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Stream: theory: category theory

Topic: non-analytic monads with arities

Tom Hirschowitz (Oct 11 2023 at 13:08):

I guess I should be answering this question rather than asking it, but here goes: are there "natural" examples of monads with arities in Weber's sense (or Berger-Melliès-Weber), which are not pointwise analytic? Concretely, in natural examples, for any $f\colon A \to TB$ , does the category of factorisations $A \xrightarrow{a} Ti \xrightarrow{Tu} TB$ through some arity $i$ have a weak initial object?

Just to be a bit more explicit, morphisms to any $A \xrightarrow{b} Tj \xrightarrow{Tv} TB$ are morphisms $k\colon i \to j$ such that
$vk=u$ and $T(k)a=b$ .

I don't know how to draw a diagonal arrow, but here's a diagram to aid intuition.

$\begin{CD} A @>b>> Tj \\ @VaVV @VVTvV \\ Ti @>>Tu> TB \end{CD}$

Let's ping @Simon Henry, who might have some clues...

James Deikun (Oct 11 2023 at 15:58):

So "pointwise analytic" is a generalization of "parametric right adjoint" where (1) the factorizations only have to exist through the specified arities and (2) the liftings for the generic factorization don't have to be unique?

If so, I think this starts to look (through the equivalence of p.r.a.s and functors with a left multi-adjoint) a lot like the [[solution set condition]] where a universal solution set is given by the arities. I could definitely believe that in natural cases of functors with arities (never mind monads with arities) that the arities form a solution set.

Tom Hirschowitz (Oct 11 2023 at 16:20):

Yes, I think your interpretation is right, up to one subtlety: generic factorisations are only considered from representable presheaves.
The point is that, because of non-uniqueness, they do not build up to all presheaves in general.

Regarding your second point, I'm not sure what you conclude. If I understand correctly, the thing is that the considered functors may not be continuous. Crucially, most monads do not preserve the terminal object. Typically, the free monoid monad maps it to $\mathbb{N}$ . Familial (=pra) functors preserve wide pullbacks, analytic functors would, I think, preserve pointwise weak pullbacks when this makes sense.

James Deikun (Oct 11 2023 at 16:25):

Yes, I definitely hope there are a lot of them that are not continuous as that is most of the interesting examples. The point is that the solution set condition is usually considered as a mild smallness condition and is implied by a lot of other mild conditions on the categories and functor involved, so it feels like the arity-bounded version would be about an equally mild condition on functors with arities.

James Deikun (Oct 11 2023 at 16:39):

Tom Hirschowitz (Oct 12 2023 at 04:46):

Right, thanks. Indeed, I guess when you throw in non-linear equations you probably lose analyticity. Idempotence probably suffices. E.g., the finite powerset functor on sets has arities in finite sets, but is not analytic. Mmmh... I should definitely prove this latter claim, but don't have time right now. Will come back to it if no one else does.

Tom Hirschowitz (Oct 12 2023 at 07:32):

Ok, a proof would be a bit tedious to write here, so I'll content with some intuition. It is relatively easy to prove that a map $1 \to 𝒫_f(n)$ can only be generic if it picks the full subset $\{ 1,…,n \}$ . To get the point, let's consider the map $1 → 𝒫_f(2)$ picking $\{1\}$ . If it was generic, then the lifting we'd find for the square

$\begin{CD} 1 @>\{1\}>> 𝒫_f(1) \\ @V\{1\}VV @VV𝒫_f(\{1\})V \\ 𝒫_f(2) @= 𝒫_f(2) \end{CD}$

would be a section of the map $1 → 2$ picking $1$ , which cannot exist.
Now, even the map $1 → 𝒫_f(1)$ picking the full $\{ 1 \}$ isn't generic, since we cannot find a lifting for the following square.

$\begin{CD} 1 @>\{1,2\}>> 𝒫_f(2) \\ @V\{1\}VV @VV𝒫_f(!)V \\ 𝒫_f(1) @= 𝒫_f(1) \end{CD}$

Indeed, for any map $k\colon 1 → 2$ , $𝒫_f(k) ∘ \{1\}$ will pick a one-element subset of $𝒫_f(2)$ , hence not $\{1,2\}$ .