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!



Product of two reflections lying in a parabolic subgroup of a Coxeter group
$\DeclareMathOperator\Im{Im}\DeclareMathOperator\Fix{Fix}$As [suggested](https://mathoverflow.net/questions/381039/product-of-two-reflections-lying-in-a-parabolic-subgroup-of-a-coxeter-group/381234#comment968750_381039) by Sam, I am posting this as an answer.


As mentioned in the comments, the above question can be solved using a result of Baumeister–Dyer–Stump–Wegener from [A note on the transitive Hurwitz action on decompositions of parabolic Coxeter elements](https://arxiv.org/abs/1402.2500), which says the following: let $t\_i$, $i=1,\dotsc, k$ be reflections in an arbitrary Coxeter group $(W,S)$. Assume that $W'$ is a (not necessarily standard) parabolic subgroup of $W$, and assume that $t\_1 t\_2\dotsm t\_k$ is a reduced reflection factorization of an element $w\in W'$. Then $t\_i\in W'$ for all $i$. Applied to the case $k=2$ and $W'=W\_I$ this answers the question.


I also think that there is a more elementary way to see this (for $k=2$). Assume that $w\mathrel{:=}t\_1 t\_2\in W\_I$. Let $V$ be the geometric representation of $W$ and let $\beta\_1$ and $\beta\_2$ be the roots corresponding to $t\_1$, respectively $t\_2$. Then $$\Im(t\_1 t\_2 - 1)\subseteq \mathbb{R} \beta\_1 \oplus \mathbb{R} \beta\_2.$$ Now since $w\in W\_I$, denoting $\alpha\_i$, $i\in I$ the simple roots attached to the simple generators in $I$ we also have $$\Im(t\_1 t\_2 - 1)=\Im(w - 1)\subseteq \bigoplus\_{i\in I} \mathbb{R} \alpha\_i.$$


**Claim :** $\mathbb{R} \beta\_1 \oplus \mathbb{R} \beta\_2\subseteq \bigoplus\_{i\in I} \mathbb{R}\alpha\_i$ (which concludes the proof, as $\beta\_i$ are then linear combinations of the simple roots in $I$, hence $t\_i\in W\_I$).


To see this, by the above inclusions, it is enough to see that $M(w)\mathrel{:=}\Im(w-1)=(w-1)(V)$ has dimension $2$. The point is that, unfortunately, this is *not* the case in general: this is related to what has already been said in the comments but for instance, this fails in type $\widetilde{A}\_1$, taking $w=st$, where $\{s,t\}=S$, then the line spanned by $\alpha\_s+\alpha\_t$ is fixed by $w$, hence $\dim(M(w))=1$. As noted in the [comments](https://mathoverflow.net/questions/381039/product-of-two-reflections-lying-in-a-parabolic-subgroup-of-a-coxeter-group#comment968583_381039), as $t\_1$ and $t\_2$ are distinct, the roots $\beta\_1$ and $\beta\_2$ are not proportional, and therefore $\Fix(t\_1t\_2)=\Fix(t\_1)\cap\Fix(t\_2)$. But these two hyperplanes may be equal, as in the example.


But I think that on can cheat as follows by taking a bigger Coxeter group $\widetilde{W}$ (with geometric representation $V'$) in which $W$ is a standard parabolic subgroup, and such that $(w-1)(V')$ has dimension two. For such an enlarged geometric representation we still have the inclusions $(w-1)(V') \subseteq \mathbb{R}\beta\_1\oplus\mathbb{R} \beta\_2$ and $(w-1)(V')\subseteq \bigoplus\_{i\in I}\mathbb{R} \alpha\_i$.


For instance, one can take $V'$ as follows: first, let $x\in W$ such that $x t\_1 x^{-1}=s\in S$, and let $w'\mathrel{:=}x w x^{-1}= s t\_2'$, with $t\_2'=xt\_2 x^{-1}$. Consider the Coxeter group $\widetilde{W}$ with one more simple generator $s'$ than $W$, such that $ss'=s's$ and $m\_{s't}=\infty$ for any $t\in S\setminus\{s\}$. Let $V'$ be its geometric representation (note that $\dim(V')=\dim(V)+1$). Then $s'$ commutes with $s$ but cannot commute with $t\_2'$ (as $s\neq t\_2'$), hence $\alpha\_{s'}\in\Fix(s)$ but $\alpha\_{s'}\notin\Fix(t\_2')$, and therefore $\Fix(s)\neq \Fix(t\_2')$. It follows that the intersection $\Fix(s)\cap \Fix(t\_2')=\Fix(st\_2')=\ker(w'-1)$ has dimension $\dim(V')-2$, and hence, that $(w'-1)(V')$ has dimension $2$. Conjugating back by $x^{-1}$ we get that $(w-1)(V')$ has dimension $2$, which concludes the proof.


