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Stream: learning: questions

Topic: Lax monoidal functors

Eric Forgy (Dec 26 2020 at 23:12):

The nLab says:

Proposition 3.1. Lax monoidal functors send monoids to monoids.

However, monoidal categories and monoids are defined in terms of morphisms and commuting diagrams and functors preserve commuting diagrams so I was under the impression that any functor would preserve monoids (thought of as one-object categories). What am I missing?

Thank you for any help :pray:

Nathanael Arkor (Dec 26 2020 at 23:14):

A monoid is defined with respect to the tensor product of the monoidal category, so you need the functor to be (lax) monoidal to send $M \otimes M$ to $F(M) \boxtimes F(M)$ .

Eric Forgy (Dec 26 2020 at 23:49):

Nathanael Arkor said:

A monoid is defined with respect to the tensor product of the monoidal category, so you need the functor to be (lax) monoidal to send $M \otimes M$ to $F(M) \boxtimes F(M)$ .

That makes sense. Thank you.

I guess I am confused because the nLab also says that any monoid can be realized as a one-object category with endomorphisms. If you realize a monoid this way, then it doesn't involve tensor product and I would think any functor should send such a one-object monoid to a one-object monoid and this should work for any category including monoidal catageories. Does that make sense? :thinking:

I'm working with the one-object monoid $\mathbb{N}$ and I was thinking any functor $\mathbb{N}\to C$ should give a one-object monoid in $C$ and I'm not sure why I would need to consider anything other than normal functors :thinking:

Nathanael Arkor (Dec 26 2020 at 23:57):

The word "monoid" is overloaded here. On the page about monoidal functor, "monoid" means "monoid object in a monoidal category". When people say that monoids are one-object categories, they mean traditional, set-theoretic monoids (which are a special case of monoid objects, namely monoid objects in the cartesian monoidal category Set). Functors between one-object categories are monoid homomorphisms, but that's not what the nLab is referring to.

Nathanael Arkor (Dec 26 2020 at 23:59):

So it depends whether you want to consider the monoid multiplication as a morphism $M \otimes M \to M$ or as the composition within a category, restricted to a single object – and these are two different (albeit related) notions of monoid.

Reid Barton (Dec 27 2020 at 00:04):

A lot of people like to say "a monoid is a one-object category" but I think it's really better to think of monoids as corresponding to one-object categories, not being them. (The same way negative integers "correspond" to positive integers, but aren't the same as them.)

Reid Barton (Dec 27 2020 at 00:05):

The thing that turns a monoid into a category is called $B$ .

Eric Forgy (Dec 27 2020 at 00:11):

Thank you :pray:

It is still not 100% clear to me, but I'll think about it :+1:

John Baez (Dec 27 2020 at 01:45):

Eric Forgy said:

It is still not 100% clear to me, but I'll think about it.

John Baez (Dec 27 2020 at 01:46):

The point is that when the nLab says "monoid" here, they don't mean monoid! - not in the sense you know, Eric:

John Baez (Dec 27 2020 at 01:46):

Proposition 3.1. Lax monoidal functors send monoids to monoids.

John Baez (Dec 27 2020 at 01:47):

When they say "monoid" here, they really mean "monoid object in a monoidal category".

John Baez (Dec 27 2020 at 01:47):

That's a different, more general concept.

John Baez (Dec 27 2020 at 01:47):

For example a monoid object in the monoidal category $(\mathsf{Vect}, \otimes)$ is what you'd call an associative algebra.

John Baez (Dec 27 2020 at 01:48):

A monoid object in the monoidal category $(\mathsf{Set}, \times)$ is what you'd call a monoid.

A monoid object in the monoidal category $(\mathsf{Set}, +)$ is what you'd call a set.

John Baez (Dec 27 2020 at 01:48):

And so on.

John Baez (Dec 27 2020 at 01:49):

But people who deal a lot with monoid objects in monoidal categories get sick of saying "monoid object in monoidal category", so they start just saying "monoid".

John Baez (Dec 27 2020 at 01:49):

E.g. if I were talking to myself I'd say "A monoid in $(\mathsf{Vect}, \otimes)$ is an associative algebra."

John Baez (Dec 27 2020 at 01:50):

But don't mix up the basic concept of monoid that you know and love with this fancy generalization!!!

John Baez (Dec 27 2020 at 01:50):

A monoid in your sense amounts to the same thing as a one-object category, but that fact is not true for this fancy generalization.

John Baez (Dec 27 2020 at 01:51):

(Luckily, some fancy generalization of this fact is true.)

Eric Forgy (Dec 27 2020 at 02:49):

John Baez said:

E.g. if I were talking to myself I'd say "A monoid in $(\mathsf{Vect}, \otimes)$ is an associative algebra."

Funny you mention this. I have seen you say "an algebra is a monoid in $\mathsf{Vect}$ " but I thought this was shorthand for saying "an algebra is $\mathsf{Hom_{Vect}}(V,V)$ (or $\mathsf{End_{Vect}}(V)$ ) for an object $V$ in $\mathsf{Vect}$ " in which case we don't even need the tensor product :thinking:

I don't know why I get hung up on this, but when I see things like $A\otimes A\to A$ , I think "I need to two objects $A$ and $A\otimes A$ to define this morphism." But then that means we also need an $A\otimes A\otimes A$ and $A^{\otimes n}$ and then

I avoid this mind explosion by just thinking of $A$ and $\mathsf{End_{Vect}}(A).$

So what is the right way to think of the monoid $\mathbb{N}$ ? I currently think of this as a one-object category. Then a functor $P: \mathbb{N}\to\mathsf{Span(Set)}$ sends this object to a set and each morphism 0,1,2,.. is sent to a span (and I don't need to think about "lax" stuff).

Fawzi Hreiki (Dec 27 2020 at 06:31):

Anyway, even if you want to define a monoid as a one object category, such a thing still has to be defined in terms of products if you want to internalise it. So functors won’t in general preserve them.

Fawzi Hreiki (Dec 27 2020 at 06:33):

Internal categories actually require finite limits since the object of composable pairs is a pullback, but in the case that the object of objects is terminal (ie the internal category is a monoid) this pullback will reduce to a product.

Fawzi Hreiki (Dec 27 2020 at 06:36):

You’re thinking of the endomorphism monoid which is attached to every object in your category already. Those are subcategories of your category, not internal categories.

John Baez (Dec 27 2020 at 16:00):

Eric Forgy said:

John Baez said:

E.g. if I were talking to myself I'd say "A monoid in $(\mathsf{Vect}, \otimes)$ is an associative algebra."

Funny you mention this. I have seen you say "an algebra is a monoid in $\mathsf{Vect}$ " but I thought this was shorthand for saying "an algebra is $\mathsf{Hom_{Vect}}(V,V)$ (or $\mathsf{End_{Vect}}(V)$ ) for an object $V$ in $\mathsf{Vect}$ " in which case we don't even need the tensor product :thinking:

Well, it's not shorthand for saying that. An algebra is not always $\mathsf{End_{Vect}}(V)$ for an object $V$ in $\mathsf{Vect}$ . There are lots of algebras that aren't an algebra of matrices!

I don't know why I get hung up on this, but when I see things like $A\otimes A\to A$ , I think "I need to two objects $A$ and $A\otimes A$ to define this morphism." But then that means we also need an $A\otimes A\otimes A$ and $A^{\otimes n}$ and then

Stop thinking that stuff. It makes no sense.

An algebra is a vector space $A$ and a bilinear associative multiplication $m: A \times A \to A$ and a unit, an element $e \in A$ . But to avoid the word "bilinear" and stay within the category of vector spaces and linear maps we can say the multiplication is a linear map $m: A \otimes A \to A$ . And to avoid the word "element" we can describe the unit as a linear map $i: k \to A$ (where $k$ is our field) with $i(1) = e$ . Now everything is happening in our monoidal category $(\mathsf{Vect} ,\otimes)$ .

So, an algebra is an object $A \in (\mathsf{Vect},\otimes)$ and morphisms $m: A \otimes A \to A$ and $i: k \to A$ obeying associativity and the left and right unit laws.

More generally, a monoid object in a monoidal category $(\mathsf{V}, \otimes)$ is an object $A \in \mathsf{V}$ and morphisms $m: A \otimes A \to A$ and $i: k \to A$ obeying associativity and the left and right unit laws.

Now if you set $(\mathsf{V}, \otimes) = (\mathsf{Set},\times)$ you'll see a monoid object in $(\mathsf{Set},\times)$ is a monoid!

I avoid this mind explosion by just thinking of $A$ and $\mathsf{End_{Vect}}$

It's better to avoid it by thinking clearly, not succumbing to weird daydreams.

So what is the right way to think of the monoid $\mathbb{N}$ ? I currently think of this as a one-object category.

There are lots of right ways to think of it. It's the free monoid on one generator. It's the category freely generated by one object and one endomorphism. Etc.

Eric Forgy (Dec 27 2020 at 19:28):

I don't know why I get hung up on this, but when I see things like $A\otimes A\to A$ , I think "I need to two objects $A$ and $A\otimes A$ to define this morphism." But then that means we also need an $A\otimes A\otimes A$ and $A^{\otimes n}$ and then

Stop thinking that stuff. It makes no sense.

Why doesn't it make sense? Let's switch hats for a second. It would be no less painful for you to think about writing code than it is for me to write proper maths so think about how you would write this up in code (I have, see TensorAlgebra.jl). A method $m: A\otimes A\to A$ takes two arguments of different types. $A\otimes A$ is merely a symbol describing a type whose instances satisfies certain properties. It is a different type than the type $A$ . So if you tell me (or you tell a computer) you have an object $A$ and a morphism $A\otimes A\to A$ it simply doesn't parse. You need to tell the computer you have a type $A$ and a type $A\otimes A$ before you can define a map between the two types.

Eric Forgy (Dec 27 2020 at 19:35):

The way to do this in code (and the way I did it) is to tell the computer how to generate new types from existing types.

Eric Forgy (Dec 27 2020 at 19:39):

I should think in code more often. It helps. When I enter the world of maths, sometimes I get lost and leave my existing knowledge at the door.

Eric Forgy (Dec 27 2020 at 19:41):

I should probably implement a monoid in code actually. It shouldn't be too difficult starting with TensorAlgebra.jl.

John Baez (Dec 27 2020 at 19:59):

Eric Forgy said:

I don't know why I get hung up on this, but when I see things like $A\otimes A\to A$ , I think "I need to two objects $A$ and $A\otimes A$ to define this morphism." But then that means we also need an $A\otimes A\otimes A$ and $A^{\otimes n}$ and then

Stop thinking that stuff. It makes no sense.

Why doesn't it make sense?

What makes no sense is this:

then that means we also need an $A\otimes A\otimes A$ and $A^{\otimes n}$ and then

Especially the part.

Yes, in a monoidal category as soon as you have an object $A$ you have objects $A \otimes A$ , $A \otimes A \otimes A$ , and so on. (Of course some or all of these objects may be the same.)

But we don't freak out over that!

It's like I told you 1+1 = 2 and you said

but that means there exists 1+1+1, 1+1+1+1 and so on! INFINITELY MANY NUMBERS!

And I guess my answer would be "yeah, math is like that, I probably found that mind-blowing too at some point, but I got used to it".

Eric Forgy (Dec 27 2020 at 20:47):

Thank you :raised_hands:

Yeah. Something clicked as soon as I started thinking in terms of code.

I started writing a Julia package Categories.jl. It is still early days (and no documentation yet). The simplest way I could think of to define a category in code was this:

struct Category{O,M}
    ob::Type{Object{O}}
    hom::Type{Morphism{O,M}}
    id::Id{O,M}

    Category(::Type{O},::Type{M}) where {O,M} = new{O,M}(Object{O},Morphism{O,M},Id{O,M}())
end

Even without knowing Julia, I think it is fairly clear. A category is a special "type" that itself consists of a type called "Object" and a type called "Morphism" where the type of Morphism depends on the type of Object, but also has its own, possibly distinct, type associated with it (which could be any type including another category).

I don't need to list every object as long as I know some way to check whether something is an object. Same with morphisms. In both cases, it is a simple check of the type (which the computer knows).

In this context, the way I would think about $\otimes$ in a monoidal category is as a way to generate new types (objects) from existing types. So just saying you have an object $A$ in a monoidal category means you have $A\otimes A$ , $A\otimes A\otimes A$ , etc. You don't need to state that explicitly because it is understood (just like "1+1+1+1" is understood) because that is what having $\otimes$ means.

Thank you again :pray:

John Baez (Dec 27 2020 at 21:03):

Okay. I'm always amazed that people find code helpful for understanding. To me it's like "It's so much easier to understand if you write it in hieroglyphics so a machine can get it!" :upside_down: But people are very different in how they think.

Eric Forgy (Dec 27 2020 at 21:17):

Getting back to the original question...

Let's say I have a functor $F: \bullet\to (C,\otimes,1)$ from the category with 1 object and one identity morphism to any mondoidal category. First, does that sentence even parse? If it does parse, then $c = F(\bullet)$ would be an object in $(C,\otimes,1)$ . What I want is $c$ , $c\otimes c$ , $c\otimes c\otimes c$ , etc. What is the correct way to write / say that?

Would I need to consider $(\bullet,\otimes,1)$ as a monoidal category with a lax monoidal functor $(\bullet,\otimes,1)\to (C,\otimes,1)$ ?

Specifically, I have a functor

$\mathbb{N}\to\mathsf{Span(Set)}$

and I want to compose this with another functor to a monoidal category

$\mathbb{N}\to\mathsf{Span(Set)}\to (C,\otimes,1).$

I think of $\mathbb{N}$ as a category with one object and morphisms labelled by natural numbers so the first step gives me a "set $V$ and a span from the set $V$ to itself (endospan) and all compositions with itself."

Going from the second to third step, my set $V$ maps to an object $c$ of $C$ , but I also want to end up somehow with $c\otimes c$ , $c\otimes c\otimes c$ (plus the morphisms). Is it possible to write that as a functor?

Eric Forgy (Dec 27 2020 at 21:18):

John Baez said:

Okay. I'm always amazed that people find code helpful for understanding. To me it's like "It's so much easier to understand if you write it in hieroglyphics so a machine can get it!" :upside_down: But people are very different in how they think.

Yes, but to 99% of the people on this planet, the maths is the hieroglyphics :blush:

Eric Forgy (Dec 27 2020 at 21:24):

I kind of think

$\mathbb{N}\to\mathsf{Span(Set)}\to (C,\otimes,1)$

should be fine because I do end up with a specific object $c\in C$ , but then (like we just said) the fact $(C,\otimes,1)$ is a monoidal category, we can generate $c\otimes c$ via $\otimes$ and we don't need to say anything special about that.

John Baez (Dec 27 2020 at 21:25):

Eric Forgy said:

Getting back to the original question...

Let's say I have a functor $F: \bullet\to (C,\otimes,1)$ from the category with 1 object and one identity morphism to any mondoidal category. First, does that sentence even parse?

It parses, but since you're saying "functor" and not "monoidal functor", you are not using anything about $C$ being a monoidal category here.

A functor of the sort you describe simply picks out an arbitrary object of $C$ , no more and no less.

For any object $c \in C$ there is a unique functor $F: \bullet\to C$ sending the one object of $\bullet$ to $c \in C$ .

Making $C$ monoidal doesn't affect this at all.

John Baez (Dec 27 2020 at 21:26):

$\bullet$ is called the terminal category, and what you've described is a useful highbrow way to talk about an object of $C$ : it's the same as a functor from the terminal category to $C$ .

Eric Forgy (Dec 27 2020 at 21:29):

Ok, but if I say "monoidal functor", then wouldn't both the domain and codomain of the functor need to be monoidal? :thinking:

Maybe what I need to say is we have a composition of lax monoidal functors

$(\mathbb{N},\otimes,1)\to (\mathsf{Span(Set)},\otimes,1)\to (C,\otimes,1).$

That would be fine if that's what I need to do, but it seems a little bloated. I wish there was a way to promote nonmonoidal category to a monoidal one by introducing a $\otimes$ or something.

Eric Forgy (Dec 27 2020 at 21:31):

Is there a forgetful functor $U: (C,\otimes,1) \to C$ ?

Eric Forgy (Dec 27 2020 at 21:34):

If so, then maybe I can write it like

$\mathbb{N}\to\mathsf{Span(Set)}\to C\to (C,\otimes,1)$

where the last functor is left adjoint to $U$ .

John Baez (Dec 27 2020 at 21:39):

Eric Forgy said:

Ok, but if I say "monoidal functor", then wouldn't both the domain and range of the functor need to be monoidal? :thinking:

Sure. But $\bullet$ is monoidal in a unique way, so there's no ambiguity.

A monoidal functor from $\bullet$ to an arbitrary monoidal category $(C,\otimes, I)$ picks out an object $c \in C$ that's isomorphic to $I$ , and some other stuff. I believe in the end it boils down to something rather boring, though it's a bit of work to show this.

John Baez (Dec 27 2020 at 21:42):

Eric Forgy said:

Is there a forgetful functor $U: (C,\otimes,1) \to C$ ?

That's a weird thing to ask, since if you just say functor then you're saying I don't give a shit about the monoidal structure of $(C,\otimes,1)$ , I'm not going to use it in this question. So you're really just saying "is there a forgetful functor $U: C \to C$ ?" And that's also a weird thing to ask.

First, the concept of "forgetful" is not precisely defined - it's a "you know it when you see it" sort of thing. So my answer would probably be "yeah, the identity functor from $C$ to $C$ is a forgetful functor, in a sort of degenerate way where you're not actually forgetting anything".

That is a long reply to a question that would make most category theorists say "Bye, I gotta go now." I'm just explaining why it would make them want to leave.

Dan Doel (Dec 27 2020 at 21:42):

I haven't been paying close attention, but I have a suspicion that some of this functor composing doesn't make a whole lot of sense. For instance, the suggestion in the other thread about $ℕ$ was that each functor $F : ℕ → \mathsf{Span}(\mathsf{Set})$ corresponds to a digraph. So, composing with such a functor is sending an entire digraph to an arrow in $C$ , which doesn't seem like it matches what your goal was.

Dan Doel (Dec 27 2020 at 22:00):

If you want to say something about the structure of a digraph, you need a setting where there are arrows between digraphs. But a particular $F : ℕ → \mathsf{Span}(\mathsf{Set})$ just picks out a single digraph that happens to be an arrow in the span category. An example of the former would be the functor category $[ℕ,\mathsf{Span}(\mathsf{Set)}]$ , but it was mentioned that its arrows aren't necessarily what you'd want for talking about digraphs.

Eric Forgy (Dec 27 2020 at 22:05):

Dan Doel said:

I haven't been paying close attention, but I have a suspicion that some of this functor composing doesn't make a whole lot of sense. For instance, the suggestion in the other thread about $ℕ$ was that each functor $F : ℕ → \mathsf{Span}(\mathsf{Set})$ corresponds to a digraph. So, composing with such a functor is sending an entire digraph to an arrow in $C$ , which doesn't seem like it matches what your goal was.

Hi Dan. Thanks for your comment :pray:

In my case, $C$ is a category with

objects $A$ -bimodules
morphisms $A$ -bimodule homomorphisms

The image of a set $V\in\mathsf{Span(Set)}$ is an algebra $A\in C$ (which is trivially an $A$ -bimodule) and the digraph $V\leftarrow E\rightarrow V$ maps to an $A$ -bimodule homomorphism $A\leftarrow\Omega^1\rightarrow A$ . The composition of the $A$ -bimodule homomorphism $A\leftarrow\Omega^1\rightarrow A$ with itself is the $A$ -bimodule homomorphism $A\leftarrow \Omega^1\otimes_A\Omega^1\to A$ , etc.

This is nice because it gives me the $A$ -bimodules I want $A$ , $\Omega^1$ , $\Omega^1\otimes_A\Omega^1$ , etc, but I also need $A\otimes_K A$ , $A\otimes_K A\otimes_K A$ , etc (which are also $A$ -bimodules).

Eric Forgy (Dec 27 2020 at 22:09):

Dan Doel said:

If you want to say something about the structure of a digraph, you need a setting where there are arrows between digraphs. But a particular $F : ℕ → \mathsf{Span}(\mathsf{Set})$ just picks out a single digraph that happens to be an arrow in the span category. An example of the former would be the functor category $[ℕ,\mathsf{Span}(\mathsf{Set)}]$ , but it was mentioned that its arrows aren't necessarily what you'd want for talking about digraphs.

Right, but I am not wanting to talk about digraphs. I want to talk about digraphs generating higher geometric structures. These higher dimensional structures can be generated by graded algebras of the form $\Omega^1$ , $\Omega^1\otimes_A\Omega^1$ , etc.

If I only wanted digraphs, then the traditional functor $X\to\mathsf{Set}$ and corresponding functor category $[X,\mathsf{Set}]$ would be sufficient, but I need the directed paths as well, which is why I am so happy Amar brought $\mathbb{N}\to\mathsf{Span(Set)}$ to my attention :pray:

Dan Doel (Dec 27 2020 at 22:13):

Oh, okay. Maybe it does make sense, then.

Dan Doel (Dec 27 2020 at 22:14):

I thought you were wanting to talk about where edges go or something.

John Baez (Dec 27 2020 at 22:14):

What kind of "maybe" is that, Dan?

Eric Forgy (Dec 27 2020 at 22:15):

Dan Doel said:

I thought you were wanting to talk about where edges go or something.

I do care about where edges go and I care about where directed paths of length $n$ go (where a directed edge is a directed path of length 1) :blush:

Eric Forgy (Dec 27 2020 at 22:16):

If you send a set to an $A$ -bimodule, then the elements of that set are a basis for the underlying algebra (vector space + product).

Dan Doel (Dec 27 2020 at 22:17):

It's the kind of maybe where I still don't really understand what's going on, but it at least looks like the right shape of things.

Eric Forgy (Dec 27 2020 at 22:17):

In this case, vertices of a directed graph generate an algebra $A$ .

Eric Forgy (Dec 27 2020 at 22:19):

Directed edges of a directed graph generate an $A$ -bimodule $\Omega^1$ .

Eric Forgy (Dec 27 2020 at 22:20):

Directed paths of length 2, generate an $A$ -bimodule $\Omega^1\otimes_A\Omega^1.$

Eric Forgy (Dec 27 2020 at 22:26):

As described in a different stream here, there is a universal derivation $\tilde d: A\to\tilde\Omega^1$ . The $A$ -bimodule $\tilde\Omega^1$ (which is the kernel of the product $m: A\otimes_K A\to A$ ) corresponds to a complete directed graph and the $A$ -bimodule $\Omega^1$ is obtained from $\tilde\Omega^1$ by removing edges from the complete graph.

Eric Forgy (Dec 27 2020 at 22:30):

In a similar way, starting with just a directed graph, we can construct a space $\Omega^2$ which is the kernel of some other map from $\Omega^1\otimes_A\Omega^1.$

Eric Forgy (Dec 27 2020 at 22:34):

This can be iterated with $\Omega^3$ being the kernel of the equalizer of maps from $\Omega^1\otimes_A\Omega^2$ and $\Omega^2\otimes_A\Omega^1.$

Eric Forgy (Dec 27 2020 at 22:35):

You end up with a cochain complex completely generated by a directed graph.

John Baez (Dec 27 2020 at 22:35):

By the way, I don't understand this stuff you're saying.

Eric Forgy (Dec 27 2020 at 22:37):

John Baez said:

By the way, I don't understand this stuff you're saying.

Any of it? Which part?

I thought you understood how to generate a diffential algebra from a directed graph. You said "Nice!" which was a highlight of the week (which is a high bar given the holidays :blush: )

John Baez (Dec 27 2020 at 22:37):

You're talking about a functor $F : \mathbb{N} \to \mathsf{Span}(\mathsf{Set)}$ , and then all of a sudden you're talking about bimodules, but I don't understand the mathematical claim you're making about how $F$ is related to something...

John Baez (Dec 27 2020 at 22:38):

I understand how to generated a differential graded algebra from a graph; I just don't understand what you're saying.

Don't take "Nice!" as equal to "I understand and believe you." If you do that, I'll have to be much more careful about making any sort of friendly remark. :upside_down:

John Baez (Dec 27 2020 at 22:38):

"Nice!" meant more like "I'm glad you think you're doing something cool."

Eric Forgy (Dec 27 2020 at 22:39):

Explaining this is the main reason for me to write the paper I'm working on. I believe there is a simple explanation for it and I'm working on it :sweat_smile:

I know it works (I can write code to implement it).

Eric Forgy (Dec 27 2020 at 22:40):

John Baez said:

"Nice!" meant more like "I'm glad you think you're doing something cool."

I am absolutely doing something cool. Just failing to explain it :pensive:

John Baez (Dec 27 2020 at 22:41):

If what you're saying makes sense, it should be possible to say a few sentences, each of which states a mathematical fact that I can verify.

Eric Forgy (Dec 27 2020 at 22:41):

Yes. That is the goal.

Eric Forgy (Dec 27 2020 at 22:42):

You sound like Sullivan :joy:

Eric Forgy (Dec 27 2020 at 22:42):

I've been working on a simple explanation for 20 years. I feel I am closer. I just need to get this last functor sorted.

John Baez (Dec 27 2020 at 22:43):

I sound like what any mathematician confronted with you is secretly thinking.

You should say something like "a functor $F: \mathbb{N} \to \mathsf{Span}(\mathsf{Set})$ is the same as..."

Eric Forgy (Dec 27 2020 at 22:46):

I don't know what it is the "same as", but I can say what it is on morphisms of $\mathbb{N}$ . $F(\bullet)$ is a set of vertices of a directed graph. $F(1)$ is a span $V\leftarrow E\rightarrow V$ , i.e. a directed graph. $F(2)$ is a like a directed graph but deals with paths of length 2. Etc.

Dan Doel (Dec 27 2020 at 22:47):

It doesn't really seem that important to talk about that part, because it's just picking out a particular digraph. The functor that sends spans of sets to bimodules seems like the complicated part. E.G. has it been said how you choose the algebra $A$ from the set $V$ ?

John Baez (Dec 27 2020 at 22:48):

Eric Forgy said:

I don't know what it is the "same as", but I can say what it is on morphisms of $\mathbb{N}$ . $F(\bullet)$ is a set of vertices of a directed graph. $F(1)$ is a span $V\leftarrow E\rightarrow V$ , i.e. a directed graph. $F(2)$ is a like a directed graph but deals with paths of length 2. Etc.

I assume you're treating $\mathbb{N}$ here as a category with one object and one endomorphism for each natural number, with composition being addition? There are a couple of famous ways of thinking of $\mathbb{N}$ as a category.

See, I don't think I even understood this much of what you're talking about! Let's clear this part up at least.

John Baez (Dec 27 2020 at 22:49):

Usually mathematicians would start by saying "Think of the additive monoid of $\mathbb{N}$ as a one-object category" before diving in and talking about functors out of $\mathbb{N}$ ... if that's what they meant.

Eric Forgy (Dec 27 2020 at 22:51):

I wish I could take credit for the functor $\mathbb{N}\to\mathsf{Span(Set)}$ , but that was Amar's idea here. I really love the idea though :+1:

John Baez (Dec 27 2020 at 22:52):

You didn't even answer my basic question.

John Baez (Dec 27 2020 at 22:53):

We're doomed....

John Baez (Dec 27 2020 at 22:53):

I don't want f------g heart symbols, I want the answer to my question!

Eric Forgy (Dec 27 2020 at 22:54):

The only question mark I see is from

I assume you're treating $\mathbb{N}$ here as a category with one object and one endomorphism for each natural number, with composition being addition?

I kind of thought that was a rhetorical question mark. If not, the answer is, "Yes. Your assumption is correct."

John Baez (Dec 27 2020 at 22:56):

I don't have time to ask "rhetorical" questions.

John Baez (Dec 27 2020 at 22:56):

Okay, good.

John Baez (Dec 27 2020 at 22:56):

So then a functor $F: \mathbb{N} \to C$ for any category $C$ is a fancy-ass way of talking about an object in $C$ and an endomorphism of that object.

Eric Forgy (Dec 27 2020 at 22:57):

I refrain from the f------ heart emoji, but it is there :joy:

Eric Forgy (Dec 27 2020 at 22:57):

John Baez said:

So then a functor $F: \mathbb{N} \to C$ for any category $C$ is a fancy-ass way of talking about an object in $C$ and an endomorphism of that object.

Yes.

John Baez (Dec 27 2020 at 22:58):

So a functor $F: \mathbb{N} \to \mathsf{Span}(\mathsf{Set})$ is completely determined by, and determines, a set and an endospan of that set - in other words, a directed graph.

John Baez (Dec 27 2020 at 22:58):

So the answer to my question "what's it the same as?" is "a directed graph".

Eric Forgy (Dec 27 2020 at 22:59):

Exactly. Yes. But more because you can compose the directed graph with itself.

John Baez (Dec 27 2020 at 22:59):

When I said this:

John Baez said:

You should say something like "a functor $F: \mathbb{N} \to \mathsf{Span}(\mathsf{Set})$ is the same as..."

you should have continued: "... a directed graph".

John Baez (Dec 27 2020 at 23:00):

Eric Forgy said:

Exactly. Yes. But more because you can compose the directed graph with itself.

No, nothing more. You can always form edge paths in any graph. This is not "extra".

John Baez (Dec 27 2020 at 23:00):

At best this way of describing a directed graph is nice because it focuses your attention on those edge paths.

Eric Forgy (Dec 27 2020 at 23:01):

It is also nice because it can send morphisms to bimodule morphisms (which is the motivation).

John Baez (Dec 27 2020 at 23:01):

"A functor $F: \mathbb{N} \to \mathsf{Span}(\mathsf{Set})$ is the same as a directed graph" is a theorem.

It's short for "the category of functors $F: \mathbb{N} \to \mathsf{Span}(\mathsf{Set})$ is equivalent to the category of directed graphs".

John Baez (Dec 27 2020 at 23:02):

If you say something like this to a category theorist they will think a while and say "yup". And you will have communicated.

Dan Doel (Dec 27 2020 at 23:02):

Well, that's probably not a theorem, because the arrows are different.

Dan Doel (Dec 27 2020 at 23:02):

Apparently.

John Baez (Dec 27 2020 at 23:03):

Eric Forgy said:

It is also nice because it can send morphisms to bimodule morphisms (which is the motivation).

Okay, now you've switched back to saying vague stuff that doesn't mean anything to me.

John Baez (Dec 27 2020 at 23:04):

Dan Doel said:

Well, that's probably not a theorem, because the arrows are different.

Okay, now we're talking. I should have said it's a potential theorem: it's something with a well-defined truth value.

John Baez (Dec 27 2020 at 23:05):

Let's see. What's a natural transformation between functors $F: N \to Span(Set)$ ? (I'm getting tired of the fancy fonts.)

Eric Forgy (Dec 27 2020 at 23:05):

For the record, I don't think it is true, but at least, like you say, it is something that can be proven (or disproven) which is apparently progress :sweat_smile:

John Baez (Dec 27 2020 at 23:06):

It's huge progress. We have entered the territory where we are actually talking about things that are either true or false.

John Baez (Dec 27 2020 at 23:06):

So let's me answer my last question....

John Baez (Dec 27 2020 at 23:07):

Say I've got a functor $F: N \to Span(Set)$ and you've got a functor $G: N \to Span(Set)$ .

John Baez (Dec 27 2020 at 23:07):

So I've got a set $X$ and an endospan on it, and you've got a set $X'$ and an endospan on it.

John Baez (Dec 27 2020 at 23:08):

Say $X \leftarrow E \to X$ and $X' \leftarrow E' \to X'$ .

John Baez (Dec 27 2020 at 23:09):

Then a natural transformation is a span $X \to A \to X'$ such that... okay yeah, this is not a map of graphs!

John Baez (Dec 27 2020 at 23:10):

I see, we need to be working in the double category of spans to describe the category of graphs this way.

John Baez (Dec 27 2020 at 23:10):

Pretend I didn't say that... I don't want to talk about it. Too much work!

Eric Forgy (Dec 27 2020 at 23:10):

I am allowed a mind exploded emoji?

John Baez (Dec 27 2020 at 23:11):

So thanks, Dan, you caught me. And I guess Eric already knew it wouldn't work.

Eric Forgy (Dec 27 2020 at 23:12):

At least we're on the same page about what $\mathbb{N}\to\mathsf{Span(Set)}$ is :+1:

John Baez (Dec 27 2020 at 23:12):

So the category of functors from $\mathbb{N}$ to $\mathsf{Span}(\mathsf{Set})$ is something cute where the objects are directed graphs but the morphisms are something more fancy than maps of directed graphs... something worth pondering for me at least....

Eric Forgy (Dec 27 2020 at 23:13):

Yes. It is cute, I think, and deserves its own name. Open to suggestions :blush:

Eric Forgy (Dec 27 2020 at 23:15):

Next, I think it is clear how to freely construct a commutative associative algebra with unit from a set.

John Baez (Dec 27 2020 at 23:16):

There are two famous ways, which agree when the set is finite.

Eric Forgy (Dec 27 2020 at 23:16):

The elements of the set map to basis elements of the algebra and the product of two bases is 1 if the bases are the same and 0 otherwise. The unit element is the sum of all basis elements.

John Baez (Dec 27 2020 at 23:17):

Okay, yes, that's one of the two ways.

Eric Forgy (Dec 27 2020 at 23:17):

I am probably fine to restrict to finite sets if it helps, but I would like to consider infinite "checkboard" graphs at some point.

John Baez (Dec 27 2020 at 23:17):

If your set is $S$ and your field is $k$ I call this algebra $k[S]$ , but sadly lots of people don't have any standard name for it.

John Baez (Dec 27 2020 at 23:18):

If your set is finite, this algebra is isomorphic to another one: $k^S$ . That's the algebra of all functions from $S$ to $k$ .

Eric Forgy (Dec 27 2020 at 23:18):

Let's call it $A$ :blush:

Eric Forgy (Dec 27 2020 at 23:18):

Yes. $K^S$ is a cool way to think about it.

John Baez (Dec 27 2020 at 23:19):

The algebra you're talking about is only the same as $k^S$ when $S$ is finite. A little point to bear in mind.

Eric Forgy (Dec 27 2020 at 23:19):

Does $K^S$ work even if $S$ is infinite?

John Baez (Dec 27 2020 at 23:20):

A typical vague question: what the hell does "work" mean?

Eric Forgy (Dec 27 2020 at 23:21):

So apparently, as you say, $K[S]$ and $K^S$ are two algebras and they are isomorphic if $S$ is finite. Can $S$ be infinite in both cases?

Eric Forgy (Dec 27 2020 at 23:23):

I am fine to choose one once and for all and not too bothered which one I use.

John Baez (Dec 27 2020 at 23:23):

Yes. I said there are two famous ways of getting a commutative associative algebra from a set $S$ : $k[S]$ and $k^S$ , and they're isomorphic when $S$ is finite.

Eric Forgy (Dec 27 2020 at 23:23):

$K^S$ is kind of nice because I call these "discrete 0-forms" anyway.

John Baez (Dec 27 2020 at 23:24):

Eric Forgy said:

I am fine to choose one once and for all and not too bothered which one I use.

That's fine if your sets $S$ are finite; otherwise you will get in trouble.

John Baez (Dec 27 2020 at 23:25):

That is, you'll get in trouble pretending $k^S$ and $k[S]$ are "more or less the same". $k^S$ is the dual of $k[S]$ - but not the other way around when $S$ is infinite.

John Baez (Dec 27 2020 at 23:27):

Anyway, we don't need to worry about it now, and if $S$ is always finite we never need to worry about it.

Eric Forgy (Dec 27 2020 at 23:27):

Oh! Ok. That makes sense. It kind of depends. $A$ should be cochains.

Eric Forgy (Dec 27 2020 at 23:27):

(Forget I said that!)

Eric Forgy (Dec 27 2020 at 23:28):

Let's go with $A := K^V$ .

Eric Forgy (Dec 27 2020 at 23:29):

So we send a set $V$ of vertices to the algebra $A := K^V$ .

Eric Forgy (Dec 27 2020 at 23:31):

The span $V\leftarrow E\rightarrow V$ gets sent to the $A$ -bimodule homomorphism $A\leftarrow \Omega^1\rightarrow A.$

John Baez (Dec 27 2020 at 23:31):

Gets sent by whom?

John Baez (Dec 27 2020 at 23:31):

Or more precisely, by what?

Eric Forgy (Dec 27 2020 at 23:34):

By our functor, but I guess before I'm allowed to talk about our functor, I need to define the category.

Consider a category $C$ with

Objects $A$ -bimodules
Morphisms $A$ -bimodule homomorphisms.

We are defining the second functor in the composition

$\mathbb{N}\to\mathsf{Span(Set)}\to C.$

Dan Doel (Dec 27 2020 at 23:34):

I don't think that's what $C$ is.

Dan Doel (Dec 27 2020 at 23:35):

Shouldn't it be objects are algebras, and arrows are bimodules?

Eric Forgy (Dec 27 2020 at 23:35):

Yes. That would be fine :+1:

Dan Doel (Dec 27 2020 at 23:36):

Does $Ω^1$ make sense in that scenario, though? You need to be able to do $K^V \leftarrow Ω^1 \rightarrow K^U$ given a span $V \leftarrow E \rightarrow U$ .

John Baez (Dec 27 2020 at 23:38):

Eric Forgy said:

Yes. That would be fine :+1:

What do you mean, "that would be fine"? Dan said something completely different than what you just said.

In math, you're not supposed to say "that would be fine" when someone says the opposite of what you just said. You're supposed to either say "you're wrong" or "whoops, I was wrong".

Eric Forgy (Dec 27 2020 at 23:38):

If our category $C$ is

0-cells are algebras
1-cells are $(A,A')$ -bimodules
2-cells are bimodule homomorphisms

then we have a bicategory (I believe), but I only care about $A$ -bimodules (i.e. $(A,A)$ -modules), in which case an algebra $A$ is already an $A$ -bimodule, so I moved things down a notch so I could deal with normal categories.

John Baez (Dec 27 2020 at 23:39):

Okay, but how you index these things completely changes what a functor $Span(Set) \to C$ does.

John Baez (Dec 27 2020 at 23:40):

In one case it sends objects of Span(Set), i.e. sets, to algebras.

In another case it sends objects of Span(Set), i.e. sets, to bimodules of some particular algebra which we should have chosen ahead of time.

John Baez (Dec 27 2020 at 23:40):

So which is it?

Eric Forgy (Dec 27 2020 at 23:41):

Yes. I would like to choose my original way if that is ok although understand there is another way to do it. The other way is probably better but requires me to think about bicategories. I think I can do this with just categories though.

Eric Forgy (Dec 27 2020 at 23:41):

There is probably some "oidification" that can happen later.

Dan Doel (Dec 27 2020 at 23:43):

But if $A$ is chosen ahead of time, then $K^V$ doesn't enter into the picture, does it?

Eric Forgy (Dec 27 2020 at 23:44):

True

Eric Forgy (Dec 27 2020 at 23:45):

Thank you. That is a good point.

Eric Forgy (Dec 27 2020 at 23:46):

So I don't really have a choice. I need to use the bicategory. Does it have a name btw?

Eric Forgy (Dec 27 2020 at 23:52):

In the paper John referenced a few days ago

Non-Commutative Geometry Indomitable

this is referred to as $\mathsf{NCAlgebra}.$ Should I go with that? Is that standard?

Eric Forgy (Dec 27 2020 at 23:54):

The 2-category of algebras $\mathsf{NCAlgebras}$ has:
A1) Objects: associative (possibly non-commutative)
algebras.
A2) Arrows: bimodules over algebras.
A3) 2-arrows: bimodule morphisms.

Dan Doel (Dec 27 2020 at 23:55):

I guess. But it seems like the problem is that the previous steps need to be made bicategorical, then.

Dan Doel (Dec 27 2020 at 23:56):

Also what about my $Ω^1$ question?

Eric Forgy (Dec 27 2020 at 23:57):

I'll go with $\mathsf{NCAlgebras}$ for now. Let's also give the cute new category $[\mathbb{N},\mathsf{Span(Set)}]$ a name. I'll just call it $P$ for now.

Eric Forgy (Dec 27 2020 at 23:57):

Dan Doel said:

I guess. But it seems like the problem is that the previous steps need to be made bicategorical, then.

Oh no.

Eric Forgy (Dec 28 2020 at 00:00):

Dan Doel said:

Does $Ω^1$ make sense in that scenario, though? You need to be able to do $K^V \leftarrow Ω^1 \rightarrow K^U$ given a span $V \leftarrow E \rightarrow U$ .

I'm not sure. I've only ever thought about $\Omega^1$ in the context of $(A,A)$ -bimodules :thinking:

Eric Forgy (Dec 28 2020 at 00:03):

I think it should be fine. It just needs a little work that I have never thought about before.

Eric Forgy (Dec 28 2020 at 00:12):

Yes. I think it makes sense.

Let $A = K^V$ and $A' = K^{V'}$ , then (following Lupercio) I'd write the span as $A\leftarrow {}_A\Omega_{A'}\rightarrow A'$ .

Similarly, I'd write $V\leftarrow {}_V E_{V'}\rightarrow V'$ in $\mathsf{Span(Set)}.$

(Not crazy about the notation ${}_A\Omega_{A'}$ . Maybe $\Omega(A,A')$ is better.)

The $(A,A')$ -bimodule ${}_A\Omega_{A'}$ is a subset of $A\otimes_K A'$ .

Dan Doel (Dec 28 2020 at 00:15):

Well, perhaps you should be saying $Ω(E)$ anyway, since it's generated by a span $E$ .

Eric Forgy (Dec 28 2020 at 00:15):

Good idea :blush:

Eric Forgy (Dec 28 2020 at 00:18):

I was wondering how to write the composition of the span

$A\leftarrow {}_A\Omega_{A'} \otimes_{A'} {}_{A'}\Omega_{A''}\rightarrow A''$

is a mess :sweat_smile:

Much better to see this as

$A\leftarrow\Omega(E) \otimes_{A'} \Omega(E') \rightarrow A''$

:+1:

Eric Forgy (Dec 28 2020 at 00:23):

I think this is a slightly nicer way to think about it in terms of spans than what is presented in Lupercio :+1:

Eric Forgy (Dec 28 2020 at 00:48):

Dan Doel said:

I guess. But it seems like the problem is that the previous steps need to be made bicategorical, then.

Is it enough to just say we will consider $\mathbb{N}$ to be a bicategory with one object, morphisms labelled by natural numbers and one identity 2-morphism? Similar, for $\mathsf{Span(Set)}$ with objects sets, morphisms spans of sets and identity 2-morphisms so that we can write down the composite

$\mathbb{N}\to\mathsf{Span(Set)}\to\mathsf{NCAlgebras}$

But then I think I need these to be "lax monoidal functors" (going full circle).

Dan Doel (Dec 28 2020 at 00:59):

I don't know. It depends what structure is important for the functors to operate on.

I don't see how lax monoidal functors are relevant, though, because these are bicategories not monoidal categories.

Eric Forgy (Dec 28 2020 at 01:01):

Maybe they are monoidal bicategories? I definitely have a tensor product around here :thinking:

Eric Forgy (Dec 28 2020 at 01:03):

If $A$ is an object of $\mathsf{NCAlgebras}$ then $A\otimes_K A$ is also an object of $\mathsf{NCAlgebras}$ :thinking:

Eric Forgy (Dec 28 2020 at 01:05):

More generally, I guess, if $A$ and $A'$ are objects, then so is $A\otimes_K A'.$

Dan Doel (Dec 28 2020 at 01:06):

Okay, but does that matter? It hasn't been mentioned yet.

Eric Forgy (Dec 28 2020 at 01:06):

Yes. I am building up the story incrementally.

Eric Forgy (Dec 28 2020 at 01:10):

Wait. Unless, $A$ and $A'$ are both commutative, I'm not sure if $A\otimes_K A'$ is an algebra. If they are commutative, then $A\otimes_K A'$ is an algebra with product

$(x\otimes x')\otimes(y\otimes_K y') \mapsto (x\,y)\otimes_K (x'\,y').$

Eric Forgy (Dec 28 2020 at 01:26):

Once I've got this composition

$\mathbb{N}\to\mathsf{Span(Set)}\to\mathsf{NCAlgebras}$

sorted, then I need to introduce some special bimodule homomorphisms so that I can generate some subspaces as kernels.

For example:

$\Omega^1$ is generated by directed edges and is a sub-bimodule of the kernel of $A\otimes_K A\to A,\quad x\otimes_K y\mapsto x\, y$
$\Omega^2$ is the kernel of a bimodule homomorphism $\Omega^1\otimes_A\Omega^1\to A\otimes_K A$
$\Omega^n$ is the kernel of a bimodule homomorphism $\Omega^{\otimes_A n}\to A^{\otimes_K n}$

I can show that the spaces $\Omega^0 = A$ , $\Omega^1$ , ..., $\Omega^n$ constitute a cochain complex with a nilpotent derivation $d:\Omega^r\to\Omega^{r+1}$ satisfying the graded Leibniz rule so I have a graded differential algebra derived from the data of a directed graph.

Eric Forgy (Dec 28 2020 at 01:31):

Then, because $\mathsf{Span(Set)}$ (and any image of it in other categories) is a $\dagger$ -category, I get not just a DGA, but a differential graded Hilbert bimodule that serves as a discrete model of spacetime for scientific computation (and possibly more).

Making all this precise is what I'm trying to do here.

Eric Forgy (Dec 28 2020 at 01:33):

Essentially I am trying to formalize / simplify and extend an old preprint I wrote with Urs Schreiber back in 2004:

Discrete Differential Geometry on Causal Graphs

John Baez (Dec 28 2020 at 01:47):

Eric Forgy said:

In the paper John referenced a few days ago

Non-Commutative Geometry Indomitable

this is referred to as $\mathsf{NCAlgebra}.$ Should I go with that? Is that standard?

It's not standard, so you have to explain it whenever you start talking to someone. But that's okay.

John Baez (Dec 28 2020 at 01:48):

Eric Forgy said:

Wait. Unless, $A$ and $A'$ are both commutative, I'm not sure if $A\otimes_K A'$ is an algebra. If they are commutative, then $A\otimes_K A'$ is an algebra with product

$(x\otimes x')\otimes(y\otimes_K y') \mapsto (x\,y)\otimes_K (x'\,y').$

You mean

$(x\otimes x') \, (y\otimes_K y') \mapsto (x\,y)\otimes_K (x'\,y').$

This product makes $A \otimes_K A'$ into an algebra works regardless of whether the algebras $A, A'$ are commutative or not. This is the usual "tensor product of algebras over the field $K$ ".

Dan Doel (Dec 28 2020 at 01:49):

I'm not really sure I understand all that. It looks like what you're describing is similar to a natural transformation from $F : ℕ → \mathsf{NCAlgebras}$ (which goes through spans) to $G : ℕ → \mathsf{NCAlgebras}$ which takes the arrow $n$ to $A^{\otimes_K n}$ . But maybe the details wouldn't work out.

Dan Doel (Dec 28 2020 at 01:50):

In that scenario, there'd be no monoidal stuff. The monoids would be part of the bicategory structure.

John Baez (Dec 28 2020 at 01:51):

Btw, my small side-remarks should not be taken to suggest that I understand what Eric is actually talking about.

Eric Forgy (Dec 28 2020 at 01:52):

John Baez said:

Btw, my small side-remarks should not be taken to suggest that I understand what Eric is actually talking about.

Understood and appreciated :+1:

Eric Forgy (Dec 28 2020 at 02:20):

Dan Doel said:

I'm not really sure I understand all that. It looks like what you're describing is similar to a natural transformation from $F : ℕ → \mathsf{NCAlgebras}$ (which goes through spans) to $G : ℕ → \mathsf{NCAlgebras}$ which takes the arrow $n$ to $A^{\otimes_K n}$ . But maybe the details wouldn't work out.

This sounds cool, but isn't obvious to me. I'll think about it. Thank you :pray:

Dan Doel (Dec 28 2020 at 02:24):

It would have to be some kind of 2-natural transformation, though, and there are potentially more than one to choose from, I think.

Eric Forgy (Dec 28 2020 at 05:06):

Dan Doel said:

I'm not really sure I understand all that. It looks like what you're describing is similar to a natural transformation from $F : ℕ → \mathsf{NCAlgebras}$ (which goes through spans) to $G : ℕ → \mathsf{NCAlgebras}$ which takes the arrow $n$ to $A^{\otimes_K n}$ . But maybe the details wouldn't work out.

I've been looking at this for a little while now and still can't quite see how we get $A^{\otimes_K n}$ from a morphim $n$ and a natural transformation.

20201227_204606.jpg

Since $F(V) = K^V = A$ and $G(V) = K^V = A$ , then F(V) = G(V) and the left and right components are equal to the "identity" bimodule homomorphism $A\leftarrow 1\rightarrow A$ . Tracing the commuting square we have

$A\leftarrow 1\otimes_A \Omega(E)^{\otimes_A n}\rightarrow A = A\leftarrow \Omega(E)^{\otimes_A n}\otimes_A 1\rightarrow A.$

Similar to how morphisms of $\mathsf{Span(Set)}$ are isomorphism classes of spans, I think morphisms in the rightmost category (so far - which Lupercio calls $\mathsf{NCAlgebras}$ ) should be isomorphism classes of bimodules so that the above commuting square is true.

Am I missing something about $A^{\otimes_K n}$ ?

I understand how we get $\Omega(E)^{\otimes_A n}$ from $n$ , but I can't see how we get $A^{\otimes_K n}$ :thinking:

John Baez (Dec 28 2020 at 05:11):

It would be great if you could describe what you're trying to do without sinking into the details of how to do it (which apparently you're still not sure about).

John Baez (Dec 28 2020 at 05:15):

Are you trying to get a dga from a graph? Something else? Maybe you already know how to do it and you're just trying to say it in a more fancy way?

Eric Forgy (Dec 28 2020 at 05:24):

Yes. I already know how to do it ("it" being construct a DGA from a graph), but it is very tedious. It takes me 5+ pages of "start with this, add this structure, then do this and add this other structure" before I can start doing calculations with it. I know it works. I can compute with it, but I believe there should be a one or two line statement about a functor that gets the message across and the rest is details.

Eric Forgy (Dec 28 2020 at 05:26):

The functor

$\mathbb{N}\to\mathsf{Span(Set)}\to\mathsf{NCAlgebras}$

helps a LOT because in one line I already get most of the basic components I need.

John Baez (Dec 28 2020 at 05:30):

Okay, you're just trying to construct a DGA from a graph in a slick way.

Eric Forgy (Dec 28 2020 at 05:31):

This gets me

Directed $n$ -paths (including directed 1-paths which are just directed graphs)
An algebra $A = K^V$
An $A$ -bimodule $\Omega(E)$
A sequence of spaces $\Omega^{\otimes_A n}$

I'm not sure yet if I get $A^{\otimes_K n}$ from this, which I need.

Eric Forgy (Dec 28 2020 at 05:31):

John Baez said:

Okay, you're just trying to construct a DGA from a graph in a slick way.

Yes

John Baez (Dec 28 2020 at 05:34):

Here's how I might try to do it.

Start with a graph G.
Apply the "free category on a graph" functor $F: \mathsf{Graph} \to \mathsf{Cat}$ to get a category $F(G)$ whose morphisms are paths in the graph $G$ .
Apply the "nerve" functor $N: \mathsf{Cat} \to \mathsf{SSet}$ , which turns categories into simplicial sets. An n-simplex in $N(F(G))$ will be a length-n path in the graph $G$ .
Apply the "cochains" functor $C: \mathsf{SSet} \to \mathsf{DGA}$ to create the dga of cochains on your simplicial set.

Eric Forgy (Dec 28 2020 at 05:35):

I get a cochain complex $\Omega^0,\cdots, \Omega^n$ by considering special maps from $\Omega^{\otimes_A n}\to A^{\otimes_K n}$ but I'd rather not get into the details yet if possible. Once I have $\Omega^{\otimes_A n}$ and $A^{\otimes_K n}$ in a slick way I'll be happy enough for now.

John Baez (Dec 28 2020 at 05:35):

All the functors I listed are really standard things to do in math. You can read about all of them on the nLab, for example.

John Baez (Dec 28 2020 at 05:36):

I may not be quite doing what you want, because there could be a few variants of what to do.

John Baez (Dec 28 2020 at 05:37):

But I think this 4-step process is pretty close to what you're talking about.

Eric Forgy (Dec 28 2020 at 05:38):

Yeah. My impression (could be wrong) is that doesn't look like what I want because nothing I do involved simplices. Simplices don't work for me (although I can't rule out that they are involved).

John Baez (Dec 28 2020 at 05:38):

It's just a fact that length-n paths in a graph are secretly n-simplices.

John Baez (Dec 28 2020 at 05:41):

There's a simplicial set where the n-simplices are length-n paths in a graph, and you get it by applying the functors I called 1. and 2.

Eric Forgy (Dec 28 2020 at 05:44):

A length $n$ -path in $\mathsf{Span(Set)}$ gets sent to a special "tensor path" that explictly looks like

$(e^1\otimes_K e^2) \otimes_A (e^2\otimes_K e^3) \otimes_A (e^3\otimes_K e^4) \otimes_A \dots$

on basis elements. This "path" might secretly be a simplex.

Then I need to construct a sub-bimodule of paths results from the kernel of certain bimodules homomorphisms.

Eric Forgy (Dec 28 2020 at 05:45):

These sub-bimodules are in the shape of "diamonds" and form a cochain complex.

Eric Forgy (Dec 28 2020 at 05:47):

Once I'm in the category with algebras $A^{\otimes_K n}$ , isomorphism classes of bimodules $\Omega^{\otimes_A n}$ and bimodule homomorphisms, then I've got it.

Eric Forgy (Dec 28 2020 at 05:51):

For example,

$(e^i\otimes_K e^{j_1})\otimes_A (e^{j_1}\otimes e^k) - (e^i\otimes_K e^{j_2})\otimes_A (e^{j_2}\otimes e^k)$

is a 2-diamond.

Eric Forgy (Dec 28 2020 at 05:54):

It is a linear combination of two length-2 "paths".

John Baez (Dec 28 2020 at 06:00):

I promise that if you learn those four functors I just mentioned you'll be better off for it.

Eric Forgy (Dec 28 2020 at 06:01):

I'm sure you're right, but I have 3 kids, a dwindling bank account and I've been unemployed for more than 2 years. I don't want a course on category theory. I want / need to finish this paper so I can get back to trying to feed my family. I am on the last step and the finish line is in sight. I don't want to start a new marathon right now.

John Baez (Dec 28 2020 at 06:02):

I meant that these constructions are likely to save you lots of time. You're probably hacking your way through the jungle when a few hundred yards to your left there's a highway.

John Baez (Dec 28 2020 at 06:04):

I may be a bit off in my recommendation, but I think it's on the right track.

However, if you're almost done with something and just want to finish it, do that. None of this stuff should be taking away time from "feeding your family".

Eric Forgy (Dec 28 2020 at 06:04):

True, I have been hacking away, but I am seriously very close to finishing. Once I get those ingredients, I can finish my paper. I actually think we could be there, I'm just not sure if I get $A^{\otimes_K n}$ from this functor.

Eric Forgy (Dec 28 2020 at 06:07):

I started this journey and I just wanted to apply for an engineering faculty position but I haven't published anything in close to 20 years so I thought if I had at least a solid preprint I could point to, it might help. In this environment, the chance I can get an engineering faculty position is probably close to zero and I'm probably too late by now anyway. I started this paper early last month.

Eric Forgy (Dec 28 2020 at 06:08):

I'm actually REALLY happy with my latest result. I just need to write it up, but I want it to be beautiful.

John Baez (Dec 28 2020 at 06:09):

Then I suggest using the tools of linear algebra that you're comfortable with, like algebras, modules, tensor products and so on, and not try to use much category theory.

Eric Forgy (Dec 28 2020 at 06:09):

Even my PhD advisor said, "Aren't you too rusty by now" when I asked for a reference letter :joy:

Maybe he is right :pensive:

John Baez (Dec 28 2020 at 06:10):

Your use of categories so far is very raw: you don't have enough concepts under your command to choose the best ones.

John Baez (Dec 28 2020 at 06:11):

It takes at least a couple years of studying category theory full-time to get really good at wielding it.

John Baez (Dec 28 2020 at 06:13):

So I think your paper will look better if you use tools you're really familiar with.

Eric Forgy (Dec 28 2020 at 06:13):

True. You and a few people here (Amar, Dan, etc) have been super helpful (and amazingly patient :pray: )

I think this guy

$\mathbb{N}\to\mathsf{Span(Set)}\to\mathsf{NCAlgebras}$

is probably what I need. Like I said, this gives me $A = K^V$ and $\Omega^{\otimes_A n}$ . Does it also give me $A^{\otimes_K n}$ or do I need to monoidify something? :sweat_smile:

John Baez (Dec 28 2020 at 06:14):

See, I think you should skip all the Span(Set) stuff and just work with the vector spaces you're talking about.

John Baez (Dec 28 2020 at 06:15):

I think if you clearly write up your construction in terms of vectors spaces, algebras, modules, tensor products and such, a category theorist could come along and make it pretty later.

Eric Forgy (Dec 28 2020 at 06:23):

A rough outline of how it goes without cute functors is:

A directed graph (without loops) $(V,E)$ gives an algebra $A = K^V$ and an $A$ -bimodule $\Omega = K^E$
We have two kinds of tensor product $\otimes_K$ and $\otimes_A$
We can build up spaces $A^{\otimes_K n}$ and $\Omega^{\otimes_A n}$ . By the way, these are $A$ -bimodules
We can define some special bimodule homomorphisms $\Omega^{\otimes_A n}\to A^{\otimes_K n}$ and use them to construct special sub-bimodules $\Omega^n$
These sub-modules form a DGA

Eric Forgy (Dec 28 2020 at 06:25):

I believe the first 3 bullets are captured by the functor

$\mathbb{N}\to\mathsf{Span(Set)}\to\mathsf{NCAlgebras}.$

Dan Doel (Dec 28 2020 at 06:28):

The reason behind what I said is the same as the other functor. A functor $G : ℕ → C$ corresponds to picking out an object $G(\bullet)$ and an endomorphism $G(1) : G(\bullet) → G(\bullet)$ , then $G(n)$ is the n-fold composition of $G(1)$ . In the previous case it was the n-fold composition of whatever $Ω(E)$ is. But instead it could be whatever that other n-fold tensor is.

However, I don't know if the natural transformation actually corresponds to the structure you want.

Dan Doel (Dec 28 2020 at 06:32):

If it's something different for each $n$ , I would guess it's not, since all the n-ary stuff is being determined by $1$ for most of this.

Eric Forgy (Dec 28 2020 at 06:32):

Dan Doel said:

The reason behind what I said is the same as the other functor. A functor $G : ℕ → C$ corresponds to picking out an object $G(\bullet)$ and an endomorphism $G(1) : G(\bullet) → G(\bullet)$ , then $G(n)$ is the n-fold composition of $G(1)$ . In the previous case it was the n-fold composition of whatever $Ω(E)$ is. But instead it could be whatever that other n-fold tensor is.

However, I don't know if the natural transformation actually corresponds to the structure you want.

Thanks Dan :pray:

How about something like this?

We have a span $K\leftarrow A\rightarrow K$ and composing this, we get $K\leftarrow A\otimes_K A\rightarrow K$ . Composing $n$ times, we get $K\leftarrow A^{\otimes_K n}\rightarrow K.$