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Stream: theory: category theory

Topic: not all monads are co-KZ

Matteo Capucci (he/him) (Jan 19 2023 at 23:58):

I recently learned from @Amar Hadzihasanovic that monads in 2-categories are classified by the delooping of the augmented simplex category, which means that a monad $t$ in $\cal K$ on an object $a$ is given by a functor $\bar t : \Delta_0 \to \cal K(a,a)$.

I was just reading [[fibration in a 2-category]] and I see this fact comes out again when they want prove that $\Phi B : B^\downarrow \rightrightarrows B$ is a colax idempotent monad (ie co-KZ afaiu) in $\bf Span(\cal K)$, for $\cal K$ a cartesian (ie finitely complete, in particular powered) 2-category.
The idea is simple: since you have powers in $\cal K$, the augmented simplex induces, for any object $B$, the aforementioned span, as well as the necessary composition operation and identity, because these are the operations $\bf \Delta_0$ describes on $\Delta[1] = {\downarrow}$, and functoriality of powers transport them to $\bf Span(\cal K)$. Cool!

Now to prove this monad is colax idempotent they argue as follows:
image.png

Matteo Capucci (he/him) (Jan 20 2023 at 00:02):

It's late here and I'm probably missing something obvious, but is this saying that in $\bf \Delta_0$, one already has the necessary adjunction $\eta T \dashv \mu$ you need to conclude a 2-monoid is colax idempotent. Then this transports along the 2-functor previously defined, and voilà you get colax idempotency on $\Phi B$. But I don't see why this argument doesn't apply to every 2-monad then, which would be given as a 2-functor $B\bf \Delta_0 \to \cal C$ (here $\cal C$ is a 3-category).

Matteo Capucci (he/him) (Jan 20 2023 at 00:03):

A probable source of confusion might arise from the 2-dimensional structure of $\bf \Delta_0$: the nLab article above states $\bf \Delta_0$ is a monoidal 2-category whose underlying 1-category is the augmented simplex, but don't seem to define the higher cells anywhere.

Matteo Capucci (he/him) (Jan 20 2023 at 00:06):

I know all the secrets are in Street's Fibrations in bicategories but maybe someone might help before I escalate

Mike Shulman (Jan 20 2023 at 00:12):

You've got it: an arbitrary 2-monad is only a map out of the augmented simplex 1-category.

Mike Shulman (Jan 20 2023 at 00:15):

For $n=1,2$, the augmented simplex $n$-category can be defined as the full sub-$n$-category of Cat (or Poset) on the finite linear orders. This definition appears on the linked page [[simplex category]], but should probably be recalled on the fibrations page. The point is that in that case, you're actually exponentiating by these linear orders themselves, so the construction as automatically 2-functorial and extends to the 2-category of them.

Matteo Capucci (he/him) (Jan 20 2023 at 09:18):

I see!

Matteo Capucci (he/him) (Jan 20 2023 at 09:21):

I added a precisation to the nLab page

Matteo Capucci (he/him) (Jan 20 2023 at 09:24):

Mike Shulman said:

You've got it: an arbitrary 2-monad is only a map out of the augmented simplex 1-category.

Actually I don't see this :thinking: I would have thought that an n-monad in $\cal C$ is given by a map from the delooped augmented n-simplex. Indeed here we get the 2-monad $\Phi B$ as a map $B\bf \Delta_0 \to \bf Span(\cal K)$ from the delooped augmented 2-simplex. Why is this not the case always?

Amar Hadzihasanovic (Jan 20 2023 at 11:51):

I don't think the n in n-simplex and the one in simplex n-category are really related...

Amar Hadzihasanovic (Jan 20 2023 at 11:55):

There happens to be a simplex 2-category just because the simplex category can be identified with a full subcategory of $\mathbf{Pos}$, so you can take the "natural transformations"/pointwise order on order-preserving maps as 2-cells.

I don't think there's such a thing as the simplex 3-category, at least not an obvious one.

Mike Shulman (Jan 21 2023 at 00:57):

Right. An $n$-monad is a map from the simplex 1-category to a hom-category of $n \rm Cat$. It is lax-idempotent if that map extends to the simplex 2-category. The 1 and 2 don't vary with $n$.

Matteo Capucci (he/him) (Jan 23 2023 at 15:01):

Oh I see!

Matteo Capucci (he/him) (Jan 23 2023 at 15:01):

Thanks a lot for the clarification :)