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Stream: learning: questions

Topic: Matrix Product From Monoidal Grothendieck

Jade Master (Oct 31 2020 at 18:23):

Does anyone recognize this matrix product? Let R be a ring equipped with a nice poset structure. You can construct a functor
$F : \mathsf{Set}^{op} \to \mathsf{MonCat}$
by:

• sending a set $X$ to the poset of matrices $\mathsf{Mat_R}(X)$ of matrices $m : X \times X \to R$,
• and sending a function $f: X \to Y$ to the monotone map
  $\mathsf{Mat_R}(Y) \to \mathsf{Mat_R}(X)$ given by
  $Y \times Y \to R$ going to $X \times X \xrightarrow{f \times f} Y \times Y \to R$.
  Moreover, each of these posets $\mathsf{Mat_R}(X)$ are monoidal under matrix product and $F$ is monoidal with respect to cartesian product of sets. Therefore you can apply the fibrewise monoidal Grothendieck construction to get a monoidal product

$\oplus : \int F \times \int F \to \int F.$

I am wondering about what this monoidal product actually is. I think I have it worked out in terms of Kronecker product $\otimes$. If $(X, m: X \times X \to R)$ and $(Y,n: Y \times Y \to R)$ are objects in $\int F$, then
$(X,m) \oplus (Y,n) = (X \times Y, (m \otimes 1(Y) ) * (n \otimes 1(X)))$
where $1(X)$ and $1(Y)$ are the square matrices on $X$ and $Y$ containing only 1's and $*$ is the ordinary matrix product.

Jade Master (Oct 31 2020 at 18:24):

One thing to note is that m and n are on the same set then
$(X,m: X \times X \to R) \oplus (X, n, X\times X \to R) = (X \times X, (m \otimes 1(X)) * (n \otimes 1(X))) = m*n \otimes 1(X) * 1(X)$
because the Kronecker product satisfies the interchange law with respect to matrix multiplication. So in this sense it starts to look like more like generalizing ordinary matrix multiplication.

Jade Master (Oct 31 2020 at 18:29):

By the way, you can find a description of the fibrewise monoidal Grothendieck construction in @Mike Shulman 's

• Framed Bicategories and Monoidal Fibrations
  Check out the proof of Theorem 12.8 halfway through page 47.

Morgan Rogers (he/him) (Nov 02 2020 at 09:50):

In the first bullet point, are you saying that the monoidal product on your poset of matrices is given by matrix multiplication? Because if so, you might have a problem with how multiplication interacts with the order :grimacing:

Morgan Rogers (he/him) (Nov 02 2020 at 09:53):

For example, I would consider $\mathbb{R}$ to be a pretty standard example of a "ring equipped with a nice poset structure", but taking $X$ to be a one-element set so that $\mathsf{Mat}_{\mathbb{R}}(X)$ is just $\mathbb{R}$ with multiplication, I have $-2 \leq -1$ and $1 \leq 3$ but $-2 \not\leq -3$, so that I can't tensor the morphisms in the poset.

Jade Master (Nov 02 2020 at 15:53):

[Mod] Morgan Rogers said:

For example, I would consider $\mathbb{R}$ to be a pretty standard example of a "ring equipped with a nice poset structure", but taking $X$ to be a one-element set so that $\mathsf{Mat}_{\mathbb{R}}(X)$ is just $\mathbb{R}$ with multiplication, I have $-2 \leq -1$ and $1 \leq 3$ but $-2 \not\leq -3$, so that I can't tensor the morphisms in the poset.

Right I see the issue. I think that I actually need R to be a rig...something which only has an additive commutative monoid rather than an additive abelian group. Probably I will end up needing R to be a quantale in the end...

Morgan Rogers (he/him) (Nov 02 2020 at 16:13):

That seems more sensible; removing negatives seems to be necessary and sufficient to fix the problem I was pointing to :+1: