Action on tensor power and "element notation" in monoidal categories

Question

Let $C$ be a symmetric monoidal category. Fix an object $X$, let $S$ denote the symmetry $X \otimes X \to X \otimes X$. Also define $X^{\otimes n}$ by induction on $n$: $X^{\otimes 0} = 1$, $X^{\otimes (n+1)} = X^{\otimes n} \otimes X$. Now it is "absolutely clear" that we have a canonical action of the symmetric group $\mathfrak{S}_n$ on $X^{\otimes n}$, i.e. a homomorphism of groups $\mathfrak{S}_n \to \text{Aut}(X^{\otimes n})$. But what is a precise and short definition?

A transposition $\sigma_i = (i,i+1)$ acts as the composite of isomorphisms $X^{\otimes n} \cong X^{\otimes (i-1)} \otimes ((X \otimes X) \otimes X^{\otimes (n-i-1)}) \stackrel{S}{\cong} X^{\otimes (i-1)} \otimes ((X \otimes X) \otimes X^{\otimes (n-i-1)}) \cong X^{\otimes n}$

Now $\mathfrak{S}_n$ is freely generated by these $\sigma_i$ modulo the relations a) $\sigma_i^2 = 1$, b) $\sigma_i \sigma_j = \sigma_j \sigma_i$ for $i \neq j \pm 1$, c) $\sigma_i \sigma_{i+1} \sigma_i = \sigma_{i+1} \sigma_i \sigma_{i+1}$. It should be possible to test these relations above, using the coherence between the symmetry and the associator on $C$, but obviously it will be tedious.

I would like to define the action in element notation via $x_1 \otimes ... \otimes x_n \mapsto x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$, but is this possible at all? Perhaps using a weak-monadic representation of monoidal categories?

In my research I often have to define such morphisms which are "obvious" for the usual symmetric monoidal categories, but are nasty to write down and manipulate in the general case. After having checked many examples, meanwhile I have "trust" in these element definitions. But I wonder if any general machinery has been established for this. Note that in my former question I asked about checking equality, but in this question about construction in monoidal categories.

Answer 1

There's a point of view on symmetric monoidal categories (dating back to Graeme Segal's "Categories and cohomology theories" and then taken up by Bertrand To\"en in "Dualit\'e de Tannaka superieure, I: Structures monoidales" to define symmetric monoidal $n$-categories) which stresses the symmetric group action from the very beginning. It goes as follows: let $\Gamma$ be the category of pointed finite sets with maps of pointed sets as morphisms. Denote by $[1]$ the set $\{0,1\}$ pointed at $0$. Then for any finite set $X$ and any element $x$ in $X$ there is a unique map $u_x:[1]\to X\coprod\{*\}$ in $\Gamma$ mapping $1$ to $x$ (where $*$ is the distinguished point of $X\coprod\{*\}$). With these notations, a symmetric monoidal category is a functor $F:\Gamma^{op}\to Categories$ such that

i) $F(\{*\})=*$ (the terminal category);

ii) the natural map $F(X\coprod\{*\})\to F([1])^{|X|}$ induced by the morphisms $u_x:[1]\to X$ and the universal property of the product is an equivalence.

The correspondence between this and the classical definition of symmetric monoidal category is (clearly) given by $F \to F[1]$. The natural action of the symmetric group $\mathfrak{S}_n$ is then induced by the isomorphism $\mathfrak{S}_n\cong \mathrm{Aut}_\Gamma([n])$, where $[n]=\{0,1,\dots,n\}$ is pointed at $0$.

Answer 2

Let $T$ be the 2-monad on $\mathbf{Cat}$ whose algebras are symmetric monoidal categories. So $T1$ is the free monoidal category on one object --say $x$,-- and any object $X$ of $C$ induces a braided monoidal functor $T1\to C$, sending $x\in T1$ to $X$. With your definition of $X^{\otimes n}$, if $x^{\otimes n}$ is defined in an analogous way, your action seems to be the function ${\frak{S}}_n\cong T1(x^{\otimes n},x^{\otimes n})\to C(X^{\otimes n},X^{\otimes n})$ which is the effect of the functor $T1\to C$ on hom-sets.

Answer 3

[Edit: I realize I should've posted it as comment, not as an answer]

You can have a look to §1.1 ("The sign rule") and §1.2 ("Categorical approach") of Chapter 1 of Quantum Fields and Strings, a Course for Mathematicians by Deligne et al.

It seems that the key phrase is "take the limit" (though it's mysterious to me what it means that "the symmetric group of $I$ acts on the family $(V_i)_{i\in I}$ " ... )