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Stream: theory: category theory

Topic: categorifying eigenvalues

Jade Master (May 01 2021 at 04:54):

I took a shot at categorifying eigenvalues and eigenvectors in this blog post:

Jade Master (May 01 2021 at 04:55):

• categorifying eigenvalues

fosco (May 01 2021 at 08:39):

Eigenvalues only make sense for endo-morphisms though. So $M : {\bf n}\times {\bf n}^o \to Set$, and there's a typo in the definition, because the domain and codomain wouldn't match otherwise.

Apart from this, as I said on twitter, I was thinking about the notion of a "general linear group": one can define the subcategory of all equivalences $C \to D$ in $\sf Prof$ to be its most proximate cognate. If you assume the categories to be Cauchy-complete then this is not really different from the bicategory having objects finite categories, equivalences (in $\sf Prof$!) between them and natural transformations of functors $C^o\times D \to Set$.

@JadeMasterMath The "general groupoid" is the subcategory of self-equivalences of objects of this category. An "invertible matrix" is an endo profunctor p over some n, which is a categorical equivalence.

- amuse-fouche (@ququ7)

I suspect this object turns out to be informative about ${\sf Prof}_F$ (profunctors of finite categories), and it has a name, somewhere.

Also, note how in this setting an "invertible matrix" can be rectangular: if $D$ is terminal, and $C$ is the walking isomorphism (so a groupoid, so Cauchy complete), then there is an equivalence in $\sf Prof$ between the two. So... vectors can be invertible matrices!

I suspect this poses a real problem when one tries to capture the notion of determinant, and before that of antisymmetric tensor. Another problem is that there seems to be no easy way to categorify the expression of the determinant

$\det A = \sum_\sigma\prod_i (-1)^{|\sigma|} A_{i\sigma i}$

(what is -1? how can you sum over permutations if A isn't square?)

Joachim Kock (May 01 2021 at 09:52):

I think the categorification maybe works better if the 'underlying vector space' of a category is defined to have as basis $\pi_0$ of the category rather that its set of objects. I confess I don't know much about the category case; I just extrapolate from groupoids. In the groupoid case there is already quite a bit of elementary linear algebra developed (see Baez-Hoffnung-Walker, Morton, Gálvez-Kock-Tonks). Regarding the minus sign in the determinant definition, it might be a good idea to split it into an odd and an even part and deal with them separately. Lawvere and Menni used this approach to give an objective account of Möbius inversion, where there is an alternating sign very similar.

Morgan Rogers (he/him) (May 01 2021 at 10:08):

When you say "we can decategorify to a matrix $\mathrm{ob}(C) \times \mathrm{ob}(D) \to \mathbb{N}$" I guess you're implicitly saying that the profunctors in FinProf are valued in finite sets?

Morgan Rogers (he/him) (May 01 2021 at 10:13):

I tend to agree with Joachim Kock that $\pi_0$ would be a better choice to use than $\mathrm{ob}$ for decategorification, since that resolves some of the issues that fosco was pointing out

Cole Comfort (May 01 2021 at 10:14):

What might help is that there is a notion of trace in Prof: namely taking the categorical trace induced by the compact closed structure. In matrices these two things are the same, and the trace is the sum of all the eigenvalues

Simon Willerton (May 01 2021 at 11:20):

Cole Comfort said:

What might help is that there is a notion of trace in Prof: namely taking the categorical trace induced by the compact closed structure. In matrices these two things are the same, and the trace is the sum of all the eigenvalues

There's at least two notions of trace in Prof for an endo-profunctor $P\colon C\to C$. As well as that one using the closed monoidal structure, you can take one using the bicategorical structure, namely the hom-object $\mathrm{Hom}(\mathrm{Id}_C, P)$.

Cole Comfort (May 01 2021 at 11:43):

I didn't know that. Although the trace induced by the compact closed structure seems to be the immediate generalization of the matrix trace.

fosco (May 01 2021 at 11:59):

I was thinking about the decomposition on connected components to split a generic profunctor into profunctors with connected domains, but it's not clear to me why it's useful to consider a single connected component as a "point in a set that parametrises a dimension"

Simon Willerton (May 01 2021 at 12:33):

Cole Comfort said:

...the trace induced by the compact closed structure seems to be the immediate generalization of the matrix trace.

That's a highly subjective statement! :grinning: It very much depends on how you think of matrix trace. I wouldn't have come down on either side. Certainly, if you think of matrix trace as being the sum of the diagonal elements then this 'diagonal trace' - as I call it - would seem to be a very natural analogue.

Anyway, if you're interested, I did give some talks on it way back, but failed to finish the paper on it. You can find slides at "Two 2-traces" in the talks section on my website.

Simon Willerton (May 01 2021 at 12:37):

The categorical trace in Prof is the coend $\int^c P(c,c)$ and the diagonal trace is the end $\int_c P(c,c)$.

Cole Comfort (May 01 2021 at 13:02):

Oh yes I saw tthese slides before but I didn't realize there were recordings on youtube!

Jade Master (May 01 2021 at 13:12):

@fosco Thanks I will correct the typo. This is super interesting about how invertible profunctors can be rectangular...maybe this is a feature but maybe not and it would be good to take $\pi_0$ as decategorification instead like @Joachim Kock said. That seems like a good suggestion because it's more standard thing to do? And also it eliminates this issue with rectangular matrices being invertible.i should definitely check out the papers referenced as I am guessing they contain a lot of ways to improve this idea. @Morgan Rogers (he/him) I am assuming that they profunctors are valued in finite sets but I didn't say that (and I should have)

Tim Campion (May 01 2021 at 13:15):

I think I once watched at least part of a talk by Ben Elias titled "categorifying eigenvalues". I think maybe he had a related paper?

fosco (May 01 2021 at 13:16):

I don't know if this was mentioned already, but the intuition for ends as a trace matches with the fact that $\text{tr}(P\circ Q)\cong \text{tr}(Q\circ P)$ fo every two composable endo-profunctors.

Jade Master (May 01 2021 at 13:20):

@Cole Comfort @Simon Willerton yeah it would be really cool if you could show that the categorified trace is (isomorphic?) to some categorified sum of eigenvalues. I'm guessing that that the coend trace is the right one because I am using coends for profunctor composition? That's just an intuition though.

Jade Master (May 01 2021 at 13:22):

@fosco That fact is true for categorified traces because you the cartesian product is symmetric monoidal? Thanks that seems like a really useful property.

Jade Master (May 01 2021 at 13:23):

@Tim Campion if you find it let me know :)

Jade Master (May 01 2021 at 13:24):

Anyway thanks so far y'all have given me lots of fun things to think about.

Tim Campion (May 01 2021 at 13:28):

@Jade Master Okay -- here we go: Elias and Hogancamp, "Categorical Diagonalization" https://arxiv.org/abs/1707.04349 I'm thinking that diagonalization is quite closely related to finding eigenvalues. I think they are focused on an abelian / linear - type setting, which might be a little different from what's going on here.

Tim Campion (May 01 2021 at 13:29):

And here's a talk by Ben : https://www.youtube.com/watch?v=y_kya4UXoaU

Tim Campion (May 01 2021 at 13:29):

A youtube search reveals a few more talks as well.

Jade Master (May 01 2021 at 13:33):

@Tim Campion Thanks this looks great

fosco (May 01 2021 at 13:49):

Jade Master said:

fosco That fact is true for categorified traces because you the cartesian product is symmetric monoidal? Thanks that seems like a really useful property.

Yes, plus Fubini rule.

$\displaystyle \text{tr}(P\circ Q) = \int^a\int^x P(a,x)\times Q(x,a) \cong \int^b\int^y Q(y,b)\times P(b,y) = \text{tr}(Q\circ P)$

Jade Master (May 01 2021 at 14:03):

Nice.

Matteo Capucci (he/him) (May 02 2021 at 22:02):

Uh really cool Jade! Skimming rapidly, I think you swapped indices in the coend formulas (lower indices indicate an end). Also, what's going on at the end? Is $\beta$ supposed to be an iso $M;v \to \lambda \cdot v$?

Matteo Capucci (he/him) (May 02 2021 at 22:03):

Still dimensions don't really match :thinking: on the left we have $M; v : m \to 1$ and on the right $\lambda \cdot v : n \to 1$

Matteo Capucci (he/him) (May 02 2021 at 22:04):

fosco said:

Eigenvalues only make sense for endo-morphisms though. So $M : {\bf n}\times {\bf n}^o \to Set$, and there's a typo in the definition, because the domain and codomain wouldn't match otherwise.

Ha, this is probably due to this

fosco (May 03 2021 at 07:24):

"How to link categories and linear algebra" is a questione that has bugged me for a long time. So, this conversation is really following in my dreams :grinning:

Unfortunately, I don't give my best in asynchronous communications: is anyone willing to discuss the matter live?

Jade Master (May 03 2021 at 13:58):

@Matteo Capucci (he/him) Thanks finally fixed the typos. I feel like coend indices should be on the bottom because that's how it works for integrals but alas. @fosco I am interested but only after I am finished with my thesis work in a month-ish give or take.

Tim Campion (May 03 2021 at 15:45):

Jade Master said:

Matteo Capucci (he/him) Thanks finally fixed the typos. I feel like coend indices should be on the bottom because that's how it works for integrals but alas. fosco I am interested but only after I am finished with my thesis work in a month-ish give or take.

Yeah, of course the convention is that it's ends which have lower indices / coends with upper indices. One could definitely argue that the opposite convention might have made more sense, since coends definitely feel more "integral-y" than ends, and are probably used explicitly more often than ends are. It might not be too late to change the convention, but one would probably want terminology / notation mavens like @Mike Shulman to weigh in before taking such a bold stance!

Tim Campion (May 03 2021 at 15:46):

One reason that coends are more "integral-like" than ends is that they are "colimit-y" in nature, which is kind of like integrals being "sum"-like in nature.

Tim Campion (May 03 2021 at 15:48):

There's also a funny thing -- the most common type of coend to take is $\int F(X) \otimes G(X)$ where $\otimes$ is a monoidal structure, whereas the most common type of end to take is $\int [F(X),G(X)]$ where $[-.-]$ is a closed structure. In both cases, one is composing some "default-like" bifunctor with some 1-variable functors, but the curious thing is that only in the first case is the bifunctor typically written with infix notation.

Tim Campion (May 03 2021 at 15:49):

The infix notation makes it feel a lot more similar to the form of an integral like $\int f(x) dx$ or more generally $\int \omega \wedge \eta$ , where there's this pairing between a function and a differential form, or just generally between different differential forms.

Tim Campion (May 03 2021 at 15:50):

Maybe what we really need to do is to convince every analyst in the world to start writing their integrals with the domain as a superscript, to match the category-theoretic convention ;).

Mike Shulman (May 03 2021 at 15:56):

Tim Campion said:

It might not be too late to change the convention, but one would probably want terminology / notation mavens like Mike Shulman to weigh in before taking such a bold stance!

Too late, I would say. Not an important enough issue to be worth the immense confusion it would cause.

Jade Master (May 04 2021 at 01:19):

Aw too bad.

Matteo Capucci (he/him) (May 04 2021 at 14:06):

fosco said:

I suspect this poses a real problem when one tries to capture the notion of determinant, and before that of antisymmetric tensor. Another problem is that there seems to be no easy way to categorify the expression of the determinant

$\det A = \sum_\sigma\prod_i (-1)^{|\sigma|} A_{i\sigma i}$

(what is -1? how can you sum over permutations if A isn't square?)

Could this work by replacing permutations with automorphisms of $C$ (the domain and codomain of $A$) and signature be replaced by some of 'character' for a representation $\phi$ of $Aut_{Cat}(C)$ on $Psh(C)$, i.e. a functor in the 'general groupoid' $Aut_{Prof}(C)$ you propose? If I generalize correctly from 0-linear algebra, a character should be a map $Aut_{Cat}(C) \to Set$ given by $Tr(\phi)$

Matteo Capucci (he/him) (May 04 2021 at 14:07):

It remains to find a suitable $\phi$... for permutations, $\phi : \sigma \to (-1)^{|\sigma|}I$ is the only non-trivial representation iirc

Matteo Capucci (he/him) (May 04 2021 at 14:11):

Conditioned on finding such $\phi$, then the determinant of an endoprofunctor could be

$$\det A = \int^{\sigma \in Aut_{Cat}(C)} \int_{i \in C} Tr(\phi(\sigma)) \times A(i, \sigma(i))$$

fosco (May 04 2021 at 14:31):

The dependence on $\sigma$ does not seem to be a functor of type (1,1)

fosco (May 04 2021 at 14:31):

But hey, I just realised that we can define tensors of higher rank :grinning: https://arxiv.org/abs/2011.13881

Matteo Capucci (he/him) (May 04 2021 at 14:36):

fosco said:

The dependence on $\sigma$ does not seem to be a functor of type (1,1)

Maybe switc $i$ with $\sigma(i)$ in $A$

Matteo Capucci (he/him) (May 04 2021 at 14:36):

fosco said:

But hey, I just realised that we can define tensors of higher rank :D https://arxiv.org/abs/2011.13881

Yep

fosco (May 04 2021 at 14:52):

Still don't see what $\sigma$ and $\phi$ are

John Baez (May 04 2021 at 14:56):

fosco said:

[...] there seems to be no easy way to categorify the expression of the determinant

$\det A = \sum_\sigma\prod_i (-1)^{|\sigma|} A_{i\sigma i}$

That formula for the determinant is hard to categorify because it's not very conceptual. This one has some advantages if you're willing to restrict attention to invertible matrices (which is okay since we all know what the determinant of a noninvertible matrix should be):

For any ring let $GL(R)$ be the group of invertible elements of that ring. The determinant is the canonical group homomorphism

$\mathrm{det} : GL(R) \to GL(R)/[GL(R),GL(R)]$

where $[GL(R),GL(R)]$ is the commutator subgroup of $GL(R)$: that is, the normal subgroup generated by all elements $ghg^{-1}h^{-1}$.

The idea is that the determinant of an invertible matrix is "all that's left if we force matrix multiplication to become commutative".

When $R$ is the ring of $n \times n$ matrices in a field $k$, $GL(R)$ is the group of invertible $n \times n$ matrices. $GL(R)/[GL(R),GL(R)]$ is canonically isomorphic to the group of invertible elements of $k$, and $\mathrm{det}$ is then the usual determinant.

More abstractly, the idea is that there's a forgetful functor from abelian groups to groups, and it has a left adjoint called abelianization sending any group $G$ to $G/[G,G]$. The counit of this adjunction is the canonical quotient map

$G \to G/[G,G]$

People use this approach to determinants in K-theory.

These slides on categorification of determinants also look interesting:

• F. Muro, On determinants (as functors).

fosco (May 04 2021 at 15:00):

Are you saying it's easier to categorify $G/G'$ (or more specifically, $GL/GL'$)?

fosco (May 04 2021 at 15:02):

Let's recap.

A proposal for GL(n) in this setting is auto-equivalences of a category with n objects; in case n is Cauchy complete these are just self-equivalences of the category itself. This is not a groupoid, as we're not considering isomorphisms, just equivalences. And the category itself will have equivalences as objects, but also non-invertible natural transformations $\alpha : f \Rightarrow f' : n \to n$

fosco (May 04 2021 at 15:04):

Now, what's your proposal for the derived subgroup of this thing which is not a group(oid)?

fosco (May 04 2021 at 15:04):

And even if it was: are there quotients for groupoids? Do they still give rise to groupoids?

fosco (May 04 2021 at 15:05):

If yes, what is an abelian groupoid, and are abelian groupoids reflective in groupoids?

fosco (May 04 2021 at 15:05):

Can you obtain the abelianisation of a groupoid as a quotient by the commutator subgroupoid?

fosco (May 04 2021 at 15:05):

None of these questions seems trivial

Oscar Cunningham (May 04 2021 at 15:06):

John Baez said:

When $R$ is the ring of $n \times n$ matrices in a field $k$, $GL(R)$ is the group of invertible $n \times n$ matrices. $GL(R)/[GL(R),GL(R)]$ is canonically isomorphic to the group of invertible elements of $k$, and $\mathrm{det}$ is then the usual determinant.

Last time this came up we decided it wasn't true for $\mathbb{F}^2_2$. https://categorytheory.zulipchat.com/#narrow/stream/266967-general.3A-mathematics/topic/Dieudonn.C3.A9.20determinant

Joe Moeller (May 04 2021 at 15:09):

fosco said:

If yes, what is an abelian groupoid, and are abelian groupoids reflective in groupoids?

Every groupoid is a coproduct of groups. So an abelian groupoid has to be one where the automorphism groups are all abelian.

John Baez (May 04 2021 at 15:10):

If I have some sort of categorified analogue of a ring $R$ - there are a number of definitions choose from - it will have a 2-group of objects that are invertible up to isomorphism. Let's call this 2-group $GL(R)$. There are two main levels of commutativity for 2-groups: braided and symmetric. I think we could take the 2-group $GL(R)$ and force it to be braided (or symmetric) using a left biadjoint to the forgetful functor from braided (or symmetric) 2-groups to 2-groups. Call the result $GL(R)/[GL(R), GL(R)]$. Then there should be a unit

$\mathrm{det}: GL(R) \to GL(R)/[GL(R),GL(R)]$.

fosco (May 04 2021 at 15:10):

I would say every groupoid is equivalent to a coproduct of groups; how do you account for invertible morphisms between non-equal objects?

fosco (May 04 2021 at 15:11):

John Baez said:

If I have some sort of categorified analogue of a ring $R$ - there are a number of definitions choose from - it will have a 2-group of objects that are invertible up to isomorphism. Let's call this 2-group $GL(R)$. There are two main levels of commutativity for 2-groups: braided and symmetric. I think we could take the 2-group $GL(R)$ and force it to be braided (or symmetric) using a left biadjoint to the forgetful functor from braided (or symmetric) 2-groups to 2-groups. Call the result $GL(R)/[GL(R), GL(R)]$. Then there should be a unit

$\mathrm{det}: GL(R) \to GL(R)/[GL(R),GL(R)]$.

I see, but how do you relate this to profunctors?

John Baez (May 04 2021 at 15:12):

I guess endoprofunctors form a 2-rig, right? You can multiply them, and add them...

John Baez (May 04 2021 at 15:13):

But to talk about this "determinant" we just need to know that endoprofunctors that are equivalences - that have an inverse up to natural isomorphism - form a 2-group.

John Baez (May 04 2021 at 15:13):

More precisely, endoprofunctors that are equivalences, and natural isomorphisms between these, form a 2-group.