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!
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.