Consider a function $f:S^2 \to \mathbb{R}$ , with $S^2$ the unit $2$-sphere in $\mathbb{R}^3$. Let's say that $f$ depends only on the polar angle $\theta$ from the north pole (e.g., $f(r,\theta,\phi) = e^{\cos(\theta)} $ which is also just $e^z$ in Cartesian cordinates). Now, we define the Laplacian on the sphere by

$$\Delta f = \frac{1}{\sin(\theta)} \frac{\partial}{\partial \theta} \sin(\theta) \frac{\partial f}{\partial \theta} + \frac{1}{\sin^2(\theta)} \frac{\partial^2 f}{\partial \phi^2}$$

where $\phi$ is the azimuthal angle in the $xy$ plane. Now, another way of looking at $f$ is saying that its value at $p \in S^2$ depends on the norm distance between the point $p_0 = (0,0,1)$ and $p$, since $||p-p_0|| = \sqrt{||p||^2 -2(p,p_0) + ||p_0||^2} = \sqrt{2-2 \cos(\theta)}$, where we just used the statement $a \cdot b = |a| |b| \cos(\theta)$. Hence, we can view $f$ as a function depending on the distance from $p$ to $p_0$. From here on, consider a chosen function $f$ whose form we know, and whose value at a point $p$ depends only on the cosine of the angle between a fixed chosen point $p_0$ (the north pole) and $p$.

With this, we can then compute $\Delta f$, and we note that all of the $\phi$ terms are automatically zero, so $\Delta f$ can be computed completely in terms of the $\theta$ derivatives.

Recall that we saw that $f$ depends on ly on $\theta$, or equivalently, $f$ depends only on $||p-p_0||$, that is $f(p) = f(||p-p_0||)$. This gives rise to many different "rotated" versions of $f$ if I move $p_0$ from the north pole to somewhere else. I want to investigate the following question: How can we take the Laplacian of $f$ if we vary $p_0$? Does it have a simple form and how does it related to the Laplacian of $f(||p-p_0||)$ when $p_0$ is the north pole?

So, let's consider this "rotated" version of $f$. That is, consider a point $p_0$ besides $(0,0,1)$ such as maybe $(1,0,0)$. Now, instead of depending on the angle between $p$ and the north pole, it depends on the angle $\theta'$ between $p$ and this shifted point $p_0$. In some sense, we are essentially rotating the sphere and assigning this new $p_0$ to be the north pole.

Here is what I want to do: I want to compuate $\Delta f$ for this new shifted $f$ where $f$ depends on this new $\theta'$. I have been told that I can just do the following: since $f(\theta')$ depends only on the polar distance, the Laplacian of $f$ is the same as taking the orignal $f$ with $p_0$ being the north pole, taking the Laplacian of that function (which only depends on $\theta$), which yields some function $g(\theta) = \Delta f$. Now, if I use a different point $p_0$ than the north pole, the claim is that $\Delta f(||p-p_0||) = g(\theta')$. That is, its the same function, just with a swap of the variable from $\theta$ to $\theta'$. Therefore, to compute $\Delta f(||p-p_0||)$ for any different $p_0$, it suffices to compute the simple case of $\Delta f(\theta)$ by

$$ g(\theta) = \Delta f(\theta) = \frac{1}{\sin(\theta)} \frac{\partial }{\partial \theta} \sin(\theta) \frac{\partial f}{\partial \theta}$$ and then swapping that $\theta$ with the different $\theta'$ depending on the new $p_0$. That seems reasonable to me, but I'm just worried that it looks like on a rotated version of the function, the azimuthal derivatives will impact it. Certainly if I view things in terms of $\theta$ and $\phi$, there will be $\phi$ components. The assertion I suppose is that in this rotated setting, $\Delta f(\theta')$ still depends only on $\theta'$ and has the same form as $\Delta f(\theta)$ as if it were centered at the north pole. Is this true?

I apologize if this is confusing or trivial!


1 Answer 1


I've ben informed that on the sphere, the Laplacian is rotation invariant. Therefore, the claim that was given to me in the original question appears to be true. I also found the following .pdf that explains in more detail : http://www.math.umn.edu/~garrett/m/mfms/notes/11_spheres.pdf See p.5 section 2 on "The Existence of the Spherical Laplacian" in these lecture notes on Harmonic Analysis on the Sphere.


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