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Given I have the most primitive spherical harmonic, $Y_{00}(\theta, \phi)=\frac{1}{2} \frac{1}{\sqrt{\pi}}$ and I look at one of the three second most primitive ones, e.g. $Y_{11}(\theta, \phi)=-\frac{1}{2} \sqrt{\frac{3}{2 \pi}} \sin \theta e^{i \phi}$, it seems to me that when I look at their plot that I did with Mathematica,

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that I would be able to approximate $Y_{11}$ with $Y_{00}$ by putting two of them next to each other.

How can I find an (analytical?) solution that lets me express the higher order spherical harmonic in terms of two lower order spherical harmonics?

ste
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1 Answers1

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Slightly too long for an answer: it definitely depends on the norm you are using for your functions. If you are using the $L^2$ norm induced by round metric on the sphere, I have very bad news: the approximation is going to be very bad in general. This is because spherical harmonics are orthogonal with respect to this norm. Let us see a baby example, for instance approximate $Y_{10}$ by $Y_{00}$, and $$\|Y_{10}-\lambda Y_{00}\|^2= \|Y_{10}\|^2+\|Y_{00}\|\lambda^2 + \lambda \langle Y_{10},Y_{00}\rangle = \|Y_{10}\|^2+\|Y_{00}\|\lambda^2$$ Minimizing with respect to $\lambda$ gives that the best approximation is... $0\times Y_{00}$.

For your general case of approximating $Y_{11}$ it will be the same, because the set $\{Y_{lm}\}$ is orthonormal!

topolosaurus
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  • Thanks for the answer! The orthogonality argument makes sense. But I wonder: Isn't it different when the $Y_{00}$ is shifted? That they have no overlap when they have the same origin makes sense, but as far as I understand it, $\lambda$ only scales it but doesn't translate it, so maybe $Y_{00}$ could be made to approximate it better by shifting it away from the origin? – ste Apr 04 '22 at 17:47
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    Note that shifting $Y_{00}$ in particular does not change too much the picture: $Y_{00}$ is already a constant! $Y_{00} + c = Y_{00} (1+c/Y_{00})$. In general, if "the constants function" are not in your linearly approximating set, then you need to be more careful, definitely. This definitely depends a lot on the context, the fact that this is a Hilbert space of functions, etc.. – topolosaurus Apr 04 '22 at 17:58
  • At the danger of sounding silly, I have to ask another bit: I just asked Mathematica to Minimize[(SphericalHarmonicY[1, 0, t, p] - l * SphericalHarmonicY[0, 0, t, p])^2, l ∈ Reals] which yielded as optimal result $l=\sqrt{3} \cos \theta$, which let's the term go to $0$. From how I misunderstand this, $l \neq 0$. – ste Apr 04 '22 at 18:18
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    Well, you should be minimizing the integral of the square of the absolute value of this difference! Otherwise Mathematica is just minimizing w.r.t. $l$ so that the values of the functions at the particular (generic) point these values are as close as possible: that is a different problem! Can you see why?

    Definitely the $l$ should not be a function if you are approximating by linear combination (constant coefficients!) of functions. Otherwise you could approximate $Y_{11}$ by $Y_{00}$ by the "naive" $\frac{Y_{11}}{Y{00}} Y_{11}$!

     – topolosaurus Apr 04 '22 at 18:26