# Do Spherical harmonics have continuous extensions to the entire sphere?

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?

• 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)"? Feb 17 at 20:39
• @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. Feb 17 at 20:46

For $$m=0$$, the term $$e^{im\phi}$$ is constant. For $$m\neq 0$$, the Legendre polynomial vanishes for $$x=\pm1$$
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.
• "So it's not a polynomial in just $\cos\theta$" - are you implying that the formula from the article is wrong? Feb 18 at 8:39
• 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 $m$. Feb 24 at 2:14