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A null normal admits no reflection of the displayed form

Example

Let $B$ be a symmetric bilinear form on a real vector space $V$ and let $a \in V$ with $a \neq 0$ and $B (a, a) = 0$ . Then $a \in \ker B (-, a)$ , and there is no linear map $r : V \to V$ with $r^{2} = {i d}_{V}$ , $r (a) = - a$ and $r (v) = v$ for every $v \in \ker B (-, a)$ : since $a \in \ker B (-, a)$ , such an $r$ would satisfy $r (a) = a$ , forcing $- a = a$ and hence $2 a = 0$ , contrary to $a \neq 0$ in a real vector space (The reals form a totally ordered field). In particular the displayed formula $r_{a} (v) = v - \frac{2 B (v, a)}{B (a, a)} a$ of The real Coxeter form, its radical, reflections, and form-preserving maps cannot be extended to normals with $B (a, a) = 0$ : the hypothesis $B (a, a) \neq 0$ is not merely a convenience of the division. For the two instantiations in $R^{2}$ below, label the coordinates by $1, 2$ : if $x$ is the function on ${0, 1}$ of The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ , write $x_{1} : = x (0)$ , $x_{2} : = x (1)$ , and let $e_{1}, e_{2}$ denote the unit vectors at $0, 1$ , respectively (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ). The instantiations are:

(i) Lorentzian plane $B (x, y) = x_{1} y_{1} - x_{2} y_{2}$ and $a = e_{1} + e_{2}$ : $B (a, a) = 0$ , while $\ker B (-, a) = R (e_{1} + e_{2}) ∋ a$ ; here $a \notin rad (B)$ , since $B (e_{1}, a) = 1 \neq 0$ (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

(ii) Radical plane $B (x, y) = x_{1} y_{1}$ and $a = e_{2}$ : $B (a, a) = 0$ and $\ker B (-, a) = V$ , because $e_{2}$ lies in the radical; in this case the only map fixing $\ker B (-, a) = V$ pointwise is the identity, which does not send $a$ to $- a$ .

Facts & Assumptions

Given: a real vector space $V$ , a symmetric bilinear form $B$ on $V$ , and $a \in V$ with $a \neq 0$ and $B (a, a) = 0$ .

[F1]

A bilinear form on $V$ is a function $V \times V \to R$ linear in each variable separately, and it is symmetric when $B (u, v) = B (v, u)$ for all $u, v \in V$ ; the set $\ker B (-, a) = {v \in V : B (v, a) = 0}$ is the kernel of the linear functional $v \mapsto B (v, a)$ (Bilinear forms, and symmetric, skew-symmetric, and alternating bilinear forms, Kernel and image of a linear map, Linear map between vector spaces over the same field).

[F2]

In any vector space over a field, $λ v = 0_{V}$ forces $λ = 0_{F}$ or $v = 0_{V}$ ; and in the totally ordered field $R$ one has $1 > 0$ , so $2 = 1 + 1 > 0$ and in particular $2 \neq 0$ (In any vector space $0_{F} v = 0_{V}$ , $λ 0_{V} = 0_{V}$ , $(- λ) v = - (λ v)$ , $(- 1_{F}) v = - v$ , and $λ v = 0_{V}$ forces $λ = 0_{F}$ or $v = 0_{V}$ , The reals form a totally ordered field).

[F3]

The displayed reflection formula $r_{a} (v) = v - \frac{2 B (v, a)}{B (a, a)} a$ of the Statement is defined only for $B (a, a) \neq 0$ ; the symbol $r_{a}$ is not defined when $B (a, a) = 0$ (The real Coxeter form, its radical, reflections, and form-preserving maps).

[F4]

The left radical is ${rad}_{L} (B) = {u : B (u, v) = 0 for every v \in V}$ , and when $B$ is symmetric $u$ lies in it exactly when the functional $B (-, u)$ is the zero functional (The matrix, left and right radicals, rank, and nondegeneracy of a bilinear form on a finite-dimensional space).

[F5]

$R^{2}$ is the function space on ${0, 1}$ ; relabel its coordinates by $x_{1} = x (0)$ , $x_{2} = x (1)$ and its standard unit vectors by $e_{1}, e_{2}$ , respectively. Thus $x = x_{1} e_{1} + x_{2} e_{2}$ (The vector space $F^{X}$ of all functions $X \to F$ with pointwise operations, and $F^{n}$ as the case $X = n = {0, 1, \dots, n - 1}$ , The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

Verification

1.1givenF1

The hypothesis $B (a, a) = 0$ says exactly that the value of the functional $v \mapsto B (v, a)$ at $v = a$ is zero, so $a \in \ker B (-, a)$ .

1.2givenF1F2algebra

Suppose $r : V \to V$ were linear with $r^{2} = {i d}_{V}$ , $r (a) = - a$ and $r (v) = v$ for every $v \in \ker B (-, a)$ . Since $B (a, a) = 0$ gives $a \in \ker B (-, a)$ , the fixed-kernel clause would give $r (a) = a$ , while the normal clause gives $r (a) = - a$ ; hence $- a = a$ , that is $2 a = 0$ . Since $a \neq 0$ , [F2] forces $2 = 0$ in $R$ , contradicting $2 \neq 0$ ; therefore no such $r$ exists, and with a null normal the three displayed requirements are already inconsistent before any question of a formula arises.

1.3givenF1F4F5algebra

Lorentzian instantiation. Take $V = R^{2}$ with basis $e_{1}, e_{2}$ and $B (x, y) = x_{1} y_{1} - x_{2} y_{2}$ , so that $B (e_{1}, e_{1}) = 1$ , $B (e_{2}, e_{2}) = - 1$ and $B (e_{1}, e_{2}) = 0$ ; let $a = e_{1} + e_{2}$ . Bilinearity gives $B (a, a) = B (e_{1}, e_{1}) + 2 B (e_{1}, e_{2}) + B (e_{2}, e_{2}) = 1 - 1 = 0$ , and $B (x, a) = x_{1} - x_{2}$ , so $\ker B (-, a) = {x : x_{1} = x_{2}} = R (e_{1} + e_{2}) ∋ a$ . Here $a$ is not in the radical: $B (e_{1}, a) = 1 \neq 0$ , so $B (-, a)$ is not the zero functional. Thus this $a$ satisfies the general hypotheses with a nonzero functional $B (-, a)$ .

1.4givenF1F2F4F5algebra

Radical-plane instantiation. Take $V = R^{2}$ with basis $e_{1}, e_{2}$ and $B (x, y) = x_{1} y_{1}$ , and let $a = e_{2}$ . Then $B (a, a) = 0$ , and $B (x, e_{2}) = x_{1} \cdot 0 = 0$ for every $x$ , so $B (-, e_{2})$ is the zero functional and $\ker B (-, e_{2}) = V$ ; in particular $e_{2}$ lies in the radical. A map $r$ fixing $\ker B (-, e_{2}) = V$ pointwise is the identity, and the identity does not send $a$ to $- a$ , since $a \neq 0$ forces $2 a \neq 0$ by [F2], that is $a \neq - a$ .

2.1givenF3step 1.2step 1.3step 1.4∎

Conclusion. Step 1.2 proves the general negative statement: for every $a \neq 0$ with $B (a, a) = 0$ there is no linear $r$ with $r^{2} = {i d}_{V}$ , $r (a) = - a$ and $r$ the identity on $\ker B (-, a)$ . Steps 1.3 and 1.4 realize the hypothesis in the two displayed planes, one with $a \notin {rad}_{L} (B)$ and $a \in \ker B (-, a)$ of dimension $1$ , the other with $\ker B (-, a) = V$ . Since the formula of [F3] is defined only for $B (a, a) \neq 0$ , the condition $B (a, a) \neq 0$ is not a removable convenience of the division: the properties required of a reflection with normal $a$ are unsatisfiable when $B (a, a) = 0$ .

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Sources

Michael W. Davis, The Geometry and Topology of Coxeter Groups (Princeton University Press; author's full institutional PDF) (standard reference, not scraped)
Anders Björner and Francesco Brenti, Combinatorics of Coxeter Groups (Graduate Texts in Mathematics 231, Springer 2005) (standard reference, not scraped)