# Laplace-Beltrami operators on differently scaled spheres: are they the same?

This question arose in the attempt of checking "chain rule" for Laplace-Beltrami operator on spheres.

Let $\Delta_{(S^{n-1}(R),g(r))}$ be the Laplace-Beltrami operator on the $(n-1)$-dimensional sphere $S^{n-1}(R)\subset \mathbb{R}^{n}$ with radius $R$, with respect to the round metric $g(r)$ with radius $r$. I often omit $g(r)$ when $R=r$.

Define $i_R\colon S^{n-1}(1)\to S^{n-1}(R)$ by $i_R(\xi):=R\xi$, and let $$(i_R^* f)(\xi):=f(i_R(\xi))$$ be the pullback of $f$ by $i_R$.

## How can we show this?

A related question Finding relationship between Laplace-Beltrami operators of two spheres starts with the formula $\Delta_{S(R)} f(x) = R^2\Delta f\left(R\frac{x}{|x|}\right)$, but I don't know where this came from (I have seen this when $R=1$ and plausible), and I would like to start with the local coordinate expression with the metric.

Context, if you are interested

I explain why I want to see if (*) is true. My goal really is to check the above question in the link: $$\Delta_{(S^{n-1}(1),g(1))}(i_R^* f)(\xi)=R^2 \Delta_{(S^{n-1}(R),g(R))}f(x)|_{x=R\xi},\tag{\star}.$$

We concentrate on the case $n=3$. Fix $R>0$ arbitrarily. First, find the local expression of the Riemannian metric for $S^{2}(R)$ inherited from $\mathbb{R}^n$ in spherical coordinates.

$S^{2}(R)$ can be parametrized by $(\theta,\phi)$ as $$x=(R\sin\theta \cos\phi,R\sin\theta\sin\phi,R\cos\theta)\in S^{2}(R),$$ at least locally. Six or so spherical caps would be enough to cover the whole sphere. Here, let $0<\theta<\pi$, $0<\phi<2\pi$.

To compute the matrix giving the Riemannian metric in this chart, we need to compute a basis $(e_1(\theta,\phi),e_2(\theta,\phi))$ of the tangent plane space $T_pS^{2}(R)$ at $p(\theta,\phi)=(R\sin\theta \cos\phi,R\sin\theta\sin\phi,R\cos\theta)$.

Let us use \begin{align} e_1(\theta,\phi)&:=\frac{\partial p}{\partial \theta} =(R\cos{\theta}\cos{\varphi}, R\cos{\theta}\sin{\varphi}, -R\sin{\theta}),\\ e_2(\theta,\phi)&:=\frac{\partial p}{\partial \phi}= ( -R\sin{\theta}\sin{\varphi} , R\sin{\theta}\cos{\varphi}, 0). \end{align} Indeed, $e_1(\theta,\phi)$ and $e_1(\theta,\phi)$ are linearly independent: we have $\|e_1\|=R^2\neq 0$ and $\langle e_1, e_2\rangle=0$, but since we avoided the north/south poles we also have $\|e_2\|=R^2\sin^2\theta\neq0$.

The metric on $T_pS^{2}(R)$ with respect to $(e_1,e_2)$ is given by $$g(R)(p)=\begin{bmatrix} R^2 & 0\\0 &R^2\sin^2\theta\end{bmatrix},$$ because $g(R)_{11}=\langle e_1,e_1\rangle=R^2$, $g(R)_{12}=g(R)_{21}=\langle e_1, e_2\rangle=0$, and $g(R)_{22}=R^2\sin^2\theta$.

Thus, $(g(R)(p))^{-1}=\begin{bmatrix} 1/R^2 & 0\\0 &1/(R^2\sin^2\theta)\end{bmatrix}$.

Now, using this chart the Laplace-Beltrami operator on $(S^{2}(R),g(R))$ is given by \begin{align} \Delta_{S^{2}(R)}&=\frac{1}{\sqrt{|g(R)|}} \bigg[\frac{\partial}{\partial \theta}\bigg(\sqrt{|g(R)|}(g(R))^{11}\frac{\partial}{\partial \phi}\bigg)+0+0+ \frac{\partial}{\partial \phi}\bigg(\sqrt{|g(R)|}(g(R))^{22}\frac{\partial}{\partial \phi}\bigg)\bigg]\\ % &=\frac{1}{R^2\sin\theta} \bigg[\frac{\partial}{\partial \theta}\bigg(R^2\sin\theta\,\frac1{R^2}\frac{\partial}{\partial \phi}\bigg)+ \frac{\partial}{\partial \phi}\bigg(R^2\sin\theta\,\frac1{R^2\sin\theta}\frac{\partial}{\partial \phi}\bigg)\bigg]\\ &=\frac{1}{R^2\sqrt{|g(1)|}} \bigg[\frac{\partial}{\partial \theta}\bigg(\sqrt{|g(1)|}(g(1))^{11}\frac{\partial}{\partial \phi}\bigg)+ \frac{\partial}{\partial \phi}\bigg(\sqrt{|g(1)|}(g(1))^{22}\frac{\partial}{\partial \phi}\bigg)\bigg]=\frac1{R^2}\Delta_{(S^{n-1}(R),g(1))},\\ \end{align} where $0$'s in the first equality comes from $(g(R))^{12}=0=(g(R))^{21}$, and $\Delta_{(S^{n-1}(R),g(1))}$ is the Laplace-Beltrami operator on $(S^{n-1}(R),g(1))$.

Thus, for any smooth function on $S^{n-1}(R)$ we have $\Delta_{(S^{n-1}(R),g(1))}f=R^2\Delta_{S^{n-1}(R)}f$.

So, if we have (*) then we obtain $$R^2\Delta_{S^{n-1}(R)}f = \Delta_{(S^{n-1}(R),g(1))}f(x)=\Delta_{S^{n-1}(1)}(i_R^*f)(\xi),$$ which is ($\star$).

I know for the Euclidean metric $\bar{g}$ we have $$(i^{-1}_R)^*\bar{g}=(d(x_1/R))^2+(d(x_2/R))^2+(d(x_3/R))^2=d\theta^2+\sin^2\theta d\phi^2=g(1),$$ and thus $i_R^{-1}\colon (S^{2}(R),g(1))\to (R^{3},\bar{g})$ is an isometry. But I am not really used to these arguments and I am not too sure if this can be/cannot be related to "$\Delta_{(S^{n-1}(R),g(1))}f(x)=\Delta_{(S^{n-1}(1)}(i_R^*f)(\xi)$".