Spherical Harmonics

Fourier decomposition transforms some function that varies over space or time ($f(t)$ , $f(x, y)$ , etc) into a sum over a set of orthogonal basis functions. For the Fourier transform, this set of basis functions are sinusoids of various frequencies. So (remembering that $e^{ix} = \cos(x) + i \sin(x)$ ) we have,

$$\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x) e^{-2 i \pi x \xi} dx$$

where $\xi$ is the frequency and $\hat{f}({\xi})$ is a measure of the amount of power in that frequency in our function $f(x)$ . If we do this for every frequency, we can then reconstruct the original function $f(x)$ by summing over the basis functions, multiplied by the power in each,

$$f(x) = \int_{-\infty}^{\infty} \hat{f}(\xi) e^{2 i \pi x \xi} d\xi$$

The core idea here is that we trade in our function defined in space or time, $f(x)$ , for a function defined in frequency, $f(\xi)$ , that contains the same information (or vice-versa). This allows us to choose the most convenient representation depending on what we want to do.

While the Fourier transform works well for functions defined in Euclidean space, it does not necessarily work for ones defined on a surface (in particular, a sphere). Is it possible to define a set of basis functions (analogous to the sinusoids of the Fourier transform) that allow us to decompose a signal defined over a sphere, $f(\theta, \phi)$ , into its harmonic components?

It is! One possible set of functions are the spherical harmonics often denoted $Y_l^m$ . Here, $l$ is the degree of the harmonic (analogous to the frequency in a sinusoid – large $l$ implies high frequency), and $m$ is the order of the harmonic (which has no direct analog in the Fourier transform). As with the Fourier transform, we can take any function defined over the surface of a sphere and instead define it as a sum over these basis functions,

$$f(\theta, \phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^{l} a_{lm} Y_l^m(\theta, \phi)$$

where $a_{lm}$ has taken the place of $f(\xi)$ , the coefficient that represents the amount of power in a given harmonic (degree and order).

The basis functions are defined as,

$$Y_l^m(\theta, \phi) = C_l^m P_l^m (\cos(\theta)) e^{im\phi}$$

where $C$ is some constant that depends on $l$ and $m$ , and $P_l^m$ is an associated Legendre Polynomial. Before digging more into any math, let’s just look at these $Y_l^m$ and their two components.

First, let’s address the slightly odd behavior of the $\theta$ property. This is not the latitude which would run from $\pi/2$ in the North to $-\pi/2$ in the South, but rather the colatitude defined as $\pi/2 - \text{latitude}$ . The longitude, $\phi$ , runs from $-\pi$ to $\pi$ as normal. The effects of varying these two components can fortunately be treated separately with the results multiplied together. This makes understanding the overall behavior far simpler.

The effect of varying longitude is relatively simple - it is a sinusoid with frequency of $m$ . The imaginary component is a $\sin$ function and the real a $\cos$ . As $\sin(x) = -\sin(-x)$ the imaginary component changes sign when $m$ does, and as $\cos(x) = -\cos(x)$ the real component is unchanged. Note that for $m = 0$ , the imaginary component is $0$ and so $Y_l^m = 0$ , and the real component is constant and $Y_l^m$ consists of bands running from East-West.

The effect of varying the (co)latitude is a bit more complicated. $\cos(\theta)$ varies from $1$ to $-1$ as we move from the North pole to the South. We evaluate the associated Legendre polynomial (ALP), $P_l^m$ , at the result of this $x = \cos(\theta)$ .

As you might guess from the name, the ALP of degree $l$ is a polynomial where the largest power of $x$ is $l$ ¹. Concretely, the first few (with $m = 0$ for now) are,

$$\begin{align} P_0^0 &= 1 \\ P_1^0 &= x \\ P_2^0 &= \frac{1}{2}(3x^2 - 1) \\ \end{align}$$

The number of zeros of a non-degenerate polynomial (which all of these are) is equal to its degree. As all of these zeros are between $-1$ and $1$ , when moving from North to South $Y_l^0 = 0$ at $l$ values of $\theta$ .

The zeroth order ($m = 0$ ) ALP is actually just the Legendre polynomial of the same degree. The ALPs are defined as,

$$P_l^m = (-1)^m (1-x^2)^{m/2} \frac{d^m}{dx^m}(P_l^m(x))$$

This also shows how we can generate the ALPs of other orders. So for example,

$$\begin{align} P_2^0 &= \frac{1}{2}(3x^2 - 1) \\ P_2^1 &= -3x(1 - x^2)^{1/2} \\ P_2^2 &= 3(1 - x^2) \\ \end{align}$$

Ignoring the $-1^m$ , we now have two components; the derivative of $P_l^0$ and $(1 - x^2)^{m/2}$ . The first of these has $l - m$ zeros (as the m’th derivative of a polynomial of degree $l$ has degree $l - m$ ). The second term has zeros only at the endpoints. So,

• Increasing $m$ by 1 reduces the number of zeros (ignore the endpoints) 1.
• Moving from $m = 0$ to $m = 1$ results in zeros at the endpoints.
• Moving to higher $m$ results in larger areas near zero near the poles (as raising $1 - x^2$ which is $< 1$ to a larger power results in an value closer to zero).

Thus we get the following intuitive explanation for the effect of $m$ . At fixed $l$ , increasing $m$ reduces the number of zeros along the North-South axis by one but increases the number of wavelengths along the East-West axis by 1 (or increases the number of zeros by 2). Thus increasing $m$ results in a function that varies more from East to West and less from North to South.

Finally, what happens to $P_l^m$ when $m$ is negative? Without getting into too many details, while $P_l^m \ne P_l^{-m}$ , the $C_l^m$ mostly cancel any difference other than a possible factor of $-1$ ,

$$P_l^m C_l^m = (-1)^m P_l^{-m} C_l^{-m}$$

The combination of the East-West and North-South effects of negating $m$ is therefore for even $m$ , $$Re(Y_l^m) = Re(Y_l^{-m}) \quad Im(Y_l^m) = -Im(Y_l^{-m})$$

and for odd $m$ , $$Re(Y_l^m) = -Re(Y_l^{-m}) \quad Im(Y_l^m) = Im(Y_l^{-m})$$

A few other useful resources with different/better/more detailed explanations are,

• Fourier Transforms: If 3Blue1Brown has a video on something, it is probably the best intuitive explanation around.
• Interactive Fourier Transforms: Draw a picture and see how it can be decomposed. Also the inspiration for the power spectrum drawing later in this series.
• Spherical Harmonics: More of the math behind spherical harmonics.