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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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


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.


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