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Stream: learning: questions

Topic: Relationship between spans and bimodules

Eric Forgy (Jan 14 2021 at 00:28):

Hello. I've been going in circles trying to understand this and can use some help :pray:

Although scary to admit (because the general topic is over my head), I'm trying to study bicategories $\mathsf{Span(C)}$ and $\mathsf{Bim}.$ Not in their full generality, but a special case.

In particular, when I look at a span $E: V\to V'$ and a span $E': V'\to V''$, their composite is a span $E\times_{V'} E':V\to V''$.

Similarly, when I look at a bimodule $\Omega:A\to A'$ and a bimodule $\Omega':A'\to A''$, their composite is a bimodule $\Omega\otimes_{A'}\Omega':A\to A''.$

The similarities between the two notations cannot be a coincidence. I'm trying to understand this.

The notation $\Omega\otimes_{A'}\Omega'$ makes me want to think of bimodules as spans (or cospans), where $\Omega\otimes_{A'}\Omega'$ is a pullback (or push out) somehow.

More specifically, I'm thinking of endomorphisms $E:V\to V$ in $\mathsf{Span(Set)}$, a.k.a. directed graphs, and endomorphisms $\Omega: A\to A$ in $\mathsf{Bim}.$

An endomorphism span $E:V\to V$ composes with itself $n$ times giving the span $E^{\times_V n}:V\to V.$

An endomorphism bimodule $\Omega:A\to A$ composes with itself $n$ times giving the bimodule $\Omega^{\otimes_A n}:A\to A.$

Given a set $S$, there two ways to obtain algebras:
${}$

1. $K[S],$ i.e. the formal linear combinations of elements of $S$.
2. $K^S,$ i.e. functions from $S$ to $K$ with product defined pointwise.

${}$
In this comment, John nicely (even I understood :sweat_smile:) explained that with $K^S$, we have a functor
${}$
$\mathsf{Span(Set)}\to\mathsf{Cospan(Alg)}$
${}$
so we are getting cospans instead of spans. In a prior comment, he also showed how to construct a functor
${}$
$\mathsf{Bim\to Cospan(Alg)}.$
${}$
(Note: Wow. Writing all this down actually gets me some way toward understanding what I'm actually trying to do :sweat_smile: )
Now, if my CT kung fu were a little better, I should be able to make the last connection and relate
${}$
$\mathsf{Span(Set)\leftrightarrows Bim}.$
${}$
My gut tells me there should be some forgetful functor
${}$
$\mathsf{Bim\to Span(Set)}.$
${}$
If so, there should be functors (free to the right and forgetful to the left)
${}$
$\mathsf{Span(Set)\leftrightarrows Cospan(Alg)\leftrightarrows Bim}.$
${}$
Unfortunately, this is a little beyond what I'm capable of figuring out on my own at the moment and I need this step to make progress on an applied / engineering paper I'm writing.

Any ideas? :pray:

PS: I came across this (unpublished) note from Urs:

• 2-Hilbert Spaces and Bimodules from Spans

It looks helpful, but it ends abrutly just when it was getting interesting :sweat_smile:

Fawzi Hreiki (Jan 14 2021 at 00:40):

Just for clarity, what exactly is the category $\text{Bim}$?

Eric Forgy (Jan 14 2021 at 00:46):

Hi Fawzi :wave:

$\mathsf{Bim}$ is a bicategory with

• Algebras (with 1)
• Bimodules
• Bimodule homomorphisms

Cole Comfort (Jan 14 2021 at 00:49):

This does not answer your question but (small) profunctors are bimodules in spans of sets

Eric Forgy (Jan 14 2021 at 00:53):

Cole Comfort said:

This does not answer your question but (small) profunctors are bimodules in spans of sets

Yeah. I'm trying to read up on profunctors, but it is pushing the limits of my poor brain :sweat_smile:

I also found this on the nLab:

Let $V=\mathsf{Set}$ and let $C=D$. Then the hom functor $C(-,-):C^{op}\times C\to\mathsf{Set}$ is a bimodule. Bimodules can be thought of as a kind of generalized hom, giving a set of morphisms (or object of $V$) between an object of $C$ and an object of $D$.

Eric Forgy (Jan 14 2021 at 00:56):

Cole Comfort said:

This does not answer your question but (small) profunctors are bimodules in spans of sets

If I could understand this statement, it would certainly help. Care to elaborate a little? :pray:

Eric Forgy (Jan 14 2021 at 01:18):

A found this fact to be pretty cool...

A category internal to some category $C$ is the same as a monad in $\mathsf{Span(C)}$.

So anywhere I see "category internal to $C$", I can think "monad in $\mathsf{Span(C)}$"

Eric Forgy (Jan 14 2021 at 01:26):

From the nLab:

Here is the slick definition: let $S$ be a category with pullbacks. Then the bicategory $\mathsf{Prof}(S)$ of internal categories, profunctors and transformations in $S$ is defined to be $\mathsf{Mod(Span}(S))$, the category of monads and bimodules in $\mathsf{Span}(S)$, the bicategory of spans in $S$.

So an internal profunctor $C⇸D$ between internal categories $C$ and $D$ is a bimodule from $C$ to $D$. An internal presheaf on $C$ is a right $C$-module, or equivalently a bimodule $1⇸C$, where $1$ is the discrete category on the terminal object of $S$ (as long as $S$ has one, of course).

Fawzi Hreiki (Jan 14 2021 at 01:27):

Regarding profunctors, I'll just point you to this.

Fawzi Hreiki (Jan 14 2021 at 01:28):

The basic idea is that, the same way 2 plays the role of truth values on Set, Set plays the role of truth values in CAT.

John Baez (Jan 14 2021 at 01:31):

Eric Forgy said:

Hello. I've been going in circles trying to understand this and can use some help :pray:

Although scary to admit (because the general topic is over my head), I'm trying to study bicategories $\mathsf{Span(C)}$ and $\mathsf{Bim}.$ Not in their full generality, but a special case.

In particular, when I look at a span $E: V\to V'$ and a span $E': V'\to V''$, their composite is a span $E\times_{V'} E':V\to V''$.

Similarly, when I look at a bimodule $\Omega:A\to A'$ and a bimodule $\Omega':A'\to A''$, their composite is a bimodule $\Omega\otimes_{A'}\Omega':A\to A''.$

The similarities between the two notations cannot be a coincidence. I'm trying to understand this.

The notation $\Omega\otimes_{A'}\Omega'$ makes me want to think of bimodules as spans (or cospans)...

I believe you're stuck because bimodules are not a special case of spans. Spans (of sets, for example) are a special case of bimodules.

I explained it here:

• John Baez, Spans versus bimodules, August 4, 2008.

When you understand this you'll be happy. I tried to explain it here on Zulip a while back, but you probably weren't desperate enough back then.

Fawzi Hreiki (Jan 14 2021 at 01:33):

At the same time though, bimodules of sets (aka binary relations) are a special case of spans of sets.

John Baez (Jan 14 2021 at 01:34):

Anyway, Eric needs to understand what I wrote. I was quite excited when I wrote it:

I always thought bimodules and spans should be related, but only recently did I learn exactly how, thanks to Paul-André Melliès.

John Baez (Jan 14 2021 at 01:35):

So I was exactly where Eric is now, until Paul-André explained what's going on here.

Eric Forgy (Jan 15 2021 at 20:02):

Hi :wave:

I'm coming back to this stream after putting a lot of elbow grease into:

Bimodules vs Spans

• Puzzle 1
• Puzzle 2
• Puzzle 3

There, we learned that in $(\mathsf{Set^{op}},\times,*)$:

• A monoid object $A$ is a set with multiplication given by the diagonal map.
• An $A$-module $X$ is a function $X\to A.$
• An $(A,B)$-module $X$ is a span $A\leftarrow X\rightarrow B.$
• Tensoring an $(A,B)$-bimodule $X$ with a $(B,C)$-bimodule $Y$ is an $(A,C)$-bimodule $X\times_B Y$ corresponding to the composition of the respective spans

John Baez (Jan 15 2021 at 20:59):

Yes, that's a nice summary!

So: spans are a special case of the general theory of bimodules.

Then you can have some fun as follows:

Given monoidal categories $C$ and $D$, and a monoidal functor $F: C \to D$, $F$ maps monoid objects in $C$ to monoid objects in $D$ in an obvious way, and bimodules to bimodules in an obvious way, preserving composition up to isomorphism. (Details left as an exercise.)

So, we can use a monoidal functor

$F: (\mathsf{Set}^{\rm op}, \times) \to (\mathsf{Vect}, \otimes)$

to turn monoid objects in $(\mathsf{Set}^{\rm op}, \times)$, which are just sets, to monoid objects in $(\mathsf{Vect}, \otimes)$, which are just algebras, and turn bimodules in $(\mathsf{Set}^{\rm op}, \times)$, which are just spans of sets, to monoid objects in $(\mathsf{Vect}, \otimes)$, which are just bimodules, in a manner preserving composition.

I'm a little confused here, but I think it works better to use finite sets, since I think I know a very nice monoidal functor

$(\mathsf{FinSet}^{\rm op}, \times) \to (\mathsf{Vect}, \otimes)$

which does not extend to infinite sets. This monoidal functor sends any finite set $X$ to $k^X$, where $k$ is the field we're using to define $\mathsf{Vect}$, and it sends a function $f: X \to Y$ to the pullback linear map $f^\ast: k^Y \to k^X$. Note this contravariance is what we need, to deal with the "op".

John Baez (Jan 15 2021 at 21:01):

I believe this is a very nice way to turn spans of finite sets into bimodules, in such a way that composition is preserved up to isomorphism. I think you want to use this in your work.

John Baez (Jan 15 2021 at 21:04):

If you need to use infinite sets I think things get a bit tricky.

Eric Forgy (Jan 15 2021 at 21:20):

Cool! That is so awesome! :smiley:

Eric Forgy (Jan 15 2021 at 21:28):

In terms of size, I'd like to be able to work with denumerable sets like $\mathbb{Z}^n.$ Is that possible?

Eric Forgy (Jan 15 2021 at 21:37):

I think your (lax) functor should be fine with sets indexed by $\mathbb{Z}.$

Eric Forgy (Jan 15 2021 at 21:41):

An element $f\in k^\mathbb{Z}$ can be written as an infinite sum:
${}$
$\begin{aligned}f = \sum_{i\in\mathbb{Z}} f(i) e^i.\end{aligned}$
${}$
I think that is nicely enough behaved so that we can write
${}$
$\begin{aligned}f\otimes_k g = \sum_{i,j\in\mathbb{Z}} f(i)\, g(j)\, e^i\otimes_k e^j\end{aligned}$
${}$
right? I hope so :sweat_smile: :pray:

Eric Forgy (Jan 15 2021 at 21:42):

If not, I might need some kind of "finite support" condition or something.

John Baez (Jan 15 2021 at 21:52):

I said finite and meant finite; countably infinite doesn't work the same.

Eric Forgy (Jan 15 2021 at 21:57):

Ok. Then I think I might need to get into some of the tricky stuff :thinking:

John Baez (Jan 15 2021 at 22:00):

The point is this: for finite sets we have

$k^{X \times Y} \cong k^X \otimes k^Y$

but if both $X$ and $Y$ are infinite this is false.

Thus, one can show the contravariant functor sending a finite set $X$ to $k^X$ and a function $f: X \to Y$ to $f^\ast : k^Y \to k^X$ is monoidal, giving a monoidal functor

$(\mathsf{FinSet}^{\text{op}}, \times) \to (\mathsf{Vect}, \otimes)$

but this procedure does not give a monoidal functor

$(\mathsf{Set}^{\text{op}}, \times) \to (\mathsf{Vect}, \otimes)$

Various attempted workarounds seem to have various problems, which I could explain.

Eric Forgy (Jan 15 2021 at 22:05):

You've explained to me a couple times how there are two ways to get a vector space from a set, i.e. $k^S$ and $k[S]$. Does the other way, i.e. $k[S]$ suffer the same restriction to finite sets? In that case, we want $k[X\times Y] \cong k[X]\otimes k[Y].$

John Baez (Jan 15 2021 at 22:07):

There is a covariant functor sending any set $X$ to $k[X]$, the free vector space on $X$ and any function $f: X \to Y$ to a certain obvious linear map $f_\ast: k[X] \to k[Y]$. This construction obeys

$k[X \times Y] \cong k[X] \otimes k[Y]$

for any pair of sets $X$ or $Y$, so we get a monoidal functor

$(\mathsf{Set}, \times) \to (\mathsf{Vect}, \otimes)$

but note: the "op" is not there now! So this sends monoids in $(\mathsf{Set}, \times)$ - that is, ordinary monoids - to algebras.

We can also think of the above functor as a monoidal functor

$(\mathsf{Set}^{\text{op}}, \times) \to (\mathsf{Vect}^{\text{op}}, \otimes)$

This sends sets (monoid objects in $(\mathsf{Set}^{\text{op}}, \times)$) to coalgebras (monoid objects in $(\mathsf{Vect}^{\text{op}}, \otimes)$).

Eric Forgy (Jan 15 2021 at 22:08):

I see :thinking: Thank you :pray:

John Baez (Jan 15 2021 at 22:11):

There are also other tricks one can play, but I'm getting tired and none of them instantly stand out as a panacea.

Eric Forgy (Jan 15 2021 at 22:11):

Sure. Understandable. Thank you. This is already a huge help :pray:

John Baez (Jan 15 2021 at 22:13):

I will say this: a lot of times when people naively think they need algebras, it turns out coalgebras would work. Finite-dimensional algebras are practically the same as finite-dimensional coalgebras, since the dual of the former is the latter and vice versa, and dualizing twice gets you back where you started. That fails in the infinite-dimensional case. So the need for coalgebras becomes more visible in the infinite-dimensional case, which is exactly the problem we're running up against now.

Eric Forgy (Jan 15 2021 at 22:17):

I'm thinking that starting with sets (not restricted to finite sets), we can obtain vectors / coalgebras following the second approach with $k[S]$ and then I can look at the dual space of that. This is analogous to the vectors / coalgebras being the "space" and then I look at "differential forms" on those spaces. I know that is vague and imprecise, but I'm just trying to convey a vague idea.

John Baez (Jan 15 2021 at 22:18):

Yeah, that was one of the "other tricks one can play". But you have to play a few moves of that game before you see if you win or get check-mated.

Eric Forgy (Jan 15 2021 at 22:18):

Roughly, I'm thinking along the lines of sets -> chains -> cochains.

John Baez (Jan 15 2021 at 22:18):

Those are good thoughts to have. Ultimately it should all make loads of sense.

Eric Forgy (Jan 15 2021 at 22:19):

John Baez said:

Yeah, that was one of the "other tricks one can play". But you have to play a few moves of that game before you see if you win or get check-mated.

Yeah. I can imagine. The dual space will need some niceness properties (I've seen the word "Fredholm" that seems relevant).

John Baez (Jan 15 2021 at 22:19):

No, "Fredholm" is not relevant, that's more about analysis, like Hilbert spaces.

John Baez (Jan 15 2021 at 22:19):

This is just algebra here.

Eric Forgy (Jan 15 2021 at 22:20):

Well, I do want Hilbert spaces out of this at the end of the day.

John Baez (Jan 15 2021 at 22:20):

Ugh, well, that's a whole other level of math.

Eric Forgy (Jan 15 2021 at 22:20):

Yeah. One step at a time :sweat_smile:

John Baez (Jan 15 2021 at 22:21):

The good news is, nothing in math fails that deserves to work. And nothing in math works that deserves to fail.

Eric Forgy (Jan 15 2021 at 22:22):

That is why I liked working with the 1-category $\mathsf{Span(Set)}$. It is a dagger category, which makes the road ahead to Hilbert spaces more clear.

Eric Forgy (Jan 15 2021 at 22:23):

I think most of the "Bimodules vs Span" discussion can be converted to a discussion about $\mathsf{Span(Set)}$ (or maybe $\mathsf{Span(Set^{op})}$ or something).

Eric Forgy (Jan 15 2021 at 22:24):

A bimodule is a morphism in $\mathsf{Span(Set^{op})}$ (guessing :sweat_smile: )

John Baez (Jan 15 2021 at 22:27):

Why guess stuff? I thought we just showed that a bimodule in $(\mathsf{Set}^{\text{op}},\times)$ is the same as as a morphism in the bicategory $\mathsf{Span}(\mathsf{Set})$.

John Baez (Jan 15 2021 at 22:27):

(Note if you want a mere category of spans you need to take isomorphism classes of spans, which will correspond to isomorphism classes of bimodules. I prefer to go bicategorical and avoid the "isomorphism class" stuff.)

Eric Forgy (Jan 15 2021 at 22:29):

True, but in this case I kind of think of it like $A\otimes (B\otimes C)$ and $(A\otimes B)\otimes C$ are the "same" :thinking:

John Baez (Jan 15 2021 at 22:29):

Btw, you might like this talk I gave, advocating the use of spans in quantum mechanics:

• John Baez, Quantum quandaries: A category-theoretic perspective, Philosophical and Formal Foundations of Modern Physics, Fondation des Treilles, April 2007.

Eric Forgy (Jan 15 2021 at 22:29):

Of course. I love all your stuff :blush:

Eric Forgy (Jan 15 2021 at 22:31):

Eric Forgy said:

True, but in this case I kind of think of it like $A\otimes (B\otimes C)$ and $(A\otimes B)\otimes C$ are the "same" :thinking:

So I think I am actually ok with thinking of isomorphism classes of bimodules, but I might change my mind when I get a little more enlightened :blush:

Eric Forgy (Jan 15 2021 at 22:32):

This is awesome too:

• The *-Category of Hilbert Spaces

John Baez (Jan 15 2021 at 22:35):

It's okay with working with isomorphism classes of bimodules until you want to talk about something like an element of a bimodule, or a homomorphism between bimodules.

There's no such thing as an element of an isomorphism classes of bimodules. There's no such thing as a homomorphism between isomorphism classes of bimodules. For these you need actual bimodules.

An isomorphism class is like a slippery greased pig twisting around so fast you can't grab any individual part of it. All you can say is "yup, it's a pig".

Eric Forgy (Jan 15 2021 at 22:43):

I'll try to do it the right way :muscle: (and forget isomorphism classes)

Eric Forgy (Jan 15 2021 at 22:45):

Through out all of this, I still have the functor $\mathbb{N}\to\mathsf{Span(Set)}$ in mind, so I have to tie this together somehow :thinking:

Eric Forgy (Jan 15 2021 at 22:46):

So I will start treating $\mathsf{Span(Set)}$ as the bicategory that it is.

Eric Forgy (Jan 15 2021 at 23:10):

Eric Forgy said:

This is awesome too:

• The *-Category of Hilbert Spaces

This is also awesome:

• Spans in Quantum Theory

John Baez (Jan 16 2021 at 02:16):

Thanks, I was actually looking for that one when I found the des Treilles talk.

Eric Forgy (Jan 17 2021 at 19:10):

Good morning :coffee:

Unlearning things you thought you knew is tough business :hurt:

Some things I've (un)learned recently:

• The unit of a monoidal category is not always terminal :face_palm:
• The unit of a cartesian category is always terminal :cool:
• $k$ is not terminal in $\mathsf{Vect}_k$ (related to above, I knew $k$ is the unit of $\mathsf{Vect}_k$) :face_palm:
• Tensor product of modules is actually a coproduct Tensor product of modules is not a product :face_palm:

Thanks to John's puzzles, I now know one important relationship between spans and bimodules.

Some definitions:
${}$
$\mathsf{Span}(C,\times)$ is a bicategory where

• $(C,\times)$ is a cartesian monoidal category with pullbacks (i.e. is finitely complete)
• Objects are objects of $(C,\times).$
• Morphisms are spans in $(C,\times).$
• 2-morphisms are maps between spans.

${}$
$\mathsf{Bim}(C^\mathsf{op},\times)$ is a bicategory where

• $(C,\times)$ is a cartesian monoidal category with pullbacks (i.e. is finitely complete)
• Objects are monoid objects of $(C^\mathsf{op},\times).$
• Morphisms are bimodules in $(C^\mathsf{op},\times).$
• 2-morphisms are bimodule morphisms.

Then,

$\mathsf{Span}(C,\times) \cong \mathsf{Bim}(C^\mathsf{op},\times).$

In other words:

• Every object $c\in (C,\times)$ is a monoid object in $(C^\mathsf{op},\times).$
• Every span in $(C,\times)$ is a bimodule in $(C^\mathsf{op},\times).$

${}$
(Note: We never discussed 2-morphisms or bicategories, but I have enough faith in the beauty of the universe to feel fairly confident this has to be the case.)

Todd Trimble (Jan 17 2021 at 20:23):

Some corrections: "the tensor product of modules is a coproduct". It's true that (for example) the tensor product of commutative rings has, for its underlying additive abelian group, the tensor product of the additive groups of the given rings. And similarly for commutative $k$-algebras over a commutative ring $k$: you take the tensor product of their underlying (additive) $k$-modules. But that's saying something else.

On bicategories of spans: the input data needs to be a category $C$ with pullbacks. Those are what allow you to compose morphisms = 1-cells. But if you want to do all the things you mention, you will need finite products as well. So you may as well assume $C$ is finitely complete.

There's a pretty big iceberg of stuff that lies beyond...

I'm probably only going to respond intermittently (if at all). John has been heroically guiding you through this -- I don't have the time myself.

Eric Forgy (Jan 17 2021 at 21:36):

Todd Trimble said:

I'm probably only going to respond intermittently (if at all). John has been heroically guiding you through this -- I don't have the time myself.

Thanks Todd :pray: Understood. For what its worth, I consider any intermittent comment from you as golden and very much welcome / appreciated :raised_hands:

Todd Trimble said:

On bicategories of spans: the input data needs to be a category $C$ with pullbacks. Those are what allow you to compose morphisms = 1-cells. But if you want to do all the things you mention, you will need finite products as well. So you may as well assume $C$ is finitely complete.

What I defined above, i.e. $\mathsf{Span}(C,\times),$ is a little more restrictive than the usual $\mathsf{Span}(C)$ because I require $(C,\times)$ to be cartesian ( Note, the first bullet in my definition. ). Is that sufficient?

Todd Trimble (Jan 17 2021 at 21:38):

How do you define the composition of spans?

Eric Forgy (Jan 17 2021 at 21:54):

I think I see. Thanks. If I require $C$ to be finitely complete, then it has finite products AND pullbacks.

Eric Forgy (Jan 17 2021 at 22:01):

[Edit: From the nLab, I see that if a category has a terminal object and pullbacks, then it is finitely complete. Since a cartesian monoidal category has a terminal object, then a cartesian monoidal category with pullbacks is finitely complete :+1: ]

I revised the definition to say "cartesian monoidal category with pullbacks (i.e. finitely complete)". Thank you for pointing out my mistake :pray:

Todd Trimble (Jan 17 2021 at 22:26):

Well, I could try to play John here and probe further why pullbacks are important here, and is composition associative, and a whole bunch of other things, but you were in the middle of talking about something else I think.

Eric Forgy (Jan 17 2021 at 22:27):

I can kind of see how some ambiguity around the term "cartesian category" arises. If a category has a terminal object and pullbacks, then it is cartesian monoidal, but more because it has all finite limits. A cartesian monoidal category has a terminal object and finite products, but might not have pullbacks and hence may not be finitely complete.

Eric Forgy (Jan 17 2021 at 22:33):

Todd Trimble said:

Well, I could try to play John here and probe further why pullbacks are important here, and is composition associative, and a whole bunch of other things, but you were in the middle of talking about something else I think.

Yeah, I was hoping to summarize (correctly - so thank you for your corrections :pray: ) what I learned in John's puzzles so I can move on to something else related that I'm thinking about. Hopefully what I said about $\mathsf{Span}(C,\times)$ and $\mathsf{Bim}(C^\mathsf{op},\times)$ is correct now :pray:

Todd Trimble (Jan 17 2021 at 23:19):

Eric Forgy said:

I can kind of see how some ambiguity around the term "cartesian category" arises. If a category has a terminal object and pullbacks, then it is cartesian monoidal, but more because it has all finite limits. A cartesian monoidal category has a terminal object and finite products, but might not have pullbacks and hence may not be finitely complete.

Incidentally, do you know how to show a category is finitely complete if it has pullbacks and a terminal object? (If it's not an immediate yes, then it's not a bad idea to see whether you can form products and equalizers and establish their universal properties, without going to the nLab or anything else.)

Eric Forgy (Jan 17 2021 at 23:22):

It is not immediately obvious to me and that does sound like good exercise to try :+1: I will think about it. Thanks :pray:

[Edit: I tried it here.]

Eric Forgy (Jan 19 2021 at 19:00):

Getting back to this :pray:

Some definitions:
${}$
$\mathsf{Span}(C,\times)$ is a bicategory where

• $(C,\times)$ is a cartesian monoidal category with pullbacks (i.e. is finitely complete)
• Objects are objects of $(C,\times).$
• Morphisms are spans in $(C,\times).$
• 2-morphisms are maps between spans.

${}$
$\mathsf{Bim}(C^\mathsf{op},\times)$ is a bicategory where

• $(C,\times)$ is a cartesian monoidal category with pullbacks (i.e. is finitely complete)
• Objects are monoid objects of $(C^\mathsf{op},\times).$
• Morphisms are bimodules in $(C^\mathsf{op},\times).$
• 2-morphisms are bimodule morphisms.

${}$
Then,

$\mathsf{Span}(C,\times) \cong \mathsf{Bim}(C^\mathsf{op},\times).$

This is pretty cool and gives us important relationships between spans and bimodules, but there are additional relationships I need to understand and those will probably turn out to be related to these :point_of_information:

Todd Trimble (Jan 19 2021 at 19:15):

"Bimodules" are a lot lot lot more general than is apparent from this context. Really pursuing this would lead to profunctors. But, all in good time.

Eric Forgy (Jan 19 2021 at 19:20):

Some background info...

Back in November, I decided to try to rewrite my old preprint with Urs:

• Discrete Differential Geometry on Causal Graphs

(as part of a desire to apply for engineering faculty positions, but I'm pretty sure I am too late for that by now :pensive:) in the language of CT in an attempt to simplify the presentation, add some exposition and extend it with some recent new results that I'm excited about.

Back in November, I roughly - very roughly - knew what a bimodule was. I thought of it as a slight generalization of vector spaces with left and right actions by commutative algebras. The bimodules I was working with were clearly related to directed graphs, but I had no idea how to make that connection precise although I knew it was there. I had heard the word "span", but had never knowingly used it in any work I was doing.

I've learned a lot these past couple months thanks to John, Todd and several others. Thank you :pray:

Now, I understand a directed graph (for my purposes) to be an endomorphism in $\mathsf{Span(Set)}$ and I understand that $\mathsf{Span(C)}$ and $\mathsf{Bim}$ are both important examples of bicategories, so if I want to express my work in the language of CT, it will involve bicategories. That is real progress :muscle:

Eric Forgy (Jan 19 2021 at 19:34):

Initially, I wanted to treat $\mathsf{Span(Set)}$ as a 1-category and I was happy with my new favorite functor
${}$
$G: \mathbb{N}\to\mathsf{Span(Set)}$
${}$
where $\mathbb{N}$ is a category with one object and morphisms labelled by natural numbers with composition = addition, i.e. $m\circ n = m+n.$ This is cool because
${}$

• $G(\bullet) = V$ (set of vertices)
• $G(0) = V:V\to V$ (identity span)
• $G(1) = E:V\to V$ (directed graph)
• $G(n) = E^{\times_V n}: V\to V$ (directed paths of length $n$)

${}$
This seems like it is related to "nerve", but is more suitable to my purposes.

Eric Forgy (Jan 19 2021 at 19:35):

Now, John has convinced me to NOT do that so I am rethinking things keeping $\mathsf{Span(Set)}$ as a bicategory, but then that forces me to rethink my favorite functor :thinking:

Eric Forgy (Jan 19 2021 at 19:37):

What I'm thinking about now is closer to $\mathsf{Span}(\mathbb{N}),$ i.e. functors
${}$
$G:\mathsf{Span}(\mathbb{N})\to \mathsf{Span(Set)},$
${}$
but I still need to think it through. Work in progress.

Eric Forgy (Jan 19 2021 at 19:59):

What I'm trying to think about now is the relationships among $\mathsf{Span(Set)},$ $\mathsf{Cospan(Vect)},$ and $\mathsf{Span(Vect)}.$

For example, we have a contravariant functor
${}$
$K^{-}: \mathsf{Span(Set)\to Copan(Vect)}$
${}$
given by $S\mapsto K^S,$ on objects and on spans gives:
${}$

image.png

${}$
and we have a covariant functor
${}$
$K[-]: \mathsf{Span(Set)\to Span(Vect)}$
${}$
given by $S\mapsto K[S]$ on objects and on spans gives:
${}$

image.png

Eric Forgy (Jan 19 2021 at 23:40):

I see. This is cool :heart_eyes: $\mathsf{Set}$ is cartesian, but $\mathsf{Vect}$ is cocartesian so we have
${}$
$\mathsf{Span(Set,\times)\cong Cospan(Set^{op},\times)\cong Bim(Set^{op},\times)}$
${}$
and
${}$
$\mathsf{Span(Vect^{op},\otimes)\cong Cospan(Vect,\otimes)\cong Bim(Vect,\otimes)}$
${}$
so that
${}$
$K^{\bullet}: \mathsf{Span(Set,\times)\to Cospan(Vect,\otimes) \cong Bim(Vect,\otimes)}$
${}$
and
${}$
$K[\bullet]: \mathsf{Span(Set^{op},\times)\to Cospan(Vect,\otimes)\cong Bim(Vect,\otimes)}$
[Edit: Was missing the $\mathsf{op}$ when I first posted this.]
${}$
so the second diagram is better thought of as
Covariant

${}$
This gives some solid justification for what I knew should somehow be true :+1:

It is actually quite beautiful. Thank you for nudging (dragging?) me through the puzzles. It was worth it :pray:

John Baez (Jan 19 2021 at 23:50):

Great! Yes, this was the goal of my remarks.

Eric Forgy (Jan 19 2021 at 23:51):

I won't apologize for the :heart: emoji. Thank you. Really :blush:

John Baez (Jan 19 2021 at 23:52):

However, I think there's a mistake in here, connected to contravariance. $K^\bullet$ is a contravariant functor and I think you neglected the effect of that contravariance on what you were doing. But $K[\bullet]$ is covariant.

John Baez (Jan 19 2021 at 23:53):

I'd have to think about this again to see exactly what's going on...

Eric Forgy (Jan 20 2021 at 00:08):

For what it's worth. I didn't neglect contravariance, but it is completely possible that I made a mistake keeping track of it.

Eric Forgy (Jan 20 2021 at 00:21):

Yep. I believe you are right. I was missing the $\mathsf{op}$ in $K[\bullet].$ I editted it now :+1:

John Baez (Jan 20 2021 at 00:45):

By the way, $(\mathsf{FinVect}^{\rm op}, \otimes)$ is equivalent to $(\mathsf{FinVect}, \otimes)$, where $\mathsf{FinVect}$ means the category of finite-dimensional vector spaces.

John Baez (Jan 20 2021 at 00:46):

So if you restrict to finite sets and finite-dimensional vector spaces, certain things get nicer.

Eric Forgy (Jan 20 2021 at 00:47):

John Baez said:

So if you restrict to finite sets and finite-dimensional vector spaces, certain things get nicer.

Ok, but I don't want to do that. The simplest example I want to work with is $\mathbb{Z}^n$ :thinking:

John Baez (Jan 20 2021 at 00:49):

It's amazing that you can count up to infinity.

Eric Forgy (Jan 20 2021 at 08:12):

I spent a few minutes looking at the bicategory $\mathsf{Cospan}(\mathbb{N,+})$. It is kind of cool.

It is defined as follows:

• There is one object $\bullet$
• Morphisms are cospans with legs labelled with natural numbers, e.g. $\bullet\overset{m}{\rightarrow}\bullet\overset{n}{\leftarrow}\bullet.$
• 2-morphisms are maps between cospans

${}$
Composition of cospans is as usual via pushout

image.png

${}$
2-morphisms

image.png

${}$
are defined only if $m = m_2-m_1 = n_2-n_1\ge 0.$
${}$

Just like my favorite functor
${}$
$\mathbb{N}\to C$
${}$
is a classifier of endomorphisms in a category $C$,
${}$
$\mathsf{Cospan}(\mathbb{N},+)\to\mathsf{Cospan}(C,\sqcup)$
${}$
seems to play a similar role for bicategories.
${}$
$(\mathbb{N},+)$ is cocartesian monoidal so we have
${}$
$\mathsf{Cospan}(\mathbb{N},+)\cong \mathsf{Span}(\mathbb{N}^\mathsf{op},+)\cong \mathsf{Bim}(\mathbb{N},+)$
${}$
and my favorite functor becomes
${}$
$G: \mathsf{Span}(\mathbb{N}^\mathsf{op},+)\to \mathsf{Span(Set,\times)},$
${}$

• $G(\bullet) = V\quad$ (set of vertices)
• $G(\bullet\overset{0}\rightarrow \bullet \overset{0}\leftarrow \bullet) = V: V\to V\quad$ (identity span)
• $G(\bullet\overset{0}\rightarrow\bullet \overset{1}\leftarrow \bullet) = E: V\to V\quad$ (directed graph)
• $G(\bullet\overset{1}\rightarrow\bullet \overset{0}\leftarrow \bullet) = E^\dagger: V\to V\quad$ (same directed graph with reversed arrows, i.e. the adjoint graph)
• $G(\bullet\overset{1}\rightarrow\bullet \overset{1}\leftarrow \bullet) = E^\dagger\circ E = E\circ E^\dagger: V\to V\quad$ (directed graph composed with its adjoint)
• $G(\bullet\overset{m}\rightarrow\bullet \overset{n}\leftarrow \bullet) = {E^\dagger}^{\times_V m}\circ E^{\times_V n} = E^{\times_V n}\circ {E^\dagger}^{\times_V m}: V\to V\quad$ (adjoint paths of length $m$ composed with paths of length $n$)

${}$
This definitely feels like I'm on track :+1:

Eric Forgy (Jan 20 2021 at 08:39):

Now, I think I can pose a question I raised above more correctly:
${}$
Is it possible that
${}$
$[\mathsf{Span}(\mathbb{N}^\mathsf{op},+),\mathsf{Span(Set,\times)}] \cong \mathsf{Span}([(\mathbb{N}^\mathsf{op},+),(\mathsf{Set},\times)],\times)$
${}$
?

Eric Forgy (Jan 20 2021 at 09:29):

Eric Forgy said:

Now, I think I can pose a question I raised above more correctly:
${}$
Is it possible that
${}$
$[\mathsf{Span}(\mathbb{N}^\mathsf{op},+),\mathsf{Span(Set,\times)}] \cong \mathsf{Span}([(\mathbb{N}^\mathsf{op},+),(\mathsf{Set},\times)],\times)$
${}$
?

Between shifus John and Todd, I think I might have learned enough CT kung fu that I might actually be able to take at stab at this :muscle:

Consider the following diagram

image.png

${}$
The action of a functor
${}$
$\mathsf{Span}(\mathbb{N}^\mathsf{op},+)\to\mathsf{Span(Set,\times)}$
${}$
is shown on the left and on the right sides of the commuting box. The horizontal arrows connecting the left and right sides are components of a natural transformation.
${}$
The action of a functor
${}$
$(\mathbb{N}^\mathsf{op},+)\to (\mathsf{Set},\times)$
${}$
is again shown on the left and on the right sides of the commuting box. The horizontal arrows connecting the left and right sides are components of a natural transformation.

However, we can insert another functor resulting in the following diagram

image.png

${}$
This looks like a span of functors, i.e. one functor in the middle with functors on either side and natural transformations connecting them, so I think it might actually be possible that
${}$
$[\mathsf{Span}(\mathbb{N}^\mathsf{op},+),\mathsf{Span(Set,\times)}] \cong \mathsf{Span}([(\mathbb{N}^\mathsf{op},+),(\mathsf{Set},\times)],\times).$
${}$
I know this doesn't count as a proof, but I think it might be a reasonable sketch of the beginning of a proof (with possible modifications) :blush:

Eric Forgy (Jan 20 2021 at 09:41):

Note: In the diagram, I introduced a new shorthand notation
${}$
$E^{m,n} := {E^\dagger}^{\times_V m}\circ E^{\times_V n} = E^{\times_V n}\circ {E^\dagger}^{\times_V m}$
${}$
so that
${}$
$E^{0,0} = V,$
${}$
$E^{0,1} = E,$
${}$
and
${}$
$E^{1,0} = E^\dagger.$

Eric Forgy (Jan 21 2021 at 18:25):

This is exactly what I needed to make CT contact with discrete differential geometry :+1:

This also helps justify defining a set
${}$
$\begin{aligned} E^{(r)} = \coprod_{r=n-m} E^{m,n} = \coprod_{r=n-m} {E^\dagger}^{\times_V m}\circ E^{\times_V (m-r)}\end{aligned}$
${}$
which, I think (still need to verify, but its looking good) gets mapped to bimodules
${}$
$\begin{aligned} \Omega^{(r)} = \bigoplus_{r=n-m} \Omega^{m,n} = \bigoplus_{r=n-m} {\Omega^\dagger}^{\times_V m}\circ \Omega^{\times_V (m-r)}.\end{aligned}$
${}$
I like it :blush: