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This recent MSE question started a conversation with the OP of that post about what are some categorical notions casted in the category of metric spaces, regarded as enriched categories over $[0,\infty]^{op}$.

Doing some digging here on MO, I found this thread particularly enlightening. Let me rephrase it slightly more generally.

Let $\cal V$ be a quantale, whose operation is denoted $\otimes$.

Then, the category of $\cal V$-categories has

  • objects the sets $X,Y,\dots$ equipped with a reflexive, transitive $\cal V$-relation;
  • 1-cells the "maps" of $\cal V$-categories, i.e. the right adjoint $\cal V$-relations;
  • 2-cells the natural transformations (families of elements $d(fx,gx)$ of the quantale, one for every object $x\in X$ such that... well, I don't want to write the definition).

What I'm doing here is taking the bicategory ${\cal V}\text{-Rel}$ of $\cal V$-profunctors/relations as primitive and working backwards to retrieve categories inside ${\cal V}\text{-Rel}$. This point of view is central in -for example- monoidal topology and -I suspect- has a wide intersection with quantaloid theory, because the standard example of a quantaloid is precisely V-Rel.

The rationale is that I'd like to be as agnostic as possible regarding where exactly the properties I want to instantiate are invoked. I'm familiar with the idea that some categorical constructions are more natural when seen as profunctorial. :-)

This level of generality is good enough to try seeing some general pattern: $\cal V$-categories, when $\cal V$ is the quantale $([0,\infty]^{op},+)$, are precisely Lawvere metric spaces, so we really are doing an abstract version of those here.

Now, one can turn ${\cal V}\text{-Cat}$ into a monoidal structure in essentially two ways: either with the cartesian structure on $\cal V$ (which is assumed to have arbitrary meets, which for $[0,\infty]^{op}$ means arbitrary sup), or through the quantale operation.

Following Leinster's answer back on MO, using the cartesian structure does not seem to carry some interesting notion of pseudomonoid in ${\cal V}\text{-Cat}$, whereas the quantale structure yields a more enticing notion of pseudomonoid: in the case of metric spaces, given $[0,\infty]^{op}$-categories $A,B$, the point-set of $A \otimes B$ is the product of the point-sets of $A$ and $B$. The distance is given by $$ d((a, b), (a', b')) = d(a, a') + d(b, b'). $$ In other words, it's the '$1$-metric', also known as the "taxicab metric". I am then allowed to call a pseudomonoid wrt this structure a "taxicab monoid" in V-Cat.

Now, what exactly is a taxicab closed pseudomonoid in metric spaces? Are there nice examples of such a thing? Theorems that say there are no examples? What is the geometric intuition to have, when the adjunction property is asking the distance from $ab$ to $c$ is equal to the distance from $a$ to $c^b$?


Surprisingly, I could find no explicit description of the adjunction identities stated in the case of $[0,\infty]^{op}$-enriched categories, nor a precise definition of what exactly is a pair of adjoint nonexpansive maps regarded as functors.

If I'm not wrong, a $[0,\infty]^{op}$-enriched natural transformation consists of the following thing:

  • the proposition $0\ge d(fx,gx)$, i.e. that it is zero (we have a family of maps);
  • the proposition $\forall x,x'.d(x,x')\ge d(fx,gx')$ (it is natural).

How should I picture this? I find suggestive how a natural transformation between nonexpansive maps is the statement that the metric cannot distinguish between $f$ and $g$, and that the distance between any two images $fx,gx'$ cannot exceed the distance $d(x,x')$.

How should I picture this more precisely?

Now, given this, two nonexpansive maps $f :X \rightleftarrows Y : g$, are adjoint maps if $d(x,gfx)=0=d(fgy,y)$ (i.e., we have unit and counit) and

$$ \begin{cases} \forall x,x'.d(x,x') \ge (x, gfx')\\ \forall y,y'.d(y,y') \ge (fgy, y') \end{cases}$$

In the specific case of a monoidal closed $[0,\infty]^{op}$-category $X$, we're after an adjunction $\_\cdot b\dashv (\_)^b$ for a given point $b\in X$, i.e.

  1. there is an equality of real numbers $d(a\cdot b,c)=d(a, c^b)$;
  2. $d(y^b\cdot b,y)=0$ and $d(x,(x\cdot b)^b)=0$, and these "satisfy the zig-zag identities".

What can we deduce from this? I think a fruitful intuition come from the example of this situation I didn't mention I already have:

Let $G$ be a topological group with a metrizable topology. Then, regarded as a metric space, i.e. as a $[0,\infty]^{op}$-enriched category, $G$ is monoidal closed if

  1. the monoidal product is the group operation, $(x,y)\mapsto x.y$
  2. the internal hom is given by multiplying by $x^{-1}$, $y^x = x^{-1}y$.

I don't think this is the only class of examples: to be more precise, I believe one can cook up a generalised metric space $M$, which is a "metric monoid", where $(\_)^b$ exists even if $b$ is not invertible w.r.t. the monoid operation in $M$. (note to self to expand further: take a measure space $(X,\Sigma,\mu)$ and the metric space $\Sigma$ --the set of events of the measurable space-- with metric $d(U,V)=\mu(U\triangle V)$).

I'm guided in this by the intuition that $(\_)^b$ really behaves like multiplying by the inverse of $b$: consider again the two equations $$\begin{cases} d(y^b\cdot b,y)=0 \\ d(x,(x\cdot b)^b)=0 \end{cases}$$ If in the first $y=1$ (the identity of the monoid operation) then $d(1^b\cdot b,1)=0$, so $1^b=[b,1]$ behaves like the dual of $b$ because the metric believes that $1^b\cdot b$ is $1$; if in the second $x=1$ then $d(1,(1\cdot b)^b)=d(1, b^b)=0$, so again the metric can't distinguish $b^b$ and $1=bb^{-1}$.

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  • $\begingroup$ If your metric monoid is a group, where each left or right multiplication is an isometry, then it is closed by exponential being given by $[x,y] = x^{-1}y$ or $yx^{-1}$ depending if we are talking of left or right exponentials. It seems to be that it is essentially a neccessary and sufficient condition for having all exponential... $\endgroup$ Apr 29, 2022 at 15:55
  • $\begingroup$ For e.g. if you require the distance to be symmetric, then $0 = d([a,b] ,[a,b]) = d([a,b] \otimes a ,b)$ shows that $[a,b] \otimes a = b $ so if this is true for all $a$ and $b$ you have $[a,b] = b a^{-1} $. $\endgroup$ Apr 29, 2022 at 16:09
  • $\begingroup$ I do not require the distance to be symmetric, nor that $d(x,y)=0$ implies $x=y$. (But thanks! I guess I could have mentioned that I'm aware the example of a metric group is probably the most interesting one: is it the only one?) $\endgroup$
    – fosco
    Apr 29, 2022 at 17:11
  • $\begingroup$ Ok, Then that's probably only one type of examples $\endgroup$ Apr 29, 2022 at 17:13

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