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Stream: learning: questions

Topic: Mimicking categorical products in enriched categories

Peva Blanchard (Nov 20 2024 at 22:13):

I've continued to play with enriched categories, and I stumbled upon the following issue.

In usual category theory ($\text{Set}$-enriched), the categorical product $a \times b$ can be defined by asking that the presheaf ${\cal C}(\_,a) \times {\cal C}(\_, b)$ be representable.

Let's try to do the same when ${\cal C}$ is a ${\cal V}$-enriched category, with $({\cal V}, \otimes, I)$ a closed monoidal category.
I.e., I'm defining $a \otimes b$ as a representative of the enriched presheaf ${\cal C}(\_, a) \otimes {\cal C}(\_, b)$

$${\cal C}(z, a \otimes b) \cong {\cal C}(z, a) \otimes {\cal C}(z,b)$$

Of course, it may not exist.

I have two questions:

1. How does this relate to weighted (co)limits? (I'm aware of the definition of weighted (co)limits)
2. Is it possible to "freely $\otimes$-complete" the category ${\cal C}$?

Mike Shulman (Nov 20 2024 at 22:20):

If $\mathcal{V}$ isn't cartesian monoidal, then $\mathcal{C}(-,a) \otimes \mathcal{C}(-,b)$ may not be an enriched presheaf: to define the functorial action, you need diagonals on hom-objects.

Peva Blanchard (Nov 20 2024 at 22:38):

Oh, I see, I fell in the trap of not checking functoriality.

I'll spell it out to make it more clear. For $F = {\cal C}(\_, a) \otimes {\cal C}(\_, b)$ to be functorial, we need, for every $x,y$, an arrow

$${\cal C}(x,y) \to [Fy, Fx]$$

where $[\_,\_]$ is the internal hom in ${\cal V}$. And these arrows should be compatible with composition and identities.

Using the tensor hom adjunction, this amounts to define an arrow

$${\cal C}(x,y) \otimes {\cal C}(y,a) \otimes {\cal C}(y,b) \to {\cal C}(x,a) \otimes {\cal C}(x,b)$$

If we had diagonals

$${\cal C}(x,y) \to {\cal C}(x,y) \otimes {\cal C}(x,y)$$

and if $\otimes$ were symmetric, we could build the above arrow via

$$\begin{align*} &{\cal C}(x,y) \otimes {\cal C}(y,a) \otimes {\cal C}(y,b) \\ &\to {\cal C}(x,y) \otimes {\cal C}(x,y) \otimes {\cal C}(y,a) \otimes {\cal C}(y,b) \\ &\to ({\cal C}(x,y) \otimes {\cal C}(y,a)) \otimes ({\cal C}(x,y) \otimes {\cal C}(y,b)) \\ &\to {\cal C}(x,a) \otimes {\cal C}(x,b) \end{align*}$$

But indeed, without the diagonals and the symmetry I can't do that.

Peva Blanchard (Nov 21 2024 at 00:30):

Actually, my original context was a bit more involved, and my question above tried to take a short-cut (in vain).
Fortunately, I think my original context doesn't suffer from the flaw above: I do get a functor it seems.
Here it is.

Category $\text{Cost}$

First, I use Lawvere-like category $\text{Cost} = ((-\infty,\infty], \ge, +, 0)$.
Following John's blog post, I also define another tensor product

$$x \wedge y = - \frac{1}{\beta}\ln \Big(e^{-\beta x} + e^{-\beta y}\Big)$$

where $\beta > 0$ is some constant.

It turns out $+$ distributes over $\wedge$

$$x + (y \wedge z) = (x + y) \wedge (x + z)$$

Category $\text{ESet}$ of energetic sets

This is a slight variant (I think) of John's energetic sets from a previous thread.

• objects are sets $X$ with an energy function $E : X \to \text{Cost}$
• morphisms are functions $f : X \to Y$ such that $E(x) \ge E(f(x))$ (energy-decreasing)

Tensor product $\otimes$

I define a first symmetric tensor product

$$X \otimes Y \triangleq (X \times Y, (x,y) \mapsto E(x) + E(y))$$

with the unit $I = (\{\bullet\}, \bullet \mapsto 0)$.
I write $x \otimes y$ the pair $(x,y)$ considered as an element of $X \otimes Y$.

This tensor product is closed.
The internal hom $[Y,Z]$ is the set of functions $f : Y \to Z$ with energy

$$E(f) = \sup_y ~ (E(f(y)) - E(y))$$

Tensor product $\wedge$

I define a second symmetric tensor product

$$X \wedge Y \triangleq (X \times Y, (x,y) \mapsto E(x) \wedge E(y))$$

with the unit $\infty = ({\bullet}, \bullet \mapsto \infty)$.
I write $x \wedge y$ the pair $(x,y)$ considered as an element of $X \wedge Y$.

It turns out the $\wedge$-diagonals $X \to X \wedge X$ exist. It maps $x \mapsto x \wedge x$.
The energy inequality is satisfied

$$\begin{align*} E(x \wedge x) &= E(x) \wedge E(x) \\ &= E(x) - \frac{\ln 2}{\beta} \\ &\le E(x) \end{align*}$$

Almost distributive

Because both $\otimes$ and $\wedge$ have as underlying set the cartesian product of the respective sets, $\otimes$ does not distribute over $\wedge$.

But, thanks to the diagonal, we have a morphism

$$X \otimes (Y \wedge Z) \to (X \otimes Y) \wedge (X \otimes Z)$$

with energy equality $E(x \otimes (y \wedge z)) = E((x \otimes y) \wedge (x \otimes z))$.

Finally

I consider $\text{ESet}$ with the monoidal structure $(\otimes, I)$ as an enriching category.
Let ${\cal C}$ be a $\text{ESet}$-enriched category.
Roughly speaking, ${\cal C}$ is like a usual locally small category, but each arrow has "an energy cost", and the costs add up when composition.

Now, I can use the second tensor product to define

$$F = {\cal C}(z,a) \wedge {\cal C}(z,b)$$

Thanks to the properties above, I think $F$ is indeed a $\text{ESet}$-enriched presheaf.

And I was wondering about the representability of $F$,

$${\cal C}(z, a \wedge b) \cong {\cal C}(z,a) \wedge {\cal C}(z,b)$$

I have to say, I'm having a good share of fun in trying to interpret this weird operation $\wedge$, but I'll stop here for now.

Morgan Rogers (he/him) (Nov 23 2024 at 11:58):

That's a lot to post in one go!

Peva Blanchard (Nov 23 2024 at 15:04):

Yes sorry about that. It was late in the night when I replied, and, as the joke says, "I didn't have time to make it shorter" :upside_down:

Here is a shorter reformulation.

Let $({\cal V}, \otimes, I)$ be a closed symmetric monoidal category, and ${\cal C}$ a ${\cal V}$-enriched category.
Assume there is another symmetric monoidal structure $({\cal V}, \wedge, \infty)$ on ${\cal V}$ which has diagonals $X \to X \wedge X$, and such that we have "distributivity natural morphisms" (not iso)

$$X \otimes (Y \wedge Z) \to (X \otimes Y) \wedge (X \otimes Z)$$

Then,

$$F = {\cal C}(\_,a) \wedge {\cal C}(\_, b)$$

defines (I think) an enriched presheaf. If $F$ were representable, there would exist some object $a \wedge b$ such that ${\cal C}(\_, a \wedge b) \cong {\cal C}(\_, a) \wedge {\cal C}(\_, b)$.

My original questions were:

• How does this "limit object" $a \wedge b$ relate to weighted (co)limits?
• Is it possible to "freely $\wedge$-complete" ${\cal C}$ so that this functor becomes representable?

Mike Shulman (Nov 23 2024 at 15:55):

I don't think this object is going to be a weighted limit of any sort.

Mike Shulman (Nov 23 2024 at 15:56):

To freely complete, you could probably do some iterative construction.