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This article contains the following formula for the spherical harmonics: $$Y_l^m(\theta,\phi) = \sqrt{\frac{(2l+1)}{4\pi}\frac{(l-m)!}{(l+m)!}} P_l^m(\cos \theta) e^{im\phi}$$ Now let $S$ be the unit sphere in three dimensions. Clearly, the function\begin{align}F\colon\mathbb R\times\mathbb R&\to S\\(\theta,\phi)&\mapsto\begin{pmatrix}\sin\theta\cos\phi\\\sin\theta\sin\phi\\\cos\theta\end{pmatrix}\end{align} to the sphere is surjective, but it is necessary that $F(\theta,\phi)=F(x,y)$ implies that $Y(\theta,\phi)=Y(x,y)$ so that we can define $Y$ on $S$.

For example, consider the north pole $N=(0,0,1)$. Clearly, $F(0,\phi)=N$ for all $\phi\in\mathbb R$, but in general the function $$\mathbb R\ni\phi\mapsto Y(0,\phi)$$is not constant, is it?

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  • $\begingroup$ Could you elaborate more on what you mean by "$\Phi(\theta,\phi)=\Phi(x,y)$ does not imply that $Y(\theta,\phi)=Y(x,y)$ (consider the north pole and the south pole of the sphere)"? $\endgroup$ Feb 17 at 20:39
  • $\begingroup$ @JackyChong Let's consider the north pole $N=(0,0,1)$. Clearly, $\Phi(0,\phi)=N$ for all $\phi\in\mathbb R$, but the function $\mathbb R\ni\phi\mapsto Y(0,\phi)$ is not constant. Thus, there is no obvious way to define $Y$ at the north pole and the south pole. Not only that, but I think there is no continuous extension to the entire sphere. $\endgroup$
    – Filippo
    Feb 17 at 20:46

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For $m=0$, the term $e^{im\phi}$ is constant. For $m\neq 0$, the Legendre polynomial vanishes for $x=\pm1$

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If we look at the formulae for spherical harmonics given in Wikipedia, we see that whenever the $m$ parameter is nonzero, the $\theta$ dependence contains factors of $\sin\theta$ (so it's not a polynomial in just $\cos\theta$), which multiply the nonconstant $\exp(im\phi)$ factor by zero at the poles. Thus no discontinuity exists.

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  • $\begingroup$ "So it's not a polynomial in just $\cos\theta$" - are you implying that the formula from the article is wrong? $\endgroup$
    – Filippo
    Feb 18 at 8:39
  • $\begingroup$ The formula is fine, but calling it a polynomial is not strictly correct. If we render $x=\cos\theta$, then the expression as a function of $x$ contains a factor $\sin^n\theta=(1-x^2)^{m/2}$ which does not fot the definition of a polynomial for all whole numbers $m$. $\endgroup$ Feb 24 at 2:14
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    $\begingroup$ That's because not all associated Legendre polynomials are polynomials, isn't it? $\endgroup$
    – Filippo
    Feb 24 at 8:25

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