Let $(C,\otimes,I)$ be a symmetric monoidal category with coequalizers and directed colimits.

Fix some object $X$ and morphism $\tau\colon I\to X.$

Using $\tau$ one can construct a sequence of morphisms:

$$ I\rightarrow^{\tau}X\rightarrow^{\tau\circ\rho_X^{-1}} X\otimes X \rightarrow X^{\otimes 3} \ldots $$

Let $(X,\tau)^{\infty}$ be a colimit of that diagram.

Question: is it true(under which assumptions it is true?) that there exists a monomorphism $\eta\colon \Sigma_{\infty}\to Aut((X,\tau)^{\infty})$ induced by braidings on terms in the diagram?

Here $\Sigma_{\infty}$ is a group of all permutations with finite support, we also assume that braiding $B_{X,X}$ is non-identity.

Why symmetric products?

Let us consider following diagram:

$$ \begin{array}{l} I&\longrightarrow^{\tau}&X&\rightarrow^{\tau\circ\rho_X^{-1}} &X\otimes X& \longrightarrow& X^{\otimes 3}& \ldots\\ \downarrow_{id}&&\downarrow_{id}&&\downarrow_{Coeq(id,B_{X,X})}&&\downarrow_{Coeq(\Sigma_3)}\\ I&\longrightarrow^{\tau}&X&\rightarrow &X^{(2)}& \longrightarrow& X^{(3)}& \ldots\\ \end{array} $$

Denote colimit of that diagram $(X,\tau)^{(\infty)}.$

Then we have canonical morphism $\zeta_{(X,\tau)}\colon (X,\tau)^{\infty} \to (X,\tau)^{(\infty)}.$

If $((X,\tau)^{(\infty)},\zeta_{(X,\tau)})$ is a coequalizer of $\eta(\Sigma_{\infty})$ then one can naturally call that pair an infinite symmetric product of $(X,\tau).$

UPDATE: Why it is non-trivial? Take a category with objects from the first diagram and terminal object. Then terminal object is colimit, but its automorphism group is trivial, therefore it does not contain $\Sigma_{\infty}.$

  • 2
    $\begingroup$ I don't see any reason to expect a monomorphism. The natural map $S_n \to \text{Aut}(X^{\otimes n})$ is often not a monomorphism, for example. $\endgroup$ – Qiaochu Yuan Jun 16 '15 at 19:21
  • $\begingroup$ Of course, it is not always a monomorphism, but more interesting cases for me are those that are closer to topological spaces, for example. $\endgroup$ – probably Jun 16 '15 at 19:28
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    $\begingroup$ Even for things like topological spaces, injectivity may fail (consider when $X$ is just a point). I don't know of any natural conditions you can put that will guarantee that $\eta$ is injective that do not include just directly assuming that $\Sigma_n$ acts faithfully on $X^{\otimes n}$ for each $n$ (or at least for $n=3$). In many particular examples, the $X$ for which this fails are just a small collection of "trivial" examples, but I don't know of any nice general characterization. $\endgroup$ – Eric Wofsey Jun 17 '15 at 2:02

This seems entirely straightforward unless I'm missing something. For any $n\in\mathbb{N}$, $\Sigma_n$ acts on $X^{\otimes m}$ for any $m\geq n$ (on the first $n$ coordinates), and this action commutes with the maps in the colimit diagram. Thus $\Sigma_n$ acts on the colimit of $X^{\otimes n}\to X^{\otimes (n+1)}\to\dots$, which is the same as $(X,\tau)^\infty$ because you've just taken a cofinal subdiagram. Furthermore, these actions over different $n$ are compatible, so they glue together to give an action of $\Sigma_\infty$.

To put it another way, let $D$ be the category of diagrams in $C$ of the form $X_0\to X_1\to X_2\to\dots$, except that only the "tail" of the diagrams are required to be defined. That is, for any $n\in\mathbb{N}$, a diagram of the form $X_n\to X_{n+1}\to X_{n+2}\to\dots$ also counts as an object of $D$, and a morphism between objects of $D$ need only be "eventually" defined (and parallel morphisms that eventually agree are identified). It is straightforward to see that taking the colimit of the diagram gives a well-defined functor from $D$ to $C$. Now observe that the diagram whose colimit is $(X,\tau)^\infty$ has an action of $\Sigma_\infty$ as an object of $D$, and hence so does its colimit.

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    $\begingroup$ Let us throw away all objects in category $C$ except objects from a diagram and terminal object. Then terminal object is a colimit of that diagram, but its automorphism group will not contain $\Sigma_{\infty}.$ $\endgroup$ – probably Jun 16 '15 at 7:12
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    $\begingroup$ So the definition of morphism $\eta$ (which is not contained in your comment) should use not only the fact that there exists a series of agreed actions, but something more. $\endgroup$ – probably Jun 16 '15 at 7:19
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    $\begingroup$ Oops, sorry, I missed that you asked for $\eta$ to be injective. Why do you care that it is injective though? There always exists a canonical homomorphism, and it will sometimes be injective, but I don't know of anything you can do when it is injective that you can't do just from its existence (but maybe you do, which is why I'm asking). $\endgroup$ – Eric Wofsey Jun 16 '15 at 8:44
  • $\begingroup$ Also, I didn't address this because you didn't ask explicitly, and I haven't checked all the details, but it seems like it should be straightforward that your $\zeta$ is the coequalizer of the action of $\Sigma_\infty$. When you write down what it means to give a map out of $(X,\tau)^{(\infty)}$ and a map out of $(X,\tau)^\infty/\Sigma_\infty$, you get equivalent data. $\endgroup$ – Eric Wofsey Jun 16 '15 at 8:50

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