# Balls in Lawvere metric spaces

Let $V$ be the monoidal category $[0,\infty)$ (as a poset) with $+$ and $0$. Lawvere shows that $V$-enriched categories are a more natural generalisation of the notion of a metric space (note no symmetry). Where it turns out many classical theorems about metric spaces (and similar structures like ultrametric spaces) are simply special cases of certain theorems in enriched category theory.

Now one of the objects one uses when working with metric spaces are balls. Let $C$ be a $V$-enriched category.

Choosing some $v\in V$ we can define the ball centered at some point $x\in C$. As the 'set' of points in $C$ such that there is a morphism $\mathrm{Hom}(x,y)\to v$. This however is not really an ideal definition. It feels very artificial and not very categorical as I would like.

What would be a nice categorical way to define an 'open ball' of a Lawvere metric space?

Edit:

I have confused the direction of the arrows in $V$, this means my second the last previous paragraph should be rephrased:

Given a $v\in V$, the ball centered at $x \in C$ is the set of points in $C$ such that $v \to \mathrm{Hom}(x,y)$.

This does feel a bit more categorical but it is not quite there yet.

• I should note that you may want to add $\infty$ to $V$ in order for certain limits and colimits to exist. – CatInTheBag Aug 13 '18 at 16:34
• I wouldn't say that the definition you give is "not really categorical": sometimes, you do precisely this (for $Set$-categorie the hom-object "being empty" is precisely asking that there are morphisms $\hom(x,y)\to \varnothing$). Aside from this, I see your point, but I would phrase the question more as "does the set of all objects $x,y$ such that there is a morphism $[x,y]\to v$ has a universal property of some sort?" – Fosco Loregian Aug 13 '18 at 16:57
• @FoscoLoregian I suppose thats a more sensible way to phrase it. I haven't made any assumptions on $C$ being small either so I have tried to tread carefully and not reference size. – CatInTheBag Aug 13 '18 at 17:05
• Well of course, a ball in this setting can be a class :-) hmmm, will it be compact? – Fosco Loregian Aug 13 '18 at 17:40
• No!! The triangle inequality is $\hom(x, y) + \hom(y, z) \geq \hom(x, z)$. – Todd Trimble Aug 13 '18 at 19:51

Luckily for you, Lawvere has already considered this in Taking categories seriously.

On page 18, he defines the family: $$\mathcal V^{\mathrm op}\times A\xrightarrow{B} \mathcal{V}^{A^{\mathrm{op}}}$$ defined by $$B(r,c)(a)=\mathcal V(r,A(a,c))$$ Where $B(r,c)$ reads the closed ball of given radius and center, since $$0 \ge B(r,c)(a)\iff r\ge A(a,c)$$ This is quite a cool construction because you can consider closed balls in any $\mathcal V$-enriched category.

• What does this construction correspond to in the case of $\mathcal{V} =$ Set? – Ivan Di Liberti Aug 18 '18 at 19:06
• @IvanDiLiberti balls of radius $\mathbf 2$ correspond to the powerset of a hom set of a category. I am not sure what this means generally. – Ali Caglayan Aug 18 '18 at 19:31
• A very strange creature. I would take a safari into this wilderness. – Ivan Di Liberti Aug 18 '18 at 19:37
• Great! This agrees with what I wrote. Probably I was remembering from Lawvere's paper. – Tim Campion Aug 18 '18 at 19:47

I would probably define balls as follows (maybe with left and right transposed):

• The left ball of radius $r$ centered at $x$ is the enriched presheaf $d(-,x)-r: X^{op} \to V$

• The right ball of radius $r$ centered at $x$ is the enriched copresheaf $d(x,-) - r: X \to V$

Here $a-b = \mathrm{max}(a-b,0)$ is the internal hom in $V$.

The idea is that subsets typically categorify to (co)presheaves. The set of $y$ such that $d(y,x)-r = 0$ is exactly the set of $y$ such that $d(y,x) \leq r$, and dually.

As a presheaf, we can ask when the left ball is representable by an object $r \ast x$. In fact, there is a standard name in enriched category theory for such a representing object: $r\ast x$ is the tensor of $x$ by $r$. Dually, a corepresenting object for the right ball is the cotensor $\{r,x\}$.

Another way of saying this is that the left ball is the cotensor $\{r,d(-,x)\}$ in the presheaf space $Fun^V(X^{op},V)$, and right ball is the cotensor $\{r,d(x,-)\}$ in the copresheaf space $Fun^V(X,V)$.

• So tensors and cotensors in some $X$, if they exist, must satisfy $d(y,x)\le r+s$ iff $d(r*y,x)\le s$ iff $d(y,\{r,x\})\le s$; in particular, in an ordinary metric space this would imply $d(y,x)*y=x$. Are there any nontrivial metric spaces with this property? – მამუკა ჯიბლაძე Aug 14 '18 at 19:25
• @მამუკაჯიბლაძე I don't think there are very many interesting (symmetric) metric spaces with a lot of (co)tensors, but there are interesting (nonsymmetric) quasimetric spaces with (co)tensors -- for example, $V$ itself and spaces of presheaves, which are enriched-complete and enriched-cocomplete. A metric space having (co)tensors feels to me a bit like a topological space being sober -- the (co)tensors are sort of "generic points". This feels similar to the subject of continuous logic to me, where a generic point is an "incomplete type", but I'm not sure how precise the correspondence is. – Tim Campion Aug 14 '18 at 19:28
• nLab likes to use power (=cotensor) and copower (=tensor) for those constructions as powers are (weighted) limits and copowers are (weighted) colimits. Calling these objects cotensor and tensor can get very confusing as the term is very overused in category theory. – Ali Caglayan Aug 18 '18 at 14:59