Proof. Let $x\in W$ such that $x^2=1$ and $P=P(x)$ . In particular, x has finite order. By [Reference Brink and Howlett8, Proposition 1.3], there are $I \subseteq S$ and $w\in W$ such that $W_I$ is finite and $x\in w W_I w^{-1}$ . Since $w W_I w^{-1}$ is a parabolic subgroup containing x, we have $P \subseteq w W_I w^{-1}$ . The latter being finite, we deduce that P is finite.

Proof. Assume that $t\leq _T u$ . Then, $V^u \subseteq H_t$ . Thus, $\mathbb {R} \alpha _t =(H_t)^\perp \subseteq (V^u)^\perp $ . However, since u is involutive, $(V^u)^\perp $ is precisely the $(-1)$ -eigenspace of u. Thus, $u(\alpha _t)=-\alpha _t$ and hence, $t\in N(u)$ . Moreover, since $\mathbb {R} \alpha _t$ is stable by u, we see that $H_t=(\mathbb {R} \alpha _t)^\perp $ is also stable by u. Since $\mathbb {R} \alpha _t$ and $H_t$ are the eigenspaces of t, it follows that $tu=ut$ .

Conversely, assume that $tu=ut$ and that $t\in N(u)$ . Since t and u commute with each other, they are simultaneously diagonalisable. It follows that u preserves $\mathbb {R} \alpha _t$ and $H_t$ , which are the eigenspaces of t, so that $u(\alpha _t)\in \{\pm \alpha _t\}$ . Since $t\in N(u)$ , this forces $u(\alpha _t)=-\alpha _t$ . Writing $H_t= E_1 \oplus E_{-1}$ , where u acts by $\mathrm {Id}_V$ on $E_1$ and by $-1$ on $E_{-1}$ ,

$$ \begin{align*}V= E_1 \oplus E_{-1} \oplus \mathbb{R} \alpha_t,\end{align*} $$

where $u t$ acts by $\mathrm {Id}_V$ on $E_1 \oplus \mathbb {R} \alpha _t$ and by $-\mathrm {Id}_V$ on $E_{-1}$ . Now, $V^{u}=E_1$ while ${V^{u t}=E_1 \oplus \mathbb {R} \alpha _t}$ . Hence, $\ell _T(u)=\ell _T(u t)+1$ , which concludes the proof.

Proof. The corollary follows immediately from the previous proposition together with the fact that $N(w_0)=T$ .

Proof. Assume that u is an involution. Let $t\in P(u)$ . By Proposition 2.3(3), $t\leq _T u$ . By Proposition 3.3, $t\in N(u)$ . It follows that every reflection in $P(u)$ is an inversion of u, which happens if and only if u is the longest element of $P(u)$ . The converse is immediate as longest elements of finite Coxeter groups are always involutive.

Proof. We begin by showing the second statement. Since $v \leq _T u$ , we have $V^u \subseteq V^v$ . Hence, v preserves $V^u$ and, thus, it also preserves $(V^u)^\perp =\mathrm {Mov}(u)$ . Since $V^u$ and $\mathrm {Mov}(u)$ are the eigenspaces of u, it follows that $uv=vu$ .

For the first statement, we begin by observing that

(3.1) $$ \begin{align} V= \mathrm{Mov}(u)\oplus V^u= \mathrm{Mov}(v)\oplus \mathrm{Mov}(v^{-1}u) \oplus V^u,\end{align} $$

where the second equality holds by [Reference Brady and Watt6, Corollary 1]. Since u is involutive, it acts by $-\mathrm {Id}_V$ on $(V^u)^\perp =\mathrm {Mov}(u)=\mathrm {Mov}(v)\oplus \mathrm {Mov}(v^{-1}u)$ . It suffices to show that v acts by $-\mathrm {Id}_V$ on $\mathrm {Mov}(v)$ to conclude that v is involutive. Hence, let $x\in \mathrm {Mov}(v)$ . We have

$$ \begin{align*}v^{-1}(v(x)+x)= v^{-1}(v(x)-u(x))=-(v^{-1}u(x)-x),\end{align*} $$

but $v^{-1}(v(x){\kern-1.2pt}+{\kern-1.2pt}x) {\kern-1.2pt}\in{\kern-1.2pt} \mathrm {Mov}(v)$ since $v(\mathrm {Mov}(v)){\kern-1.2pt}={\kern-1.2pt}\mathrm {Mov}(v)$ , while ${-(v^{-1}u(x){\kern-1.2pt}-{\kern-1.2pt}x){\kern-1.2pt}\in{\kern-1.2pt} \mathrm {Mov}(v^{-1}u)}$ . The decomposition (3.1) forces $v^{-1}(v(x)+x)=0$ and hence, $v(x)=-x$ .

Proof. Let $1 \leq i < j \leq k$ . Since $u = t_i t_j {}^{t_j t_i}( t_1 \cdots t_{i-1}) {}^{t_j}(t_{i+1} \cdots t_{j-1}) (t_{j+1} \cdots t_k)$ and the conjugate of a reflection is again a reflection, we have $t_i t_j \leq _T u$ . We deduce from the previous proposition that $t_i t_j$ is an involution and hence, $t_i t_j = t_j t_i$ .

Proof. Suppose that condition $(1)$ holds. We deduce condition $(2)$ and also the last statement using Propositions 3.6 and 2.3(2).

Assume condition $(2)$ . Since $v\in P(u)=C_W(V^u)$ , we have $V^u \subseteq V^v$ , which gives condition $(3)$ .

Assume condition $(3)$ . We have $V^u \subseteq V^v$ and hence, $V^v = V^u \oplus (V^v \cap \mathrm {Mov}(u))$ yielding

$$ \begin{align*}V= V^v \oplus \mathrm{Mov}(v)= (V^u \oplus (V^v \cap \mathrm{Mov}(u))) \oplus \mathrm{Mov}(v).\end{align*} $$

Since u and v are involutive and $\mathrm {Mov}(v)= (V^v)^\perp \subseteq (V^u)^\perp = \mathrm {Mov}(u)$ , we see that u (respectively v) acts by $1$ on $V^u$ (respectively by $1$ ), by $-1$ on $V^v \cap \mathrm {Mov}(u)$ (respectively by $1$ ) and by $-1$ on $\mathrm {Mov}(v)$ (respectively by $-1$ ). Thus, $v^{-1}u$ is involutive with ${V^{v^{-1}u} = V^u \oplus \mathrm {Mov}(v)}$ and $\mathrm {Mov}(v^{-1}u)= V^v \cap \mathrm {Mov}(u)$ . We deduce that

$$ \begin{align*}\ell_T(v)+ \ell_T(v^{-1} u)&=\dim(\mathrm{Mov}(v))+\dim(V^v \cap \mathrm{Mov}(u))= \dim(V) - \dim(V^u) \\&=\dim(\mathrm{Mov}(u))=\ell_T(u),\end{align*} $$

which shows that $v\leq _T u$ , yielding condition $(1)$ .

Proof. The first statement is an immediate consequence of the equivalence between points $(1)$ and $(2)$ of Proposition 3.9.

We now prove the second statement. Since $P(u)$ is parabolic, the parabolic subgroups of $P(u)$ are exactly the parabolic subgroups of W included in $P(u)$ (this is easily seen, for instance, using the description of parabolic subgroups as centralisers of subspaces in the geometric representation). Let $v_1, v_2\in [1,u]_T$ , which by the already proven statement are involutions. If $\varphi (v_1) = \varphi (v_2)$ , then $v_1 \in P(v_2)$ and $v_2 \in P(v_1)$ . Using the equivalence between Proposition 3.9(1) and (2), we obtain $v_1 \leq _T v_2$ and $v_2 \leq _T v_1$ , and hence, $v_1 = v_2$ . The map is thus injective. It is surjective since, if P is an involutive parabolic subgroup of $P(u)$ , then there is an involution $v\in P(u)$ such that $P=P(v)$ . By Proposition 3.9, again we deduce that $v \leq _T u$ and hence, $v\in [1,v]_T$ . Finally, the map $\varphi $ is an isomorphism of posets since, using again Proposition 3.9, we have

$$ \begin{align*} v \leq_T u \Leftrightarrow v \in P(u) \Leftrightarrow P(v) \subseteq P(u).\\[-34pt] \end{align*} $$

Proof. Assume that the involutive parabolic subgroups of $P(u)$ are stable by intersection. It suffices to show the existence of meets since the poset is finite and has a maximal element $P(u)$ (see [Reference Stanley28, Proposition 3.3.1]). Hence, let $v, w\in [1,u]_T$ . Then, by Proposition 3.9, $v, w\in P(u)$ and $v^2=w^2=1$ . In particular, the parabolic subgroups $P(v)$ and $P(w)$ are involutive, and hence by assumption, the same holds for ${P=P(v)\cap P(w)}$ . Thus, there is $z\in P$ such that $P=P(z)$ and $z^2=1$ . By Proposition 3.9, $z \leq _T v, w$ . Now, let $z'\in W$ be such that $z' \leq _T v$ , $z' \leq _T w$ . In particular, ${z'\in P(w)\cap P(y)= P(z)}$ . Since $z\in [1,u]_T$ , by Proposition 3.6, it is an involution. Using Proposition 3.9, we deduce that $z' \leq _T z$ .

Conversely, assume that the involutive parabolic subgroups of $P(u)$ are not stable by intersection. Then, there are two involutions $v, w\in P(u)$ such that $P:=P(v) \cap P(w)$ is not involutive. Consider the longest element $z_1=w_{0,P}$ of P. Since it is an involution, by Proposition 3.9, we have $z_1 \leq _T v$ and $z_1 \leq _T w$ . Since P is not involutive, we have $P(z_1) \subsetneq P$ . Since both $P(z_1)$ and P are generated by reflections, there is $t\in P\cap T$ such that $t\notin P(z_1)$ . Hence, $t \not \leq _T z_1$ , and by Corollary 3.4 applied to the Coxeter group P, it follows that $t z_1 \neq z_1 t$ . Setting $z_2 := t z_1 t$ , we see that $z_2$ is an involution lying in P and distinct from $z_1$ , with $\ell _T(z_1)=\ell _T(z_2)$ . By Proposition 3.9, $z_2 \leq _T v, w$ . We can now easily deduce that v and w cannot have a meet for $\leq _T$ . Assume for a contradiction that $Z\in [1,u]_T$ is a meet of v and w. Then, Z is an involution, and $Z\leq _T v$ and $Z\leq _T w$ ; hence, $Z \in P(v)\cap P(w)=P$ and $z_1, z_2 \leq _T Z$ . Since $\ell _T(z_1) = \ell _T(z_2)$ and $z_1 \neq z_2$ , we have $z_1 <_T Z$ . Hence, there is $t\in P \cap T$ such that $z_1 \leq _T z_1 t \leq _T Z$ . Since Z is an involution, by Proposition 3.6, $z_1 t$ is also an involution. We thus have $z_1 t = t z_1$ and hence, $t \leq _T z_1$ by Corollary 3.4, which contradicts $z_1 \leq _T z_1 t$ .

The last statement follows immediately from the isomorphism of posets in Proposition 3.10, since the meet in the lattice of involutive parabolic subgroups is clearly given by the intersection.

Proof. $(1) \Leftrightarrow (2)$ . Since T generates W, this equivalence is obtained by using Proposition 2.3(3).

$(2) \Leftrightarrow (3)$ . Assume statement $(2)$ . Then, $t \leq _T u$ for all $t\in T$ . By Proposition 3.3, we deduce that $tu=ut$ for all $t\in T$ , and hence that $u\in Z(W)$ since T generates W. Conversely, assume that $u \in Z(W)$ . Then, $ut$ is an involution for all $t\in T$ . Let ${T= T_1 \cup T_2}$ , where $T_1 =\{t\in T \ \vert \ t \leq _T u\}$ and $T_2 =\{t\in T \ \vert \ t \not \leq _T u\}$ . Since $u\neq 1$ , we have $T_1 \neq \emptyset $ . Assume that $T_2 \neq \emptyset $ and let $t_i \in T_i$ , $i=1, 2$ . Then, $u t_1 \leq _T u \leq _T u t_2$ and, hence, $u t_2$ is an involution such that $t_1 t_2 \leq _T u t_2$ . By Proposition 3.6, $t_1 t_2$ is an involution and hence, $t_1 t_2 = t_2 t_1$ . The reflections in $T_1$ therefore commute with the reflections in $T_2$ . We thus obtain a decomposition as a direct product $W= \langle T_1 \rangle \times \langle T_2 \rangle $ , which is impossible since W is irreducible. Hence, $T_2 = \emptyset $ and $T_1=T$ , yielding statement $(2)$ .

The equivalence between statements $(1)$ and $(4)$ follows from Proposition 2.3(1).

Proof. The fact that $P(u)$ is finite is given by Lemma 3.2. Since $P(u)$ is a parabolic subgroup, it is a Coxeter group and we can decompose it into irreducible components $P_1, P_2, \ldots , P_k$ , which are themselves parabolic, and this decomposition is unique up to the order of the factors. We thus have $P(u) = P_1 \times P_2 \times \cdots P_k$ , and u decomposes uniquely according to this decomposition as $u = u_1 u_2 \cdots u_k$ , where $u_i \in P_i$ and $u_i$ is involutive for all i. It remains to show that $P_i = P(u_i)$ . If not, there is i such that $P(u_i) \subsetneq P_i$ . However, then, u lies in $P_1 \times \cdots \times P_{i-1} \times P(u_i) \times P_{i+1} \times \cdots \times P_k$ , which is strictly included in $P(u)$ and parabolic, which is a contradiction. We thus get the existence of the decomposition, and the uniqueness follows from the uniqueness of the decomposition of $P(u)$ into irreducible components and Corollary 3.5.

The last statement follows from the decomposition $P(u) = P(u_1) \times \cdots \times P(u_k)$ and the fact that $\ell _T(u) = \sum _{i=1}^k \ell _T(u_i)$ ; this last claim holds since by Proposition 2.3(1),

$$ \begin{align*} \ell_T(u) =\mathrm{rank}(P(u)) = \sum_{i=1}^k \mathrm{rank}(P(u_i)) = \sum_{i=1}^k \ell_T(u_i).\\[-42pt] \end{align*} $$

Proof. We go through the list given in (3.2) and check for each family whether involutive parabolic subgroups are stable by intersection or not; applying Theorem 3.11 then completes the proof.

We first note that the cases $B_n \ (n\geq 3)$ and $D_{2k} \ (k\geq 2)$ have already been treated by Kallipoliti in [Reference Kallipoliti21, Theorems 7 and 8], who characterised all the elements w in types $B_n$ and $D_n$ such that $[1,w]_T$ is a lattice. We can also reobtain it using the framework developed in this paper by checking that involutive parabolic subgroups are stable by intersection in these cases: in type $B_n$ , using the list in (3.2), we see that involutive parabolic subgroups are of type $B_k \times (A_1)^\ell $ and the intersection of subgroups of such type is still of this type; in $D_4$ , proper involutive parabolic subgroups are of the form $(A_1)^\ell $ for $\ell \leq 3$ and hence, also stable by intersection. In type $D_{2k}$ with $k \geq 3$ , one can consider a standard parabolic subgroup $P_1$ of type $D_4$ and a simple reflection $s\in D_{2k} \backslash P_1$ commuting with every simple reflection in $P_1$ except one. Then, ${P_2:= s P_1 s}$ is again an involutive parabolic subgroup and $P_1 \cap P_2$ has type $A_3$ ; hence, is not involutive. For instance, in $D_6$ , labelling the Dynkin diagram as in Figure 1, we set $P_1= \langle s_1, s_2, s_3, s_4 \rangle $ , $P_2= s_5 P_1 s_5 = \langle s_1, s_2, s_3, s_5 s_4 s_5 \rangle $ and $P_1 \cap P_2= \langle s_1, s_2, s_3 \rangle $ : indeed $P_1 \cap P_2$ is parabolic, of rank at most three since $P_1$ and $P_2$ have rank four and are distinct, and $\langle s_1, s_2, s_3 \rangle $ is a rank-three parabolic subgroup included in $P_1 \cap P_2$ . Note that the argument also works for types $E_7$ and $E_8$ .

Figure 1

Labelling of the Dynkin diagram of type $D_6$ .

The case $A_1$ is trivial, and in type $I_2(2k)$ with $k\geq 2$ , proper involutive parabolic subgroups are of type $A_1$ ; hence, stability by intersection is clear (but in this case, every element w of the group has reflection length at most $2$ and hence, $[1,w]_T$ is necessarily a lattice).

The case $H_3$ is already covered by our previously obtained result [Reference Gobet19, Theorem 2.2] since $\ell _T(w_0)=3$ in this case. Nevertheless, we can reobtain it easily since proper involutive parabolic subgroups in type $H_3$ are of type $(A_1)^\ell $ for $\ell \leq 2$ ; hence, stability by intersection is clear.

For types $F_4$ and $H_4$ , we can use an argument similar to that used for $D_{2k}$ ( $k\geq 3$ ). Consider a standard parabolic subgroup $P_1$ of type $B_3$ (respectively $H_3$ ) in $F_4$ (respectively $H_4$ ). Let s be the unique simple reflection which remains, that is, not lying in $P_1$ . Then, $P_2:= s P_1 s$ has the same type as $P_1$ and hence, is involutive, but $P_1 \cap P_2$ has type $A_2$ (respectively $I_2(5)$ ); hence, involutive parabolic subgroups are not stable by intersection.