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Stream: learning: questions

Topic: two partial orders at once

David Egolf (Feb 05 2022 at 03:38):

Say I have a set $X$, and I want to consider two partial orders on it at once. What kind of category would model this situation?

David Egolf (Feb 05 2022 at 03:39):

(what I have in mind: a topological space where we order open sets by inclusion, and where we also have the image of some net mapping to the open sets which allows us to say some open sets are "after" others)

David Egolf (Feb 05 2022 at 03:46):

It seems like a situation where we have different kinds of morphisms - the "is a subset" kind, and the "comes after" kind. I don't think we want to be able to compose morphisms of different kinds (?).

Mike Shulman (Feb 05 2022 at 04:03):

Any chance it is a double category?

David Egolf (Feb 05 2022 at 04:19):

Unfortunately, I don't know what a double category is (and all the definitions I can find go over my head currently). However, if a double category does model this kind of setting - with multiple kinds of relationships present at once - that's really interesting and maybe a reason to work on understanding what a double category is!

David Egolf (Feb 05 2022 at 04:20):

The presence of "vertical" and "horizontal" morphisms does make it sounds like a double category allows us to talk about different kinds of morphisms at once.

John Baez (Feb 05 2022 at 06:22):

Yes, it does.

A preorder is the same as a category where for any pair of objects $x,y$ there's at most one morphism from $x$ to $y$ . If there's a morphism from $x$ to $y$ we write $x \le y$.

We can build a double category from a pair of preorders on the same set, say $\le$ and $\triangleleft$.

John Baez (Feb 05 2022 at 06:26):

We say there's a single vertical morphism from $x$ to $y$ iff $x \le y$, and there's a single horizontal morphism from $x$ to $y$ iff $x \triangleleft y$.

David Egolf (Feb 05 2022 at 06:45):

We also have "two-morphisms" in a double category, which in this case seem to describe a sort of interaction between the two preorders (?). I'm not sure why we really need that, but presumably it's important.

John Baez (Feb 05 2022 at 06:52):

In general 2-morphisms are the most interesting part of a double category, just as morphisms are the most interesting part of a category. But we can make up examples where they are not so interesting... like the one we're talking about now.

If I have two perorders on the same set, we can make up a double category where the vertical and horizontal morphisms are as above, and we have a 2-morphism from the horizontal arrow $x \triangleleft y$ to the horizontal arrow $x' \triangleleft y'$ whenever we have a vertical arrow $x \le x'$ and a vertical arrow $y \le y'$.

John Baez (Feb 05 2022 at 06:53):

(I would draw this as a square if I had the energy.)

David Egolf (Feb 05 2022 at 17:28):

2-morphism

David Egolf (Feb 05 2022 at 17:30):

I remember the idea that the composition of morphisms characterizes the objects.
Maybe the composition of 2-morphisms characterizes the morphisms (and maybe, consequently, also the objects)?
That would explain why 2-morphisms could be the most interesting part of a double category.

David Egolf (Feb 05 2022 at 17:38):

I guess a 2-morphism relates four 1-morphisms. If we take the philosophy that the meaningful attributes of things corresponds to how they compare to other things, a 2-morphism can be viewed as saying something about the 1-morphisms.

David Egolf (Feb 05 2022 at 17:41):

As a perhaps related thought.... I was thinking about sequences of real numbers, viewed as maps $f: \mathbb{N} \to \mathbb{R}$ from a preorder. I know we care a lot about sequences, but thinking of a sequence in this way just makes it sound like a "redecorating" of $\mathbb{N}$, corresponding to a preorder isomorphic to the one we started with. But maybe we care about these sequences because of how they relate to a standard ordering (or topology?) already present on $\mathbb{R}$. In particular, we maybe care about how the preorder induced by a sequence compares to other structure on the real numbers we wish to consider simultaneously.

Fawzi Hreiki (Feb 05 2022 at 17:47):

Yes. The 2-morphisms totally capture the 1 and 0 morphisms. This is basically by induction.

David Egolf (Feb 05 2022 at 18:36):

Fawzi Hreiki said:

Yes. The 2-morphisms totally capture the 1 and 0 morphisms. This is basically by induction.

Cool! It makes me wonder if sometimes it is handy to do all one's reasoning in terms of 2-morphisms, as a result. Maybe this can sometimes allow for a clean restating of arguments involving various horizontal and vertical morphisms.

John Baez (Feb 05 2022 at 23:59):

I've proved a lot of theorems and done a lot of calculations involving double categories - they get used a lot in my theory of open networks - and I find it useful to talk explicitly about objects, vertical and horizontal 1-morphisms, and 2-morphisms.