Proposition: No finite-event Lorentz generation

For a finite set of experiment identifiers, the resulting quotient carrier is finite. Any faithful generated proper cone representing that finite carrier has finitely many extreme rays and cannot be linearly isomorphic to

Proof. The finite carrier gives finitely many nonzero comparable differences $G=\{g_1,\dots ,g_s\}$. If it generates a proper cone, $G\ne \varnothing$, and pointedness implies $0\notin \mathrm{conv}G$. Otherwise a nontrivial nonnegative combination summing to zero would put the negative of a nonzero generator in the same cone.

Choose an auxiliary Euclidean inner product and let $z\ne 0$ be the closest point of the compact set $\mathrm{conv}G$ to zero. Minimization gives

For any convergent sequence ${\sum }_j{\lambda }_{n,j}{g}_j$, ${\lambda }_{n,j}\ge 0$, these inequalities bound all coefficients. A convergent coefficient subsequence shows that $\mathrm{cone}G$ is closed. Thus the generated closed cone is $\mathrm{cone}G$.

Every extreme ray contains one of the finitely many generators: a generating decomposition of a nonzero vector on an extreme ray forces each nonzero summand onto that ray. There are therefore finitely many extreme rays.

By contrast, every $(1,u)$, $\Vert u{\Vert }_2=1$, generates an extreme ray of ${\mathcal{L}}_m$, as follows from equality in the triangle inequality in any cone decomposition of that vector. There are infinitely many such rays for $m\ge 3$. Invertible linear maps preserve extreme rays.