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Stream: learning: questions

Topic: monads with all algebras free

James Deikun (Jan 07 2023 at 23:11):

It's pretty hard to google this up because of all the talk about free monads that comes up. Are there any interesting things we can say about a monad where all algebras are free algebras, particularly if there's a nice equivalence (coreflective maybe) between the Eilenberg-Moore and Kleisli categories?

fosco (Jan 07 2023 at 23:12):

@Ivan Di Liberti

Nathanael Arkor (Jan 07 2023 at 23:22):

There's been some discussion on MathOverflow about this in the past, e.g. Varieties where every algebra is free.

Nathanael Arkor (Jan 07 2023 at 23:23):

Every idempotent monad is an example.

Nathanael Arkor (Jan 07 2023 at 23:23):

particularly if there's a nice equivalence (coreflective maybe) between the Eilenberg-Moore and Kleisli categories

Note that every equivalence can be strengthened to an [[adjoint equivalence]], so we don't gain anything by restricting to "nice" equivalences.

Ivan Di Liberti (Jan 07 2023 at 23:33):

I gave a talk on this topic at some point of my life.

James Deikun (Jan 07 2023 at 23:34):

It's nice when the nice equivalence is the easily available one.

James Deikun (Jan 07 2023 at 23:36):

Nathanael Arkor said:

There's been some discussion on MathOverflow about this in the past, e.g. Varieties where every algebra is free.

Wow, that is some seriously strong result! My monad is not going to be on Set though. (The prototypical cases are on "interesting" presheaf categories.)

James Deikun (Jan 08 2023 at 00:00):

The talk is very interesting and even though it also deals with stable monads on Set it makes the underlying ideas more approachable.

James Deikun (Jan 08 2023 at 00:10):

My own examples are completion under degeneracies in a presheaf category, for example, completion from semisimplicial to simplicial sets. Aside from being stable they are "essentially unary" in that the action of the monad on the subobject lattice preserves meets and joins.

James Deikun (Jan 14 2023 at 15:50):

Say you have a left adjoint monad $T$, which is also Cartesian. Call the right adjoint $D$. Consider an algebra $TA \xrightarrow{\overline{a}} A$ and its adjunct coalgebra $A \xrightarrow[\underline{a}]{} DA$. Is it possible to prove that $\overline{a}$ is free on a subobject of $A$? If not, what if $T$ is strongly Cartesian?

James Deikun (Jan 14 2023 at 16:18):

Never mind, I found a counterexample. It comes down to the fact that every degenerate cell is a degeneracy of a unique cell but not necessarily in a unique way.

James Deikun (Jan 16 2023 at 02:18):

Take $B$ to be the pushout of the kernel pair of $\overline{a}$. This has a natural comparison morphism $\chi$ into $A$ that gives two factorizations of $\overline{a}$. Consider the free algebra on this object, $TTB \xrightarrow{\mu} TB$. Since $T$ preserves pullbacks and pushouts, $TB$ is constructed from $T\overline{a}$ in the same way $B$ is constructed from $\overline{a}$, and it has a comparison morphism to $TA$ which has to be $T\chi$. Take $\overline{a}\circ{}T\chi$ as a comparison between $\mu_{B}$ and $\overline{a}$. $\overline{a}\circ{}T(\overline{a}\circ{}T\chi) = \overline{a}\circ{}T\overline{a}\circ{}TT\chi = \overline{a}\circ{}\mu_A\circ{}TT\chi = \overline{a}\circ{}T\chi\circ\mu_B$, so it is a legit morphism. Now let's build it an inverse.

Morphisms into $TB$ are not easy to build since it is a pushout, but there is already $\delta : TA \times_A TA \to B$ that's pretty canonical. What's more, $\chi\circ\delta$ is the common diagonal of the pushout square involving $A$. So $\overline{a}\circ{}T\chi\circ{}T\delta = \overline{a}\circ{}T\overline{a}\circ{}Tp_1 = \overline{a}\circ\mu_A\circ{}Tp_1$. Still need to invert that, so take the morphism $\tilde{\eta}$ that factors $\eta_A$ through $TA \times_A TA$. $\overline{a}\circ\mu_A\circ{}Tp_1\circ{}T\tilde{\eta}\circ\eta_A = \overline{a}\circ\mu_A\circ{}T\eta_A\circ\eta_A = \overline{a}\circ\eta_A = \mathrm{id}_A$.

I can't seem to prove it's more than a semi-inverse though. It feels like the right thing because it glues together everything that can become equal through operations in $T$, but I'm not quite getting it.