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Oct 25, 2021 at 7:32 comment added Emily Oh sorry, absolutely! I had a misunderstanding about the classical case (I thought the set of conjugacy classes of an abelian group was just a point, and it's precisely the opposite :P). Thanks!
Oct 25, 2021 at 7:03 comment added Maxime Ramzi (Look at the uncategorified version : the set of conjugacy classes of an abelian group is not punctual)
Oct 25, 2021 at 7:02 comment added Maxime Ramzi For a presentable symmetric monoidal $C$, an approptiate notion seems to be $C\otimes_{C\otimes C} C$, in other words the Hovhschild homology of $C$ as an object of $Pr^L$. This always has a SM functor to $C$ given by multiplication (as you'd expect) and so is never punctual unless $C$ is
Oct 25, 2021 at 6:58 comment added Maxime Ramzi Emily : Why would this imply that $C/\sim_3$ is punctual ? On the contrary, if anything it says that it is close to $C$. Consider e.g. $C$ a groupoid and the functor $C\to \pi_0(C)$. This sends $A\otimes B$ and $B\otimes A$ to the same element so it should factor through your $C/\sim_3$ and be surjective.
Oct 25, 2021 at 4:14 comment added Emily ($\newcommand{\C}{\mathcal{C}}$This definition is also in a sense dual to that of the Drinfeld centre of $\C$, which is the pseudo biend $\int^{\mathsf{ps}}_{\bullet\in\mathrm{B}\C}\mathsf{Hom}_{\mathrm{B}\C}(\bullet,\bullet)$; the uncategorified version of this being that for a monoid $A$, the set $\int_{\bullet\in\mathrm{B}A}\mathrm{Hom}_{\mathrm{B}A}(\bullet,\bullet)$ is the centre of $A$ and $\int^{\bullet\in\mathrm{B}A}\mathrm{Hom}_{\mathrm{B}A}(\bullet,\bullet)$ is its set of conjugacy classes)
Oct 25, 2021 at 4:12 comment added Emily $\newcommand{\C}{\mathcal{C}}$This way the conjugacy relation takes into account the monoidal structure, with the $g\circ f\sim f\circ g$ relation now being taken on the $1$-morphisms of $\mathrm{B}\C$, and looking like $A\otimes_\C B\sim B\otimes_\C A$. We also have now "conjugacies between conjugacies", with $\C/{\sim}_3$ being a category instead of a set. I think under this definition the category $\C/{\sim}_3$ should be punctual (or something similar) if $\C$ is symmetric monoidal, as then we have isos $A\otimes B\cong B\otimes A$.
Oct 25, 2021 at 4:11 comment added Emily @MaximeRamzi $\newcommand{\C}{\mathcal{C}}$I think the appropriate notion of "$\sim_3$-conjugacy classes" for monoidal categories should be different than the one for plain categories: instead of directly taking the coend $\int^{X\in\C}\mathrm{Hom}_{\C}(X,X)$, we should first pass to the delooping of $\C$ and then take the (pseudo bi)coend $\int^{\bullet\in\mathrm{B}\C}_{\mathsf{ps}}\mathsf{Hom}_{\mathrm{B}\C}(\bullet,\bullet)\overset{\mathrm{def}}{=}\C/\sim_3$.
Oct 24, 2021 at 9:10 comment added Maxime Ramzi Emily : your statement about SMCs is wrong I think, finite sets being a couterexample ! (They are symmetric monoidal in two ways at least)
Oct 24, 2021 at 9:06 comment added Maxime Ramzi Emily : yes, it is not too hard an exercise to show that this coend is literally the trace of the presheaf category over that category in presentable categories with the Lurie tensor product. I agree that it's good to leave the question open - even the answer for finite sets if you take the colimit in $\infty$-groupoids could be interesting (e.g. is it just the same as this one without $\pi_0$ ?).
Oct 24, 2021 at 9:02 comment added Maxime Ramzi @BenjaminSteinberg : Ah sorry I hadn't seen that particular comment ! Good to know I'm not wrong then :)
Oct 24, 2021 at 5:13 comment added Emily (Talking about the other cases, a related question which seems fun to think about is how these conjugacy classes might play with $E_n$-structures: the ($\sim_3$-)conjugacy classes of a commutative monoid are all equal, and similarly the category of $\sim_3$-conjugacy classes of a SMC $C$ is punctual (probably; I haven't checked this carefully yet). But what if $C$ were just braided, or more generally an $E_n$-monoidal $\infty$-category?)
Oct 24, 2021 at 5:12 comment added Emily I'll leave the question open for a little bit longer since maybe someone else might have something to say about the other cases, though let me mention again that I find your answer really fantastic!
Oct 24, 2021 at 5:08 comment added Emily Thanks, Maxime (and Alexander and everyone else), this is very cool! I didn't know about the relation with traces, and searching for it led me to the nLab page "trace of a category". There the trace of a category $C$ is defined to be precisely the coend $\int^X\mathrm{Hom}_C(X,X)$, and Section 3 gives exactly the same argument you worked out, finishing with an identification of $\mathsf{FinSets}/{\sim}_3$ with the "class of all finite Young diagrams"!
Oct 24, 2021 at 1:01 comment added Benjamin Steinberg I.mentioned the answer for finite sets which is well known from semigroup representation on the comment deleted answer
Oct 23, 2021 at 22:56 history edited Maxime Ramzi CC BY-SA 4.0
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Oct 23, 2021 at 22:49 history answered Maxime Ramzi CC BY-SA 4.0