1

Consider usual local coordinates $(\theta, \varphi)$ em $S^2 \subset \mathbb{R}^3$ defined by the parametrization $\phi:(0, \pi) \times (0, 2 \pi) \rightarrow \mathbb{R}^3$ given by $$\phi(\theta, \varphi) = (\sin \theta \cos \varphi, \sin \theta \sin \varphi, \cos \theta)$$ Using those coordinates determine the expression of the riemannian metric induced over $S^2$ by the euclidian metric of $\mathbb{R}^3$.

Conceptually the riemannian metric will associate each point $p \in S^2$ to a product $\langle \cdot,\cdot \rangle_p$ in its tangent space $T_pM$. I think that what the question means by metric induced by $\mathbb{R}^3$ is taking $\langle \cdot,\cdot \rangle_p$ to be the usual internal product in $\mathbb{R}^3$. I'm having trouble with getting an explicit expression for the metric.

José
  • 953
  • Have you tried expressing the basis vectors $\partial_{\varphi},\partial_{\theta}$ in terms of $\partial_{x},\partial_{y},\partial_{z}$? – Johnny El Curvas Feb 14 '21 at 18:32
  • Doing that I get $\partial_\varphi = -(\sin \theta \sin \varphi) \partial_x + (\sin \theta \cos \varphi) \partial_y$ and $\partial_\theta = (\cos \theta \cos \varphi) \partial_x + (\cos \theta \sin \varphi) \partial_y - (\sin \theta) \partial_z$. I have no idea where I should go from here. – José Feb 14 '21 at 18:44
  • You're almost there! Just evaluate the metric using bilinearity. – Johnny El Curvas Feb 14 '21 at 18:45
  • Do you mean I should evaluate the product $\langle \partial_\theta, \partial_\varphi \rangle$ over the $\partial_x, \partial_y, \partial_z$ coordinates? – José Feb 14 '21 at 18:54
  • 1
    Yes. Formally, you would write $g(\partial_{\theta},\partial_{\phi})=g_{0}(dj(\partial_{\theta}),dj(\partial_{\phi}))$, where $g$ is the metric in the sphere and $g_{0}$ is the Euclidean metric. – Johnny El Curvas Feb 14 '21 at 18:56

2 Answers2

2

Essentially, the metric will have the form $${\rm d}s^2 = E(\theta,\varphi){\rm d}\theta^2 +2F(\theta,\varphi){\rm d}\theta\,{\rm d}\varphi + G(\theta,\varphi){\rm d}\varphi^2,$$where $$E(\theta,\phi)=\left\langle\frac{\partial\phi}{\partial \theta}(\theta,\varphi),\frac{\partial\phi}{\partial \theta}(\theta,\varphi)\right\rangle,\quad F(\theta,\varphi)=\left\langle\frac{\partial\phi}{\partial \theta}(\theta,\varphi),\frac{\partial\phi}{\partial \varphi}(\theta,\varphi)\right\rangle,\quad\mbox{and}\quad G(\theta,\varphi)=\left\langle\frac{\partial\phi}{\partial \varphi}(\theta,\varphi),\frac{\partial\phi}{\partial \varphi}(\theta,\varphi)\right\rangle,$$where these inner products are computed with the ambient metric. This is a general mechanism: if $(M^n,g)$ is a Riemannian manifold, $\iota:S \to M$ is a submanifold, and $(u^1,...,u^k)$ are coordinates for $S$, then $\iota^*g$ is described in these coordinates as $$\sum_{i,j=1}^k a_{ij}\,{\rm d}u^i\,{\rm d}u^j,$$where $$a_{ij} = g\left({\rm d}\iota\left(\frac{\partial}{\partial u^i}\right),{\rm d}\iota\left(\frac{\partial}{\partial u^j}\right)\right).$$

Ivo Terek
  • 77,665
  • +1 for writing the final answer in terms of the parametrization itself. Unrelated question: how do you put that straight $"d"$ for the differential? – Johnny El Curvas Feb 14 '21 at 19:00
  • '{\rm d}' renders as ${\rm d}$. Supposedly this is a bad LaTeX practice and one should write '\mathrm{d}' instead, but I don't have the patience for that nor to create a macro just for the answer. – Ivo Terek Feb 14 '21 at 19:29
1

Let $(\theta,\varphi)$ be the coordinates defined by $\phi^{-1}$, and $(x,y,z)$ be the standard coordinates in $\mathbb{R}^{3}$, so that the euclidean metric is

$$g_{0}=dx\otimes dx+dy\otimes dy+dz\otimes dz$$

We identify, for $p\in \mathbb{S}^{2}$, $T_{p}\mathbb{S}^{2}$ as a subspace of $T_{p}\mathbb{R}^{3}$ (via the differential $dj_{p}$ of the inclusion). Now if $U$ is the domain of $(\theta,\varphi)$, and $p\in U$, we have

$$ \dfrac{\partial}{\partial \theta}=\dfrac{\partial x}{\partial \theta}\dfrac{\partial}{\partial x}+\dfrac{\partial y}{\partial \theta}\dfrac{\partial}{\partial y}+\dfrac{\partial z}{\partial \theta}\dfrac{\partial}{\partial z} \\ \dfrac{\partial}{\partial \varphi}=\dfrac{\partial x}{\partial \varphi}\dfrac{\partial}{\partial x}+\dfrac{\partial y}{\partial \varphi}\dfrac{\partial}{\partial y}+\dfrac{\partial z}{\partial \varphi}\dfrac{\partial}{\partial z} $$

Since $x=\sin \theta \cos \varphi$, $y=\sin \theta \sin \varphi$, $z=\cos \theta$, we get:

$$ \dfrac{\partial}{\partial \theta}=\cos \theta \cos \varphi\dfrac{\partial}{\partial x}+\cos \theta \sin \varphi\dfrac{\partial}{\partial y}-\sin \theta\dfrac{\partial}{\partial z} \\ \dfrac{\partial}{\partial \varphi}=-\sin \theta \sin \varphi\dfrac{\partial}{\partial x}+\sin \theta \cos \varphi\dfrac{\partial}{\partial y} $$

Finally, $g(v,w)=g_{0}(v,w)$ for $v,w\in T_{p}\mathbb{S}^{2}$. Since the basis vectors of $T_{p}\mathbb{R}^{3}$ are orthonormal:

$$ g_{\theta \theta}=\cos^{2}\theta \cos^{2}\varphi+\cos^{2}\theta \sin^{2}\varphi+\sin^{2}\theta=1 \\ g_{\theta \varphi}=g_{\varphi \theta}=-\cos \theta \cos \varphi \sin \theta \sin \varphi+\cos \theta \cos \varphi \sin \theta \sin \varphi=0 \\ g_{\varphi \varphi}=\sin^{2}\theta\sin^{2}\varphi+\sin^{2}\theta\cos^{2}\varphi=\sin^{2}\theta. $$

Hope this helps!

Johnny El Curvas
  • 1,592