Theorem: Representation from cone order and

Let $K$ be a proper cone in $V$, with $m=\dim V\ge 3$. If $\mathsf{S}$ holds, there is a linear isomorphism $T:V\to \mathbb{R}\times {\mathbb{R}}^{m-1}$ such that

Proof. Homogeneity and self-duality make $K$ a symmetric cone. The Koecher–Vinberg correspondence identifies symmetric cones with cones of squares in Euclidean Jordan algebras Koecher,Vinberg,FK. Irreducibility makes the algebra simple. The classification lists the one-dimensional algebra, the Hermitian matrix algebras over $\mathbb{R},\mathbb{C},\mathbb{H}$, spin factors, and the exceptional algebra of $3\times 3$ Hermitian octonionic matrices FK (Chapter V); see also JNW.

We now apply the strict-convexity hypothesis. In each matrix algebra of size $n\ge 3$, let $a=\mathrm{diag}(1,0,\dots ,0)$ and $b=\mathrm{diag}(0,1,0,\dots ,0)$. They are nonproportional positive elements, but $a+b$ has a zero eigenvalue and is on the boundary of the positive cone. The same diagonal elements in the exceptional rank-three algebra have a boundary sum, by the Jordan spectral criterion for positivity FK (Chapter III). These cases violate (eq:strict).

The one-dimensional case is excluded by $m\ge 3$. The remaining cases are spin factors, including the $2\times 2$ Hermitian cases. Explicitly,

identifies their positive cones with $t\ge \sqrt{s^2+|z{|}^2}$, treating $z$ in real coordinates.

For completeness, the spin factor on $\mathbb{R}\oplus {\mathbb{R}}^{m-1}$ has product

A square is

which belongs to $L_m$. Conversely, if $a\ge \Vert b{\Vert }_2$, put

For $b\ne 0$, the element

squares to $(a,b)$; the case $b=0$ is immediate. Thus the cone of squares is exactly $L_m$.