TITLE: Product of two reflections lying in a parabolic subgroup of a Coxeter group
QUESTION [10 upvotes]: Let $(W,S)$ be a Coxeter group, $I\subseteq S$ a subset of simple reflections, and $W_I \subseteq W$ the corresponding parabolic subgroup (we could also assume $|W_I|<\infty$, if needed).

Let also $t_1,t_2\in W$ be two reflections (i.e. elements in $W$ conjugated to some $s_1,s_2 \in S$ respectively) such that $t_1t_2\in W_I$ but $t_i\notin W_I$, $i=1,2$.
Is it then true that the only possibility is $t_1t_2=e$ (i.e. $t_1=t_2$)?
If I look at the geometric picture with alcoves and reflection hyperplanes this seems to me to be true, but I don't know how to prove it in full generality via geometric arguments. I rather tried with some combinatoric method (e.g. using the strong exchange condition or some result about parabolic double cosets) but I have succeeded only in the case $t_i \in S$ for at least one $i$.
Thanks a lot for any comment about that!

REPLY [4 votes]: $\DeclareMathOperator\Fix{Fix}\DeclareMathOperator\Stab{Stab}$I want to share a possible solution that I found just developing the idea contained in Nathan's deleted answer. His argument just needed a representation $V$ with the following two properties:
(i) $\Fix(t_1t_2)=\Fix(t_1)\cap \Fix(t_2)$, if $t_1$, $t_2$ are distinct reflections in $W$.
(ii) $\exists v_0\in V$ such that $\Stab(v_0)=W_I$
With such a representation in hand one can easily conclude as follows (assuming $t_1\neq t_2$):
$t_1t_2\in W_I \Rightarrow v_0\in \Fix(t_1t_2)=\Fix(t_1)\cap \Fix(t_2) \Rightarrow t_1,t_2 \in W_I$.
As discussed in previous comments, the geometric representation of $W$ satisfies (i) but fails (ii), while its dual satisfies (ii) but fails (i).
However, a certain enlargement of the geometric representation (having the dual of it as a quotient) does the job:
Claim: Let $V$ be a finite dimensional real vector space with given linearly independent vectors $\{e_s\}_{s\in S}\subseteq V$ and given linearly independent linear forms $\{e_s^\vee\}_{s\in S}\subseteq V^*$ such that $\langle e_t,e_s^\vee\rangle=-2\cos(\frac{\pi}{m_{st}})$ $\forall s,t\in S$. Assume also that $V$ has the smallest possible dimension to satisfy these requests. Then $s\cdot v\mathrel{:=}v-\langle v,e_s^\vee\rangle e_s$ turns $V$ into a representation of $W$ that satisfies (i) and (ii).
The fact that such a vector space $V$ always exists is straightforward, while the fact that the above formula actually defines a representation of $W$ can be found in Soergel's article Kazhdan–Lusztig polynomials and indecomposable bimodules over polynomial rings (Proposition 2.1) (actually the original article is in German, the linked one is just an English preprint). The same Proposition also shows that this representation is reflection faithful: this in particular means that $\Fix(w)$ is a hyperplane if and only if $w$ is a reflection in $W$ and that distinct reflections fix different hyperplanes, which implies at once (i).
For (ii), we use the fact (contained in the proof of the above mentioned Proposition) that, if $K\mathrel{:=}\{v\in V \mathrel\vert \langle v, e_s^\vee\rangle=0\ \forall s\in S\}$ (so $K$ is a subrepresentation of $V$, where $W$ acts trivially), then $V/K$ is isomorphic to the dual of the geometric representation of $W$, and so it contains a vector with stabilizer equal to $W_I$. As a consequece, for any $v_0\in V$ lying in the preimage of such a vector one has $\Stab(v_0)\subseteq W_I$. To show the other inclusion, let $s \in I$: one has then $s\cdot v_0\in v_0+K$, but by definition one also has $s\cdot v_0=v_0-\langle v_0, e_s^\vee\rangle e_s$. Since $e_s\notin K$ ($\langle e_s,e_s^\vee\rangle=2$) it has to be $\langle v_0,e_s^\vee\rangle=0$, i.e. $s\in \Stab(v_0)$, as desired.