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In $\mathbb{R}^n$, in terms of polar coordinates $(r,\theta)$ where $r>0$ and $\theta\in S^{n-1}$, we have the following formula: $$\Delta_{\mathbb{R}^n}=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\Delta_{S^{n-1}}.$$ To prove it, you can first try to prove it when $n=2$: When $n=2$, $(x,y)=(r\cos\theta, r\sin\theta)$...I think you can fill out the details. So the answer to your question is yes when $g$ is Euclidean. |
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In $\mathbb{R}^n$, in terms of polar coordinates $(r,\theta)$ where $r>0$ and $\theta\in S^{n-1}$, we have the following formula: $$\Delta_{\mathbb{R}^n}=\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\Delta_{S^{n-1}}.$$ To prove it, you can first try to prove it when $n=2$: When $n=2$, $(x,y)=(r\cos\theta, r\sin\theta)$...I think you can fill out the details. So the answer to your question is yes. |
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