# Action on tensor power and “element notation” in monoidal categories

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.

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Can't you use the coherence theorem to reduce to the strict case? Then everything should be clear. – Qiaochu Yuan Jul 9 '11 at 14:04
You absolutely must read "Notes on Supersymmetry", by Deligne and Morgan, in Quantum Fields and Strings. They develop precisely the categorical machinery necessary to write formulas like the one you want. – Theo Johnson-Freyd Jul 9 '11 at 15:01
A very clean treatment is given in Ezra Getzler's "Operads revisited", right in the beginning of section 2. – Dan Petersen Jul 9 '11 at 16:08
@Theo: Do they really do it? It seems to me that they use element notation, where only the first examples are backened by their morphism definiton (and it is clear how to do it in any case). No general machinery is developed. So basically this is the same which I have done so far. – Martin Brandenburg Jul 9 '11 at 20:06
@Dan: This is the "n-ary tensor product" definition I was aiming at! Could you post this as an answer? – Martin Brandenburg Jul 9 '11 at 20:09

## 3 Answers

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 $0$.

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Probably you mean "is an equivalence" in (ii)? I have found another reference, a paper by Tom Leinster arxiv.org/PS_cache/math/pdf/0002/0002180v1.pdf , 3.3. – Martin Brandenburg Jul 9 '11 at 12:44
Yes, that sentence should have ended (and now it does) with "is an equivalence." I don't know how that remained unfinished (my fault, I should have checked more carefully my answer before posting). Concerning Tom Leistner paper, that's a very good reference. – domenico fiorenza Jul 9 '11 at 13:03

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.

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Yes, but isn't this cheating? For me, $T1$ is defined to be as follows: The objects are natural numbers, writen as $x^{\otimes n}$ instead of $n$, with the obvious tensor product. There is only a morphism $x^{\otimes n} \to x^{\otimes m}$ if $n=m$, and then they are all isomorphisms and $Aut(x^{\otimes n})=\mathfrak{S}_n$. But now in order to show that this is actually the free symmetric monoidal category on one object, we have to show that for every object $X \in C$ and every $\sigma \in Aut(x^{\otimes n})$, there is an induced automorphism of $X^{\otimes n}$ - this is exactly my question. – Martin Brandenburg Jul 9 '11 at 12:08
On other hand, if the existence of $T1$ is guaranteed by more abstract means, it should be then proved that $\mathfrak{S}_n \cong Aut(x^{\otimes n})$ in the free case. – Martin Brandenburg Jul 9 '11 at 12:09
@Martin the category you describe as $T1$ is the free symmetric strict monoidal category on $1$, that is not exactly the one I was referring to, although it's equivalent. The 2-monad $T$ is described somewhere in the literature, probably using clubs a la Max Kelly. That description yields $T1(x^{\otimes n},x^{\otimes n})\cong\frak{S}_n$. Alternatively, if $S$ is the 2-monad with algebras symmetric strict monoidal categories, by general theory in Blackwell-Kelly-Power, $S\simeq T$, so $S1\simeq T1$. But $S1$ is the category you describe in your comment above, with homs $\frak{S}_n$. – Nacho Lopez Jul 9 '11 at 16:37
This is exactly right, but it is also exactly right that one needs to prove something to conclude that the automorphisms in the free symmetric monoidal category are, in fact, the symmetric groups (regardless of which half of the theorem one takes as the "definition"). And as Qiaochu said in a comment above, this is essentially the content of the coherence theorem for symmetric monoidal categories. – Mike Shulman Jul 10 '11 at 6:57
Yes and therefore this answer is just blowing up the question without answering it. – Martin Brandenburg Jul 10 '11 at 12:30

[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 $(V_i)_{i\in I}$ " ... )

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@who downvoted: could you please explain why? I like to learn something from my mistakes/imprecisions/misunderstandings. :) – Qfwfq Jul 11 '11 at 11:49