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Stream: learning: questions

Topic: Distributive laws for monoids


view this post on Zulip Ralph Sarkis (Jan 25 2021 at 19:20):

I have learned that monads are monoids in the monoidal category of endofunctors and that monad distributive laws are useful to compose monads and they are very well studied. It seems to me that distributive laws could be defined between monoids in any monoidal categories, has this been studied? If not (I didn't find anything online), is there a particular reason why?

view this post on Zulip Reid Barton (Jan 25 2021 at 19:38):

In a symmetric monoidal category you can "compose" monoid objects just by tensoring them--if you like, because the "swap" isomorphism defines a distributive law. And looking at endomorphisms is one of the prototypical ways to get something noncommutative/nonsymmetric.

view this post on Zulip Reid Barton (Jan 25 2021 at 19:38):

That's not to say there couldn't be interesting examples in the symmetric case as well of course.

view this post on Zulip Reid Barton (Jan 25 2021 at 19:40):

I think things like semidirect products could be obtained from distributive laws

view this post on Zulip Amar Hadzihasanovic (Jan 25 2021 at 20:28):

Reid Barton said:

I think things like semidirect products could be obtained from distributive laws

Never thought of that before but it sounds quite plausible.

view this post on Zulip Ralph Sarkis (Jan 25 2021 at 20:47):

Thanks Reid! Does this work in braided monoidal categories as well (ie: do we need the swap to be invertible)?

I didn't think of this link between swaps and distributive laws but according to this MO question, the converse direction should be rare.

view this post on Zulip Amar Hadzihasanovic (Jan 25 2021 at 20:58):

Braided is fine. In fact you need even less: just a not necessarily invertible natural transformation XYYXX \otimes Y \to Y \otimes X gives you a distributive law between any pair of monoids.

view this post on Zulip Amar Hadzihasanovic (Jan 25 2021 at 21:02):

Actually it should also satisfy the hexagon equation. But the point is you don't need invertibility.