June 30th: Maxwell’s Theorem and Spherical Functions

Maxwell’s Theorem: In 3-dim., all spherical functions of degree d can be obtained by successive differentiations of 1/r along d directions.

The derivative of a harmonic function along a constant vector field is harmonic. Why?

Suppose the constant vector field is given by $\textbf{v}$ , then the corresponding directional derivative of a function f is given by $D_{v} f = \textbf{v} \cdot \nabla{f}$ . Since $\nabla^2 f = 0$ , it follows that:

$\nabla^2 (\textbf{v} \cdot \nabla{f}) = (\nabla^2 \textbf{v}) \nabla{f} + \textbf{v} \cdot (\nabla^2 \cdot \nabla{f}) = \textbf{v} \cdot \nabla (\nabla^2 f) = 0$ (follows from the fact that f is smooth (*)).

Since $D_{v} f$ is a harmonic function, when restricted to the sphere, it is an eigenfunction of the spherical Laplacian. Claim: $D_{v} f|_{S^2}$ coincides with a spherical function, i.e. a harmonic homogeneous polynomial.

Suppose $D_{v} f|_{S^2}$ to be an eigenfunction corresponding to eigenvalue $- \Lambda = -(k^2 + k(n-2)$ . Then, we can extend this to the entire space (with 0 removed, why?) as a harmonic homogeneous function of degree k. We now show that 0 is a removable singularity of our function.

Since k is positive and a harmonic function is defined to satisfy $h(r^k*q) = r^k*h(q)$ where $q \in S^2$ , it follows that the function is bounded near the origin, because as r tends to 0, the function value also tends to 0 (we know that the function is well-defined on the entire $S^2$ ).

Theorem (to be proven in the following post): a bounded harmonic function in a neighbourhood of 0 is still harmonic when continuously extended to 0.

It follows by continuity that our function should be defined to equal 0 at 0.

Arnold claimed that since this is a harmonic function, it is smooth (note that (*) also used the fact that harmonic implies smooth). This is true in 2-dim., since a harmonic function defined on a simply connected domain is the real/imaginary part of a holomorphic function, but why is it true in 3-dim. as well?

If we accept his claim for now, then our harmonic function is infinitely differentiable (i.e. we can express it by its Taylor series) and also homogeneous of degree k.

It follows that the harmonic function must be a homogeneous polynomial of degree k.

Question: so now that we know that differentiating 1/r (a harmonic function when n=3) along d constant vector fields give us a harmonic function that when restricted to the sphere coincides with a spherical function, how do we know that this spherical function is degree d?

Claim: show that after n differentiations, the numerator is a n degree homogeneous polynomial and the denominator is $r^{2n+1}$ . From above, we know that this function is harmonic and it is homogeneous of degree $n - (2n+1) = -n-1$ . When restricted to the sphere, we get an eigenfunction for the eigenvalue $- \Lambda = - (-n-1)^2 - (n-1 + 1)^2 = - (n+1)^2 - n^2$ . It is easy to see that the other k-value that gives $-k^2 - (k+1)^2 = - \Lambda$ is $k = n$ , and so, we can extend this eigenfunction as a homogeneous polynomial of degree n to the entire space. Since the numerator is a $n$ degree homogeneous polynomial and coincides with the eigenfunction on the unit sphere, it follows that the polynomial in the numerator is exactly the harmonic homogeneous polynomial that we are looking for.

We prove the above claim using induction. Suppose after m differentiations we obtain $F/r^{2m+1}$ , where F is a homogeneous polynomial of degree m. Then when we differentiate this function along a constant vector field, we obtain:

$\nabla_{v} \frac{F}{r^{2m+1}} = \frac{(\nabla_{v} F)r^2 - a F(v, r)}{r^{2m+2 + 2}}$ .

Since $\nabla_{v}$ preserves homogeneity but decreases the degree by 1, it follows that the numerator is a homogeneous polynomial of degree m+1.

It is easy to see that the base case for when m =1 holds. Take for example differentiating along a unit vector in the z-direction. Then, $\frac{\partial 1/r}{\partial z} = \frac{z}{r^3}$ .

Thus, by this method, we obtain a space of harmonic polynomials (to show that it is a vector space, we would need to do a lot more work; this will be done in a following post). If we want to show that we indeed obtain the entire space of harmonic polynomials, we would need to show that no harmonic polynomial is invariant under the action of SO(3). Since by construction, the space of harmonic polynomials that we obtain is invariant under SO(3), it follows that this must be the entire space.

Claim: the (2k+1)-dimensional spherical function of degree k can be generated by a linear combination of a single spherical function and its rotations.

Let’s take a zonal function, i.e. a spherical function depending only on z, to be the one generating the entire space of functions. The first question that we need to ask ourselves: does there exist a zonal function of any degree k (where k is a non-negative integer)?

Yes. When k = 0, the function 1/r restricted to the sphere gives 1. When k = 1, we can take the function $\frac{\partial 1/r}{\partial z} = \frac{z}{r^3}$ restricted to the unit sphere, which gives $z$ . For k = m, suppose we have the function $\frac{F(z, r)}{r^{2n+1}}$ , which when restricted to the unit sphere gives $f(z) = F(z, 1)$ . Then, for k = m+1, we can take:

$\frac{\partial \frac{F(z, r)}{r^{2n+1}}}{\partial z}$

$= \frac{ \frac{\partial F}{\partial z} r^{2n+1} + (2n+1)F r^{2n} \frac{\partial r}{\partial z}}{r^{2(2n+1)}}$

$= \frac{\frac{\partial F}{\partial z} r^{2} + (2n+1)zF}{r^{2n+3}}$ .

When restricted to $S^2$ , this gives a function of z only. Hence, in the space of spherical functions of any non-negative degree, there exists a function depending only on z. These polynomials that we obtain when such zonal functions are restricted to the unit sphere are called the Legendre polynomials.

Interesting: the Legendre polynomials are mostly not harmonic functions, but when restricted to the unit sphere, they coincide with a harmonic homogeneous polynomial.

Okay, so how do we generate the other spherical functions? The key is that we can obtain a basis of functions by “rotating” the zonal function along the x- or y-axis (rotating around the z-axis does nothing since the function is independent of x and y).

To be honest, I am still not sure how such rotated functions are related to the associated zonal functions that Arnold introduced.

June 30th: Maxwell’s Theorem and Spherical Functions

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