# Differential operators on the sphere

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. – Giuseppe Negro Oct 16 '14 at 12:29

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.
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
@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