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The Wall form of an orthogonal operator, subspace restriction, and the interval structure of the orthogonal reflection-length order

Statement

Let $(V, B)$ be a finite-dimensional real inner product space (Real and complex inner-product spaces and their induced length); on this page $V = R^{S}$ with the positive definite Coxeter form $B$ of a Coxeter system of finite type (The real Coxeter form, its radical, reflections, and form-preserving maps, Finiteness criterion: W is finite exactly when the Coxeter form is positive definite), and $O (V)$ is the group of $B$ -preserving invertible linear maps (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces). Let $M$ , $F$ and $\leq_{O}$ be as in Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator. Then:

(1) Basic identities. For every $A \in O (V)$ one has $F (A) = F (A^{- 1})$ , $M (A) = M (A^{- 1})$ , $M (A) = F (A)^{⊥}$ , $V = M (A) \oplus F (A)$ , $A (M (A)) = M (A)$ , $A (F (A)) = F (A)$ , and $(A - i d) ∣_{M (A)} : M (A) \to M (A)$ is a bijection (In finite dimension, $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$ , Rank-nullity: $\dim_{F} V = nullity T + rank T$ ). For all $X, Y \in O (V)$ , $M (X Y) \subseteq M (X) + X M (Y)$ and hence $\dim M (X Y) \leq \dim M (X) + \dim M (Y)$ .

(2) The Wall form. For $A \in O (V)$ put

$χ_{A} (u, v) : = B ((A - i d) ∣_{M (A)}^{- 1} u, v) (u, v \in M (A)) .$

Then $χ_{A}$ is a bilinear form on $M (A)$ satisfying

$χ_{A} (u, v) + χ_{A} (v, u) = - B (u, v) (u, v \in M (A));$

in particular $χ_{A}$ is nondegenerate and its symmetric part is $- \frac{1}{2} B ∣_{M (A)}$ (The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ , Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ ).

(3) Subspace restriction. Let $A \in O (V)$ , let $U \subseteq M (A)$ be a subspace and let $Π_{U} : V \to U$ be the orthogonal projection (The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$ , For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ ). There is a unique $H_{U} \in End (U)$ with $B (H_{U} u, v) = χ_{A} (u, v)$ for all $u, v \in U$ , and it satisfies $H_{U} + H_{U}^{*} = - {i d}_{U}$ , so that $H_{U}$ is invertible. Define $A_{U} \in O (V)$ by $A_{U} u = u + H_{U}^{- 1} u$ for $u \in U$ and $A_{U} = i d$ on $U^{⊥}$ . Then $M (A_{U}) = U$ and

$(A_{U} - i d) ∣_{U}^{- 1} = H_{U}, H_{U} u = Π_{U} ((A - i d) ∣_{M (A)})^{- 1} u (u \in U) .$

An element $A \in O (V)$ is a reflection (that is, $\dim M (A) = 1$ ) if and only if $A = i d - 2 Π_{L}$ for the line $L = M (A)$ ; in particular every line $L \subseteq V$ is the moved space of exactly one reflection of $O (V)$ (For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and $T^{*} T = I$ are equivalent).

(4) The restriction theorem. For every subspace $U \subseteq M (A)$ one has $A_{U} \leq_{O} A$ , that is $\dim M (A) = \dim U + \dim M (A_{U}^{- 1} A)$ ; and for $U \subseteq U^{'} \subseteq M (A)$ one has $(A_{U^{'}})_{U} = A_{U}$ , hence $A_{U} \leq_{O} A_{U^{'}}$ . Conversely every $B \in O (V)$ with $B \leq_{O} A$ satisfies $M (B) \subseteq M (A)$ , $χ_{A} ∣_{M (B)} = χ_{B}$ and $B = A_{M (B)}$ . Consequently the assignment $U \mapsto A_{U}$ is a bijection from the set of subspaces of $M (A)$ onto ${B \in O (V) : B \leq_{O} A}$ , with inverse $B \mapsto M (B)$ , and it is an order isomorphism for inclusion of subspaces and $\leq_{O}$ .

(5) Rank-length equality and prefixes. Let $A \in O (V)$ and let $r_{1}, \dots, r_{k} \in O (V)$ be reflections with $A = r_{1} r_{2} \dots r_{k}$ . Then $k \geq \dim M (A)$ , and $A$ is a product of exactly $\dim M (A)$ reflections. Moreover $B \leq_{O} A$ if and only if there are a shortest factorization $A = r_{1} \dots r_{m}$ (so $m = \dim M (A)$ ) and an index $k$ with $B = r_{1} \dots r_{k}$ .

Facts & Assumptions

Given: A finite-dimensional real inner product space $(V, B)$ with the positive definite Coxeter form of a finite-type Coxeter system, an element $A \in O (V)$ , and the moved space $M (A) = im (A - i d)$ , the fixed space $F (A) = \ker (A - i d)$ and the relation $\leq_{O}$ of Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator.

[F1]

For a linear map $T : V \to W$ with $V$ finite-dimensional, $\dim V = \dim \ker T + \dim im T$ ; in particular an injective endomorphism of a finite-dimensional vector space is bijective. Rank-nullity: $\dim_{F} V = nullity T + rank T$

[F2]

For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ , $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$ . For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ In finite dimension, $W^{⊥ ⊥} = W$ and $\dim W + \dim W^{⊥} = \dim V$

[F3]

The orthogonal projection $P_{W} : V \to W$ sends $v = w + z$ with $w \in W$ , $z \in W^{⊥}$ to $P_{W} v = w$ ; equivalently $P_{W} v$ is the unique vector of $W$ with $v - P_{W} v \in W^{⊥}$ , and $P_{W} w = w$ for every $w \in W$ . The projection is linear: combining the unique decompositions of $x$ and $y$ gives $P_{W} (a x + b y) = a P_{W} x + b P_{W} y$ . The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$

[F4]

$A \in O (V)$ means that $A$ is invertible and $B (A u, A v) = B (u, v)$ for all $u, v \in V$ ; equivalently $B (A u, v) = B (u, A^{- 1} v)$ for all $u, v \in V$ . Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and $T^{*} T = I$ are equivalent

[F5]

$B$ is symmetric and bilinear, and positive definite: $B (u, u) \geq 0$ with $B (u, u) = 0$ only for $u = 0$ ; consequently $B$ is nondegenerate, so $B (x, y) = 0$ for all $y \in V$ implies $x = 0$ , and the same holds for the restriction of $B$ to any subspace. Real and complex inner-product spaces and their induced length Finiteness criterion: W is finite exactly when the Coxeter form is positive definite

[F6]

$M (A) = im (A - i d)$ , $F (A) = \ker (A - i d)$ and $B \leq_{O} A$ holds if and only if $\dim M (A) = \dim M (B) + \dim M (B^{- 1} A)$ ; no order property of $\leq_{O}$ is asserted by the definition. Reflection length, the absolute order on a finite Coxeter group, and the moved and fixed spaces of an orthogonal operator

Proof

technique · direct

1.1F2F4F5algebra

For $y \in V$ one has $B ((A - i d) x, y) = B (x, (A^{- 1} - i d) y)$ for all $x \in V$ by [F4], so $y \in M (A)^{⊥}$ holds exactly when $(A^{- 1} - i d) y = 0$ and hence exactly when $A^{- 1} y = y$ , that is, when $y \in F (A^{- 1})$ ; thus $M (A)^{⊥} = F (A^{- 1})$ . Since $A^{- 1} y = y$ holds exactly when $A y = y$ , one has $F (A^{- 1}) = F (A)$ , and $A^{- 1} - i d = - A^{- 1} (A - i d)$ shows $M (A^{- 1}) = M (A)$ . Consequently $M (A)^{⊥} = F (A)$ , and [F2] gives $M (A) = M (A)^{⊥ ⊥} = F (A)^{⊥}$ , $V = M (A) \oplus F (A)$ and $\dim M (A) + \dim F (A) = \dim V$ . Moreover $A (A - i d) = (A - i d) A$ , so $A (M (A)) = M (A)$ ; and $A (F (A)) \subseteq F (A)$ with equality because $A$ is injective.

1.2F1F4algebra

For $X, Y \in O (V)$ the identity $X Y - i d = (X - i d) + X (Y - i d)$ gives $M (X Y) \subseteq M (X) + X M (Y)$ ; the invertible $X$ restricts to an injective map $M (Y) \to X M (Y)$ , whose image therefore has dimension $\dim M (Y)$ by [F1]; consequently $\dim M (X Y) \leq \dim M (X) + \dim M (Y)$ .

2.1step 1.1F1

By step 1.1 the direct sum $V = M (A) \oplus F (A)$ gives $F (A) \cap M (A) = {0}$ , so the restriction $(A - i d) ∣_{M (A)} : M (A) \to M (A)$ , whose image lies in the $A$ -stable space $M (A)$ , has trivial kernel and is therefore injective; by [F1] it is bijective, and $S_{A} : = ((A - i d) ∣_{M (A)})^{- 1} \in G L (M (A))$ is defined.

2.2step 1.1F3F4F5algebra

Let $A \in O (V)$ with $\dim M (A) = 1$ , put $L : = M (A)$ and let $Π_{L}$ be the orthogonal projection onto $L$ [F3]. By step 1.1 the line $L$ and the space $F (A) = L^{⊥}$ are $A$ -stable, so $A$ fixes $L^{⊥}$ pointwise; for $0 \neq u \in L$ one has $A u = c u$ with $c \in R$ , and since $A$ preserves $B$ by [F4] and $B (u, u) \neq 0$ by [F5], $c^{2} B (u, u) = B (A u, A u) = B (u, u)$ forces $c^{2} = 1$ , while $c = 1$ would put $u \in L \cap F (A) = {0}$ ; hence $c = - 1$ and $A = i d - 2 Π_{L}$ . Conversely if $L$ is a line then $A^{'} : = i d - 2 Π_{L}$ acts as $- 1$ on $L$ and as the identity on $L^{⊥}$ , so it preserves $B$ and is invertible, and $M (A^{'}) = L$ has dimension one; if $A^{'} \in O (V)$ is a reflection with $M (A^{'}) = L$ , the first part applied to $A^{'}$ gives $A^{'} = i d - 2 Π_{L}$ . Hence every line $L \subseteq V$ is the moved space of exactly one reflection of $O (V)$ , namely $i d - 2 Π_{L}$ .

3.1step 1.1step 2.1F4F5algebra

Applying step 2.1 to the orthogonal element $A^{- 1}$ , whose moved space is $M (A)$ by step 1.1, gives the bijection $T_{A} : = ((A^{- 1} - i d) ∣_{M (A)})^{- 1} : M (A) \to M (A)$ . For $u, v \in M (A)$ put $w : = S_{A} u$ and $z : = T_{A} v$ , so that $(A - i d) w = u$ and $(A^{- 1} - i d) z = v$ ; then $B (S_{A} u, v) = B (w, (A^{- 1} - i d) z) = B (w, A^{- 1} z) - B (w, z) = B (A w, z) - B (w, z) = B ((A - i d) w, z) = B (u, T_{A} v)$ , using [F4] and symmetry of $B$ . Moreover $A^{- 1} - i d = - A^{- 1} (A - i d)$ on the $A$ -stable space $M (A)$ , so the inverse there is $T_{A} = - S_{A} A$ and $S_{A} + T_{A} = S_{A} (i d - A) = - ((A - i d) ∣_{M (A)})^{- 1} (A - i d) ∣_{M (A)} = - {i d}_{M (A)}$ .

4.1step 1.1step 3.1F4F5algebra

The Wall form $χ_{A} (u, v) = B (S_{A} u, v)$ is bilinear on $M (A)$ , and for $u, v \in M (A)$ the transpose identity of step 3.1 gives $B (S_{A} v, u) = B (v, T_{A} u) = B (T_{A} u, v)$ , so $χ_{A} (u, v) + χ_{A} (v, u) = B (S_{A} u, v) + B (T_{A} u, v) = B ((S_{A} + T_{A}) u, v) = - B (u, v)$ . If $χ_{A} (u, v) = 0$ for all $v \in M (A)$ , then $B (S_{A} u, \cdot)$ vanishes on $M (A)$ and, since $S_{A} u \in M (A) = F (A)^{⊥}$ , also on $F (A)$ , hence on $V$ ; by [F5] $S_{A} u = 0$ , and $S_{A}$ is injective, so $u = 0$ : the form $χ_{A}$ is nondegenerate, and its symmetric part is $\frac{1}{2} (χ_{A} (u, v) + χ_{A} (v, u)) = - \frac{1}{2} B (u, v)$ on $M (A)$ .

5.1step 3.1step 4.1F1F3F5algebra

Let $U \subseteq M (A)$ and define $H_{U} : = Π_{U} S_{A} ∣_{U} : U \to U$ . For $u, v \in U$ , [F3] gives $B (H_{U} u, v) = B (S_{A} u, v) = χ_{A} (u, v)$ ; if another operator has these pairings, its difference from $H_{U}$ pairs to zero with every $v \in U$ , so it equals $H_{U}$ by [F5]. Define $H_{U}^{t} : = Π_{U} T_{A} ∣_{U}$ , with $T_{A}$ from step 3.1. That step and [F3] give $B (H_{U}^{t} u, v) = B (T_{A} u, v) = B (u, S_{A} v) = B (u, H_{U} v)$ , so $H_{U}^{t}$ is the adjoint $H_{U}^{*}$ of The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ . Since $S_{A} + T_{A} = - {i d}_{M (A)}$ , compression to $U$ gives $H_{U} + H_{U}^{t} = - {i d}_{U}$ . If $H_{U} u = 0$ , then $0 = B (H_{U} u, u) = χ_{A} (u, u) = - \frac{1}{2} B (u, u)$ by step 4.1, hence $u = 0$ ; thus $H_{U}$ is injective and invertible by [F1], including when $U = 0$ .

6.1step 5.1F1F2F5algebra

Define $A_{U} \in End (V)$ by $A_{U} u = u + H_{U}^{- 1} u$ for $u \in U$ and $A_{U} = i d$ on $U^{⊥}$ ; this is well defined and linear because $V = U \oplus U^{⊥}$ [F2]. Then $A_{U} - i d = H_{U}^{- 1} Π_{U}$ has image $U$ , so $M (A_{U}) = U$ and $(A_{U} - i d) ∣_{U}^{- 1} = H_{U}$ . Put $T : = H_{U}^{- 1} \in G L (U)$ and $T^{t} : = (H_{U}^{t})^{- 1}$ ; inverting the transpose relation of step 5.1 shows that $T^{t}$ is the transpose of $T$ , that is, $B (T x, y) = B (x, T^{t} y)$ for all $x, y \in U$ . Multiplying $H_{U} + H_{U}^{t} = - {i d}_{U}$ on the left by $T$ and on the right by $T^{t}$ , and also on the left by $T^{t}$ and on the right by $T$ , gives $T + T^{t} = - T T^{t} = - T^{t} T$ , hence $(i d + T) (i d + T^{t}) = {i d}_{U} = (i d + T^{t}) (i d + T)$ . Therefore $B (A_{U} u, A_{U} v) = B ((i d + T^{t}) (i d + T) u, v) = B (u, v)$ for all $u, v \in U$ , the transpose of $i d + T$ being $i d + T^{t}$ ; on $U^{⊥}$ the operator $A_{U}$ is the identity, and $U ⊥ U^{⊥}$ , so $A_{U}$ preserves $B$ on $V$ . If $A_{U} x = 0$ , then $B (x, y) = B (A_{U} x, A_{U} y) = 0$ for every $y \in V$ , so $x = 0$ by [F5] and $A_{U}$ is injective, hence invertible by [F1]: thus $A_{U} \in O (V)$ .

7.1step 1.1step 5.1step 6.1F1F6algebra

Fix $U \subseteq M (A)$ and let $A_{U}$ be as in step 6.1. In the direct sum $V = M (A) \oplus F (A)$ of step 1.1 write $x = m + f$ ; then $(A - A_{U}) x = (A - i d) m - H_{U}^{- 1} Π_{U} m$ , so $x \in \ker (A - A_{U})$ exactly when $(A - i d) m = H_{U}^{- 1} Π_{U} m$ . With $u : = Π_{U} m \in U$ this equation reads $m = S_{A} H_{U}^{- 1} u$ , and it is consistent because $Π_{U} S_{A} Π_{U} = H_{U}$ is step 5.1: the solutions are exactly the $x = S_{A} H_{U}^{- 1} u + f$ with $u \in U$ and $f \in F (A)$ arbitrary. Hence $\ker (A - A_{U}) = S_{A} H_{U}^{- 1} (U) \oplus F (A)$ has dimension $\dim U + \dim F (A)$ , so $rank (A - A_{U}) = \dim M (A) - \dim U$ . Since $A_{U}^{- 1} A - i d = A_{U}^{- 1} (A - A_{U})$ , the space $M (A_{U}^{- 1} A)$ is the image of $A - A_{U}$ under $A_{U}^{- 1}$ and has the same dimension $rank (A - A_{U})$ ; therefore $\dim M (A) = \dim U + \dim M (A_{U}^{- 1} A)$ , that is, $A_{U} \leq_{O} A$ by [F6].

7.2step 1.1step 5.1step 6.1F6algebra

Let $B \leq_{O} A$ and put $C : = B^{- 1} A$ , so that $\dim M (A) = \dim M (B) + \dim M (C)$ by [F6]. Since $A - i d = (B - i d) C + (C - i d)$ , one has $M (A) \subseteq M (B) + M (C)$ , and comparing dimensions gives $M (A) = M (B) \oplus M (C)$ ; in particular $M (B) \subseteq M (A)$ . Fix $w \in U : = M (B)$ and write $m : = S_{A} w = m_{B} + m_{C}$ with $m_{B} \in M (B)$ and $m_{C} \in M (C)$ ; then $w = (A - i d) m = (B - i d) (C m) + (C - i d) m$ with $(B - i d) (C m) \in M (B)$ and $(C - i d) m \in M (C)$ , so comparing the two direct summands gives $C m = m$ and $(B - i d) m = w$ . Hence $m - S_{B} w \in F (B) = M (B)^{⊥} = U^{⊥}$ , where the first equality uses step 1.1 applied to $B$ ; therefore for all $u, v \in U$ one has $χ_{A} (u, v) = B (S_{A} u, v) = B (S_{B} u, v) = χ_{B} (u, v)$ , because $S_{A} u - S_{B} u \in U^{⊥}$ . Thus the operator defined by $χ_{A}$ on $U$ in step 5.1 is $H_{U} = S_{B}$ , and the construction of step 6.1 for $A$ and $U$ gives $A_{M (B)} = i d + S_{B}^{- 1}$ on $U$ and $i d$ on $U^{⊥}$ , while $B = i d + S_{B}^{- 1}$ on $U$ and $B = i d$ on $U^{⊥} = F (B)$ by step 1.1, so $B = A_{M (B)}$ .

8.1step 5.1step 6.1step 7.1algebra

Let $U \subseteq U^{'} \subseteq M (A)$ and apply step 6.1 to $A_{U^{'}}$ , whose moved space is $M (A_{U^{'}}) = U^{'}$ . Since $(A_{U^{'}} - i d) ∣_{U^{'}}^{- 1} = H_{U^{'}}$ , the Wall form of $A_{U^{'}}$ on $U^{'}$ is $χ_{A_{U^{'}}} (x, y) = B (H_{U^{'}} x, y)$ ; for $x, y \in U$ this equals $B (Π_{U^{'}} S_{A} Π_{U^{'}} x, y) = B (S_{A} x, y) = χ_{A} (x, y)$ by step 5.1, because $Π_{U^{'}} x = x$ and $y \in U \subseteq U^{'}$ . Hence the operator attached to $A_{U^{'}}$ and $U$ by step 5.1 is again $H_{U}$ , and the construction of step 6.1 gives $(A_{U^{'}})_{U} = A_{U}$ ; applying step 7.1 with $A_{U^{'}}$ in place of $A$ and the subspace $U \subseteq M (A_{U^{'}}) = U^{'}$ then gives $A_{U} \leq_{O} A_{U^{'}}$ .

8.2step 1.2step 2.2step 7.1

Let $r_{1}, \dots, r_{k} \in O (V)$ be reflections and $A = r_{1} r_{2} \dots r_{k}$ ; since $\dim M (r_{i}) = 1$ by step 2.2, the subadditivity of step 1.2 gives $\dim M (A) \leq k$ . Conversely every $A \in O (V)$ is a product of exactly $\dim M (A)$ reflections: if $M (A) = 0$ then $A = i d$ is the empty product, and otherwise one picks a line $L \subseteq M (A)$ and applies step 7.1 to $U : = L$ , obtaining $\dim M (A_{L}^{- 1} A) = \dim M (A) - 1$ , so by induction on $\dim M (A)$ the element $A_{L}^{- 1} A$ is a product of $\dim M (A) - 1$ reflections and $A = A_{L} \cdot (A_{L}^{- 1} A)$ is a product of $\dim M (A)$ of them.

9.1step 6.1step 7.2step 8.1

The assignment $U \mapsto A_{U}$ from subspaces of $M (A)$ to ${X \in O (V) : X \leq_{O} A}$ is injective, because $A_{U} = A_{U^{'}}$ forces $U = M (A_{U}) = M (A_{U^{'}}) = U^{'}$ by step 6.1, and surjective by step 7.2; with inverse $X \mapsto M (X)$ it is a bijection. It is an order isomorphism: if $U \subseteq U^{'}$ then $A_{U} \leq_{O} A_{U^{'}}$ by step 8.1, and conversely $A_{U} \leq_{O} A_{U^{'}}$ gives $U = M (A_{U}) \subseteq M (A_{U^{'}}) = U^{'}$ by the inclusion clause of step 7.2.

9.2step 1.2step 2.2step 8.2F6algebra

Suppose $A = r_{1} \dots r_{m}$ with $m = \dim M (A)$ and $B = r_{1} \dots r_{k}$ ; the $r_{i}$ are involutions by step 2.2, so $B^{- 1} A = r_{k + 1} \dots r_{m}$ , whence $\dim M (B) \leq k$ and $\dim M (B^{- 1} A) \leq m - k$ by step 8.2, while $\dim M (A) \leq \dim M (B) + \dim M (B^{- 1} A)$ by step 1.2; both inequalities are therefore equalities and $B \leq_{O} A$ by [F6].

10.1step 8.2F6∎

Conversely, if $B \leq_{O} A$ , write $B = r_{1} \dots r_{k}$ with $k = \dim M (B)$ and $B^{- 1} A = s_{1} \dots s_{l}$ with $l = \dim M (B^{- 1} A)$ , both by the factorization clause of step 8.2; then $A = r_{1} \dots r_{k} s_{1} \dots s_{l}$ is a product of $k + l = \dim M (A)$ reflections by [F6], that is, a shortest factorization of $A$ by step 8.2, of which $B$ is the prefix of length $k$ .

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Sources

T. Brady and C. Watt, Lattices in finite real reflection groups (arXiv:math/0501502) (standard reference, not scraped)
R. W. Carter, Conjugacy classes in the Weyl group, Compositio Mathematica 25 (1972) 1-59 (Numdam full text) (standard reference, not scraped)