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In $n$-dimensional spherical coordinates, the gradient of a real valued function $f : \mathbb{R}^n \rightarrow \mathbb{R}$ can be represented by $\mathrm{grad} f = \left( \dfrac{\partial f}{\partial r}, \dfrac{1}{r} \nabla_\theta f \right) $, where $$ \nabla_\theta f = \dfrac{\partial f}{\partial \theta_1} \vec{\theta_1} + \dfrac{1}{\sin \theta_1} \dfrac{\partial f}{\partial \theta_2} \vec{\theta_2} + \dfrac{1}{\sin \theta_1 \sin \theta_2} \dfrac{\partial f}{\partial \theta_3} \vec{\theta_3} + \cdots + \dfrac{1}{\sin \theta_1 \cdots \sin \theta_{n-2}} \dfrac{\partial f}{\partial \theta_{n-1}} \vec{\theta_{n-1}}.$$

On the other hand, let us consider the unit sphere $S^{n-1} \subset \mathbb{R}^n$ with the usual metric. (Pullback of the Euclidean metric on $\mathbb{R}^n$.) I guess that $\nabla_\theta f$ is the gradient of a restricted function $f \vert_{S^{n-1}}$ on the sphere, but I do not know how to check it. Please give any advice.

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  • $\begingroup$ Could you please give a reference for this formula for $\nabla_{\mathbb{S}^{n-1}}$. Thanks. $\endgroup$
    – Medo
    Dec 28, 2019 at 6:14

1 Answer 1

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Let $(g_{ij})$ denote the usual metric on $\mathbb{R}^{n+1}$. Consider the parameterization of $\mathbb{R}^{n+1}\setminus\{0\}$ given by $\vec{x}:\mathbb{R}^+\times\mathbb{S}^n\to\mathbb{R}^{n+1}$, $\vec{x}(r,\sigma)=r\sigma$. In these polar coordinates, we have that $$(g_{ij})=\renewcommand{\arraystretch}{1.5} \left(\begin{array}{@{}c|c@{}} \,\,1 & \begin{matrix} 0 & \cdots & 0 \end{matrix} \\ \hline \begin{matrix} 0 \\ \vdots \\ 0 \end{matrix} & \begin{matrix} r^2\left(g_{ij}^{\mathbb{S}^n}\right) \end{matrix} \end{array}\right),$$ where $\left(g_{ij}^{\mathbb{S}^n}\right)$ is the metric on $\mathbb{S}^{n}\subseteq\mathbb{R}^{n+1}.$ You should show this if you haven't shown this before! Note that the inverse matrix is given by $$(g^{ij})=\renewcommand{\arraystretch}{1.5} \left(\begin{array}{@{}c|c@{}} \,\,1 & \begin{matrix} 0 & \cdots & 0 \end{matrix} \\ \hline \begin{matrix} 0 \\ \vdots \\ 0 \end{matrix} & \begin{matrix} \frac{1}{r^2}\left(g^{ij}_{\mathbb{S}^n}\right) \end{matrix} \end{array}\right),$$ where $\left(g^{ij}_{\mathbb{S}^n}\right)$ is the inverse of the metric of $\mathbb{S}^n$.

Now recall that the gradient is defined as $\nabla f = (df)^\sharp$. In local coordinates, that is $$\nabla f = g^{ij}\frac{\partial f}{\partial x^j}\frac{\partial}{\partial x^i},$$ where in the above I used the Einstein summation notation. Now we compute the gradient of $\mathbb{R}^{n+1}$ in polar coordinates: \begin{align*} \nabla_{\mathbb{R}^{n+1}}f = \frac{\partial f}{\partial r}\frac{\partial}{\partial r}+\frac{1}{r^2}\sum_{i=1}^{n}\sum_{j=1}^{n}g^{ij}_{\mathbb{S}^n}\frac{\partial f}{\partial \theta_j}\frac{\partial}{\partial\theta_i} = \frac{\partial f}{\partial r}\frac{\partial}{\partial r} + \nabla_{\mathbb{S}^n(r)}f. \end{align*} Note that my result is different from yours by a factor of $\frac{1}{r^2}$, this is because the surface gradient on $\mathbb{S}^n(r)$ differs from the surface gradient on $\mathbb{S}^n(1)$ by this factor. If we were to write the surface gradient in coordinates, and normalize our tangent vectors, we would obtain the exact same representation you have in your question.

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  • $\begingroup$ Then are you assuming that $\mathbb{S}^n$ is a sphere with radius $r$? $\endgroup$
    – MKim
    Jul 20, 2017 at 1:43
  • $\begingroup$ Yes, I should have written it in terms of the gradient on $\mathbb{S}^n(r)$. I made a mistake in my last equality since $\frac{\partial}{\partial\theta_i}$ is not constant as a function of $r$, and so I can't simply write $\nabla_{\mathbb{S}^n}f$. When I write it in terms of the gradient on $\mathbb{S}^n(r)$, the factor of $\frac{1}{r^2}$ gets absorbed into the surface gradient. I'm just about to edit my answer. $\endgroup$ Jul 20, 2017 at 2:17
  • $\begingroup$ Great, I'm glad to hear it :) $\endgroup$ Jul 20, 2017 at 5:42

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