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The sphere $\mathbb{S}^2$ is a Riemannian submanifold of the Euclidean space $\mathbb{R}^3$ and as such comes equipped with an array of differential operators, particularly gradient, divergence and Laplace-Beltrami. Can we compute them in terms of the corresponding Euclidean operators? Specifically:

Let $f$ be a smooth function and $\mathbb{A}$ a smooth vector field on the unit sphere. Denote $\tilde{f}, \tilde{\mathbf{A}}$ the smooth function and vector field on $\mathbb{R}^3 \setminus \{O\}$ defined by the identity

$$\tilde{f}(x)=f\left(\frac{x}{\lvert x \rvert}\right),\ \tilde{\mathbf{A}}(x)=\mathbf{A}\left( \frac{x}{\lvert x \rvert}\right).$$

Is it true that

  1. $\mathrm{grad}_{\mathbb{S}^2} f(y)=\mathrm{grad}_{\mathbb{R}^3} \tilde{f}(y)$;
  2. $\mathrm{div}_{\mathbb{S}^2}\mathbf{A}(y)=\mathrm{div}_{\mathbb{R}^3} \tilde{\mathbf{A}}(y)$;
  3. $\Delta_{\mathbb{S}^2}f(y)=\Delta_{\mathbb{R}^3}\tilde{f}(y)$;

for all $y \in \mathbb{S}^2$?


(secondary)

More generally, if $\mathbf{T}$ is a tensor field on $\mathbb{S}^2$ and $\tilde{\mathbf{T}}(x)=\mathbf{T}(x/\lvert x \rvert)$ is the corresponding tensor field on $\mathbb{R}^3\setminus\{O\}$, is there any relationship similar to the ones above between the covariant derivative $\nabla^{(\mathbb{S}^2)}_X \mathbf{T}$ and the Euclidean derivative of $\tilde{\mathbf{T}}$?

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Generally, considering f in coordinates $f=f(r,\xi)$, where $r$ is a radius and $\xi$ a point on the unit sphere $$(\Delta_{\mathbb{R}^{n+1}}\widetilde{f})|_{S^n}=\Delta_{S^n}f-\frac{\partial^2 \widetilde{f}}{\partial r^2}|_{S^n}-n\frac{\partial \widetilde{f}}{\partial r}|_{S^n}$$

In your case $\widetilde{f}(r,\xi)=f(\frac{r\xi}{\xi})=f(\xi)$. So the equality for the Laplacian should hold. I am not sure about the others.

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Also here: en.wikipedia.org/wiki/… –  user31373 Jun 26 '12 at 16:49
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Works for the gradient too, because in general the gradient can be found by taking an arbitrary extension of $f$ and then projecting the gradient of extension back to the tangent space. Here the extension is such that its gradient is already in the tangent space... The divergence is dual to gradient via integration by parts, so it works as well. –  user31373 Jun 26 '12 at 16:53
    
Thank you both for the illuminating answers and references. @LeonidKovalev: Do you mind elaborating a bit on the following statement? "The divergence is dual to gradient via integration by parts, so it works as well". I cannot understand how you could use integration to help us here. –  Giuseppe Negro Jun 26 '12 at 17:06
    
@LeonidKovalev: (I'm rewriting this comment only because I suspect the previous one wasn't reported to you.) Do you mind elaborating a bit on the statement of yours "The divergence is dual to gradient via integration by parts, so it works as well"? If you do that in a separate answer I could also upvote it. –  Giuseppe Negro Jun 27 '12 at 9:58
    
@GiuseppeNegro I'm not sure that integration by parts is really a good way to prove this. Perhaps it's best to calculate $\mathrm{div}_{\mathbb R^3}$ in spherical coordinates and observe that the normal direction does not contribute. –  user31373 Jun 27 '12 at 13:33
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