# Is there a natural notion of completion of a Coxeter system?

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.

• What about affine and hyperbolic Coxeter groups? – Will Sawin Mar 5 '14 at 4:56
• @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? – Sam Hopkins Mar 5 '14 at 4:59
• 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. – Will Sawin Mar 5 '14 at 5:06
• 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. – Allen Knutson Mar 5 '14 at 7:37
• @AllenKnutson: see edit above. – Sam Hopkins Mar 5 '14 at 16:54

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.
• @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 $GL(E)$). – Sam Hopkins Mar 19 '14 at 12:49
• 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 $W = S_{\infty}$. – Sam Hopkins Mar 19 '14 at 12:51