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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?

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  • $\begingroup$ Seeing as this is still unanswered, I could probably find a proof for this. Is that sufficient, or do you want a published reference? $\endgroup$ – Matt Samuel Nov 4 '18 at 11:50
  • $\begingroup$ @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.) $\endgroup$ – Jim Humphreys Nov 4 '18 at 14:54
  • $\begingroup$ 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 $W$. $\endgroup$ – Jim Humphreys Nov 4 '18 at 15:00
  • $\begingroup$ @MattSamuel: I'd be happy with both! $\endgroup$ – ouimerci Nov 4 '18 at 15:02
  • $\begingroup$ @JimHumphreys: I'm not sure I understand the remark; is some sort of rewriting of the question in order? $\endgroup$ – ouimerci Nov 4 '18 at 15:06
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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.

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  • $\begingroup$ 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!) $\endgroup$ – ouimerci Nov 4 '18 at 16:23
  • $\begingroup$ @ouimerci That's right, I prove that such a word exists. $\endgroup$ – Matt Samuel Nov 4 '18 at 16:24
  • $\begingroup$ Well, I think this answers my question then! $\endgroup$ – ouimerci Nov 4 '18 at 16:30
  • $\begingroup$ @oui I finished the proof for completeness. It doesn't use symmetry of the word. $\endgroup$ – Matt Samuel Nov 4 '18 at 17:32
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In response to the reference request: The existence of a symmetric reduced word is Exercise 10 (page 22) in Chapter 1 of the Björner-Brenti book.

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