It's been a long time since I posted the following question on stackexchange.

Now I think it's better to adress it to you, in the hope I will reach the right audience: Martin's comment, albeit useful, didn't help me to satisfy my curiosity.


A functor $F\colon \bf Sets\to Sets$ is said to be analytic if it results from the left Kan extension of a functor $f\colon \mathbf{Bij}(\mathbb N)\to \bf Sets$ (the "species" of the functors $F$) along the natural inclusion $\mathbf{Bij}(\mathbb N)\to \bf Sets$, where $\mathbf{Bij}(\mathbb N)$ is the category having objects natural numbers and where $\mathbf{Bij}(\mathbb N)(m,n)$ are the bijective functions $\{1,\dots,m\}\to \{1,\dots,n\}$ (empty if $n\neq m$). Representing a left Kan extension as a coend it means that $$ F(T)\cong \int^n T^n\times f(n) $$ (the most of you will recognize the fact that a functor is "anaytic" if it can be written in Taylor form, and the coend is in a suitable sense exactly that Taylor series) This can be expressed replacing $\bf Sets$ with any symmetric monoidally cocomplete category: there is a functor $\mathbf{Bij}(\mathbb N)^\text{op}\times \mathcal V\to \mathcal V\colon (n,V)\mapsto V\otimes\dots\otimes V=V^{\otimes n}$, which allows to define $$ \int^n V^{\otimes n}\otimes f(n) $$ for any "species" $f\colon \mathbf{Bij}(\mathbb N)\to \mathcal V$.

In the case of $\bf Sets$, it seems to be possible to extend the definition of an analytic functor to the case of an arbitrary cardinal $\kappa$:

  1. the category $\mathbf{Bij}(\kappa)$ is defined to be the category having objects cardinals $<\kappa$ and $\mathbf{Bij}(\kappa)(\mu,\nu)$ bijections between those cardinals (hence empty if $\mu\neq\nu$).
  2. I say that a functor $F\colon\bf Sets\to Sets$ is $\kappa$-analytic if it results from $\text{Lan}_jf$ in the diagram $$ \begin{array}{ccc} \mathbf{Bij}(\kappa) &\to& \bf Sets \\ \downarrow&\nearrow_F&\\ \bf Sets & \end{array} $$ $F(T)\cong \int^{\mu}T^\mu\times f(\mu)$ where $T^\mu$ is the product of $\mu$ copies of $T$, in the obvious sense.

Now my question: can I define something analogous in the case of a generic monoidal(ly cocomplete) category, maybe imposing some additional property? In the end all boils down to the possibility of defining a $\kappa$-fold tensor product $\otimes\colon \mathcal V^\kappa\to \mathcal V$, for any cardinal $\kappa$. Modules over a ring $R$ seem to have this "extension property", together with sets.

Being these two very explicit and natural examples of a monoidal category where "tensoring has an arbitrarily large arity", I think the problem has its own interest, and I would be surprised if the topic was new.


As I pointed out in the comments on math.SE, I was motivated by Martin's topic about k-ary tensors in $Mod_R$ to pose the question in the form of a "k-ary tensor product". I think there is no hope to recover "usual properties"; it's better to think about an axiomatization of the structure I need, but for the moment I don't have the slightest idea. I repeat myself: on the one side I find quite astounding the lack of a theory of those monoidal categories where tensoring has an arbitrarily large arity, and on the other, it's obviously due to the absence of "good" properties for such a structure.


1 Answer 1


Not sure this will meet your expectations, but maybe Mark Weber's work on functors with arities is relevant. See Remark 2.12 therein, in particular.

The upshot, if I'm correct, is that any analytic endofunctor $F$ obtained from $f$ as above may be recovered from a polynomial diagram along the lines of

$$1 \leftarrow D \to \int^{n} f(n) \to 1,$$

where $1 \leftarrow D \to \int^{n} f(n)$ is the comma cone from the Yoneda embedding $1 \to Sets$ to the functor $\int^{n} f(n) \to Sets$ mapping $(n,x)$ to the finite ordinal $n$.

The functor $Sets \to Sets$ associated to such a diagram takes any set $X$, viewed as a functor $1^{op} \to Sets$, and

  • restricts it along (the opposite of) $1 \leftarrow D$, then
  • right extends it along (...) $D \to \int^{n} f(n)$, and finally
  • left extends it along (...) $\int^{n} f(n) \to 1.$

In our case, this should map $X$ to $\int^p X^p \times {f(p)}$, as desired.

So we obtain $F$ as a so-called parametric right adjoint functor (I think Mark now calls these local right adjoints). There is an even more general notion of functor with arities, defined in the particular case of monads in the above paper, and made more explicit in Berger, Melliès, and Weber.


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