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Let $\mathcal{C}$ and $\mathcal{D}$ be symmetric monoidal categories and assume that the symmetric monoidal product $\otimes_{\mathcal{D}}$ on $\mathcal{D}$ preserves colimits in both variables. Then the Day convolution $*$ defines a symmetric monoidal structure on the category of functors $Fun(\mathcal{C},\mathcal{D})$.

(The obligatory nlab link and here a treatment for $\infty$-categories. In fact nlab only considers the case that $\mathcal{D}$ is the category we're enriched in, so $\mathit{Set}$ for the purpose of this question.)

There is also a (much easier to define) symmetric monoidal structure on $Fun(\mathcal{C},\mathcal{D})$, given by 'point-wise multiplication' using $\otimes_\mathcal{D}$. (This does not use the symmetric monoidal structure on $\mathcal{C}$.)

I have two questions about these structures:

  1. Does $\otimes_{\mathcal{D}}$ distribute over $*$, i.e. do we have $F\otimes (G* H) \cong (F\otimes G) * (F \otimes H)$ ?

More precisely, does $Fun(\mathcal{C},\mathcal{D})$ have the structure of a bimonoidal category/rig category, where $*$ is '$+$' and $\otimes$ is '$\times$'? If not, are they compatible in some other sense?

  1. Is the Day convolution of two symmetric monoidal functors again symmetric monoidal?

This is easy for the other product $\otimes_{\mathcal{D}}$, which in fact induces a symmetric monoidal structure on the category of symmetric monoidal functors $Fun^{\otimes}(\mathcal{C}, \mathcal{D})$. Combined the question would be whether this extends to a bimonoidal structure using the Day convolution.

I believe both of the answers to be yes when $\mathcal{C}$ is the category of oriented $1$-bordisms between $0$-manifolds (aka the free symmetric monoidal category on one dualisable object). So I was wondering if this in general gives a bimonoidal structure on the category of TFTs. This would be nice because the definition of the direct sum of two TFTs is kind of awkward and written out it looks a lot like the Day convolution. (One defines $(F\oplus G)(M) := F(M) \oplus G(M)$ for connected objects $M$ and then uses the fact that every manifold decomposes uniquely as the disjoint union of its connected components.) On the other hand it might only be this nice property of the bordism category, namely having a 'unique prime decomposition', that makes the result work in this case.

Edit: There was a mistake concerning which way the distributivity is supposed to go, I hope this is more clear now.

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  • $\begingroup$ I think the original paper (web.math.rochester.edu/people/faculty/doug/otherpapers/…) deals with the case $Fun(\mathcal{C},\mathcal{V})$ for $\mathcal{C}$ a $\mathcal{V}$-enriched category. However, as far I can see, he does not consider the case $\mathcal{D} \neq \mathcal{V}$ and also does not mention the 'point-wise' product. (I might well be wrong, particularly about the latter.) $\endgroup$
    – Jan Steinebrunner
    Mar 15, 2018 at 12:33

2 Answers 2

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I don't know in general whether $\otimes_D$ distributes over $\ast$. However, I think it is unlikely, because there is in general a different compatibility between them: they are a duoidal category. This is stated without proof on the nLab page, probably as a generalization from the special case $[V_\kappa,V]$ discussed in arXiv:1507.08710.

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  • $\begingroup$ This is interesting, but I am confused about the duoidal relation $(F \otimes G) * (H\otimes K) \cong (F * H) \otimes (G * K)$. Doesn't this contradict the idea that $*$ behaves like the convolution product on the group ring $\mathbb{Z}[G]$, whereas $\otimes$ behaves like the point-wise multiplication? $\endgroup$
    – Jan Steinebrunner
    Mar 15, 2018 at 16:18
  • $\begingroup$ More explicitly, what happens in the following example: Say we have the discrete category $\mathcal{C}=\mathbb{N}$ with $+$ as s.m. product. Then, if I'm not mistaken, $(F*G)(1) \cong F(1) \amalg G(1)$ for $\amalg$ the coproduct, and $(F\otimes G)(1) \cong F(1) \otimes G(1)$. Then, by evaluation on $1$ we can show that the duoidal relation is not satisfied for e.g. $\mathcal{D}=Vect_\mathbb{C}$. $\endgroup$
    – Jan Steinebrunner
    Mar 15, 2018 at 16:19
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    $\begingroup$ @J.Steinebrunner The duoidal structure is not an isomorphism, but a directed transformation. $\endgroup$
    – Mike Shulman
    Mar 15, 2018 at 16:21
  • $\begingroup$ @J.Steinebrunner In your example, I would have thought that $(F\ast G)(1) = (F(1) \otimes G(0)) \amalg (F(0) \otimes G(1))$. So the duoidal morphism $(F\otimes G) \ast (H \otimes K) \to (F\ast H) \otimes (G \ast K)$ at $1$ is $(F1 \otimes G1 \otimes H0 \otimes K0) \amalg (F0 \otimes G0 \otimes H1 \otimes K1) \to ((F1 \otimes H0) \amalg (F0 \otimes H1)) \otimes ((G1 \otimes K0) \amalg (G0 \otimes K1))$. $\endgroup$
    – Mike Shulman
    Mar 15, 2018 at 16:42
  • $\begingroup$ Since $\otimes$ in $D$ distributes over $\amalg$, this codomain is a coproduct of 4 summands, one of which is $F1\otimes H0 \otimes G1 \otimes K0$ and another of which is $F0 \otimes H1 \otimes G0 \otimes K1$. The duoidal morphism is then the inclusion of these two summands, plus symmetry of $\otimes$ to switch $G$ and $H$ in the middle. $\endgroup$
    – Mike Shulman
    Mar 15, 2018 at 16:42
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This distributivity doesn't seem right to me:

in the case of presheaves, the convolution product on $[(C,\oplus),(V,\otimes)]$ of two representables $\hom(a,-),\hom(b,-)$ is the representable on $a\oplus b$ (so that the convolution "restricts" to the $\oplus$ product on $C$), and it seems false to me that the convolution $$\hom(a,-) * \big( \hom(b,-)\otimes\hom(c,-) \big)$$ is the presheaf $\hom(a\oplus b,-)\otimes \hom(a\oplus c,-)$... it would imply that (in the same notation) $$ \int^q \hom(b,q)\otimes\hom(c,q)\otimes \hom(a\oplus q,\_)\cong \hom(a\oplus b,\_)\otimes\hom(a\oplus c,\_) $$ which doesn't seem right.

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  • $\begingroup$ I'm sorry, I made a mistake in the question. I meant to say ' Does $\otimes_\mathcal{D}$ distribute over $*$?'. I'll fix it quickly. Does a similar argument to the one you gave above still work to disprove that claim? $\endgroup$
    – Jan Steinebrunner
    Mar 15, 2018 at 15:34

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