Structure of Column Sums of Real Orthonormal Matrices

Question

Suppose I have a square real orthonormal matrix $A \in O(D)$. I'd like to understand what structure exists in the set of column sums of $A$.

For instance, $O(2)$ can be parameterized by a single scalar. To see why, consider $A = \begin{bmatrix} a & b\\ c & d \end{bmatrix}$. Since the first column must have unit norm, $c = \sqrt{1 - a^2}$. Since the second column must be orthogonal to the first column and must also have unit norm, $b = -c$ and $d = a$. Consequently, $A = \begin{bmatrix} a & -\sqrt{1 - a^2}\\ \sqrt{1 - a^2} & a \end{bmatrix}$ and the column sums are $a + \sqrt{1 - a^2}$ and $a - \sqrt{1 - a^2}$. When I plot the column sums as a function of $a$, I observe these nice curves:

My question is: how does this structure generalize to $O(D)$? Is some quantity conserved? If I order the column sums in decreasing order, does some relationship between them hold?

Maybe what I'd like is some theorem that states "if the previous columns' sums were $A, B, C,...$ then the next column's sum is is equal to $Z$ / bounded between $[-X, Y]$"

Answer 1

Knowing that the set of all possible column-sum-vectors is a sphere essentially answers all possible questions you could want to ask about such vectors. Specifically, we have:

Let $S(n)$ be the set of column-sum-vectors of orthogonal matrices in $O(n)$. Then $S(n)$ is equal to the sphere of radius $\sqrt n$ centered at the origin.

From the comments:

can I say anything beyond that? Since the vectors are orthonormal, it suggests that fixing one (or several) severely limits what remaining points on the sphere can be chosen.

Bringing in the hypothesis that the vectors are orthonormal cannot possibly get you any stronger results, since that hypothesis is embedded in the theorem that the set of all column-sum-vectors is a sphere. So yes, fixing one or several coordinates restricts the others - but it restricts them only and precisely in that they must be chosen so that the resulting point ends up on a sphere. There is no point trying to get any further restrictions, as the result is that $S(n)$ is equal to a sphere - not a subset of it, and not a superset of it, but equal. Therefore the restriction is as tight as it gets.

For example:

1. You can parameterize $S(n)$, using any standard parameterization of a sphere.

2. Yes, if you fix the first $k$ coordinates, this restricts the remaining coordinates since the whole vector must end up on a sphere. Specifically, the remaining coordinates $a_{k+1}, ..., a_n$ must be chosen so that $$a_{k+1}^2+...+a_n^2=n-(a_1^2+...+a_k^2)$$ In other words, if $r^2=a_1^2+...+a_k^2$, the remaining coordinates must be chosen from a sphere of radius $\sqrt{n-r^2}$ in $(n-k)$-dimensional space.