A shared column of an independent orthogonal mixing must be Gaussian

Let ${\phi }_i(u)=\mathbb{E}{e}^{iu{S}_i}$ and ${\theta }_j(v)=\mathbb{E}{e}^{iv{T}_j}$. Each $T_j$ has a finite second moment, since $({\sum }_i{O}_{ji}{S}_i{)}^2\le ({\sum }_i{O}_{ji}^2)({\sum }_i{S}_i^2)$ pointwise. Thus [F2] makes all these functions $C^2$, and each is one at zero. Choose $\delta >0$ such that none vanishes on $(-\delta ,\delta )$; this is possible by continuity and finiteness of the families. Define ${\psi }_i(u)={\int }_0^u\frac{{\phi }_i^{\mathrm{\prime }}(s)}{{\phi }_i(s)}ds,{\chi }_j(v)={\int }_0^v\frac{{\theta }_j^{\mathrm{\prime }}(s)}{{\theta }_j(s)}ds\mathrm{.}$ By [F5] applied on every compact subinterval, these are complex-valued $C^2$ functions on that interval, equal to zero at zero. Differentiation shows $(e^{-{\psi }_i}{\phi }_i{)}^{\mathrm{\prime }}=0$, so $e^{{\psi }_i}={\phi }_i$; likewise $e^{{\chi }_j}={\theta }_j$. Moreover ${\psi }_i^{\mathrm{\prime }}(0)=0$ and ${\psi }_i^{\mathrm{\prime }\mathrm{\prime }}(0)={\phi }_i^{\mathrm{\prime }\mathrm{\prime }}(0)-{\phi }_i^{\mathrm{\prime }}(0{)}^2=-1$. This constructs the needed local logarithms without any global nonvanishing hypothesis.

For $\parallel t\parallel <\delta$, also $\parallel O^{T}t\parallel <\delta$. Factoring the characteristic function of $t^{T}T=(O^{T}t{)}^{T}S$ in both independent coordinate systems by [F3] gives ${\prod }_i{\phi }_i((O^{T}t{)}_i)={\prod }_j{\theta }_j(t_j)\mathrm{.}$ Consequently $D(t):={\sum }_i{\psi }_i((O^{T}t{)}_i)-{\sum }_j{\chi }_j(t_j)$ has $e^{D(t)}=1$. It is continuous on the ball, is zero at zero, and takes values in the discrete set $2\pi i\mathbb{Z}$. Along each line segment $s\mapsto st$ the intermediate value property of its imaginary part makes it constant; hence $D(t)=0$ throughout the ball. Differentiating this $C^2$ identity with respect to distinct coordinates $t_j,t_k$ gives ${\sum }_i{O}_{ji}{O}_{ki}{\psi }_i^{\mathrm{\prime }\mathrm{\prime }}((O^{T}t{)}_i)=0(j\mathrm{\ne }k)\mathrm{.}$

Fix $i$ and two distinct rows $j,k$ with $O_{ji}{O}_{ki}\mathrm{\ne }0$. Orthogonality maps the ball onto itself, so in the last identity one may set $O^{T}t=u$ for every $\parallel u\parallel <\delta$. First take $u=v{e}_i$, $\mathrm{\mid }v\mathrm{\mid }<\delta$, and then $u=0$. Subtraction cancels all summands except the $i$th, giving $O_{ji}{O}_{ki}({\psi }_i^{\mathrm{\prime }\mathrm{\prime }}(v)-{\psi }_i^{\mathrm{\prime }\mathrm{\prime }}(0))=0$. Thus ${\psi }_i^{\mathrm{\prime }\mathrm{\prime }}(v)=-1$ throughout $(-\delta ,\delta )$. With ${\psi }_i^{\mathrm{\prime }}(0)={\psi }_i(0)=0$, [F6] applied successively to ${\psi }_i\mathrm{\prime }(v)+v$ and ${\psi }_i(v)+v^2\mathrm{/}2$ gives ${\psi }_i(v)=-v^2\mathrm{/}2$ and ${\phi }_i(v)=e^{-v^2\mathrm{/}2}$ there. By [F4], $S_i$ is standard Gaussian.

The contrapositive of step 3.1 says that a non-Gaussian source column has at most one nonzero entry. Its squared norm is one because $O^{T}O=I$, so it has exactly one entry, whose square is one. This proves both assertions.