Proposition: Finite-event obstruction

A finite event carrier cannot generate a cone linearly isomorphic to ${\mathcal{L}}_m$, $m\ge 3$, under (eq:generated).

Proof. There are finitely many nonzero comparable differences $G=\{g_1,\dots ,g_s\}$. If they generate a proper cone, pointedness gives $0\notin \mathrm{conv}G$. Otherwise a nontrivial nonnegative sum equal to zero would put the negative of a nonzero generator in the cone.

Choose an auxiliary Euclidean inner product and the closest point $z\ne 0$ of $\mathrm{conv}G$ to zero. Then

In a convergent sequence ${\sum }_j{\lambda }_{n,j}{g}_j$, ${\lambda }_{n,j}\ge 0$, these inequalities bound all coefficients. A coefficient subsequence proves $\mathrm{cone}G$ is closed.

Every extreme ray of $\mathrm{cone}G$ contains a generator, hence there are finitely many extreme rays. The Lorentz cone has an extreme ray ${\mathbb{R}}_{\ge 0}(1,u)$ for every unit $u$; extremality follows from equality in the triangle inequality. There are infinitely many such rays for $m\ge 3$, and linear isomorphisms preserve extreme rays.