2
$\begingroup$

Background

It is known that the spherical harmonics $Y_n^m$ with order $n$ and degree $m$ (such that $n \ge 0, -n \le m \le n$) are functions on the sphere which form a complete, orthogonal infinite set in $L^2$ space on the sphere surface $S^2$.

Given any arbitrary square-integrable function $f(\theta, \phi) \in L^2(S^2)$, the spherical Fourier series with coefficients for order $n \le N$, denoted $S_N(f)$ converges absolutely in the 2-norm:

$$ \lim_{N \to \infty} \lVert f - S_N(f) \rVert_2 = 0 $$

For a finite, $N$-th order limited set of spherical harmonics, what functions can be exactly represented (in the norm convergence sense)?

Consider the spherical function: $$ f(\theta, \phi) = \cos(2 \phi) \sin \theta $$

The function f in a colatitude spherical coordinate system

Due to the identities: $$ \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2}, \quad \sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i} $$ it appears that $f(\theta, \phi)$ is a 2nd-order trigonometric polynomial, as one would conclude for a 1-dimensional function $g(x)$ with a traditional Fourier series (source): $$ S_\infty (g) = \sum_{n=-\infty}^\infty c_n e^{i n x} $$ which is equal to: $$ c_0 + (c_1 e^{ix} + c_{-1} e^{-ix}) + \ldots + (c_n e^{inx} + c_{-n} e^{-inx}) + \ldots $$ and we know that $e^{\pm i n \theta} = \cos (n \theta) \pm i \sin (n \theta)$. I think this expresses a relation between the trigonometric functions and the complex exponential, where order is the multiple of the dependent variable. I would assert that: the SHs are just this process in two dimensions. Therefore, the SHs of order 2 can exactly represent trig polynomials of order 2, regardless of whether it is in $\theta$ or $\phi$.

But I must be missing some crucial theory. Numerically, I can determine that there are SH coefficients that are non-zero beyond the 2nd-order:

Spherical harmonic coefficients using ACN numbering

Coefficients corresponding to order 2 and below are to the left of the red dashed line. (The ACN channel numbering is from Ambisonics to identify the specific SH coefficient according to $n^2 + n + m$.)

So the questions are: why can't the spherical harmonics of order $n \le 2$ represent $f(\theta, \phi)$? And furthermore, what exactly can they represent in terms of orders of polynomials? What math theory am I missing?

$\endgroup$
9
  • 1
    $\begingroup$ What are polynomials on $S^2$? Restrictions of polynomials on the ambient $\mathbb R^3$ to the unit sphere? $\endgroup$
    – doetoe
    Sep 13, 2021 at 23:52
  • 1
    $\begingroup$ But the spherical harmonic you list above, a pretty typical one, is not a polynomial over $\theta$ and $\phi$, so it's not clear what you mean. $\endgroup$
    – jwimberley
    Sep 14, 2021 at 0:50
  • 1
    $\begingroup$ Some clarification is necessary. In your comment above, you said that $Y_4^3$ is a polynomial in $\theta$ and $\phi$, but really, it's a polynomial in $\cos\theta$, and $e^\phi$. Is that what you mean? Because if that's the case, then I think the answer is a pretty clear (to me) yes (because the $Y^lm$'s are just associated Legendre polynomials of $\cos\theta$ mutiplied by powers of $e^i\phi$). But if you actually mean polynomials of $\theta$ and $\phi$, then $Y^m_n$ is not an example of such a polynomial, and in general my guess it that the answer is no. $\endgroup$
    – march
    Sep 14, 2021 at 3:08
  • 1
    $\begingroup$ No, I mean, when I hear "polynomials of $\theta$ and $\phi$", I picture something like $\theta^3\phi^2 + 3\theta+1$, but what I think you mean is polynomials $x^3y^2+x+1$, where $x=\cos\theta$ and $y=e^{i\phi}$. Is that right? That's the clarification I'm talking about. But I think your post edit clarified that. $\endgroup$
    – march
    Sep 14, 2021 at 15:13
  • 1
    $\begingroup$ I think the problem with $\cos(2\phi)\sin(\theta)$ is that it corresponds to a state with $m>l$ ($m=2$, $l=1$), but the spherical harmonics restrict to $m\leq l$. It seems, then, that the spherical harmonics are not the generalizations of trigonometric polynomials that you have in mind. $\endgroup$
    – march
    Sep 14, 2021 at 15:55

1 Answer 1

2
$\begingroup$

Instead of polar, let us work in Cartesian coordinates $(x, y, z)\in\mathbb R^3$, with the constraint that $x^2+y^2+z^2=1$. You are representing functions on the sphere as $f=f(\theta, \phi)$, but we can equivalently represent them as functions of $(x, y, z)$ via the following change of coordinates formulas (physicist's convention): $$\tag{1} \begin{cases} x=\sin\theta \cos \phi, \\ y=\sin\theta\sin\phi, \\ z=\cos \theta. \end{cases}$$ The usefulness of this lies in the following.

Fact. The function $Y=Y(\theta, \phi)$ is a spherical harmonic of degree $n$ if and only if, letting $$ Y(\theta, \phi)=H(x, y, z), $$ the function $H$ is a homogeneous polynomial of degree $n$ and moreover $$ \frac{\partial^2 H}{\partial x^2} + \frac{\partial^2 H}{\partial y^2}+ \frac{\partial^2 H}{\partial z^2} =0,\qquad \forall (x, y, z)\in \mathbb R^3.$$ Notice that this last equation must hold for all $(x, y, z)\in\mathbb R^3$, not just on the sphere. In other words, $H$ must be a harmonic homogeneous polynomial.

Here there is a proof of this fact.

Examples of homogeneous harmonic polynomials $H=H(x, y, z)$ of various degrees are $$ x,\quad y,\quad z,\quad x^2-z^2, $$ which, using (1), yield the following spherical harmonics in polar coordinates: $$ \sin\theta \cos\phi,\quad \sin\theta\sin\phi,\quad \cos\theta,\quad\sin^2\theta\cos^2\phi-\cos^2\theta.$$

We conclude from all this that, in particular, every linear combination of spherical harmonics of degree up to $N$ must be a harmonic polynomial in the Cartesian coordinates of degree up to $N$.


Now let us consider the example given above, namely $$ f(\theta, \phi)=\cos(2\phi)\sin(\theta).$$ Using (1), we see that $$ \begin{split} f(\theta, \phi)&=(\cos^2\phi-\sin^2\phi)\sqrt{1-\cos^2\theta} \\ &=\frac{x^2-y^2}{x^2+y^2}\sqrt{1-z^2}\\ &=\frac{x^2-y^2}{\sqrt{1-z^2}}. \end{split}$$ This is not a polynomial. Therefore, it is not a finite linear combination of spherical harmonics. It must possess nonzero spherical harmonics coefficients of arbitrarily high order, as observed numerically.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.