Question

It is known that if you choose n point at random on S¹ = [0,1], the nearest neighbor spacings between the points are exponentially distributed with mean 1.

For example, two of our n points could be x, x + α/n Then with probability (1 - α/n)^n-2 -> e^-α The remaining n-2 points avoid the interval [x, x + α/n].

For the sequence kα (mod 1) does the nearest neighbor spacings of the first n elements approach a limit? I've heard it does when α is the golden ratio $(\sqrt{5}-1)/2$, but not in general. So maybe this is only true for specific α irrational.

Any readings on this problem would be appreciated, too.

EDIT: Apparently this sequence arises in the energy spacings of the two dimensional quantum harmonic oscillator. The equation would be $ - \frac{\hbar^2}{2m}\left( \frac{d^2 \psi}{dx^2} + \frac{d^2 \psi}{dx^2} \right) + \frac{1}{2}m[(\omega_1 x)^2 + (\omega_2y)^2]= E \psi$ and the spectrum would be $\frac{\hbar}{2}(m \omega_1 + n \omega_2 )$. This comes from Pavel Bleher's The energy level spacing for two harmonic oscillators with golden mean ratio of frequencies

Answer 1

This is called the 3-gap theorem: the set of gaps in the points α, 2α, 3α, ..., Nα (all taken modulo 1) has cardinality 1, 2 or 3, and if it has cardinality 3 then the largest gap is the sum of the other 2. If it has cardinality 1, then α is rational. If it has cardinality 2, then N is the denominator of a convergent to the continued fraction of α.

Vera Sos had the earliest proof, as far as I know. A more recent reference is Allesandri & Berth´e.

Answer 2

This reminds me of the following cute statement (which I believe has been recently generalized to arbitrary manifolds, but I couldn't find the paper): Given the sequence of points $k\alpha (\mod 1)$, there are at most 3 different distances between nearest neighbors.

Actually, I first heard it stated as $\alpha^k$ on the unit circle. To prove it, notice that the first cycle of points around the circle all have the same length between them, and then there is the remainder at the end.

When the wrap occurs, this is the remainder just shifted (because the distance between two consecutive points is always the same), and it keeps appearing between previous points in the same way at each step (and then there are the remaining points it has yet to appear between). This still leaves only 3 lengths (the two from the "cut" it produced in the previous length, and the previous lengths it has yet to cut).

I hope this helps.