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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

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