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I want to show $g_r=r^2g_1$ where $g_1$ is the (Riemannian) metric in the unit sphere induced by its inclusion in $\mathbb{R}^n$ and $g_r$ is the metric in the sphere of radius $r$ also induced by its inclusion in $\mathbb{R}^n$. I gave the following argument but it seems rather unelegant (maybe wrong?): $$ g_r = \sum_{i=1}^n \left.dx^i \otimes dx^i \right|_{\lVert x \rVert = r} = \sum_{i=1}^n r^2 \left. (x^i/r) \otimes (dx^i/r) \right|_{\lVert x \rVert = r} = \sum_{i=1}^nr^2 \left. dy_i\otimes dy_i \right|_{\lVert y \rVert=1} =r^2g_1$$ Is this an acceptable answer? What would be a more elegant way?

Note: the equality $g_r=r^2g_1$ is taken in the sense that an element of the sphere of radius $r$ and of radius 1 is determined only by its polar coordinates (by a direction coming from the origin). So a vector orthogonal to radial unit vector in $\mathbb{R}^n$ defines a vector in every tangent space of every sphere. That is, if we use polar coordinates and we write a vector field in $\mathbb{R}^n\setminus \{0\}$ as $v=v_r\partial_r + v_{\bar{\theta}} \partial _\bar{\theta}$ the quantity $v_{\bar{\theta}} \partial _\bar{\theta}$ defines a vector field in every sphere.

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Technically, $g_{1}$ is a metric on the unit sphere $S^{n-1}(1)$ and $g_{r}$ is a metric on the sphere $S^{n-1}(r)$ of radius $r$, so they're not immediately comparable.

What you're proven is: If $\phi:S^{n-1}(1) \to S^{n-1}(r)$ is radial scaling by a factor of $r$ (i.e., $\phi(x) = rx$), and if $g_{r}$ is the metric induced on $S^{n-1}(r)$ by the ambient Euclidean metric, then $\phi^{*}g_{r} = r^{2} g_{1}$. (Note that both of these metrics "live" on the unit sphere.)

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  • $\begingroup$ If I understand your explanation, then under the same logic, $g_r=(\phi^{-1})^*g_1$ right? Is there a better way to prove this fact? $\endgroup$ – Mauricio G Tec May 26 '15 at 15:38
  • $\begingroup$ 1. Yes, $g_{r} = r^{2} (\phi^{-1})^{*}g_{1}$. 2. It may be easier to avoid using a coframe field, and simply to note (using angle brackets for the ambient metric and putting $\psi = \phi^{-1}$) that if $x$, $y$ are tangent vectors to $S^{n-1}(r)$, then$$\psi^{*}g_{1}(x, y) = g_{1}(\psi_{*}x, \psi_{*}y) = \langle x/r, y/r\rangle = \tfrac{1}{r^{2}}\langle x, y\rangle = \tfrac{1}{r^{2}} g_{r}(x, y).$$ $\endgroup$ – Andrew D. Hwang May 26 '15 at 15:56

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