15
$\begingroup$

Let $(W,S)$ be a Coxeter system. Then any element of $W$ can be written as a finite products of elements of $S$. I want some notion of a "completion" of $W$, call it $\hat{W}$, whose elements are (suitably nice) infinite products of $S$. For instance, let $S$ be the set of adjacent transpositions of $\mathbb{Z}$. $W$ is therefore the infinite symmetric group; i.e, group of permutations $\mathbb{Z} \to \mathbb{Z}$ that fix all but finitely many numbers. I would want $\hat{W}$ to be the group of all permutations $\mathbb{Z} \to \mathbb{Z}$.

EDIT:

Okay, here is the motivation; I heard this in a class of George Lusztig. Let $(W,S)$ be an affine Coxeter system. Let $s \in S$, and let $W_s$ be the set of elements of $W$ that have a unique reduced expression which starts with $s$. (So $W_s$ is a right cell.) We can put a graph on $W_s$ by joining two elements by an edge if one is obtained from the other by multiplying on the right by an element of $S$. It turns out that this graph will always be the Coxeter diagram of a finite or affine Coxeter group, and in particular, the graph will be a tree and simply-laced. Let $(W',S')$ be this output. Then there is embedding of $(W,S)$ into $(W',S')$, where what I do is map $s_i \in S$ to the product of all the vertices of the graph of $(W',S')$ whose label in the above scheme ends with $s_i$.

However, what I just said is slightly a lie because it turns out that the diagram of $(W',S')$ can end up being infinite (although still always a tree and simply-laced). For instance, consider what happens in the case of $\widetilde{A}_{n-1}$. In this case, the graph will look like:

... n-1 - n - 1 - 2 - 3 - ... - n-1 - n - ...

(Here in the labeling I only write the last letter in the reduced word because that's all that's relevant.)

So then $W' = S_{\infty}$ is the infinite symmetric group. However, in carrying out the procedure to embed $\widetilde{A}_{n-1}$ inside $S_{\infty}$, I need to multiply infinitely many transpositions. And so I can't actually embed into $S_{\infty}$; I will end up inside what I want to call $\widehat{S}_{\infty}$: the group of all permutations $\mathbb{Z} \to \mathbb{Z}$. If I can do this, I get the realization of $\widetilde{A}_{n-1}$ as affine permutations, which is nice.

$\endgroup$
7
  • $\begingroup$ What about affine and hyperbolic Coxeter groups? $\endgroup$
    – Will Sawin
    Mar 5, 2014 at 4:56
  • $\begingroup$ @WillSawin: I don't understand your question. Certainly these (affine/hyperbolic Coxeter groups) are included as possibilities for $(W,S)$. My question is: how do you complete them? $\endgroup$ Mar 5, 2014 at 4:59
  • $\begingroup$ I was just wondering if there is a natural way to complete them. If not, it might not bode well for a good way to complete in general. $\endgroup$
    – Will Sawin
    Mar 5, 2014 at 5:06
  • 1
    $\begingroup$ Can you motivate your choice of $\hat W$? There are lots of intermediate groups, like the permutations that have a uniform bound on how far elements can move. That looks related to infinite words in the generators that only use any letter finitely many times. $\endgroup$ Mar 5, 2014 at 7:37
  • $\begingroup$ @AllenKnutson: see edit above. $\endgroup$ Mar 5, 2014 at 16:54

2 Answers 2

5
$\begingroup$

Check out the paper by Thomas Lam and Anne Thomas: "Infinite reduced words and the Tits boundary of a Coxeter group" arxiv: 1301.0873. There is also, I think, an earlier paper by Lam and Pylyavskyy.

$\endgroup$
1
1
$\begingroup$

The following is what Lusztig told me about the situation; I thought it was worth recording. Consider the geometric representation of $(W,S)$ as a reflection group in $GL(E)$ for some vector space $E$. We can let the completion $\hat{W}$ be the subgroup of $GL(E)$ generated by $\{ \prod_{s \in S'} s\}$ where $S' \subset S$ ranges over all subsets of pairwise commuting elements. These (potentially infinite) products make sense as elements of $GL(E)$ precisely because of the pairwise commuting property. This works to explain the $S_{\infty}$ example, but somehow I was imagining something more general than this.

$\endgroup$
4
  • $\begingroup$ @ Sam: I'm confused about what you mean. I'm assuming S is the finite set of "simple" generators of W, so as you define it, hat(W) is generated by certain products in W, and can't be bigger than W. (I guess I don't see how these products are "potentially infinite". Each is the product of elements of a subset of a finite set.) Please clarify. $\endgroup$ Mar 19, 2014 at 12:47
  • $\begingroup$ @NathanReading: yes, $\hat{W}$ is generated by products of generators of $W$, but these products are allowed to be infinite (if all terms pairwise commute). This infinite product does not belong to $W$ already (but it does make sense in $GL(E)$). $\endgroup$ Mar 19, 2014 at 12:49
  • $\begingroup$ Ah right, but I should point out that this is only meaningful when the set $S$ of generators itself is infinite, as in the case of $W = S_{\infty}$. $\endgroup$ Mar 19, 2014 at 12:51
  • $\begingroup$ Ah, I see. I'm used to assuming S is finite. The Lam-Pylyavskyy reference in my answer is for S finite. $\endgroup$ Mar 22, 2014 at 23:13

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.