Let $W$ be a Coxeter Group with generators $s_1, \cdots, s_n$ and $l:W \to \mathbb N$ be the lenght function, and $T \subset W$ the subgroup of reflections. Then $$l(w)<l(tw) \iff l(w)<l(wt)$$ for every $w \in W$, $t \in T$.

Thoughts: I have tried using that $l(x)= l(x^{-1 })$ and that we can write $t$ explicitally i.e., if $w=s_1s_2\cdots s_k$ is a reduced form for $w$ and $l(tw)<l(w)$ then $t= s_1s_2\cdots s_i\cdots s_2s_1$ for some unique $i$.


If I'm not mistaken, you are not able to prove the claim, since it is not true.

Consider the Coxeter group $\tilde A_1=I_2(\infty)$ with the Coxeter diagram $\circ\underset{\infty}{--}\circ$ and generators $S=\{\,s,t\,\}$. Let's take the element $w=st$, which has $\ell(w)=2$. But $tw=tst$ has $\ell(tw)=3$, where $\ell(wt)=\ell(s)=1$.

Therefore we found $w\in W, t\in T$ (even $t\in S$) with $$\ell(w)<\ell(tw) ~\text{ but }~ \ell(w)>\ell(wt)$$

Where did you get the claim from? Is there a restriction to the Coxeter group? Or do you want to use a length function with respect to some other generating set?

In retrospect, I don't know why I used $\tilde{A}_1$. This should work with every Coxeter group.

  • $\begingroup$ Maybe Im trying to prove the wrong thing. I must prove that the bruhat order defined with the lenght function as I did is the same using the left or the right multiplication! $\endgroup$ – Maffred Aug 27 '18 at 8:19
  • $\begingroup$ I think you are using the wrong quantifiers. If $w\overset{t}{\rightarrow}_R wt$ in the standard bruhat order, which means that $\ell(w)<\ell(wt)$, then you have to show that $w <_L wt$ in the alternative order. So you have to find reflections $t'_1,\dots,t'_k$ such that $w \overset{t'_1}{\rightarrow}_R \dots \overset{t'_k}{\rightarrow}_R wt$. Note that the $t'$ won't have to coincide with $t$. So in your claim you should not assume that the reflections on both sides are the same. $\endgroup$ – Babelfish Aug 27 '18 at 8:32
  • $\begingroup$ So you mean that if $w<ws_{\alpha}$, we look for $t$ such that $ws_{\alpha}=tw \implies t=ws_aw^{-1}=s_{w\alpha}$? $\endgroup$ – Maffred Aug 28 '18 at 2:29
  • $\begingroup$ Yes, and you already found $t$. Since $\ell(w)<\ell(tw)$, you got $w\overset{t}{\rightarrow}_L ws_\alpha$ //// I had one error in my last expression in the above comment, it should of course be $w \overset{t'_1}{\rightarrow}_L \dots \overset{t'_k}{\rightarrow}_L wt$. $\endgroup$ – Babelfish Aug 28 '18 at 7:04
  • $\begingroup$ @Maffred Do you need more help with your question? $\endgroup$ – Babelfish Aug 30 '18 at 7:51

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