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Has anyone ever seen a Monad that is very much like the List Monad but is also a co-monad, and also a Frobenius monad? In this paper they give examples of List-like monads called Containers and they give one that is a Comonad, namely Trees. The comonad axiom takes each node in the tree and labels it with the tree rooted at that node. In the comments, there is an answer to a previous question of mine where someone suggests a bimonad, so comonad and monad that does not satisfy the frobenius property.

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    $\begingroup$ Tom Leinster in an answer to your question mathoverflow.net/a/237967/41291 has described structures of monad and comonad on $L^+$ (nonempty lists). He however calls this bimonad rather than Frobenius monad. $\endgroup$ – მამუკა ჯიბლაძე Jul 25 '16 at 19:14
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    $\begingroup$ A "Frobenius monad" and a "bimonad" would be two different possible ways of combining a monad and comonad structure, satisfying axioms analogous to those of a Frobenius algebra and a bialgebra. (In particular, having a monad and comonad doesn't immediately imply either one; there are extra compatibility axioms to check, different ones in each case.) So I expect that Tom knew what he was talking about when choosing one terminology rather than the other. $\endgroup$ – Mike Shulman Jul 26 '16 at 3:28
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    $\begingroup$ @MikeShulman The part I don't understand is "and hence". Is just a Frobenius monad structure on a list-like endofunctor meant, or something more (or less)? $\endgroup$ – მამუკა ჯიბლაძე Jul 30 '16 at 9:47
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    $\begingroup$ Right, that was exactly my point: having a monad that is "also a comonad" is not sufficient to have a "Frobenius monad", because there is an extra condition. So you can't say "and hence". $\endgroup$ – Mike Shulman Jul 31 '16 at 6:59
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I claim the only Frobenius monad on $\mathrm{Set}$ is the trivial monad given by the identity functor (which I guess needless to say isn't "very much like" the List monad).

Frobenius monads on a category $C$ are essentially the same as monads on $C$ whose underlying endofunctor is left adjoint to itself. For details on this assertion, see the nLab article on Frobenius monads.

Endofunctors on $\mathrm{Set}$ that have right adjoints, i.e., that are left adjoints, are very easily described: they are precisely the endofunctors of the form $X \mapsto S \cdot X$ where $S$ is a set and $S \cdot X$ denotes the coproduct of an $S$-indexed family of copies of $X$; it can be identified with $S \times X$. Natural transformations between such functors are exactly the ones induced by functions $S \to T$. The right adjoint to such a functor is $X \mapsto X^S$. The only case where we have an isomorphism (natural in $X$) $S \cdot X \cong X^S$ is where $S = 1$; this is immediate from the component at $X = 1$. This $S = 1$ corresponds to the identity endofunctor.

Thus the only endofunctor on $\mathrm{Set}$ adjoint to itself is the identity functor, and this has exactly one monad structure which is the one who unit and multiplication are identity transformations.

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  • $\begingroup$ Hi Todd. That sucks for my research....What about FinSet? $\endgroup$ – Ben Sprott Aug 30 '17 at 21:45
  • $\begingroup$ Sorry, ignore the question in my comment. It's just a bit of a let down. I will wait a few days to see if anyone else has anything to add, but I guess this settles the matter. $\endgroup$ – Ben Sprott Aug 30 '17 at 21:57
  • $\begingroup$ @BenSprott Not knowing what are your restrictions - nontrivial such guys might be f.d. vector spaces with a nondegenerate bilinear form (more generally, Hilbert spaces too) $\endgroup$ – მამუკა ჯიბლაძე Aug 31 '17 at 6:42
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    $\begingroup$ Ben, if I had a better idea what you are trying to do in your research with regard to Frobenius, I might be able to say something more hopeful. I got the impression though you were looking at monads on $\mathrm{Set}$. $\endgroup$ – Todd Trimble Sep 1 '17 at 0:15
  • $\begingroup$ HI Todd, Yes, I was thinking Containers as in the paper, and I believe they are all Monads on Set. In my research, I might switch away from either Set or the Frobenius requirement as there are a few ways to combine a monad and comonad structure.. $\endgroup$ – Ben Sprott Sep 5 '17 at 1:40

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