# Each $w\in W$ can be expressed as product of distinct reflections?

For a Weyl group $W$, I would like to know whether each $w\in W$ can be expressed as $w=s_{\alpha_1}s_{\alpha_2}\cdots s_{\alpha_k}$ for some distinct positive roots $\{\alpha_1, \alpha_2, \cdots, \alpha_k\}\subseteq \Phi^+$.

I know for type A, the above is true.

Since $W(A_n)\cong S_{n+1}$ with the map $s_{e_i-e_j}\mapsto (i \ j)$. For $w\in W(A_n)$, let $\sigma\in S_{n+1}$ be the image of $w$ under the map. By the fact that every permutation can be decomposed as a product of disjoint cycles in the form of $(a_1\cdots a_l)$. And also the fact that $(a_1\cdots a_l)=(a_1 \ a_l)\cdots (a_1 \ a_2)$. We get $\sigma$ is a product of distinct transpositions. And distinct transpositions correspond to distinct reflections of the form $s_{e_i-e_j}$. Therefore, the claim follows.

I would like to know whether the above statement is also true for other types?

• This would imply that the longest element is the product, in some order, of the reflections in all the positive roots. Have you checked this for any non-$A_n$ types? – LSpice Sep 5 '18 at 14:57
• As a consequence of Thm 1.8 in Humphreys "Reflection Groups and Coxeter Groups" (CUP), pages 15 and 16, there exists a longest element in $W$ whose order is $|\Phi^+|$ (or in his notation $|\Pi|$. This is precisely LSpice previous remark. – F Zaldivar Sep 5 '18 at 17:07
• @FZaldivar The order of the longest element is 2. I think you mean "length", not order. – Johannes Hahn Sep 9 '18 at 17:08
• Indeed, I meant length. Thanks, Johannes! – F Zaldivar Sep 11 '18 at 3:57
• Expanding on Jim Humphreys' comment on one of the answers below, this is completely general, not even requiring finiteness. (As usual, we define "reflections" to be the elements that are conjugate to simple reflections. That is perfectly reasonable, because there is a standard way to represent $W$ so that these are precisely the reflections in a geometric sense. This is in Humphreys' book.) A reference: Combine Lemma 1.3.1 of the Bjorner-Brenti book with equation (1.12) right before the Lemma. James Cheung's proof seems to do the same thing. – Nathan Reading Nov 29 '18 at 15:39

The answer is "yes" and there is a geometric explanation.

Let $\mathcal{H}$ denote the set of hyperplanes corresponding to the reflections $s_\alpha$ with $\alpha\in\Phi^+$ (note that $s_\alpha=s_{-\alpha}$), and let $\Sigma$ denote the connected components of $V\setminus\bigcup_{H\in\mathcal{H}}H$ (where $V$ is the vector space where $\Phi$ lives). The elements of $\Sigma$ are the chambers of the Coxeter complex of $(V,\Phi)$, and it is well-known that $W$ acts simply transitively on $\Sigma$.

Fix some $C_0\in\Sigma$, let $w\in W$, and put $C_1=w(C_0)$. Then you can "walk" from the chamber $C_0$ to $C_1$ by crossing through the walls of the chambers (here, a wall means a codimension-$1$ face of a chamber). Every time you cross from $C$ into an adjacent chamber $C'$, you walk though some hyperplane $H\in \mathcal{H}$, and in that case you have $C'=s_\alpha C$, where $\alpha\in \Phi^+$ corresponds to $H$. Since you can walk from $C_0$ to $C_1=wC_0$ without crossing the same hyperplane $H$ twice (Edit: e.g. take a walk minimizing the number of chambers visited, or go along a straight line in general position w.r.t. $\mathcal{H}$), it follows that $C_1=s_{\alpha_t}\cdots s_{\alpha_1}C_0$, where $\alpha_t,\dots,\alpha_1\in\Phi^+$ are distinct and correspond to the hyperplanes you crossed on your way from $C_0$ to $C_1$. Thus, $w=s_{\alpha_t}\cdots s_{\alpha_1}$ with distinct $\alpha_t,\dots,\alpha_1\in \Phi^+$.

• I think you have to be a little bit careful: a naïve walk might cross several hyperplanes simultaneously, and a naïve genericity adustment to fix this might cross the same hyperplane several times. I have no doubt that this can be fixed, but it would need to be argued. – LSpice Sep 5 '18 at 18:18
• @LSpice I was a bit sloppy there. You can take a straight line in general position with respect to the (finitely many) hyperplanes in $\mathcal{H}$ and their intersections, which always exists. Alternatively, you can take a minimal walk (gallery) from $C_0$ to $C_1$. Indeed, if you cross the same hyperplane $H$ more than once, say at $C'$ and $C''$, then you can replace everything between $C'$ and $C''$ by their reflection along $H$ and remove $C'$ and $C''$, resulting in a shorter walk. – Uriya First Sep 5 '18 at 19:19
• @Uriya: It would help a lot to give a couple of examples, such as the longest element of a Weyl group (at least for $S_3$). However, the usual objective is to express elements of $W$ as products of simple reflections for some fixed simple system of roots: this is considerably more efficient than using the entire collection of reflections for all positive roots. – Jim Humphreys Sep 8 '18 at 17:52
• P.S. Note too that the arguement you give is valid for any finite Coxeter group (including those which are not Weyl groups). This indicates that the question has nothing to do with Lie theory, but onlly with the geometry of finite real reflection groups. – Jim Humphreys Sep 8 '18 at 17:57
• @Jim Humphreys I agree to both. Let me add that, personally, I find the argument using the non-simple reflections conceptually easier to understand. (I also feel that I should acknowledge that my first introduction to reflection groups was through your book on the subject.) – Uriya First Sep 9 '18 at 7:20

After reading your comments, I come up with the following proof. I would like to know whether my proof is correct or not.

For weyl group $W$, each $w\in W$ can be expressed as $w=s_{\beta_l}\cdots s_{\beta_{2}} s_{\beta_1}$ for some distinct positive roots $\{\beta_1, \beta_2, \cdots, \beta_l\}\subseteq \Phi^+$.

My proof:

Let $w=s_{i_1}s_{i_2}\cdots s_{i_l}$ be a reduced expression with $s_{i_k}$ is the simple reflection of root $\alpha_{i_k}\in\Delta$.

Let $w_k=s_{i_1}s_{i_2}\cdots s_{i_k}$, $w_0=1$, $C_k=w_{k}C_+$ and $C_0=C_+$. By direct computation, $C_k=w_{k}C_+ =w_{k-1}s_{i_k}C_+ =w_{k-1}s_{i_k}w_{k-1}^{-1}w_{k-1}C_+ =w_{k-1}s_{i_k}w_{k-1}^{-1}C_{k-1} =s_{\beta_k}C_{k-1},$ where $\beta_k:=w_{k-1}\alpha_{i_k}=s_{i_1}s_{i_2}\cdots s_{i_{k-1}}(\alpha_{i_k})$ and $\beta_1=\alpha_{i_1}$.

Note that $wC_+ =C_l =s_{\beta_l}C_{l-1} =s_{\beta_l}s_{\beta_{l-1}}C_{l-2}=\cdots =s_{\beta_l}s_{\beta_{l-1}}\cdots s_{\beta_2}C_1 =s_{\beta_l}s_{\beta_{l-1}}\cdots s_{\beta_1}C_+.$ By the simply-transitivity of $W$, we get $w=s_{\beta_l}s_{\beta_{l-1}}\cdots s_{\beta_1}$.

Since $s_\alpha=s_{-\alpha}$ for all $\alpha\in\Phi^+$, we may WLOG assume $\beta_k\in\Phi^+$.

Suppose $\beta_k=\beta_j$ for some $j<k$, then $s_{\beta_k}=s_{\beta_j}$. Then $w_{k-1}s_{i_k}w_{k-1}^{-1}=w_{j-1}s_{i_j}w_{j-1}^{-1}.$ $s_{i_j}=(s_{i_j}\cdots s_{i_{k-1}})s_{i_k}(s_{i_j}\cdots s_{i_{k-1}})^{-1},$ from which we obtain, upon right-multiplying by $s_{i_j}\cdots s_{i_k}$. $s_{i_{j+1}}\cdots s_{i_{k-1}} =s_{i_{j}}\cdots s_{i_{k}}.$ Then $w=s_{i_1}\cdots s_{i_l}=s_{i_1}\cdots\hat{s_{i_j}}\cdots\hat{s_{i_k}}\cdots s_{i_l},$ a contradiction to the fact that $w=s_{i_1}s_{i_2}\cdots s_{i_l}$ is a reduced expression. Therefore, the claim follows.

• Your proof appears to be correct. In fact, I think it is an "algebraic" rewriting of the the "geometric" argument above (a minimal gallery is essentially the same as a reduced expression). – Uriya First Sep 9 '18 at 7:30
• I agree with your viewpoint. I just use your idea and try to write an "algebraic" proof for that. Thank a lots. – James Cheung Sep 9 '18 at 12:35