

• Are there noncartesian monoidal categories with A⊗B=A×B?

John Baez (Jul 24 2024 at 22:54):

I think I've answered it: yes. Let me test out the answer on you folks.

First think about the category of affine spaces over some field. This famously has both a cartesian monoidal structure $\times$ and a semicartesian monoidal structure $\otimes$: the latter is explained here.

John Baez (Jul 24 2024 at 22:57):

Let's look at the full subcategory $\mathsf{C}$ of affine spaces that are either terminal (the one-element affine space) or of countably infinite dimension, by which I mean they're a coproduct of a countably infinite number of copies of the terminal affine space. I claim that for all objects $A, B \in \mathsf{C}$ we have

$A \otimes B \cong A \times B$

John Baez (Jul 24 2024 at 22:58):

I also claim $\mathsf{C}$ is a monoidal subcategory with respect to either $\times$ or $\otimes$: for example, the cartesian or tensor product of two affine spaces of countably infinite dimension again has countably infinite dimension.

John Baez (Jul 24 2024 at 23:01):

Now take a skeleton $\mathsf{S}$ of $\mathsf{C}$. We can transfer both the $\times$ and $\otimes$ monoidal structures from $\mathsf{C}$ to this skeleton, and since we had

$A \otimes B \cong A \times B$ in $\mathsf{C}$

we have

$A \otimes B = A \times B$ in $\mathsf{S}$

Concretely, $\mathsf{C}$ has just two objects, the terminal object $1$ and the affine space of countably infinite dimension, $x$. These have

$1 \times x = x \times 1 = x, x \times x = x$

and also

$1 \otimes x = x \otimes 1 = x, x \otimes x = x$

John Baez (Jul 24 2024 at 23:05):

To finish the proof we just need to check that $(S, \otimes, 1)$ is not cartesian monoidal. I need to check that more carefully. However, it would seem amazing if the non-cartesian tensor product $\otimes$ of affine spaces were cartesian on affine spaces of countably infinite dimension!

Jean-Baptiste Vienney (Jul 24 2024 at 23:53):

I don’t understand from the nLab what’s the tensor product of two affine spaces? And of two affine maps?

Jean-Baptiste Vienney (Jul 25 2024 at 00:09):

Ok, now I see that I must learn what is the tensor product of two algebras for a commutative algebraic theory and this will just be an instance of this: really nice.

John Baez (Jul 25 2024 at 06:32):

The nLab link I gave describes the tensor product of two finite-dimensional affine spaces quite explicitly, and the same basic idea works in general. We get this description using the fact that

1) every affine space is free on some set $X$, so call it $F(X)$
2) $F(X) \otimes F(Y) \cong F(X \times Y)$

This should remind you immensely of how vector spaces work. But you have to adjust your intuitions a bit, remembering that in an affine space we can only form affine combinations, not more general linear combinations, and in particular an affine space doesn't have an element called $0$.

Thus, the free vector space on $\emptyset$ has one point, while the free affine space on $\emptyset$ is empty. Similarly the free vector space on $1$ is $k$ (your chosen field), while the free affine space on $1$ is a one-element set. The free vector space on $2$ is $k^2$, while the free affine space on $2$ is a line containing the two points of $2$. The free vector space on $3$ is $k^3$, while the free affine space on $3$ is a plane containing the three points of $3$, which form the vertices of a triangle. And so on.

John Baez (Jul 25 2024 at 08:25):

Using this idea (which extends to the free affine space on an infinite set) and 2), we can understand a lot about the tensor product of affine spaces.

Martti Karvonen (Jul 25 2024 at 08:38):

Is there a reason you start from affine spaces? It seems to me that you could've done the same construction on vector spaces and still get a seemingly plausible example of what you're after.

John Baez (Jul 25 2024 at 11:18):

Affine spaces are known to have a semicartesian tensor product $\otimes$ that's not cartesian, so the unit object for $\otimes$ is the terminal object, so it's same as the unit object for $\times$.

Vector spaces don't have any semicartesian tensor product that I know of: in particular, the unit object for the usual $\otimes$ of vector spaces is not the terminal vector space.

John Baez (Jul 25 2024 at 11:27):

So, my plan was this: start with the category of affine spaces with its two tensor products $\otimes$ and $\times$. They share the same unit object. Next, try to make them equal on all objects. The two different tensor products of finite-dimensional affine spaces aren't isomorphic since they have different dimension. But:

• if $A$ and $B$ have countably infinite dimension then $A \times B \cong A \otimes B$ since they have the same dimension.

• if $1$ is the terminal affine space we have $1 \times A \cong 1 \otimes A$ since $\otimes$ is semicartesian.

• also $1 \times 1 \cong 1 \otimes 1$

John Baez (Jul 25 2024 at 11:29):

So let's consider the full subcategory $\mathsf{C}$ of the category of affine spaces consisting only of terminal and countably-infinite-dimensional affine spaces! For any objects $A, B \in \mathsf{C}$ we have

$A \times B \cong A \otimes B$

Further, $\mathsf{C}$ is a monoidal subcategory of $\mathrm{Aff}$ under either $\otimes$ or $\times$, since the product or tensor product of two affine spaces of countably infinite dimension again has countably infinite dimension.

(This is easy to see in terms of bases. To get a basis of $A \times B$ we take the disjoint union of a basis of $A$ and a basis of $B$. To get a basis of $A \otimes B$ we take the cartesian product of a basis of $A$ and a basis of $B$.)

Now let $\mathsf{S}$ be a skeleton of $\mathsf{C}$. Now isomorphic objects become equal, so we get

$A \times B = A \otimes B$

for all objects of $\mathsf{S}$.

Martti Karvonen (Jul 25 2024 at 13:02):

John Baez said:

Affine spaces are known to have a semicartesian tensor product $\otimes$ that's not cartesian, so the unit object for $\otimes$ is the terminal object, so it's same as the unit object for $\times$.

Vector spaces don't have any semicartesian tensor product that I know of: in particular, the unit object for the usual $\otimes$ of vector spaces is not the terminal vector space.

Oh yeah of course, I was just thinking of the two monoidal products coinciding in the countable case but forgot about being semicartesian.