# How to Sanity Check your Spherical Harmonics Projection Code


We begin by giving a short re-introduction to SH. Let us say we have a spherical function $f(\theta,\phi)$, where $\theta$ and $\phi$ are spherical coordinates and they are in range $0 \le \theta \le \pi$ and $0 \le \phi \le 2\pi$. This is a function that assigns a value for every point $(\theta,\phi)$ on the surface of the unit sphere. Spherical harmonics allows us to cheaply approximate and encode such a function. In graphics, we use the following nine spherical harmonics functions: (Taken from Patapom's blog.) \begin{align*} Y_{0}^{0} &= \frac{1}{2} \sqrt{\frac{1}{\pi}} \\ Y_{1}^{-1} &= \frac{1}{2} \sqrt{\frac{3}{\pi}} y \\ Y_{1}^{0} &= \frac{1}{2} \sqrt{\frac{3}{\pi}} z \\ Y_{1}^{1} &= \frac{1}{2} \sqrt{\frac{3}{\pi}} x \\ Y_{2}^{-2} &= \frac{1}{2} \sqrt{\frac{15}{\pi}} xy \\ Y_{2}^{-1} &= \frac{1}{2} \sqrt{\frac{15}{\pi}} yz \\ Y_{2}^{0} &= \frac{1}{4} \sqrt{\frac{5}{\pi}} (3z^2 - 1) \\ Y_{2}^{1} &= \frac{1}{2} \sqrt{\frac{15}{\pi}} xz \\ Y_{2}^{2} &= \frac{1}{4} \sqrt{\frac{15}{\pi}} (x^2 - y^2) \\ \end{align*} Where \begin{align*} x &= \sin(\theta)\cos(\phi) \\ y &= \sin(\theta)\sin(\phi) \\ z &= \cos(\theta) \\ \end{align*} There are actually an infinite number of SH functions, but for real-time rendering purposes, we in practice rarely use more than the nine above. We use SH to approximate $f$ as follow: \begin{align*} f(\theta,\phi) \approx C_{0}^{0} Y_{0}^{0}(\theta,\phi) + C_{1}^{-1} Y_{1}^{-1}(\theta,\phi) + C_{1}^{0} Y_{1}^{0}(\theta,\phi) + C_{1}^{1} Y_{1}^{1}(\theta,\phi) + \\ C_{2}^{-2} Y_{2}^{-2}(\theta,\phi) + C_{2}^{-1} Y_{2}^{-1}(\theta,\phi) + C_{2}^{0} Y_{2}^{0}(\theta,\phi) + C_{2}^{1} Y_{2}^{1}(\theta,\phi) + C_{2}^{2} Y_{2}^{2}(\theta,\phi) \end{align*} Let us give a cute analogy: Essentially, these 9 SH functions are like LEGO bricks that we can use to build spherical functions. We combine together these LEGO bricks into a sum, and this sum forms our approximation for the spherical function $f$. The constants $C_{0}^{0}, C_{1}^{-1}, C_{1}^{0},\dots$ specifies how much of each LEGO brick we will use. If a constant is zero, then its corresponding LEGO brick is not used at all, and if it is non-zero, then the brick is used in the sum that builds $f$. Also, do note the use of $\approx$ instead of $=$. This is because using only nine SH functions results in a very rough approximation of $f$; this approximation is poor for capturing high frequency detail in $f$, but for the purposes of diffuse indirect lighting, this is not a large drawback, and thus it is a useful approximation in practice. We can determine the values of the constants by convolving the SH-functions with $f$, one by one. \begin{align*} C_{0}^{0} &= \int_0^{2\pi} \int_0^{\pi} f(\theta,\phi) Y_{0}^{0} \sin(\theta) d\theta\ d\phi \\ C_{1}^{-1} &= \int_0^{2\pi} \int_0^{\pi} f(\theta,\phi) Y_{1}^{-1} \sin(\theta) d\theta\ d\phi \\ &\dots \\ &\dots \end{align*} To perform this convolution in practice, we need to perform a numerical integration, as is described at Patapom's blog. This kind of calculation is often called a projection in some literature. After projecting all the nine SH-functions, we will have our nine constants. I think that this projection calculation can be easy to get wrong, so it is good if we can sanity check our implementation. What I often do, is that I give a spherical function defined as $f(\theta,\phi)=1$ to the projection code. So the function is 1 everywhere on the unit sphere. If we have our function $f$ encoded as a cubemap, we would just give an all-white cubemap to the projection code. For this spherical function $f(\theta,\phi)=1$, we can analytically perform the convolution. \begin{align*} C_{0}^{0} &= \int_0^{2\pi} \int_0^{\pi} f(\theta,\phi) Y_{0}^{0} \sin(\theta) d\theta\ d\phi = \\ &= \int_0^{2\pi} \int_0^{\pi} \frac{1}{2} \sqrt{\frac{1}{\pi}} \sin(\theta) d\theta\ d\phi = 2\sqrt{\pi}\\ \end{align*} We got this analytic answer by plugging it into Wolfram Alpha. Calculating the next constant is also easy \begin{align*} C_{1}^{-1} &= \int_0^{2\pi} \int_0^{\pi} \frac{1}{2} \sqrt{\frac{3}{\pi}} \sin(\theta)\sin(\phi) \sin(\theta) d\theta\ d\phi = 0\\ \end{align*} If we do all the integrations, it turns out that $C_{0}^{0} = 2\sqrt{\pi}$ and $C_{1}^{-1} = C_{1}^{0} = C_{1}^{1} = C_{2}^{-2} = C_{2}^{-1} = C_{2}^{0} = C_{2}^{1} = C_{2}^{2} = 0$. Note also that the obtained constants perfectly describe the input function $f(\theta,\phi)=1$, since \begin{align*} &C_{0}^{0} Y_{0}^{0}(\theta,\phi) + C_{1}^{-1} Y_{1}^{-1}(\theta,\phi) + C_{1}^{0} Y_{1}^{0}(\theta,\phi) +\\ &C_{1}^{1} Y_{1}^{1}(\theta,\phi) + C_{2}^{-2} Y_{2}^{-2}(\theta,\phi) + C_{2}^{-1} Y_{2}^{-1}(\theta,\phi) +\\ &C_{2}^{0} Y_{2}^{0}(\theta,\phi) + C_{2}^{1} Y_{2}^{1}(\theta,\phi) + C_{2}^{2} Y_{2}^{2}(\theta,\phi) = \\ &C_{0}^{0} Y_{0}^{0}(\theta,\phi) = \\&(2\sqrt{\pi}) (\frac{1}{2} \sqrt{\frac{1}{\pi}}) = 1 = f(\theta,\phi) \end{align*}

So this is my approach to sanity checking my SH projection code: Give a spherical function with a constant value of $1$ to the code. If the first constant has a value very close to $2\sqrt{\pi}$(due to floating point imprecision issues, you should not check for exact equality!), and the remaining constants are all close to zero, then the projection code is very likely correctly implemented. Because the above calculations shows that the constants must assume these values for that input spherical function.