# Product of two reflections lying in a parabolic subgroup of a Coxeter group

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!

• @LSpice: that's not quite right. That's a criterion for being conjugate to an element that lies in a proper parabolic subgroup. Jan 12, 2021 at 18:59
• @SamHopkins, I am using the term "parabolic subgroup" to mean "conjugate of a standard parabolic subgroup", with hopefully obvious meaning. If $t_1 t_2$ lies in a proper parabolic subgroup in this broader sense, then some conjugate of it lies in a standard proper parabolic subgroup, and then we could just conjugate $t_1$ and $t_2$ by the same element so that their product lies in that standard parabolic. Jan 12, 2021 at 19:37
• @LSpice: Fair enough! Jan 12, 2021 at 19:40
• More generally, reduced reflection factorizations of an element lying in a (not necessarily standard) parabolic subgroup of a Coxeter group always have all their factors in the parabolic subgroup : this is Theorem 1.4 of arxiv.org/pdf/1402.2500.pdf Jan 13, 2021 at 10:08
• I think there is a proof based on the theorem in Section 5.13 of Humphreys' book "Reflection groups and Coxeter groups" and indeed I posted an answer to that effect here. I think such a proof can probably be made to work, but in trying to address some objections in the comments, I see that I am not able to easily prove something that I had thought was obvious, and I've given up for now and deleted the answer so as not to leave an incorrect argument up. Thanks to the commenters who made me aware of my errors and saved me from permanently publishing something wrong. Jan 14, 2021 at 3:14

$$\DeclareMathOperator\Im{Im}\DeclareMathOperator\Fix{Fix}$$As suggested 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, 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, 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.

• Where does the Baumeister–Dyer–Stump–Wegener result appear? EDIT: Aha, I found the remark in the comments. I will edit it in. Jan 14, 2021 at 23:33
• No, $s$ was already defined above. I edited. Jan 14, 2021 at 23:44
• Oops, sorry! I will delete my wrong comment. Jan 14, 2021 at 23:50
• The cited result says for reduced reflection factorizations $w = t_1 \dots t_k \in W$ that $w \in W_I \Rightarrow t_1,\dots,t_k \in W_I$ for any parabolic $W_I$. Applying this to $W_I = W^{Fix(w)} = \{ \tau \in W \mid Fix(\tau) \supseteq Fix(w) \}$ yields $Fix(w) = Fix(t_1) \cap \dots \cap Fix(t_k)$ as you intend to prove for $k=2$. So there ought to be a geometric argument proving this in general. Jan 15, 2021 at 10:57
• Thanks for the misprint, I'll edit. About your question : let $s_1 s_2\cdots s_k$ be a reduced expr. of $t_2'$. As $t_2'\in W$ and $W$ is a stand. parabolic of $\widetilde{W}$, then all the $s_i$'s lie in $W$. It follows that $s' s_1 s_2 \cdots s_k$ is reduced. Now if $s'$ and $t_2'$ commute, then $s' t_2'=s' s_1 s_2 \cdots s_k$ has a reduced expression ending by $s'$. It is not possible, as at least one of the $s_i$'s must be in $S\backslash \{ s\}$, hence has $m_{s' s_i}=\infty$ : you cannot move $s'$ to the right of the first occurrence of $s_i$ in $s_1 s_2 \cdots s_k$ by a braid move. Jan 15, 2021 at 13:36

$$\DeclareMathOperator\Fix{Fix}\DeclareMathOperator\Stab{Stab}$$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.

• Interesting: @ThomasGobet's answer enlarges $W$, but yours keeps $W$ the same while enlarging the dual of its reflection representation. Jan 16, 2021 at 0:45