Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2008 / 2

The functional $\ell (u)=⟨u,v⟩$ is linear and nonzero because $\ell (v)=⟨v,v⟩>0$. It therefore has rank one, and rank-nullity theorem givesNo Gram–Schmidt argument is needed. Every vector decomposes uniquely as $u=w+av$, where $w\perp v$ and $a=⟨u,v⟩/⟨v,v⟩$. Substitution gives $S_v(w+av)=w-av$. Thus the eigenspace for $-1$ is $\mathbb{R}v$, and that for $+1$ is $v^{\perp }$. The latter is an actual eigenvalue when $n\ge 2$; when $n=1$ its eigenspace is zero and only $-1$ occurs. In the plane the map is reflection in the line perpendicular to $v$.

For equal-length crystallographic root configurations in a plane, taking $u=v$ in the closure condition gives $S_v(v)=-v$, so every vector occurs with its negative. For fixed $v\in \Phi$, a vector $u\in \Phi$ has $⟨u,v⟩\in \{-1,-1/2,0,1/2,1\}$, because both vectors are unit vectors. Thus its angle relative to $v$ belongs to the finite list $0,\pm \pi /3,\pm \pi /2,\pm 2\pi /3,\pi$.

If all vectors are collinear, the set is exactly $\{v,-v\}$, giving two opposite points of the unit circle.

If an orthogonal vector $w$ occurs, closure supplies $\pm v,\pm w$. A unit vector making an angle of $\pi /3$ or $2\pi /3$ with $v$ would have inner product $\pm \sqrt{3}/2$ with $w$, contradicting integrality of twice that inner product. Thus these four points are the whole set: the vertices of a square.

In the remaining case a noncollinear pair makes an angle of $\pi /3$ or $2\pi /3$. Changing a sign, take their angles as $0,\pi /3$. Reflecting the second in the line perpendicular to the first supplies the angle $2\pi /3$, and negation supplies the other three hexagon vertices. No perpendicular vector can occur, by the preceding inner-product obstruction. The initial finite list now leaves only these six points. Conversely the opposite pair, square and regular hexagon each satisfy both reflection closure and the integrality condition. Hence