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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1$\begingroup$ 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 ჯიბლაძეCommented Jul 25, 2016 at 19:14
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4$\begingroup$ 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 ShulmanCommented Jul 26, 2016 at 3:28
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1$\begingroup$ @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 ჯიბლაძეCommented Jul 30, 2016 at 9:47
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3$\begingroup$ 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 ShulmanCommented Jul 31, 2016 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.
answered Aug 30, 2017 at 20:58
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$\begingroup$ Hi Todd. That sucks for my research....What about FinSet? $\endgroup$ Commented Aug 30, 2017 at 21:45
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$\begingroup$ 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. $\endgroup$ Commented Aug 30, 2017 at 21:57
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$\begingroup$ @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) $\endgroup$ Commented Aug 31, 2017 at 6:42
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1$\begingroup$ 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 $\mathrm{Set}$. $\endgroup$ Commented Sep 1, 2017 at 0:15
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$\begingroup$ 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.. $\endgroup$ Commented Sep 5, 2017 at 1:40