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Stream: learning: questions

Topic: Monoids and left adjoint to U: Vect -> Set

Eric Forgy (Dec 12 2020 at 17:34):

I thought I understood free vector spaces, but I am questioning myself and could use some help :pray:

As a former physicist / engineer, I am comfortable with sentences like "The vector space $V$ spanned by the set $S$."

By this, I always understood it to mean that if $v\in V$, then $v$ can be written as a formal linear combination of elements of $S$, i.e.

$v = \sum_i v_i s_i,$

where $v_i\in K$, $s_i \in S$ for some field $K$ (usually $\mathbb{R}$ or $\mathbb{C}).$

I always assumed the free functor $F: \text{Set}\to\text{Vect}$ did exactly that, i.e.

$F^{(1)}_{Idea}(S) =$ "The vector space spanned by $S$."

Now, I am not so sure it is that simple :thinking:

The issue is that $\text{Vect}$ is monoidal so if $V$ is an object in $\text{Vect}$, then so is $V\otimes V$. So we actually get a tower of vector spaces $V^{\otimes n}.$ So if $V_S$ is the "vector space of formal linear combinations of elements of $S$", then I'd be tempted to say that

$F^{(2)}_{Idea}(S) = \bigoplus_{r\ge 0} V_S^{\otimes r},$

where $V_S^{\otimes 0} = K.$

But then I don't think that is the full picture either because if $V$ is an object in $\text{Vect}$, then so is the dual space $V^*.$ But then if $V^*$ is an object of $\text{Vect}$, then so is $V^*\otimes V^*$ and the story above repeats. So I am tempted to denote $V^*$ by $V^{-1}$ so that what I am thinking now is that

$F^{(3)}_{Idea}(S) = \bigoplus_{r\in\mathbb{Z}} V_S^{\otimes r}.$

But even this might not be the end of the story because now if $F(S)$ is an object of $\text{Vect}$, then so are $F(S)\otimes F(S)$ and $F(S)^*$. If I squint my eyes, I can kind of believe that

$F(S)\otimes F(S) \simeq F(S)$

and

$F(S)^* \simeq F(S).$

So maybe what we really want is something like

$F^{(4)}_{Idea}(S) = \left[\bigoplus_{r\in\mathbb{Z}} V_S^{\otimes r}\right],$

i.e. equivalence classes of vector spaces of that form :thinking:

That is a LOT different than "the vector space spanned by linear combinations of elements of $S$"

This train of thought came about while thinking about monoids.

If I take Wikipedia's definition for example:

In category theory, a branch of mathematics, a monoid (or monoid object) $(M, μ, η)$ in a monoidal category $(C, ⊗, I)$ is an object $M$ together with two morphisms

$μ: M ⊗ M → M$ called multiplication,

$η: I → M$ called unit, [snip]

I look at this and say, "Hey! Wait a minute!"

If all you have is an object $M$, how can you have a map $M\otimes M\to M$??? You need both $M$ AND $M\otimes M$ and for associativity you need $M\otimes M\otimes M$ so wouldn't it be more correct to say,

In category theory, a branch of mathematics, a monoid (or monoid object) $(M, μ, η)$ in a monoidal category $(C, ⊗, I)$ is an object

$M = \sum_{r\ge 0} M^{\otimes r},$

where $M^{\otimes 0} = I$ together with two morphisms

$μ: M ⊗ M → M$ called multiplication,

$η: I → M$ called unit, [snip]

?

Nathanael Arkor (Dec 12 2020 at 17:38):

The definition of monoidal category means that for any object $M$, you also have an object $M \otimes M$ (and $(M \otimes M) \otimes M$, etc.) by applying the tensor product to two copies of $M$.

Eric Forgy (Dec 12 2020 at 17:39):

A punchline to the above is that $F(S)$ is a monoid in $\text{Vect}.$ Is that correct?

Eric Forgy (Dec 12 2020 at 17:43):

Nathanael Arkor said:

The definition of monoidal category means that for any object $M$, you also have an object $M \otimes M$ (and $(M \otimes M) \otimes M$, etc.) by applying the tensor product to two copies of $M$.

Hi Nathanael :wave:

Yeah. That is my point. I'm more than likely just confused, but it seems you can never really have an "object $M$ in a monoidal category" unless $M$ is an infinite tower of tensor powers of something so that $M\otimes M = M$. So a morphism $M\otimes M\to M$ is really an endormorphism $M\to M.$

Dan Doel (Dec 12 2020 at 17:46):

There is a different presentation of monoids where instead of $η$ and $μ$ you use $φ : (\sum_{r \geq 0} M^{\otimes r}) → M$. I don't see why you would assume $M$ is equal to that sum, though.

Nathanael Arkor (Dec 12 2020 at 17:47):

You start with a category $\mathscr V$ equipped with a tensor product $\otimes : \mathscr V \times \mathscr V \to \mathscr V$ subject to the various axioms for a monoidal category. A monoid in $\mathscr V$ is then a choice of some object $M \in \mathrm{ob}(\mathscr V)$. You don't need to mention $M \otimes M$ explicitly, because as soon as you pick the $M$, its tensors with other objects are determined by the structure of $\otimes$ on $\mathscr V$. So you know that $M \otimes M$ exists (and is a specific object of $\mathscr V$) without needing to declare it to be anything else. It's not going to be equal to $M$ in general.

Nathanael Arkor (Dec 12 2020 at 17:48):

Regarding the free vector space: your initial understanding looks to me to be correct. You send each set $S$ to the set of formal sums, equipped with the obvious vector space structure.

Eric Forgy (Dec 12 2020 at 17:52):

Yes. I'm sorry. I struggle to make myself clear sometimes :sweat_smile:

My point is, as you say, if $C$ is monoidal and $c$ is an object of $C$, then $c\otimes c$ is also an object of $C$ and you cannot escape that, so anything that maps into a monoidal category gives you an infinite tower of objects, but that itself is an object so $c$ itself should be that infinite tower.

Dan Doel (Dec 12 2020 at 17:55):

You know the category of sets is monoidal, right?

Nathanael Arkor (Dec 12 2020 at 17:55):

Defining a $c$ does define an infinite set in that sense, yes, but the set itself is not an object of the monoidal category. For instance, the set contains $(c \otimes c) \otimes c$ but also $c \otimes (c \otimes c)$. Even if you just focus on left-associated tensor products, you can't take an infinite tensor product in a monoidal category. There's also no way to recover $c$ from $c \otimes c$ in general, so even if you could do this, it would lose information over just considering $c$ itself.

Nathanael Arkor (Dec 12 2020 at 17:57):

It's also arbitrary to consider the set of tensor products of $c$ with itself: why not tensor products with other objects of $C$ as well?

John Baez (Dec 12 2020 at 17:58):

Eric Forgy said:

I thought I understood free vector spaces, but I am questioning myself and could use some help :pray:

As a former physicist / engineer, I am comfortable with sentences like "The vector space $V$ spanned by the set $S$."

By this, I always understood it to mean that if $v\in V$, then $v$ can be written as a formal linear combination of elements of $S$, i.e.

$v = \sum_i v_i s_i,$

where $v_i\in K$, $s_i \in S$ for some field $K$ (usually $\mathbb{R}$ or $\mathbb{C}).$

I always assumed the free functor $F: \text{Set}\to\text{Vect}$ did exactly that, i.e.

$F^{(1)}_{Idea}(S) =$ "The vector space spanned by $S$."

Yes, that's correct.

$F$ is a left adjoint of the forgetful functor $U : \text{Vect} \to \text{Set}$, and you can use the definition of left adjoint to check that the vector space spanned by $S$ indeed behaves in the correct way to be the left adjoint. It's essential to do this, to understand what's going on here.

If I squint my eyes, I can kind of believe that

$F(S)\otimes F(S) \simeq F(S)$

Don't squint your eyes then. It's false. If $S$ has 3 elements, $F(S)$ is a 3-dimensional vector space while $F(S) \otimes F(S)$ is 9-dimensional, so they're not isomorphic.

I don't know why you're bringing tensor products into this game, since they're not relevant to the construction of the functor $F$. They obey some interesting properties connected to $F$, e.g.

$F(X \times Y) \cong F(X) \otimes F(Y)$

but if you're trying to understand $F$ itself you should not be thinking about tensor products of vector spaces. I have no clue what you're doing with them here, but it doesn't look good.

John Baez (Dec 12 2020 at 18:12):

(Btw, in case anyone is wondering why I'm talking so rudely to Eric, it's because we're old friends and I've decided long ago that when he goes off the rails it's good for me to tell him very clearly exactly where he went off the rails - there's no need to be shy about it, he can take it.)

Eric Forgy (Dec 12 2020 at 18:37):

John Baez said:

If I squint my eyes, I can kind of believe that
$F(S)\otimes F(S) \simeq F(S)$

Don't squint your eyes then. It's false. If $S$ has 3 elements, $F(S)$ is a 3-dimensional vector space while $F(S) \otimes F(S)$ is 9-dimensional, so they're not isomorphic.

If $S$ has 3 elements, then if $F(3)$ is three-dimensional as it would be if $F(S)$ is "linear combinations of elements of $S$" as I originally understood it (and I guess is the correct way to think about it), then I was confused because $F(S)\otimes F(S)$ is also in $\text{Vect}$ and we can't escape that, so I thought one way to pick up those extra elements could be to form that infinite tower so that both $F(S)$ and $F(S)\otimes F(S)$ are infinite dimensional. I guess that would not be great :sweat_smile:

But like Dan reminded me, $\text{S}$ is also monoidal so if $S$ is an object of $\text{Set}$, then so is $S\times S$ and as you say

$F(S\times S) \cong F(S)\otimes F(S).$

So this clears things up for "free vector spaces", so now I just need to unstick the way I think of $\text{Set}$ :sweat_smile:

When I'm trying to understand a concept, I try to reduce it to the simplest case possible. Unfortunately, sometimes, like now, I consider cases that are simpler than possible :face_palm:

I was thinking of a "subcategory of $\text{Set}$ with just one object $S$". However, since $\text{Set}$ is monoidal, there is no one-object subcategory of $\text{Set}$ because if $S$ is an object of $\text{Set}$ then you have at least two objects because $S\times S$ will ALSO be in that category. But then we actually have an infinite number of objects. Is it true that any subcategory of $\text{Set}$ has an infinite number of objects?

If so, then I have learned something :tada:

Thank you guys for your help and your patience :raised_hands: (except John who gets thanks for help but not patience (just kidding - you always have amazing patience with me) :joy::hug:)

Fawzi Hreiki (Dec 12 2020 at 18:43):

There's no reason that a subcategory of $\text{Set}$ should be closed under products.

Eric Forgy (Dec 12 2020 at 18:47):

Thanks Fawzi. Yeah. Is there a term to distinguish just a general subcategory of $\text{Set}$ from a subcategory that has the same structure, e.g. closed under products?

Fawzi Hreiki (Dec 12 2020 at 18:48):

What you're talking about are monoidal subcategories of $(\text{Set}, \times, 1)$, i.e. subobjects in the category $\text{MonCat}$

Fawzi Hreiki (Dec 12 2020 at 18:49):

'Subcategory' unqualified would refer instead to a subobject in $\text{Cat}$

Eric Forgy (Dec 12 2020 at 18:54):

Just to clarify, would a monoidal subcategory of $(\text{Set},\times, 1)$ have an infinite number of objects? (I think yes, but a ":+1:" for confirmation or something would be great :pray:

Reid Barton (Dec 12 2020 at 18:55):

Eric Forgy said:

so I thought one way to pick up those extra elements could be to form that infinite tower so that both $F(S)$ and $F(S)\otimes F(S)$ are infinite dimensional. I guess that would not be great :sweat_smile:

It seems like you might be imagining that given some "formal vectors" $v_1$, $v_2$, $v_3$, in addition to formal linear combinations like $3v_1 - 2v_2$, you could also write down formal tensor products like $v_2 \otimes v_3$. But these formal tensor products aren't elements of $F(S)$.

Reid Barton (Dec 12 2020 at 18:55):

And the reason is just that if $v_1$ and $v_2$ are elements of a vector space $V$, then (in general) $v_1 \otimes v_2$ doesn't make sense as an element of $V$. It's an element of $V \otimes V$ instead.

Reid Barton (Dec 12 2020 at 18:56):

On the other hand $3v_1 - 2v_2$ is again an element of $V$.

Reid Barton (Dec 12 2020 at 18:56):

Eric Forgy said:

Just to clarify, would a monoidal subcategory of $(\text{Set},\times, 1)$ have an infinite number of objects? (I think yes, but a ":+1:" for confirmation or something would be great :pray:

This isn't true--it could consist of just the unit object, for example.

Eric Forgy (Dec 12 2020 at 18:56):

Yeah. Thanks Reid. I know that, I just managed to confused myself into thinking I might have been wrong the last 20 years :sweat_smile:

Eric Forgy (Dec 12 2020 at 19:01):

Reid Barton said:

Eric Forgy said:

Just to clarify, would a monoidal subcategory of $(\text{Set},\times, 1)$ have an infinite number of objects? (I think yes, but a ":+1:" for confirmation or something would be great :pray:

This isn't true--it could consist of just the unit object, for example.

Ok. I was wondering about that case. So $1\times 1$ is considered the same object as $1$? Other than that case, if a monoidal subcategory has at least one non-unit element, it has an infinite number of objects?

Reid Barton (Dec 12 2020 at 19:01):

No, that's also not true...

Fawzi Hreiki (Dec 12 2020 at 19:19):

E.g. $0 \times 0 = 0$.

Fawzi Hreiki (Dec 12 2020 at 19:21):

So the full subcategory on $\{0, 1\}$ is closed under finite products.

Eric Forgy (Dec 12 2020 at 19:35):

Fawzi Hreiki said:

So the full subcategory on $\{0, 1\}$ is closed under finite products.

Sorry :pray:

Is $1\times 1 = 1$ in $\text{Set}$? :thinking:

I think I can see that $0\times 0 = 0$ and even $0\times 1 = 1\times 0 = 0$ (if $0$ is the empty set), but I don't quite see how $1\times 1=1$ (although I can see that $1\times 1\cong 1$) :thinking:

So I'm struggling to see how the smallest monoidal subcategory of $(\text{Set},\times, 1)$ isn't infinite

$\{1,1\times 1, (1\times 1)\times 1, 1\times(1\times 1),\cdots\}$

although each object is equivalent :thinking:

Fawzi Hreiki (Dec 12 2020 at 19:40):

Just think of basic arithmetic. More generally, in any category with finite products, the terminal object is unital (i.e. neutral) with respect to the product.

Fawzi Hreiki (Dec 12 2020 at 19:42):

You seem to be getting the objects and the symbols (presentations) for the objects confused.

Eric Forgy (Dec 12 2020 at 20:04):

Fawzi Hreiki said:

You seem to be getting the objects and the symbols (presentations) for the objects confused.

That wouldn't surprise me at all (although I don't really understand what that means :sweat_smile: )

If I look at the nLab:

Initial object:

An initial object in a category $C$ is an object $∅$ such that for any object $x$ of $C$, there is a unique morphism $!:∅→x$. An initial object, if it exists, is unique up to unique isomorphism, so we speak of the initial object.

Terminal object:

A terminal object in a category $C$ is an object $1$ of $C$ satisfying the following universal property:

for every object $x$ of $C$, there exists a unique morphism $!:x→1$. The terminal object of any category, if it exists, is unique up to unique isomorphism. If the terminal object is also initial, it is called a zero object.

So if I look at

$\{1,1\times 1, (1\times 1)\times 1, 1\times(1\times 1),\cdots\}$

I see that each object is isomorphic to $1$ so each object is the terminal object so you can say they are the same thing I guess, but aren't they really different objects? :thinking:

Reid Barton (Dec 12 2020 at 20:07):

They might be the same object or different objects, it doesn't really matter

Eric Forgy (Dec 12 2020 at 20:21):

I guess the only reason it would matter is if you are trying to count objects. If they are the same object, then $\{1\}$ has one object and is a monoidal subcategory (of any category I guess). If they are different, then the smallest monoidal subcategory has an infinite number of (ismorphic) objects.

Dan Doel (Dec 12 2020 at 20:22):

That isn't something you should be doing, though.

Dan Doel (Dec 12 2020 at 20:23):

It's "evil".

Eric Forgy (Dec 12 2020 at 20:24):

image.png

:joy:

Eric Forgy (Dec 12 2020 at 20:32):

Does that mean that to not be evil, we should think in terms of equivalence classes so that

$[1] = [1\times 1] = [(1\times 1)\times 1] = [1\times(1\times 1)]$

so it takes you a few seconds before you get tired of writing brackets and just understand that $1$, $1\times 1$, $1\times(1\times 1)$ etc are the same object for practical purposes?

Dan Doel (Dec 12 2020 at 20:32):

No, I think equivalence classes are not a good way to think about it.

Fawzi Hreiki (Dec 12 2020 at 20:34):

To say that the full subcategory on 0 and 1 is closed under products is to say that the inclusion functor preserves products. But when we say that a functor $F: C \rightarrow D$ preserves products, we mean that it preserves products up to isomorphism.

Eric Forgy (Dec 12 2020 at 20:35):

How can I make progress if I don't even understand $\text{Set}$? :pensive:

Fawzi Hreiki (Dec 12 2020 at 20:36):

I don't think this is a case of not understanding sets - more so not understanding equivalence.

Dan Doel (Dec 12 2020 at 20:38):

One thread of how to better think about it is to ditch sets at the fundamental building blocks, and say that a category has a groupoid of objects, where groupoids are a (more) fundamental building block. And isomorphisms between objects mean that the objects are equivalent in that groupoid.

Eric Forgy (Dec 12 2020 at 20:39):

I would say I don't fully understand $\text{Set}$ if I don't understand monoidal subcategories of $\text{Set}$, but yeah, I have little intuition when it comes to equivalence (despite trying).

Dan Doel (Dec 12 2020 at 20:42):

And a groupoid with infinitely many 'points' that are all equivalent behaves the same as a groupoid with just one point, even if in some technical sense they don't have the same number of points. Because all constructions automatically respect equivalences, and can't distinguish between equivalent things in a meaningful way.

John Baez (Dec 12 2020 at 20:43):

Eric Forgy said:

Just to clarify, would a monoidal subcategory of $(\text{Set},\times, 1)$ have an infinite number of objects?

It either has an infinite number of isomorphism classes of objects, or just one.

Here's a simpler but extremely related fact: a set of natural numbers containing $1$ and closed under multiplication either contains infinitely many numbers or just one.

John Baez (Dec 12 2020 at 20:45):

And notice, the phrase "just one" is a pun: you can read that in two ways, and the sentence is true either way.

Shea Levy (Dec 12 2020 at 20:45):

"all constructions automatically respect equivalences" is the key thing here. Any reasoning you can do categorically about one object carries over mechanically to any other object isomorphic to it.

Also might be helpful to distinguish a pre-mathematical notion of sets, where the elements matter, from the category of sets, which IMO is better thought of as "the category of the empty algebraic theory and its homomorphisms." (thanks to Chris Brown for this formulation)

John Baez (Dec 12 2020 at 20:46):

I think that sort of remark is mainly gonna make Eric more confused, Shea. We're talking about some really basic stuff here; "the empty algebraic theory" is about 10 times harder to understand than this stuff.

Fawzi Hreiki (Dec 12 2020 at 20:46):

Hold on, surely the full subcategory on $0$ and $1$ has two isomorphism classes.

Dan Doel (Dec 12 2020 at 20:46):

If you try to model this with equivalence classes, I think you get into trouble because things can be equivalent/isomorphic in inequivalent ways, and set theoretic equivalence classes start forcing you to do messy stuff like choosing particular isomorphisms and such.

John Baez (Dec 12 2020 at 20:48):

Fawzi Hreiki said:

Hold on, surely the full subcategory on $0$ and $1$ has two isomorphism classes.

Whoops, I was wrong! Okay: a monoidal subcategory of $(\text{Set},\times,1)$ has either infinitely many isomorphism classes of objects, or just two, or just one. This time I swear I'm right. :upside_down:

John Baez (Dec 12 2020 at 20:50):

And the other thing I said was wrong too. A submonoid of $(\mathbb{N},\times,1)$ has either infinitely many elements, or just two, or just one.

Eric Forgy (Dec 12 2020 at 20:52):

Here's a simpler but extremely related fact: a set of natural numbers containing $1$ and closed under multiplication either contains infinitely many numbers or just one.

Yeah. I can see that, so how do we reconcile? :blush:

It is the same thing. It is either "just one" $\{1\}$ or it is $\{1,1\times 1,1\times(1\times 1),\cdots\}$ together with a map $m$ :sweat_smile:

Dan Doel (Dec 12 2020 at 20:54):

I don't think those are the only choices of technical set theoretic details.

John Baez (Dec 12 2020 at 20:55):

All those sets $1$, $1 \times 1$, etc. are isomorphic. That's why I said this (after correction):

John Baez said:

a monoidal subcategory of $(\text{Set},\times,1)$ has either infinitely many isomorphism classes of objects, or just two, or just one.

Dan Doel (Dec 12 2020 at 20:55):

Like, you could have $\{1, 1×1, 1×(1×1), (1×1)×1\}$, and then all further products are set-theoretic-equal to something else in that set, maybe.

Dan Doel (Dec 12 2020 at 20:57):

Like, what if it's doing something with the elements of the singleton set mod-3?

Dan Doel (Dec 12 2020 at 20:57):

That doesn't matter for the categorical structure, but it's 'technically not the same set'.

Eric Forgy (Dec 12 2020 at 20:59):

So is there even a difference between $\text{Set}$ and natural numbers $\mathbb{N}$ categorically thinking? Since two sets with the same number of elements are isomorphic?

Dan Doel (Dec 12 2020 at 20:59):

You can have way more than $ℕ$ many elements of a set.

John Baez (Dec 12 2020 at 21:00):

If you do category theory correctly, it never makes a bit of difference how many objects you have in a category. Category theorists never care about this. All that can matter is how many isomorphism classes of objects.

There are occasional exceptions to this rule of thumb, but if you're a beginner and you try to claim you're in one of these exceptional cases, there's a 99% chance you're doing something dumb.

Whenever you start trying to count objects in a category, a big alarm bell should start ringing, and big red warning light should start flashing.

John Baez (Dec 12 2020 at 21:00):

Eric Forgy said:

So is there even a difference between $\text{Set}$ and natural numbers $\mathbb{N}$ categorically thinking?

Yes, because there are things called "infinite sets", but not "infinite natural numbers".

Eric Forgy (Dec 12 2020 at 21:01):

How about $\text{FinSet}$ and $\mathbb{N}$? Those should be the same categorically, right?

Dan Doel (Dec 12 2020 at 21:02):

Depends what you mean by $ℕ$, but those are more similar, yeah.

John Baez (Dec 12 2020 at 21:03):

Don't humor him so much, Dan.

Eric Forgy (Dec 12 2020 at 21:03):

So the moral is: Don't be evil. If you are going to count something, only count isomorphism classes.

John Baez (Dec 12 2020 at 21:03):

$\text{FinSet}$ is a category: it has objects and morphisms.

$\mathbb{N}$ is a set: it just has elements.

They're different.

John Baez (Dec 12 2020 at 21:04):

Eric Forgy said:

So the moral is: Don't be evil. If you are going to count something, only count isomorphism classes.

Right.

Eric Forgy (Dec 12 2020 at 21:05):

That is going to be hard for me, but I will try. Thank you :raised_hands:

I suck in counting, but one thing I suck in even more is understanding isomorphism classes (I know how they are defined but lack any intuition).

John Baez (Dec 12 2020 at 21:06):

A set of five penguins is isomorphic to a set of five politicians because you can draw a little line connecting each penguin to its own politician: a one-to-one correspondence.

John Baez (Dec 12 2020 at 21:07):

In category theory we say "the five-element set" because we don't give a shit about the difference between penguins and politicians.

John Baez (Dec 12 2020 at 21:08):

So: the correct statement, that you were working towards, is this:

Isomorphism classes of objects in FinSet are in 1-1 correspondence with natural numbers.

John Baez (Dec 12 2020 at 21:09):

This sentence means the exact same thing:

Decategorifying $\text{FinSet}$, we get $\mathbb{N}$.

John Baez (Dec 12 2020 at 21:10):

All this is a fancy way to say:

We can tell if two finite sets are isomorphic by counting them and seeing if they have the same number of elements.

John Baez (Dec 12 2020 at 21:10):

This is the reason they invented counting.

Dan Doel (Dec 12 2020 at 21:11):

The point isn't just, "don't be evil," though. The point is that categorical stuff actually works up to equivalences, so if you think about things in an "evil" way, and say, "well, actually there are infinitely many of these things, not one," you'll be missing the way things actually behave. And sometimes it'll be like, "well, there is technically a proper class of these things," but categorically the class will act like a small set.

Eric Forgy (Dec 12 2020 at 21:11):

So after all that, maybe I can reframe my original question.

If $S$ is a non-unital object in $\text{Set}$, then the smallest monoidal subcategory of $(\text{Set},\times, 1)$ having $S$ as an object has an infinite number of isomorphism classes of objects. Namely $S$, $S\times S$, $S\times S\times S$ (where $S\times(S\times S) \cong (S\times S)\times S$), etc.

Is that right?

John Baez (Dec 12 2020 at 21:12):

No. You forgot about the empty set, just like I did.

Eric Forgy (Dec 12 2020 at 21:12):

At least I'm in good company :smiley:

John Baez (Dec 12 2020 at 21:13):

Yeah. It's really fun to slap you down for making the exact same mistake I got corrected on 2 minutes ago.

John Baez (Dec 12 2020 at 21:14):

It almost makes that mistake seem worthwhile.

Eric Forgy (Dec 12 2020 at 21:14):

Take two...

If $S$ is a non-unital object in $\text{Set}$, then the smallest monoidal subcategory of $(\text{Set},\times, 1)$ having $S$ as an object has an infinite number of isomorphism classes of objects. Namely $0$, $S$, $S\times S$, $S\times S\times S$ (where $S\times(S\times S) \cong (S\times S)\times S$), etc.

Is that right?

John Baez (Dec 12 2020 at 21:15):

No.

Eric Forgy (Dec 12 2020 at 21:15):

The smallest monoidal subcategory of $(\text{Set},\times,1)$ would be $\{0,1\}$?

John Baez (Dec 12 2020 at 21:16):

No.

Eric Forgy (Dec 12 2020 at 21:16):

:face_palm:

John Baez (Dec 12 2020 at 21:16):

It's the subcategory containing just a 1-element set and its identity morphism.

Eric Forgy (Dec 12 2020 at 21:17):

Ah! Because in that case $0\cong 1$?

John Baez (Dec 12 2020 at 21:18):

Huh??? No, the empty set is not isomorphic to the one-element set.

Eric Forgy (Dec 12 2020 at 21:18):

You have to admit, it is impressive the number of ways I can be wrong :joy:

John Baez (Dec 12 2020 at 21:18):

I think if you just answered randomly you'd have a better batting average.

John Baez (Dec 12 2020 at 21:20):

Wrong:

If $S$ is a non-unital object in $\text{Set}$, then the smallest monoidal subcategory of $(\text{Set},\times, 1)$ having $S$ as an object has an infinite number of isomorphism classes of objects. Namely $0$, $S$, $S\times S$, $S\times S\times S$ (where $S\times(S\times S) \cong (S\times S)\times S$), etc.

Right:

If $S$ is a non-unital object in $\text{Set}$, then the smallest monoidal subcategory of $(\text{Set},\times, 1)$ having $S$ as an object has these isomorphism classes: $1$, $S$, $S\times S$, $S\times S\times S$ (where $S\times(S\times S) \cong (S\times S)\times S$), etc.

Note: we need to make two changes to correct the answer.

Eric Forgy (Dec 12 2020 at 21:20):

I think I see where I messed up...

Take three...

If $S$ is a non-unital object in $\text{Set}$, then the smallest monoidal subcategory of $(\text{Set},\times, 1)$ having $S$ as an object has an infinite number of isomorphism classes of objects. Namely $1$, $S$, $S\times S$, $S\times S\times S$ (where $S\times(S\times S) \cong (S\times S)\times S$), etc.

Is that right?

Eric Forgy (Dec 12 2020 at 21:20):

Checking...

John Baez (Dec 12 2020 at 21:20):

No, it's not right.

John Baez (Dec 12 2020 at 21:20):

It's not right when $S$ is empty.

Eric Forgy (Dec 12 2020 at 21:20):

Checking for the difference...

John Baez (Dec 12 2020 at 21:21):

When $S$ is empty you don't get infinitely many isomorphism classes, just two.

Eric Forgy (Dec 12 2020 at 21:21):

I see

Eric Forgy (Dec 12 2020 at 21:24):

Attempt four...

If $S$ is a non-initial and non-terminal object in $\text{Set}$, then the smallest monoidal subcategory of $(\text{Set},\times, 1)$ having $S$ as an object has an infinite number of isomorphism classes of objects. Namely $1$, $S$, $S\times S$, $S\times S\times S$ (where $S\times(S\times S) \cong (S\times S)\times S$), etc.

If $S$ is the empty set, the smallest monoidal subcategory of $(\text{Set},\times, 1)$ is $\{0,1\}$.

Is that right?

Eric Forgy (Dec 12 2020 at 21:26):

Or better...

If $S$ is a non-empty non-unital object in $\text{Set}$, then the smallest monoidal subcategory of $(\text{Set},\times, 1)$ having $S$ as an object has an infinite number of isomorphism classes of objects. Namely $1$, $S$, $S\times S$, $S\times S\times S$ (where $S\times(S\times S) \cong (S\times S)\times S$), etc.

If $S$ is the empty set, the smallest monoidal subcategory of $(\text{Set},\times, 1)$ is $\{0,1\}$.

Is that right?

Eric Forgy (Dec 12 2020 at 21:26):

Typo! Fixed.

Reid Barton (Dec 12 2020 at 21:27):

This is still false actually

Reid Barton (Dec 12 2020 at 21:28):

But, it's at least something that could be true. :upside_down:

Reid Barton (Dec 12 2020 at 21:28):

$S$ could be an infinite set.

Eric Forgy (Dec 12 2020 at 21:28):

Reid Barton said:

But, it's at least something that could be true. :upside_down:

I'll take that as a win :joy:

Eric Forgy (Dec 12 2020 at 21:40):

Thank you everyone :raised_hands:

Eric Forgy (Dec 12 2020 at 21:42):

And at the risk of being wrong yet again, I think the "free functor" $F: \text{Set}\to\text{Vect}$ is monoidal, i.e. preserves products.

Eric Forgy (Dec 12 2020 at 21:46):

So if I want to study that functor by looking at what it does to a finite non-empty set $S$, then I need not just $S$, but the monoidal subcategory of $\text{Set}$ whose objects include $1$, $S$ and products of $S$ with itself.

Eric Forgy (Dec 12 2020 at 22:02):

Now, I have some homework. I'm trying to understand

$X\overset{\mathcal{G}}{\to}\text{Set}\overset{\mathcal{F}}{\to}\text{Vect}$

where

$\mathcal{G}: X\to\text{Set}$

is a directed graph and my understanding of $\mathcal{F}: \text{Set}\to\text{Vect}$ was wrong, so I wasn't getting anywhere.

Thank you again :raised_hands:

John Baez (Dec 12 2020 at 22:07):

Eric Forgy said:

Or better...

If $S$ is a non-empty non-unital object in $\text{Set}$, then the smallest monoidal subcategory of $(\text{Set},\times, 1)$ having $S$ as an object has an infinite number of isomorphism classes of objects. Namely $1$, $S$, $S\times S$, $S\times S\times S$ (where $S\times(S\times S) \cong (S\times S)\times S$), etc.

If $S$ is the empty set, the smallest monoidal subcategory of $(\text{Set},\times, 1)$ is $\{0,1\}$.

Is that right?

It's false, because as Reid Barton pointed out, if $S$ is infinite then

$S \cong S \times S \cong S \times S \times S \cong \cdots$

So in this case the collection $1$, $S$, $S\times S$, $S\times S\times S, \dots$ has just two isomorphism classes of objects.

If you don't want to think about infinite sets, just say "finite sets".

Eric Forgy (Dec 12 2020 at 22:11):

If I summarize:

If $S=1$, the smallest monoidal subcategory is $\{1\}$.

If $S=0$ or $S$ is infinite, the smallest monoidal subcategory is $\{1,S\}$.

If $S$ is finite, the smallest monoidal subcategory is $\{1,S,S\times S,\cdots\}$.

John Baez (Dec 12 2020 at 22:11):

Eric Forgy said:

And at the risk of being wrong yet again, I think the "free functor" $F: \text{Set}\to\text{Vect}$ is monoidal, i.e. preserves products.

Something like that is true, but you should say it correctly.

The category $\mathsf{Vect}$ has "products" $V \times W$ and also "tensor products" $V \otimes W$. "Preserving products" would mean

$F(X \times Y) \cong F(X) \times F(Y)$

which is false. It's better to say

$F: (\text{Set}, \times) \to (\text{Vect}, \otimes)$ is monoidal.

By the way, it's also true that

$F: (\text{Set}, +) \to (\text{Vect}, \times)$ is monoidal.

John Baez (Dec 12 2020 at 22:13):

It's also true that

$F : (\text{Set}, +) \to (\text{Vect}, +)$ is monoidal

since in $\text{Vect}$ it happens that products ($\times$) are also coproducts ($+$, also written $\oplus$ in this example).

John Baez (Dec 12 2020 at 22:14):

In $\text{Set}$, $+$ and $\times$ are very different, so one of the charms of linear algebra is that in $\text{Vect}$ they are the same, and we get a new monoidal structure $\otimes$.

John Baez (Dec 12 2020 at 22:15):

I found this confusing for years, since nobody would come out say it.

Eric Forgy (Dec 12 2020 at 22:16):

In the case of left adjoint to the forgetful functor, $U: \mathsf{Vect}\to\mathsf{Set}$, which of those $F$ s are we talking about? :blush:

Eric Forgy (Dec 12 2020 at 22:17):

Or are they all the same $F$?

Eric Forgy (Dec 12 2020 at 22:18):

I think I know

Eric Forgy (Dec 12 2020 at 22:21):

Instead of saying $\text{Vect}\to\text{Set}$ , we need to be more specific, e.g. $(\text{Vect},\otimes)\to(\text{Set},\times)$.

John Baez (Dec 12 2020 at 22:25):

We're talking about just one $F$ here, the functor from $\mathsf{Set}$ to $\mathsf{Vect}$ sending each set to the vector space with that set as basis. We haven't been talking about any other $F$'s lately.

John Baez (Dec 12 2020 at 22:25):

But we can make $\mathsf{Set}$ into a monoidal category in two famous ways, and ditto for $\mathsf{Vect}$.

John Baez (Dec 12 2020 at 22:28):

So there are 2 $\times$ 2 = 4 possible ways to try to make $F$ into a monoidal functor, and I told you which two work.

John Baez (Dec 12 2020 at 22:29):

A mathematician who suddenly changed what they meant by $F$ without warning in the middle of a conversation is a mathematician who'd deserve to be severely punished.

Eric Forgy (Dec 12 2020 at 22:30):

Would it be 3 ways for $\text{Vect}$ corresponding to $(\text{Vect},\otimes)$, $(\text{Vect},\times)$ and $(\text{Vect},\oplus)$?

John Baez (Dec 12 2020 at 22:31):

I just told you that for vector spaces $+ = \times = \oplus$.

John Baez said:

It's also true that

$F : (\text{Set}, +) \to (\text{Vect}, +)$ is monoidal

since in $\text{Vect}$ it happens that products ($\times$) are also coproducts ($+$, also written $\oplus$ in this example).

Eric Forgy (Dec 12 2020 at 22:32):

Sorry. I meant are those 3 valid monoidal categories (not thinking about functors)?

John Baez (Dec 12 2020 at 22:33):

Yes, those are 3 valid monoidal categories... and two of them are the same, I keep telling you.

Eric Forgy (Dec 12 2020 at 22:34):

Interesting :+1:

John Baez (Dec 12 2020 at 22:34):

There's no difference between $V \oplus W$ and $V \times W$, for vector spaces.

Eric Forgy (Dec 12 2020 at 22:34):

That feels like a homework assignment, but I believe you :blush:

John Baez (Dec 12 2020 at 22:35):

A guy in $V \oplus W$ or $V \times W$ or $V+W$ is just an ordered pair $(v,w)$ with $v \in V$ and $w \in W$.

John Baez (Dec 12 2020 at 22:35):

This is for vector spaces, mind you.

Eric Forgy (Dec 12 2020 at 22:37):

Yeah and $(kv)\times w \ne v\times(kw)$, but $(kv)\otimes w = v\otimes(kw)$.

John Baez (Dec 12 2020 at 22:41):

I don't know what $(kv) \times w$ means - nobody says that - but if it means the ordered pair $(kv,w)$ then you're right.

John Baez (Dec 12 2020 at 22:43):

It's a good homework assignment to show that the set of ordered pairs $(v,w)$ with $v \in V, w \in W$ is both the product $V \times W$ and the coproduct $V + W$. This is part of learning what products and coproducts are like in category theory.

Eric Forgy (Dec 12 2020 at 22:44):

I'm almost ready to move on and complete my homework ($X\to\text{Set}\to\text{Vect}$).

$(Vect,\otimes,1_{\otimes})$ is monoidal and

$(Vect,\oplus,1_{\oplus})$ is monoidal.

Are the two units the same?

John Baez (Dec 12 2020 at 22:45):

Figure it out!

John Baez (Dec 12 2020 at 22:45):

Figure out what they are.

Eric Forgy (Dec 12 2020 at 22:45):

I'll try :blush:

Eric Forgy (Dec 12 2020 at 22:46):

$1_\otimes = 1_\oplus = K$ :face_palm:

Dan Doel (Dec 12 2020 at 22:47):

That doesn't seem right, but I'm no vector guy.

Eric Forgy (Dec 12 2020 at 22:48):

I am pretty sure $1_\otimes = K$. Let me think more about $1_\oplus$.

Dan Doel (Dec 12 2020 at 22:48):

Since it was mentioned that $F$ preserves coproducts, and the coproducts correspond to the direct sum(?), then $1_\oplus \cong F(0)$, I think.

Eric Forgy (Dec 12 2020 at 22:55):

Dan Doel said:

Since it was mentioned that $F$ preserves coproducts, and the coproducts correspond to the direct sum(?), then $1_\oplus \cong F(0)$, I think.

This seems right where $F(0) = 0$, i.e. "the" 1-element vector space 0.

Eric Forgy (Dec 12 2020 at 22:56):

So

$(Vect,\otimes,K)$ is monoidal and

$(Vect,\oplus,0)$ is monoidal.

John Baez (Dec 12 2020 at 22:59):

Right!

John Baez (Dec 12 2020 at 22:59):

It's a lot like how $(\mathbb{N}, \times, 1)$ is a monoid and so is $(\mathbb{N}, +, 0)$.

Eric Forgy (Dec 12 2020 at 22:59):

What would be a way to denote that $\text{Vect}$ is monoidal in two ways?

$(\text{Vect},\oplus,\otimes,0,K)$?

John Baez (Dec 12 2020 at 23:03):

Well, there are two separate monoidal categories, which you already have names for. There's not much point in just slamming them together. But Vect becomes a "bimonoidal category" or "rig category" because $\otimes$ distributes over $\oplus$, up to isomorphism.

John Baez (Dec 12 2020 at 23:03):

I've been doing lots of work on rig categories these days.

Eric Forgy (Dec 12 2020 at 23:03):

Homework to myself:

If $V$ is a finite-dimensional non-empty vector space, what is the smallest monoidal subcategory of $\text{Vect}$ containing $V$ that preserves both $\otimes$ and $\oplus$?

Eric Forgy (Dec 12 2020 at 23:07):

I'm pretty sure this has to be:

$T(V) = K\oplus V\oplus (V\otimes V)\oplus (V\otimes V\otimes V)\cdots.$

Eric Forgy (Dec 12 2020 at 23:09):

Unlike $\text{Set}$, I think that should also be true if $V$ is infinite dimensional because I'm not sure if $V\otimes V\cong V$ when $V$ is infinite dimensional :thinking:

Eric Forgy (Dec 12 2020 at 23:15):

(I'm generally happy enough if I understand finite dimensional stuff because I ultimately want to write some code and do numerical computations)

John Baez (Dec 12 2020 at 23:20):

Eric Forgy said:

Homework to myself:

If $V$ is a finite-dimensional non-empty vector space, what is the smallest monoidal subcategory of $\text{Vect}$ containing $V$ that preserves both $\otimes$ and $\oplus$?

I'm pretty sure this has to be:

$T(V) = K\oplus V\oplus (V\otimes V)\oplus (V\otimes V\otimes V)\cdots.$

That answer makes no sense: $T(V)$ is a vector space, not a monoidal subcategory of $\text{Vect}$.

A key step to rapid progress in category theory is developing this mental habit: whenever you guess an answer to question, see if the guess parses, or type-checks. That is, see if it's the sort of thing that could be an answer to the question.

If the question is asking for a category, don't guess a vector space. If someone asks you who is the president of Russia, don't say "Moscow".

If you do this religiously, your batting average will immediately improve.

John Baez (Dec 12 2020 at 23:21):

I think I can go further and say that anyone who doesn't develop this habit is doomed to fail miserably in category theory.

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

I had one grad student who was so confused he thought "type errors" meant "typographical errors" even after I explained the idea repeatedly.

Eric Forgy (Dec 13 2020 at 00:01):

Instead of saying "be", I should have said "include".

If that monster term I wrote as $T(V)$ is an object in the smallest monoidal subcategory, then each "term" should also be an object, e.g. $0$, $K$, $V$, $V\otimes V$. I misspoke, but the morale is that even if $V$ is finite, the smallest monoidal subcategory that preserves both products (product and coproduct) is pretty heinous.

John Baez (Dec 13 2020 at 00:04):

That guy $T(V)$ is an infinite direct sum, it will not appear in the smallest category containing $V$ and closed under $\oplus$ and $\otimes$, if by $\oplus$ you mean the binary direct sum, like $X \oplus Y$.

John Baez (Dec 13 2020 at 00:04):

We hadn't been talking about infinite direct sums up to now, just monoidal structures like binary direct sum.

John Baez (Dec 13 2020 at 00:06):

$T(V)$ will indeed be an object in the smallest subcategory of $\text{Vect}$ containing $V$ and closed under $\otimes$ and infinite direct sums.

fosco (Jan 30 2021 at 16:51):

I would say the answer is "polynomials in $V$ of the form $\sum_{i=0}^d V^{\otimes i}$"?

Fawzi Hreiki (Jan 30 2021 at 17:09):

It might be worthwhile to draw the diagram consisting of the adjunctions between the monoidal categories of monoids, rigs, modules, algebras, etc..

Fawzi Hreiki (Jan 30 2021 at 17:10):

Either over some base symmetric monoidal category, or over say Ab if you want to be concrete

Fawzi Hreiki (Jan 30 2021 at 17:11):

Almost all the free functors in basic algebra come from either taking free monoids or taking free actions with respect to some appropriate base category.

Fawzi Hreiki (Jan 30 2021 at 17:27):

(And taking inverse completions)

Eric Forgy (Jan 30 2021 at 19:37):

For what's its worth, looking back on this discussion, I now better understand why I got hung up.

From what I understand, people sometimes say "monoidal category $C$", sometimes they say "monoidal category $(C,\otimes,1)$" and I don't recall seeing it, but I'm sure it happens that people say "monoidal category $(C,\otimes,1,a,\lambda,\rho)".$

Then, one ingredient of the definition of a monoid object in a monoidal category $(C,\otimes)$ (see, I made up my own way now) is a morphism
${}$
$\mu: c\otimes c\to c$
${}$
satisfying some conditions.

I looked at that :point_of_information: and confused myself by thinking that morphism can only be defined if both objects $c\otimes c$ and $c$ were already present a priori in the category. I found the escape hatch thinking in terms of code. The symbol $\otimes$ is like code that takes two objects $c$ and $c'$ and returns a new object $c\otimes c'$. So just like we don't need to have every instance of a floating point number in memory to talk about multiplication, we just need to know the rule how to multiply numbers. It sounds so obvious (and it is once you get it), but we don't need to have both $c\otimes c$ and $c$ hanging around "in memory" to define $\mu$. We just need some rules defining $\otimes$ and then compose those rules with the rules for how to define $\mu$ :face_palm:

John Baez (Jan 30 2021 at 19:49):

Yeah, people say "the monoidal category $C$" when they feel like saving time, usually when the monoidal structure in question has already been specified earlier in the discussion. But what's a monoid object in $C$ depends on the monoidal structure... and as we've seen, a monoid object in $(\mathsf{Vect}, \otimes)$ is utterly different than a monoid object in $(\mathsf{Vect}, \oplus)$: the first is an algebra, the second is just a vector space.

Eric Forgy (Jan 30 2021 at 19:50):

Yes. I will get there eventually :blush:

Eric Forgy (Jan 30 2021 at 19:53):

Btw, I don't recall reading anywhere that a monoid object in $(\mathsf{Vect},\oplus)$ is a vector space, which is kind of cool albeit trivial / obvious. First time students might benefit from seeing it as an example (or homework).

John Baez (Jan 30 2021 at 19:55):

Well, you read it here.

Eric Forgy (Jan 30 2021 at 19:58):

I know I learned that from you here on this Zulip recently, but I don't remember seeing it before, but that isn't saying much because my memory is so bad :older_man: I'm not surprised you teach this example :blush: :raised_hands:

John Baez (Jan 30 2021 at 20:02):

James Dolan made me figure out what a comonoid in $(\mathsf{Set}, \times)$ is, as part of teaching me that monoids in cocartesian categories are just objects.

John Baez (Jan 30 2021 at 20:04):

This really lies near the foundations of algebraic geometry: namely, the way that commutative monoids resemble spaces, but in the opposite category.

Eric Forgy (Jan 30 2021 at 20:10):

I hope to understand that one day when I grow up. Those kinds of questions are the reason I'm trying to learn CT in the first place.

John Baez (Jan 30 2021 at 20:12):

To make that a bit more precise: we can tentatively define a generalized "space" to be an object in any category $C$ with finite products and coproducts, where products distribute over coproducts. Then every object in $C$ is a cocommutative comonoid in $(C, \times)$ in a unique way, and every morphism of $C$ becomes a homomorphism of cocommutative cmonoids. Thus, every object in $(C^{\rm op}, \times)$ is a commutative monoid object - where I'm writing $\times$ for the product in $C$, which is the coproduct in $C^{\rm op}$.

A commutative monoid object is a generalized "commutative ring".

So, we see that the opposite of a category of generalized spaces is a category of generalized commutative rings.

John Baez (Jan 30 2021 at 20:13):

Lawvere calls this something like the duality between "substance" and "quantity" - where "generalized spaces" are what he'd call"substance", and "generalized commutative rings" are what he'd call "quantity".

John Baez (Jan 30 2021 at 20:14):

He has some paper or something on "categories of substance and quantity".

John Baez (Jan 30 2021 at 20:19):

Note by the way that algebras of differential forms, or more generally supercommutative dgas, are commutative monoid objects in a certain monoidal category, so they fit into this business.

Eric Forgy (Jan 30 2021 at 20:22):

Very cool stuff :heart_eyes:

I've always felt my research in discrete differential geometry related to all these beautiful concepts (the relation to NCG is obvious), but they were just out of my mathematical reach. The last few months have helped a lot, but the road ahead is still long for me :pray:

John Baez (Jan 30 2021 at 20:23):

It's startlingly short. Everything I just said is something you know.

John Baez (Jan 30 2021 at 20:24):

That is, every mathematical fact I just said is something we've been either been talking about a lot lately, or a trivial observation. (Maybe you didn't notice that the algebra of differential forms is a commutative monoid object in the category of supervector spaces, but if I wrote this out in formulas you'd know all the formulas.)

John Baez (Jan 30 2021 at 20:24):

I was just packaging it up in some words.

Eric Forgy (Jan 30 2021 at 20:27):

That is encouraging and I'm sure I'll wake up one day and everything will just click and be completely obvious, but I'm still feeling a bit overwhelmed. Until yesterday, I thought tensor product was pushout in $\mathbf{Vect}$ :face_palm:

Eric Forgy (Jan 30 2021 at 20:36):

Tensor product is pushout in $\mathbf{CAlg}$, but that leap feels a little unsatisfactory to me so I need some meditation. Instead of jumping from
${}$
$\mathbf{Set\to CAlg}$
${}$
I might take a detour through
${}$
$\mathbf{Set\to Vect}.$

John Baez (Jan 30 2021 at 20:43):

Eric Forgy said:

That is encouraging and I'm sure I'll wake up one day and everything will just click and be completely obvious, but I'm still feeling a bit overwhelmed. Until yesterday, I thought tensor product was pushout in $\mathbf{Vect}$ :face_palm:

Oookay.... I guess you never tried an example? It's good to always look at examples.

John Baez (Jan 30 2021 at 20:44):

Whenever you think something, you should have some examples hovering in your mind. Abstraction is nice but the human mind is frail.

John Baez (Jan 30 2021 at 20:44):

I actually think you will discover that $\mathsf{CommAlg}$ is what you want, or more likely $\mathsf{CocommCoalg}$.

John Baez (Jan 30 2021 at 20:45):

The functor $K[-]: \mathsf{Set} \to \mathsf{Vect}$ that you like actually lands in $\mathsf{CocommCoalg}$.

John Baez (Jan 30 2021 at 20:45):

This follows from the fact that symmetric monoidal functors send cocommutative comonoids to cocommutative comonoids.

John Baez (Jan 30 2021 at 20:46):

A cocommutative comonoid in $(\mathsf{Set}, \times)$ is just a set.

A cocommutative comonoid in $(\mathsf{Vect}, \otimes)$ is a cocommutative coalgebra.

$K[-]: (\mathsf{Set}, \times) \to (\mathsf{Vect}, \otimes)$ is symmetric monoidal.

So, it sends sets to cocommutative coalgebras!

This is following the tao: not doing anything sneaky, just using what's there.

Eric Forgy (Jan 30 2021 at 20:46):

Sure. I know I want $\mathbf{CAlg}$ and I do have examples in mind. My situation is that I already know the bimodule I want to land in and I'm just filling in details. I know it must work at the end of the day, so I am being too sloppy.

Eric Forgy (Jan 30 2021 at 20:52):

There is a nice natural progression in terms of concepts:
${}$
$\mathbf{Cospan}(\mathbb{N},+)\cong\mathbf{Bim}(\mathbb{N},+)$
${}$
$\mathbf{Span(Set,\times)\cong Bim(Set^\mathsf{op},\times)}$
${}$
$\mathbf{Cospan(Vect,\oplus)\cong Bim(Vect,\oplus)}$
${}$
$\mathbf{Cospan(CAlg,\otimes)\cong Bim(CAlg,\otimes)}$
${}$
I'm trying to tell a nice story. I made a mistake blurring the last two (was anxious).

Eric Forgy (Jan 30 2021 at 20:56):

Btw, I defined $\mathbf{Cospan}(\mathbb{N},+)$ here.

John Baez (Jan 31 2021 at 17:44):

I agree that the right story should consist of a lot of beautiful functors between a lot of beautiful categories.

John Baez (Jan 31 2021 at 17:44):

I still think you're downplaying the importance of cocommutative coalgebras, probably because they're a bit less familiar.

John Baez (Jan 31 2021 at 17:46):

But the fact is, if you've got a set $S$, the vector space $K[S]$ is functorially a coalgebra but not an algebra, unless $S$ is finite.

John Baez (Jan 31 2021 at 17:49):

As you mentioned, this is connected to the difference between chains and cochains. If we've got a graph or simplicial set and $S$ is its set of vertices, $K[S]$ can be thought of as the vector space of 0-chains on that graph or simplicial set. This is naturally a coalgebra. Its dual $K[S]^\ast$, the space of 0-cochains, is naturally an algebra.

John Baez (Jan 31 2021 at 17:51):

You may want to check that $K[S]$ is not an algebra when $S$ is infinite, if you don't believe me.

Eric Forgy (Jan 31 2021 at 22:25):

Thank you for your notes John :pray:

John Baez said:

I still think you're downplaying the importance of cocommutative coalgebras, probably because they're a bit less familiar.

Although it is true I am less familiar with coalgebra, I wouldn't put it like that (I'm not familiar with most of this stuff and that's never stopped me). I'm just progressing painfully slow. My thinking is still along the same lines we discussed, i.e. $K[-]$ gets us closer to chains / coalgebras than cochains / algebras and then I'll look at the dual of chains to get the algebra I'm ultimately after.

Roughly, I have this nice simple bicategory $\mathbf{Cospan}(\mathbb{N},+)$ and a new favorite functor
${}$
$G: \mathbf{Cospan}(\mathbb{N},+)\to\mathbf{Span(Set,\times)}$
${}$
which gives not only the directed graphs and paths we've talked about before in a nice compact fashion, but naturally also gives the adjoint (reversed) graphs and paths
${}$
$E^\dagger: V\to V$
${}$
solving some other riddles I was putting off in the process, which makes me believe I'm on the right path. So now I want to insist all my functors respect adjoints, i.e.
${}$
$\dagger\circ F = F\circ \dagger$
${}$
just like those other papers on CT in QM.

Then, from $\mathbf{Span(Set,\times)}$, the next functorial step involving $K[-]$ should give a path (co?)algebra including adjoint / reversed paths. From there, I think I'll finally get the "Diamond complex", which is the main contribution. Diamond complexes are actually defined very similarly to Moore complexes. I was surprised when I stumbled on the definition of Moore complex by how similar the definition is to diamonds. A Moore complex is like a forward facing cone. A diamond is like a combination of a Moore complex and a reversed Moore complex (forming something like a causal set).

Once I've got the diamond complex, the DGA stuff will follow naturally along the lines of the universal differential calculus stuff I was looking at originally, but taking care to carry $\dagger$ along for the ride, which will be used to define an inner product, Hodge star, etc.

This route is harder than the treasure map you outlined. If it ends up reproducing the same thing, that would be pretty awesome. If it ends up producing something new (which I still think is the case), it will be even more awesome :blush:

This, from our paper, is the general roadmap:

image.png

Eric Forgy (Jan 31 2021 at 22:30):

I'm getting there a little different than Urs and I did originally. Back then, we postulated $\langle\cdot |\cdot\rangle$ and derived everything. I'm trying to build things up incrementally / functorially so trying to follow an approach more similar to your note:
${}$

• The *-Category of Hilbert Spaces

${}$
So, today, I'm actually wondering if the functorial step after $\mathbf{Span(Set,\times)}$ should land me in a $C^*$-(co)algebra :thinking:

In other words, the metric can be derived from $\dagger.$