Proposition: The cone determines a quadratic form up to scale

Let $T(K)=L_m$, $m\ge 3$, and set

Then $Q_T$ is nondegenerate of signature $(1,m-1)$. For every other linear isomorphism $T^{\prime }$ with $T^{\prime }(K)=L_m$, there is $c>0$ such that $Q_{T^{\prime }}=c{Q}_T$.

Proof. Only uniqueness needs proof. Work in the $T$-coordinates and write the second quadratic form as

where $C$ is symmetric. It vanishes at every $(1,u)$, $\Vert u{\Vert }_2=1$, because the two maps send the same boundary to the null boundary. Comparing $u$ and $-u$ gives $b=0$. Then $u^{\mathsf{T}}Cu=-a$ on the unit sphere, so $C=-aI$. Hence $q=(-a)(-t^2+\Vert x{\Vert }_2^2)$. The second form is negative on the cone interior, so $a<0$.