Reference Request: Length of a reflection in a Coxeter group can be achieved by symmetric word

In a given coxeter group $$(W,S)$$, a reflection is an element of $$W$$ that can be written with a symmetric word in the generators $$S$$.

In multiple sources, I found the following formula: $$\mathrm{dp}(\alpha) = \frac{1}{2}(l(t_\alpha) + 1)$$ where $$\alpha$$ is a positive root, $$t_\alpha$$ the corresponding reflection and the depth $$\mathrm{dp}(\alpha)$$ is the length of a shortest word $$w$$ such that $$w\cdot \alpha$$ is a negative root.

Assuming that any reflection has length achievable by a symmetric word, this formula is rather easy to check, but I couldn't find a proof for just this fact.

In X Fu's thesis, Lemma 1.3.19, the formula is proven but I'm looking for a more elementary proof of this fact:

Question: Is the length of a reflection in a Coxeter group achievable by a symmetric word?

• Seeing as this is still unanswered, I could probably find a proof for this. Is that sufficient, or do you want a published reference? – Matt Samuel Nov 4 '18 at 11:50
• @Matt: in your added comment, it's not quite clear what "this" refers to, but in any case a published proof is not required. However, it would probably help to give a more detailed reference to Fu's thesis or a published version of his depth formula. (Note too that a label such as (*) would be appropriate for that formula, and that you can highlight your question by typing > first.) – Jim Humphreys Nov 4 '18 at 14:54
• P.S. As background, the definition of "reflection" implies that such an element is conjugate to a "simple" reflection (one relative to an element of $S$), though you do need to specify how a "root system" comes into the picture for infinite $W$. – Jim Humphreys Nov 4 '18 at 15:00
• @MattSamuel: I'd be happy with both! – ouimerci Nov 4 '18 at 15:02
• @JimHumphreys: I'm not sure I understand the remark; is some sort of rewriting of the question in order? – ouimerci Nov 4 '18 at 15:06

Let $$\beta$$ be a positive root. Let $$u$$ be an element of length $$\mathrm{dp}(\beta)$$ such that $$u(\beta) <0$$. Then $$u(\beta)=-\alpha$$ for some simple root $$\alpha$$ (because otherwise we could multiply $$u$$ by a left descent to get an element of shorter length inverting $$\beta$$), so $$u^{-1}(\alpha)=-\beta$$ and hence $$\ell(s_\alpha u) <\ell(u)$$.
Now since $$u^{-1}s_\alpha(\alpha) = \beta$$, we have that $$(s_\alpha u)^{-1}s_\alpha (s_\alpha u)=s_\beta$$. Thus $$s_\beta$$ has a symmetric word of length $$2\mathrm{dp}(\beta)-1$$.
This proves that $$\ell(s_\beta) \leq 2\mathrm{dp}(\beta)-1$$. For the opposite inequality, let $$(s_1,\ldots,s_k)$$ be a reduced word for $$s_\beta$$. Let $$k+1-i$$ be the maximal index such that $$s_{k+1-i}\cdots s_k(\beta) =-\alpha<0$$. Then $$k+1-(k+1-i)= i\geq\mathrm{dp}(\beta)$$. Now $$s_1\cdots s_{k-i }s_{k+1-i}(\alpha)=-\beta$$, so $$s_{k+1-i}\cdots s_{2}s_{1}(\beta)=-\alpha$$, so $$k+1-i\geq\mathrm{dp}(\beta)$$. Hence $$k+1\geq 2\mathrm{dp}(\beta)$$, and the result follows.
• So, you're showing that for any root $\beta$, $s_\beta$ can be written with a symmetric word of length $2\mathrm{dp}(\beta) -1$, right? Then, to answer the question/claim in my post, you'd essentially combine that with the equality written there. I'm not sure of what you assume and what you conclude. (And thanks!) – ouimerci Nov 4 '18 at 16:23