# Is convolution in spherical harmonics equivalent to multiplication in the spatial domain?

Spherical harmonic convolution is defined as:

$$( k \star f )^l_m = \sqrt{ \frac{ 4 \pi }{2l+1} } h^l_m f^l_m$$

I have a function with RGB values for every $(\theta,\phi)$ in the spatial domain. Call this the "light function":

I project this "light function" to spherical harmonics.

Now, I want to use this lobe function:

to "select" a portion of the "light function", in the SH domain.

Note that the lobe is circularly symmetric.

In the spatial domain, this "selection" could be done by a simple multiplication: for every $(\theta,\phi)$ simply multiply. The result will be "keep anything that lines up with the lobe, and discard the rest."

Multiplication in the spatial domain should be convolution in the frequency domain.

But when I try to convolve the spherical harmonic "lobe" with the spherical harmonic light function, I am getting a "scaled" or "filtered" copy of the light map, but not in the way I expect! I expect to keep everything along the lobe, and discard the rest, as described above.

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## 2 Answers

That assumption is only true if one of the harmonics is ZONAL

I.e. every component with m not equal to 0, is 0

so then the convolution is with h being the zonal harmonic

$( k \star f )^l_m = \sqrt{ \frac{ 4 \pi }{2l+1} } h^l_0 f^l_m$

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No, convolution in spherical harmonics is not multiplication in the spatial domain. According to this paper, convolution in spherical harmonics corresponds with low pass filtering.

Just as the Fourier basis is convenient for examining the results of convolutions in the plane, similar tools exist for understanding the results of the analog of convolutions on the sphere. We now introduce these tools, and use them to show that in producing reflectance, k acts as a low-pass filter.
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But isn't low-pass filtering a form of convolution? I don't see how your point follows. – Rahul May 4 '12 at 22:19