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It is true that the restiction of the Laplace operator on $\mathbb R^n$ to functions on the sphere is the Laplacian for the round metric on the sphere. Is this true for any Riemannian metric $g$ on $\mathbb R^n$?

I mean, is it true that the restriction of $\Delta_g$ to functions on the sphere is the Laplacian on $S^{n-1}$ of the metric induced by $g$?

Thanks in advance.

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    $\begingroup$ This seems like a nice exercise for you to do yourself. $\endgroup$
    – Deane Yang
    Jul 14, 2011 at 3:17
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    $\begingroup$ Your first sentence is not correct. The Laplace operator on $R^n$ does not act on functions on the sphere, so I don't see what you mean by restriction. Once your reformulate the sentence you'll probably find that it's false, although something is true for homogeneous functions. For a general metric $g$ I don't quite see what simple statement could be correct. $\endgroup$
    – Jean-Marc Schlenker
    Jul 14, 2011 at 6:18
  • $\begingroup$ First, as Jean-Marc points out, you need to restate your question more precisely (and correctly). Second, your question applies to any hypersurface in a Riemannian manifold, and you can compare the two Laplacians by writing them with respect to local co-ordinates, where the hypersurface is the level set of one of the co-ordinates. $\endgroup$
    – Deane Yang
    Jul 14, 2011 at 11:45
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    $\begingroup$ Here's one possible statement to prove or refute: Given a hypersurface in a Riemannian manifold, consider a function that is constant along each geodesic normal to the hypersurface. The manifold Laplacian of this function, restricted to the hypersurface, is equal to the hypersurface Laplacian of the function restricted to the hypersurface. $\endgroup$
    – Deane Yang
    Jul 14, 2011 at 13:11
  • $\begingroup$ Dear Yang, I was able to prove your statement. I'm very grateful for your help. I'll try to post the solution in the following days. Thank you again. Yaiza. $\endgroup$
    – user16436
    Jul 15, 2011 at 5:08

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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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  • $\begingroup$ Hi Paul, thank you for your quick answer. I knew this already, it's what I wote in the first line. What I wanted to know is if it would hold for any other Riemannian metric on $R^n$. $\endgroup$
    – user16436
    Jul 14, 2011 at 5:19

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