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Stream: learning: questions

Topic: distributive laws


view this post on Zulip Tom Hirschowitz (Mar 23 2021 at 17:22):

Does anyone know of a reference for the following fact?

Consider an endofunctor FF and a monad TT on a sufficiently nice category, together with a distributive law δ ⁣:FTTFδ \colon FT \to TF
that commutes with the unit and multiplication of TT.
Furthermore, let FF^* denote the free monad over FF.

Then there is a unique distributive law FTTFF^*T \to TF^* commuting with both units and multiplications, and satisfying the obvious diagram relating it to δδ. I found a few variants, notably in Bartel's thesis and a paper by Lenisa, Power, and Watanabe, but not this precise one.

Thanks in advance for any hint!

view this post on Zulip Tom Hirschowitz (Mar 23 2021 at 17:28):

Funny, I tried to use unicode symbols instead of math mode, but the star didn't show...

view this post on Zulip Mike Shulman (Mar 24 2021 at 08:41):

I think this follows from the fact that such a distributive law δ\delta is equivalent to a lifting of TT to a monad on the category of FF-endofunctor-algebras, and similarly δ\delta^* is equivalent to a lifting of TT to a monad on the category of FF^*-monad algebras, while these latter two categories of algebras are equivalent (assuming FF^* is algebraically-free on FF).

view this post on Zulip Tom Hirschowitz (Mar 24 2021 at 10:14):

Thanks, @Mike Shulman, that's a neat proof, hardly longer than a mere reference.

view this post on Zulip John Baez (Mar 24 2021 at 16:23):

What does "algebraically-free" mean here - it's different than free?

view this post on Zulip Nathanael Arkor (Mar 24 2021 at 16:58):

Algebraically free means that the category of algebras for the monad coincide with the category of algebras for the endofunctor. It's a stronger condition than just being free.

view this post on Zulip John Baez (Mar 24 2021 at 18:03):

Nice! I don't have the energy to understand the proof right now, but the nLab says also that conversely, a free monad on a locally small and complete category is algebraically free. That covers most of our usual categories of mathematical gadgets.

view this post on Zulip Mike Shulman (Mar 24 2021 at 22:01):

Free monads that aren't algebraically-free are one of the pathologies of category theory that we make definitions and theorems to let us ignore, like limit-preserving functors that don't have left adjoints and Kan extensions that aren't pointwise.

view this post on Zulip Mike Shulman (Mar 24 2021 at 22:02):

(Of course, now that I've said that, someone will pop up and tell me about how important all three of those things are...)