Suppose I have a square real orthogonal matrix $A \in \mathbb{R}^D$, and I compute the element-wise sum of the $i$th column as $a_i := \sum_{d=1}^D A_{di}$. How can I describe the distribution of the $a_i$ values for the D columns?
I know that the maximum value $a_i$ can take is $\sqrt{D}$ and the minimum value is $-\sqrt{D}$. I'm wondering how fixing one $a_i$ affects the remaining $a_j$ for $j \neq i$.
Edit 1: I might be more interested in how the $a_i$ values decay if you order them such that $a_1 \geq a_2 \geq ... \geq a_D$. For instance, if $a_1 = \sqrt{D}$, what is the largest that $a_2$ can be?
The possible vectors $(a_i)_{1\leq i \leq D}$ are exactly the vectors of $\mathbb{R}^n$ with norm $\sqrt{D}$. That is, we always have
$$ \sum_{i=1}^D a_i^2 = D $$
So to answer one question, if $a_1$ has the extreme value $a_1 = \pm \sqrt{D}$, then all other $a_i$ are exactly zero. This makes sense because the first column of $A$ must be $\pm D^{-1/2} [1, \ldots, 1]$, and taking the dot product with any other column must give zero, meaning the sum of that other column's elements is zero.
To prove the claim more generally, note that calling the unit basis $\{e_i\}$, we have $A_{ij} = e_i^T A\, e_j$, so
$$ a_i = \sum_{d=1}^D A_{di} = \sum_{d=1}^D (e_d^T A\, e_i) = \left[\sum_{d=1}^D e_d^T \right] A e_i $$
The left row vector is a vector with all elements $1$. So denoting $u = [1,\ldots,1]$, that gives
$$ a_i = u^T A \, e_i = (A^T u)^T e_i $$
That is, the numbers $a_i$ are the coordinates of the vector $A^T u$. Since $A$ is orthonormal, so is $A^T$, and therefore
$$\big| (a_i)_{1 \leq i \leq D} \big| = |A^T u| = |u| = \sqrt{\sum_{d=1}^D 1^2} = \sqrt{D} $$
One way of looking at this is that each value $(a_i / \sqrt{D})$ is the cosine of the angle between $A e_i$ (the image of an original basis vector) and the constant vector $u$.
For the converse, suppose we are given values $\alpha_i$ such that $\sum \alpha_i^2 = D$. Choose an orthonormal basis $\{x_i\}$ of $\mathbb{R}^D$ with $x_1 = D^{-1/2} (\alpha_i)_{1 \leq i \leq D}$, and choose another orthonormal basis $\{y_i\}$ of $\mathbb{R}^D$ with $y_1 = D^{-1/2} u$, for example by the Gram-Schmidt algorithm. If matrix $X$ has $\{x_i\}$ as columns and matrix $Y$ has $\{y_i\}$ as columns, then the matrix $A = Y X^T$ is one example of matrix with the values $a_i = \alpha_i$ as its column sums, since
$$ a_i = u^T A e_i = \sqrt{D} \cdot y_1^T Y X^T e_i = \sqrt{D} \cdot e_1^T X^T e_i = \sqrt{D} \cdot x_1^T e_i = \alpha_i $$
So for a probability distribution on vectors $(a_i)$, we could certainly use the usual distribution for the sphere $S^{D-1}$. I would expect this is the same distribution we would get from a sensible probability distribution on orthonormal matrices $A$ (however that would be defined), just as a principle of symmetry.