Proposition: Finite event differences cannot generate a Lorentz cone

Suppose that $E_D$ is finite and that Definition def:representation holds. Then $K_D$ cannot be linearly isomorphic to ${\mathcal{L}}_m$ for $m\ge 3$.

Proof. Let $G=\{g_1,\dots ,g_s\}$ be the nonzero comparable differences appearing in (eq:generation). Properness excludes the case $G=\varnothing$ in the stated dimension.

Because $G$ lies in a pointed cone, zero is not in $\mathrm{conv}G$. Indeed, a convex combination equal to zero would put the negative of a nonzero generator in the same cone. Choose any auxiliary Euclidean inner product on $V_D$ and let $z\ne 0$ be the closest point of the compact set $\mathrm{conv}G$ to zero. The minimizing property gives

If a sequence ${\sum }_j{\lambda }_{n,j}{g}_j$, ${\lambda }_{n,j}\ge 0$, converges, these inequalities bound all its coefficients. A convergent coefficient subsequence shows that $\mathrm{cone}G$ is closed. Thus $K_D=\mathrm{cone}G$.

Every extreme ray of $\mathrm{cone}G$ contains one of the finitely many generators: decomposing a nonzero vector on an extreme ray into its generating summands forces every nonzero summand onto that ray. Hence $K_D$ has finitely many extreme rays.

The Lorentz cone has an extreme ray ${\mathbb{R}}_{\ge 0}(1,u)$ for every $\Vert u{\Vert }_2=1$. Extremality follows from equality in the triangle inequality in a decomposition of $(1,u)$ into two cone vectors. There are infinitely many such rays when $m\ge 3$. Invertible linear maps preserve extreme rays, giving a contradiction.