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Stream: learning: questions

Topic: Lipschitz functions

Joe Moeller (Sep 18 2024 at 19:07):

For categories enriched in a monoidal poset, being a V-functor is really a property of the object-function, the property of the existence of the appropriate V-morphisms between hom-objects. Functors between Lawvere metric spaces turn out to be Lipschitz continuous with constant 1, i.e. distance-shrinking functions. Lawvere points out that it's not a big deal because a Lipschitz continuous map between metric spaces with a constant $\lambda$ other than 1 still lives in $VCat$, but as a functor of the form $\lambda X \to Y$. If V is a bimonoidal/rig category, you can always multiplicatively scale a category enriched in the additive structure by an object of the enriching category by tensoring with all the hom-objects and leaving the object set alone.

This isn't satisfying to me though, because I'm currently thinking about some non-shrinking endomorphisms of a metric space. So instead I'm working with the category of Lawvere metric spaces and object-functions $f \colon ob X \to ob Y$ for which there exists a finite positive real number $\lambda$ such that f is a V-functor of the form $\lambda X \to Y$. I'm also thinking about doing analogous stuff for categories enriched in a monoidal poset, often a quantale. I think this formulation transfers just fine, but here's a few questions:

1. Has someone else worked with these maps before? I'm currently calling them "generalized V-functors", and considering "Lipschitz V-functors", but I'd rather use an established name if it exists.
2. Is this actually a case of something more standard that already exists? Like maybe profuctors already cover this in a way I didn't catch?

Martti Karvonen (Sep 19 2024 at 07:36):

Not sure if this will work out in this case, but whenever I see a definition like "generalized morphisms $X\to Y$ are ordinary maps $\lambda X\to Y$", I immediately guess that we're working in a co-Kleisli category of a comonad. Here of course you can act on an object $X$ by many different $\lambda$ s, but that just suggests we're basically doing the same over a graded comonad.

Paolo Perrone (Sep 19 2024 at 09:52):

This reminds me of something I've always struggled with:

• Given a linear map f: V -> V, the set of f-invariant vectors is the equalizer of f and the identity.
• The above can also be seen as the eigenspace of the eigenvalue 1.
• What is, categorically, the eigenspace of a generic eigenvalue x? Can we express it as a nice weighted limit, suitably enriched?

Chris Grossack (they/them) (Sep 19 2024 at 15:18):

@Paolo Perrone -- is the point that you don't want to use arbitrary elements of your base field as symbols? I'm trying to see what stops you from defining $E_\lambda$ to be the equalizer of $f$ and $\lambda \text{id}$...

If that's what you're going for, then we should win when $k = \mathbb{F}_p, \mathbb{Q}$, right? Since in that case even if your signature is only $0,1,+,-,\times,(-)^{-1}$ you can get the whole field. Iirc there's even diagrammatics for this in work of Pawel Sobocinski, but I forget the details.

I feel like I'm probably misinterpreting what you want, though.

Paolo Perrone (Sep 19 2024 at 15:21):

Maybe a way of restating my question is, can we keep the identity arrow, and express the scalar multiplication in terms of weights?

Kevin Carlson (Sep 19 2024 at 15:54):

Sure! Your diagram of $f$ and $\mathrm{id}$ is indexed by the free $k$-linear category on a parallel pair. The equalizer is weighted by the constant weight at $k$. Now consider the weight $1,\lambda:k\rightrightarrows k.$ A cylinder under this weight for the diagram $f,\mathrm{id}: V\rightrightarrows V$ is a $V$-natural transformation, so, pedantically, a $v_1$ and a $v_2$ such that $v_2=f(v_1)$ and $v_2=\lambda v_1.$

Kevin Carlson (Sep 19 2024 at 15:55):

This is a great motivating example of a weighted limit, hadn’t thought of it before! I guess this isn’t at the right level to help Joe, but maybe it’s inspiring…

Kevin Carlson (Sep 19 2024 at 16:01):

Martti Karvonen said:

Not sure if this will work out in this case, but whenever I see a definition like "generalized morphisms $X\to Y$ are ordinary maps $\lambda X\to Y$", I immediately guess that we're working in a co-Kleisli category of a comonad. Here of course you can act on an object $X$ by many different $\lambda$ s, but that just suggests we're basically doing the same over a graded comonad.

I think the scaling by $\lambda$ isn’t actually a comonad since you would need the identity to be a map $\lambda X\to \lambda^2 X$ and also one $\lambda X\to X$, but these aren’t both contractions.

Paolo Perrone (Sep 19 2024 at 16:24):

Kevin Carlson said:

Sure! Your diagram of $f$ and $\mathrm{id}$ is indexed by the free $k$-linear category on a parallel pair. The equalizer is weighted by the constant weight at $k$. Now consider the weight $1,\lambda:k\rightrightarrows k.$ A cylinder under this weight for the diagram $f,\mathrm{id}: V\rightrightarrows V$ is a $V$-natural transformation, so, pedantically, a $v_1$ and a $v_2$ such that $v_2=f(v_1)$ and $v_2=\lambda v_1.$

This sounds promising. What's the enriching category?

Kevin Carlson (Sep 19 2024 at 16:28):

Oh, I'm working in $k$-linear categories for $k$ whatever field your vector spaces are over.

Kevin Carlson (Sep 19 2024 at 16:29):

i.e. just enriching over vector spaces.

Joe Moeller (Sep 19 2024 at 16:38):

I had thought about these maps being a function equipped with a “Lipschitz constant”, but that category does not have the properties I want. For instance, the previously unique map to 1 now has infinitely many copies, one for each constant.

Kevin Carlson (Sep 19 2024 at 16:41):

Wait, so why are you unhappy with the idea of having the family of $V$-categories $\lambda X$ where $V$ is a rig category, again?

Joe Moeller (Sep 19 2024 at 16:52):

Because I want a non-shrinking endomorphism of a space to be an endomorphism.

Paolo Perrone (Sep 19 2024 at 16:52):

Kevin Carlson said:

i.e. just enriching over vector spaces.

I see, thanks.

Paolo Perrone (Sep 19 2024 at 16:55):

So, coming back to Joe's question (sorry for partially derailing the thread): could we model Lipschitz functions as morphisms of a category enriched in sets equipped with an action of R^+?

Martti Karvonen (Sep 19 2024 at 17:04):

Kevin Carlson said:

Martti Karvonen said:

Not sure if this will work out in this case, but whenever I see a definition like "generalized morphisms $X\to Y$ are ordinary maps $\lambda X\to Y$", I immediately guess that we're working in a co-Kleisli category of a comonad. Here of course you can act on an object $X$ by many different $\lambda$ s, but that just suggests we're basically doing the same over a graded comonad.

I think the scaling by $\lambda$ isn’t actually a comonad since you would need the identity to be a map $\lambda X\to \lambda^2 X$ and also one $\lambda X\to X$, but these aren’t both contractions.

I could be wrong, but I think we have a graded comonad where the grading monoidal category is the poset $[1,\infty]$ under multiplication : for each $\lambda$, scaling by it gives an endofunctor $T_\lambda$, the comultiplication $T_{\lambda\cdot\theta}\to T_\lambda \circ T_\theta$ can be taken to be the identity and so can the counits $T_\lambda \to 1$. Numbers at most one should similarly give a graded monad, but I don't see how to treat all of $[0,\infty]$ in the same picture.