Real Forms and Reflection Geometry — Examples

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 $a\in \ker B(-,a)$.

Suppose $r:V\to V$ were linear with $r^2={\mathrm{i}\mathrm{d}}_V$, $r(a)=-a$ 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 $2a=0$. Since $a\mathrm{\ne }0$, [F2] forces $2=0$ in $\mathbb{R}$, contradicting $2\mathrm{\ne }0$; therefore no such $r$ exists, and with a null normal the three displayed requirements are already inconsistent before any question of a formula arises.

Lorentzian instantiation. Take $V={\mathbb{R}}^2$ 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 $B(a,a)=B(e_1,e_1)+2B(e_1,e_2)+B(e_2,e_2)=1-1=0$, and $B(x,a)=x_1-x_2$, so $\ker B(-,a)=\{x:x_1=x_2\}=\mathbb{R}(e_1+e_2)\ni a$. Here $a$ is not in the radical: $B(e_1,a)=1\mathrm{\ne }0$, so $B(-,a)$ is not the zero functional. Thus this $a$ satisfies the general hypotheses with a nonzero functional $B(-,a)$.

Radical-plane instantiation. Take $V={\mathbb{R}}^2$ 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 $a\mathrm{\ne }0$ forces $2a\mathrm{\ne }0$ by [F2], that is $a\mathrm{\ne }-a$.

Conclusion. Step 1.2 proves the general negative statement: for every $a\mathrm{\ne }0$ with $B(a,a)=0$ there is no linear $r$ with $r^2={\mathrm{i}\mathrm{d}}_V$, $r(a)=-a$ and $r$ the identity on $\ker B(-,a)$. Steps 1.3 and 1.4 realize the hypothesis in the two displayed planes, one with $a\mathrm{\notin }{\mathrm{rad}}_L(B)$ 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 $B(a,a)\mathrm{\ne }0$, the condition $B(a,a)\mathrm{\ne }0$ is not a removable convenience of the division: the properties required of a reflection with normal $a$ are unsatisfiable when $B(a,a)=0$.

The case $m=3$. Here $c=\cos (\pi \mathrm{/}3)$. The triple-angle identity of [F3] at $x=\pi \mathrm{/}3$ gives $\cos \pi =4c^3-3c$; with $\cos \pi =-1$ this is $4c^3-3c+1=0$, that is $(c+1)(2c-1{)}^2=0$. The factor $c+1$ is nonzero: $\cos x=-1$ forces ${\sin }^{2}x=1-{\cos }^{2}x=0$, hence $\sin x=0$ and $x\in \pi \mathbb{Z}$, whereas $\pi \mathrm{/}3$ is not an integer multiple of $\pi$. Hence $2c-1=0$ and $c=\frac{1}{2}$. Substituting into [F1], $[A]=\left(\begin{array}{cc}0& -1\\ 1& -1\end{array}\right),[A{]}^2=\left(\begin{array}{cc}-1& 1\\ -1& 0\end{array}\right),[A{]}^3=[A{]}^2[A]=I_2,$ and $[A]\mathrm{\ne }{I}_2\mathrm{\ne }[A{]}^2$, so $A$ has order exactly $3=m$. Its trace is $-1$, and $2\cos (2\pi \mathrm{/}3)=2(2c^2-1)=2(\frac{1}{2}-1)=-1$, consistent with the rotation through $2\pi \mathrm{/}3$ of the positive definite plane: $A$ preserves $B{\mathrm{\mid }}_P$ and has determinant $1$.

The case $m=2$. Here $c=\cos (\pi \mathrm{/}2)=0$, so [F1] gives $[A]=\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)=-I_2$, and $A^2={\mathrm{i}\mathrm{d}}_P$ while $A\mathrm{\ne }{\mathrm{i}\mathrm{d}}_P$; thus $A$ has order $2=m$, again a rotation through $2\pi \mathrm{/}2=\pi$ of the positive definite plane.

The case $m=\mathrm{\infty }$. Here $c=1$, so [F1] gives $[A]=\left(\begin{array}{cc}3& -2\\ 2& -1\end{array}\right)=I_2+N$ with $N=\left(\begin{array}{cc}2& -2\\ 2& -2\end{array}\right)\mathrm{\ne }0$, and direct multiplication gives $N^2=0$. The binomial theorem in a ring with $N^2=0$ gives $(I_2+N{)}^k=I_2+kN$ for every $k\ge 0$, and $A^{-1}=I_2-N$ because $(I_2+N)(I_2-N)=I_2-N^2=I_2$, so $(I_2-N{)}^j=I_2+(-j)N$ for $j\ge 0$ and $A^k=I_2+kN\text{for every }k\in \mathbb{Z}\mathrm{.}$ Therefore $A^k(e_s)=e_s+kN{e}_s=e_s+2k(e_s+e_t)$, and since $kN$ has first entry $2k$, which is nonzero for $k\mathrm{\ne }0$, no nonzero power of $A$ is the identity: the product $r_s{r}_t$ has infinite order in $\mathrm{G}\mathrm{L}(V)$.

Conclusion in the abstract group. By [F2], $r_s,r_t$ are invertible involutions. If $m<\mathrm{\infty }$, [F1] gives $(r_s{r}_t{)}^m={\mathrm{i}\mathrm{d}}_V$; the reversed finite relator holds too, since $r_t{r}_s=(r_s{r}_t{)}^{-1}$. If $m=\mathrm{\infty }$, there is no finite pair relator to check. Thus [F5] supplies a homomorphism $\rho :W\to \mathrm{G}\mathrm{L}(V)$ with $\rho (s)=r_s$, $\rho (t)=r_t$, and $\rho (st)=A$. For $m=2,3$, the relator gives $(st{)}^m=1$, and steps 1.1–1.2 show that no smaller positive power can be $1$: it would map to the corresponding nonidentity power of $A$. Hence $st$ has order exactly $m$ in these finite cases. For $m=\mathrm{\infty }$, if $(st{)}^k=1$ for any nonzero integer $k$, applying $\rho$ would give $A^k={\mathrm{i}\mathrm{d}}_V$, contradicting step 1.3. Hence $st$ has infinite order. This conclusion uses the homomorphism and the computed nonidentity powers, rather than inferring element order from the absence of a relator.

General reflection identities. Put $d:=B(a,a)\mathrm{\ne }0$. Bilinearity makes $r_a$ linear, gives $B(r_a(v),a)=-B(v,a)$, and hence $r_a^2(v)=v$. The formula gives $r_a(a)=-a$ and fixes every $v$ with $B(v,a)=0$; conversely $r_a(v)=v$ forces $B(v,a)=0$, since $a\mathrm{\ne }0$ and $2\mathrm{/}d\mathrm{\ne }0$. Finally, symmetry and bilinearity give $B(r_a(u),r_a(v))=B(u,v)-4B(u,a)B(v,a)\mathrm{/}d+4B(u,a)B(v,a)\mathrm{/}d=B(u,v)$. Thus the algebraic identities hold for every symmetric $B$, without requiring it to be a Coxeter form.

Positive plane. Take $B$ the dot product, so $B(e_1,e_1)=B(e_2,e_2)=1$ and $B(e_1,e_2)=0$, and take $a=\frac{3}{5}{e}_1+\frac{4}{5}{e}_2$, for which $B(a,a)=\frac{9}{25}+\frac{16}{25}=1$. Here $B(e_1,a)=\frac{3}{5}$ and $B(e_2,a)=\frac{4}{5}$, so $r_a(e_1)=e_1-\frac{6}{5}a=\frac{7}{25}{e}_1-\frac{24}{25}{e}_2,r_a(e_2)=e_2-\frac{8}{5}a=-\frac{24}{25}{e}_1-\frac{7}{25}{e}_2,$ giving $[r_a]=\frac{1}{25}\left(\begin{array}{cc}7& -24\\ -24& -7\end{array}\right)$. Squaring, $[r_a{]}^2=\frac{1}{625}\left(\begin{array}{cc}625& 0\\ 0& 625\end{array}\right)=I_2$, and $\det [r_a]=\frac{-49-(-24)(-24)}{625}=\frac{-625}{625}=-1$. The dot product has inertia $(2,0,0)$ and $r_a$ fixes $\ker B(-,a)=\{x:\frac{3}{5}{x}_1+\frac{4}{5}{x}_2=0\}=\mathbb{R}(-4e_1+3e_2)$ pointwise, the Euclidean reflection across that line.

Lorentzian plane. Take $B(x,y)=x_1{y}_1-x_2{y}_2$, of inertia $(1,1,0)$, and $a=2e_1+e_2$, so $B(a,a)=4-1=3$, $B(e_1,a)=2$ and $B(e_2,a)=-1$. Then $r_a(e_1)=e_1-\frac{4}{3}a=-\frac{5}{3}{e}_1-\frac{4}{3}{e}_2,r_a(e_2)=e_2+\frac{2}{3}a=\frac{4}{3}{e}_1+\frac{5}{3}{e}_2,$ so $[r_a]=\frac{1}{3}\left(\begin{array}{cc}-5& 4\\ -4& 5\end{array}\right)$, whose square is $\frac{1}{9}\left(\begin{array}{cc}9& 0\\ 0& 9\end{array}\right)=I_2$ and whose determinant is $\frac{-25+16}{9}=-1$. The invariance identities hold on the basis: $B(r_a{e}_1,r_a{e}_1)=\frac{25}{9}-\frac{16}{9}=1$, $B(r_a{e}_2,r_a{e}_2)=\frac{16}{9}-\frac{25}{9}=-1$, and $B(r_a{e}_1,r_a{e}_2)=-\frac{20}{9}+\frac{20}{9}=0$, so $[r_a]$ preserves $B$ by bilinearity.

Plane with radical. Take $B(x,y)=x_1{y}_1$, whose matrix $\mathrm{diag}(1,0)$ has inertia $(1,0,1)$ and whose radical is $\{u:u_1{v}_1=0\text{ for all }v\}=\mathbb{R}{e}_2$, and take $a=e_1$, so $B(a,a)=1$ and $r_a(v)=v-2v_1{e}_1$. Hence $r_a(e_1)=-e_1$ and $r_a(e_2)=e_2$, that is $r_a=\mathrm{diag}(-1,1)$, and its fixed hyperplane is $\ker B(-,e_1)=\{v:v_1=0\}=\mathbb{R}{e}_2=\mathrm{rad}(B)$. The vector $e_2$ is $B$-null, $B(e_2,e_2)=0$, so the displayed formula assigns it no reflection.

Conclusion. In each of the three cases $a$ satisfies $B(a,a)\mathrm{\ne }0$ and the displayed matrix is $[r_a]$ in the ordered basis $(e_1,e_2)$: an involution with determinant $-1$ in the two nondegenerate cases, with the inertia readings $(2,0,0)$ and $(1,1,0)$ of the form and the fixed hyperplane $\ker B(-,a)$ computed above, and in the degenerate case the fixed hyperplane coincides with the radical $\mathbb{R}{e}_2$. These explicit numbers verify the general identities proved in step 1.1 and display why the hypothesis $B(a,a)\mathrm{\ne }0$ of [F1] is exactly what the definition of $r_a$ requires: the null vector $e_2$ of the last case is a normal for which no reflection is defined by the displayed formula.