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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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    If somebody is interested in this question, she might find interesting the Proposition 22.1 in Shubin's book "Pseudodifferential operators and Spectral Theory", 2001 edition.2014-10-16

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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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    @LeonidKovalev: Ok, done. With the aid of [this page](http://mathworld.wolfram.com/SphericalCoordinates.html) I just checked that $$div_{\mathbb{R}^3}A=\underbrace{\frac{1}{r^2}\frac{\partial r^2A_r}{\partial r}}_{\text{radial part}} + div_{r\mathbb{S}^2}P(A), $$ where $P(A)$ is orthogonal projection onto the sphere and $div_{r\mathbb{S}^2}$ has been computed in polar coordinates by means of the formula $$div B=\frac{1}{\sqrt{g}}\frac{\partial}{\partial u^i}(\sqrt{g}B^i).$$ From this, as you suggested, the claim follows. This might not be very elegant but it surely works! Thank you!2012-06-27