6
$\begingroup$

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 $\Sigma_{\infty}.$

$\endgroup$
3
  • 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. $\endgroup$ Jun 16, 2015 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, 2015 at 19:28
  • 2
    $\begingroup$ 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. $\endgroup$ Jun 17, 2015 at 2:02

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ 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, 2015 at 7:12
  • 1
    $\begingroup$ 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, 2015 at 7:19
  • 3
    $\begingroup$ 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). $\endgroup$ Jun 16, 2015 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. $\endgroup$ Jun 16, 2015 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.