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    • CommentRowNumber1.
    • CommentAuthorTodd_Trimble
    • CommentTimeMar 2nd 2012

    Started an article on monoidal monad. An earlier redirect had sent it over to Hopf monad which is something that Zoran was working on, but I think it deserves an article to itself, with discussion of the relation to commutative monads, etc. (which I have started).

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTimeMar 2nd 2012

    Excellent, thanks!

    • CommentRowNumber3.
    • CommentAuthorPaoloPerrone
    • CommentTimeJan 9th 2020

    Started page about tensor product of algebras over a commutative monad.

    diff, v8, current

    • CommentRowNumber4.
    • CommentAuthorPaoloPerrone
    • CommentTimeJan 30th 2020

    Added functoriality of correspondence.

    diff, v13, current

    • CommentRowNumber5.
    • CommentAuthorPaoloPerrone
    • CommentTimeFeb 3rd 2020

    Added monoidal structure on Kleisli category

    diff, v15, current

    • CommentRowNumber6.
    • CommentAuthorSam Staton
    • CommentTimeFeb 4th 2020

    Attempt to better summarize the situation at the beginning of the article.

    diff, v16, current

  1. It looks like the proof is actually for the second composite.

    tgeng

    diff, v17, current

    • CommentRowNumber8.
    • CommentAuthorJonasFrey
    • CommentTimeSep 6th 2021

    Does anybody know of an example of a monoidal monad on a symmetric monoidal category, where the monad itself is not symmetric, i.e. the square

    TATB T(AB) TBTA T(BA) \begin{matrix} TA \otimes TB &\to& T(A\otimes B)\\ \downarrow && \downarrow\\ TB \otimes TA &\to& T(B\otimes A)\\ \end{matrix}

    doesn’t necessarily commute?

    I thought that André Joyal mentioned such a monad at a CT talk about the Dold Kan correspondence and Tom Leinster blogged about it, but although I found the blogpost here, it doesn’t mention any such monad.

    Does anybody happen to remember the monad in question, or another non-symmetric monoidal monad on a symmetric monoidal category?

    • CommentRowNumber9.
    • CommentAuthorRichard Williamson
    • CommentTimeSep 7th 2021
    • (edited Sep 7th 2021)

    Interesting question! Do cubical sets perhaps provide an example? I’d guess that they can be viewed as algebras for some monoidal monad on the category Set \mathsf{Set}^{\mathbb{N}} (with symmetric monoidal structure given by levelwise product), or something like that (thinking of each element of X(n)X(n), for some nn \in \mathbb{N}, as a free-standing nn-cube), with the monoidalness of the monad inducing the usual tensor product of cubical sets; but this tensor product is not symmetric, and I think this would necessitate that the square you mention does not always commute.

    • CommentRowNumber10.
    • CommentAuthorJonasFrey
    • CommentTimeSep 7th 2021
    • (edited Sep 7th 2021)

    Hmm, I think the square can only be defined if the tensor product is symmetric (or at least braided) since the vertical maps are meant to be braiding maps. Anyway, which non-symmetric tensor product are you referring to? A kind of substitution tensor product?

    EDIT: I assume that you mean the convolution tensor product on presheaves on this cube category. That’s indeed not symmetric.

    • CommentRowNumber11.
    • CommentAuthorRichard Williamson
    • CommentTimeSep 8th 2021
    • (edited Sep 8th 2021)

    Yes, that is the tensor product I meant: the one induced by Day convolution from the monoidal structure of the cube category.

    Hmm, I thought that you were only using the symmetric structure of the underlying symmetric monoidal category to define that square, not the one on algebras? My feeling was that if that square commutes, it will force the monoidal structure on algebras to be symmetric, but I haven’t gone through the details.

    One thing that I think is true is that the only way to get an example would be to have left and right strengths where the right strength is not compatible with the symmetric structure of the underlying symmetric monoidal category, but where one still has the necessary compatibility between the strengths to have a commutative monad (as defined in the non-symmetric setting). I think there should be such examples (I see no reason for the compatibility between strengths to force that the right strength comes from the symmetric structure), but if my idea about looking at non-symmetric-ness of the tensor product on algebras is no good, then I’m not sure where to look for them!

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