Magic angles

The function ${u(x, y, z) = 2z^2 - x^2 - y^2}$ is notable for the following combination of properties:

1. It is a harmonic function: ${\Delta u = -2 - 2 + 4 = 0}$
2. It vanishes at the origin ${(0, 0, 0)}$ together with its gradient.
3. It is positive on the cone ${C=\{(x, y, z) : z > \sqrt{(x^2+y^2)/2}\}}$

The cone C has the opening angle ${\theta_3 = \cos^{-1}(1/\sqrt{3}) \approx 57.4^\circ}$ which is known as the magic angle in the context of NMR spectroscopy. Let us consider the mathematical side of its magic.

If C is replaced by any larger cone, the properties 1-2-3 cannot be satisfied by a harmonic function in a neighborhood of the origin. That is, C is the largest cone such that a harmonic function can have a critical point at its vertex which is also a point of its extremum on the cone. Why is that?

Let ${u}$ be a harmonic function in some neighborhood of ${(0,0,0)}$ and suppose ${u(0,0,0)=0}$, ${\nabla u(0,0,0) = 0}$, and ${u>0}$ on some cone ${C = \{(x, y, z) \colon z>c \sqrt{x^2+y^2+z^2} \}}$. Expand ${u}$ into a sum of polynomials ${p_2+p_3+\cdots }$ where each ${p_k}$ is a harmonic homogeneous polynomial of degree ${k}$. Let ${m}$ be the smallest integer such that ${p_m}$ is not identically zero. Then ${p_m}$ has the same properties as ${u}$ itself, since it dominates the other terms near the origin. We may as well replace ${u}$ by ${p_m}$: that is, ${u}$ is a spherical harmonic from now on.

Rotating ${u}$ around the ${z}$-axis preserves all the properties of interest: harmonic, positive on the cone, zero gradient at the origin. Averaging over all these rotations we get a rotationally symmetric function known as a zonal spherical harmonic. Up to a constant factor, such a function is given by ${P_m(\cos \phi)}$ where ${\phi}$ is a spherical coordinate (angle with the ${z}$-axis) and ${P_m}$ is the Legendre polynomial of degree ${m}$.

The positivity condition requires ${P_m(t) > 0}$ for ${t>c}$. In other words, the bound on ${\theta}$ comes from the greatest zero of the Legendre polynomial. As is true for orthogonal polynomials in general, the zeros are interlaced: that is, the zeros of ${P_m=0}$ appear strictly between any two consecutive zeros of ${P_{m+1}}$. It follows that the value of the greatest zero grows with ${m}$. Thus, it is smallest when ${m=2}$. Since ${P_2(t) = (3t^2-1)/2}$, the zero of interest is ${1/\sqrt{3}}$, and we conclude that ${c \ge 1/\sqrt{3}}$. Hence the magic angle.

The magic angle is easy to visualize: it is formed by an edge of a cube and its space diagonal. So, the magic cone with vertex at (0,0,0) is the one that passes through (1, 1, 1), as shown above.

In other dimensions the zonal harmonics are expressed in terms of Gegenbauer polynomials (which reduce to Chebyshev polynomials in dimensions 2 and 4). The above argument applies to them just as well. The relevant Gegenbauer polynomial of degree ${2}$ is ${nx^2-1}$ up to a constant. Thus, in ${\mathbb R^n}$ the magic angle is ${\cos^{-1}(1/\sqrt{n})}$, illustrated by the harmonic function ${u(x)=nx_n^2 - |x|^2}$.

This analysis is reminiscent of the Hopf Lemma which asserts that if a positive harmonic function in a smooth domain has boundary value 0 at some point, the normal derivative cannot vanish at that point. The smoothness requirement is typically expressed as the interior ball condition: one can touch every boundary point by a ball contained in the domain. The consideration of magic angles shows that if the function is also harmonic in a larger domain, the interior ball condition can be replaced by the interior cone condition, provided that the opening angle of the cone is greater than ${\cos^{-1}(1/\sqrt{n})}$.

The magic of cyclic sieving and q-analogs

In my previous post the 14 possible triangulations of hexagons were arranged by Whitehead moves. But if asked, most people would probably group them in this way:

Well, some may decide to put the S- and Z-shapes together since they are mirror images of each other. But I won’t. It’s easy to rotate a piece of paper with a triangulation on it; it’s not nearly as easy to fetch a mirror and study triangulations through it. So let it be recorded that the cyclic group $\mathbb Z_6$ acts on triangulations by rotating the hexagon. The orbits have lengths 6,3,3,2.

In the previous post I also mentioned that the number of triangulations of an (n+2)-gon is the Catalan number $C_n=\frac{1}{n+1}\binom{2n}{n}$. With n=4 we get

$\displaystyle C_4=\frac{1}{5}\,\frac{8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2}=\frac{8\cdot 7\cdot 6}{4\cdot 3\cdot 2} \qquad (*)$

Amazingly, the formula (*) does not just count triangulations: it also knows how they behave under rotations of the hexagons. But it will only tell us if we don’t rush to its evaluation. Instead, replace each number on the right of (*) by its q-analog

$\displaystyle [n]_q=\frac{1-q^n}{1-q} =1+q+q^2+\dots+q^{n-1}$

The result is

$\displaystyle C_4(q) = \frac{(1-q^8)(1-q^7)(1-q^6)}{(1-q^4)(1-q^3)(1-q^2)} =1+q^2+q^3+2q^4+q^5+2q^6+q^7+2q^8+q^9+q^{10}+q^{12}\qquad (**)$

The fact that we get a polynomial can be explained: since the Catalan number is an integer, every prime divisor of the denominator in (*) also appears in the numerator (to equal or higher power), and this can be translated into the order of vanishing of polynomials in (**) at the roots of unity. It’s not as easy to explain why the coefficients end up nonnegative.

Now, both the cyclic group $\mathbb Z_6$ and the roots of unity were mentioned. Let’s bring them together by introducing $\zeta=\exp(2\pi i/6)$ which generates $\mathbb Z_6$ by multiplication. The cyclic sieving phenomenon (in this special case) is the following: $C_4(\zeta^d)$ is the number of triangulations fixed by the action of $\zeta^d$. Indeed, evaluating $C_4(q)$ at $q=\zeta,\zeta^2,\dots,\zeta^6$ we get $0,2,6,2,0,14$. So, no triangulations are invariant under rotation by $\zeta$, two are invariant under rotation by $\zeta^2$, etc. It’s easy to read the entire orbit structure from this:

• there are no fixed points
• 2 elements of order 2 form a single orbit of size 2
• 6 elements of order 3 form two orbits of size 3
• there are no elements of order 4, since their number is $C_4(\zeta^4)-C_4(\zeta^2)=2-2=0$
• there are $C_4(\zeta^6)-C_4(\zeta^2)-C_4(\zeta^3)=6$ elements of order 6, which form a single orbit of size 6

And it’s not just triangulations. Consider noncrossing matchings of 2n points — these can be observed experimentally by sitting 2n people at a round table and asking them all to shake hands with someone with no arms crossing.

The number of such matchings is also the Catalan number $C_n$ (hence 14 matchings for 8 people). But the orbit structure is completely different! How can the same polynomial $C_4(q)$ work here? It can because we now act by $\mathbb Z_8$ and therefore take $\zeta=\exp(2\pi i/8)$. Evaluation of $C_4(q)$ at the powers of $\zeta$ yields $0,2,0,6,0,2,0,14$. As above, we see the orbit structure in these numbers:

• 2 elements of order 2 form a single orbit
• 4 elements of order 4 form a single orbit
• 8 elements of order 8 form a single orbit

And it’s not just the Catalan numbers. Suppose that among those 8 guests at the round table are 2 men and 6 women. The number of seating arrangements is $\displaystyle \binom{8}{2}=\frac{8\cdot 7}{2}$. Don’t evaluate to 28 just yet. Replace each integer with its q-analog and you’ll get the polynomial $\displaystyle 1+q+2q^2+2q^3+3q^4+3q^5+4q^6+3q^7+3q^8+2q^9+2q^{10}+q^{11}+q^{12}$, which will tell you that under rotation, 4 arrangements form an orbit of size 4, while the other 24 form 3 orbits of size 8.

And there is much more. The cyclic sieving phenomenon was established recently (in 2004) by three mathematicians from Minnesota (Reiner, Stanton, and White), and there are still paths waiting to be explored.