Iterating projections to random halfspaces Consider the following process:

*

*Start with a set $S = \mathbb R^n$. Repeat $L$ times: choose a random orthonormal basis $u_1, \ldots, u_n$, and consider the cone $C = \{ \sum \alpha_i u_i : \alpha_i \in \mathbb R_{\ge 0} \}$. Project $S$ to this cone.

What is the minimum dot product between two unit vectors in $S$, after $L$ iterations? Can it be bounded away from zero, say with high probability in $n$?
See the illustration of the process for $n = 2$. The red area outlines the set $S$. After projecting to the cones defined by the blue, green, and orange bases, the set $S$ becomes contained in a line.
The regime of interest is ideally $L = O(1)$, but $L = o(n)$ could also work.
 A: This addresses only the easy part.
The unit $(n-1)$-sphere in $\mathbb{R}^n$ is covered by finitely many disks (sectors) each portending an acute cone from the origin. Each projection to a half-space of an acute cone becomes the union of (possibly) a subset of the intersecting hyperplane with (possibly) a sub-cone of the cone, also acute. Eventually, with probability $1$, each cone (or sub-cone) will lie entirely outside a half-space and therefore will be projected into a hyperplane. Since there are only finitely many cones, all of $\mathbb{R}^n$ will eventually project into a finite union of subsets the hyperplanes defining the half-spaces. This has measure $0$.
For the harder question, I did some experimenting.
First notice that a ray (semi-line from the origin) is always projected to a ray. Then I ran some simulation in $\mathbb{R}^3$, and noticed that starting from 2 rays, no matter what the angle between them is, eventually I always end up with an angle between them of less than pi/180, almost always within less than 100 iterations, and typically within less than 30. It would be tedious but not hard to run such simulations for other dimensions and to collect some data. Though I wonder if floating point precision could quickly become an issue.
In $\mathbb{R}^2$ it's pretty obvious that we'll eventually hit, with probability 1, a pair of projections such that first brings two starting rays to within an acute angle of each other, and the second projects both of those rays into a single one. So eventually all of $\mathbb{R}^2$ has to end up in a single ray, as the OP was speculating.
