Riemannian metric on the sphere $S^2$ using a parametrization

Question

How does one go about deriving the Riemannian metric on the sphere $S^2$? I was reading Lee's book and he explains quite clearly that if we consider $S^2$ embedded in $\Bbb R^3$ we can use the inclusion $\iota : S^2 \to \Bbb R^3$ to pullback the Euclidean metric $g_E$ to get a metric $g = \iota^\ast g_E$ on $S^2$, but I could not find an explicit formulae for this.

It is later in the book noted that computations on a submanifold $M \subset \widetilde{M}$ are usually carried out most conveniently in terms of a smooth local parametrization $X : U \to \widetilde{M}$ where $X(U)$ is open in $M$. He denotes $\varphi = X^{-1}:X(U) \to U$ and states that the metric $g$ is then $$X^*g = X^*\iota^*\widetilde{g}=(\iota \circ X)^*\widetilde{g}$$ where $\widetilde{g}$ is the metric on $\widetilde{M}$.

Say I want to use this to find a metric for $S^2$. I can parametrize $S^2$ with $$X(\theta,\varphi) = (\cos\theta\sin\varphi,\sin\theta\sin\varphi,\cos\varphi)$$ where $0\le \theta \le 2\pi$ and $0\le \varphi \le \pi$.

It is not entirely clear to me now how do I find the metric given by this parametrization?

Answer 1

Whenever you have a tensor field $T$ on a manifold $M$ (let's say twice-covariant, for simplicity), and coordinates $(x^1,\ldots, x^n)$ on some open subset of $M$, the components of $T$ relative to this coordinate system are defined by $$T_{ij} = T\left(\frac{\partial}{\partial x^i} ,\frac{\partial}{\partial x^j}\right),\qquad \mbox{ for }i,j=1,\ldots,n,$$and we have $$T = \sum_{i,j=1}^n T_{ij}\,{\rm d}x^i\otimes {\rm d}x^j.$$

Here, $M = \Bbb S^2$, $T = \iota^*g_E$, and the coordinates $(\theta,\varphi)$ are given by the inverse of the parametrization $X$. By taking derivatives, we see that $$\iota_*\left(\frac{\partial}{\partial \theta}\bigg|_{X(\theta,\varphi)}\right) = \frac{\partial X}{\partial\theta}(\theta,\varphi) = -\sin\theta\sin\varphi \frac{\partial}{\partial x}\bigg|_{X(\theta,\varphi)} + \cos\theta\sin\varphi\frac{\partial}{\partial y}\bigg|_{X(\theta,\varphi)}, $$and so $$(\iota^*g_E)\left(\frac{\partial}{\partial\theta},\frac{\partial}{\partial\theta}\right) = g_E\left(\frac{\partial X}{\partial\theta},\frac{\partial X}{\partial \theta}\right) = \sin^2\varphi.$$This means that $$\iota^*g_E = \sin^2\varphi\,{\rm d}\theta \otimes {\rm d}\theta + \cdots.$$

You can compute $\iota_* (\partial/\partial \varphi) = \partial X/\partial\varphi$ and the remaining components and inner products like above.

Answer 2

The other answer is great, but here's a pedestrian version of the underlying nuts-and-bolts of the calculation since you asked "How do I find the metric given by this parametrization?" This is how I would find it.

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With $\iota \circ X = (x, y, z)$, i.e. $$ \left\{ \begin{aligned} x &= \sin\varphi\cos\theta \\ y &= \sin\varphi\sin\theta \\ z &= \cos\varphi, \end{aligned} \right. $$ we can calculate differentials: $$ \left\{ \begin{aligned} \mathrm{d}x &= \cos\varphi\cos\theta\,\mathrm{d}\varphi - \sin\varphi\sin\theta\,\mathrm{d}\theta \\ \mathrm{d}y &= \cos\varphi\sin\theta\,\mathrm{d}\varphi + \sin\varphi\cos\theta\,\mathrm{d}\theta \\ \mathrm{d}z &= -\sin\varphi\,\mathrm{d}\varphi. \end{aligned} \right. $$

Hence, $$ \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2 = \cdots (\text{trig identities}) \cdots = \sin^2 \varphi\,\mathrm{d}\theta + \mathrm{d}\varphi^2, $$ which is the pullback of the euclidean metric $(\iota \circ X)^* g_E$ that you asked about, i.e. the symmetric rank $2$ tensor $$ \begin{pmatrix} \sin^2 \varphi & 0 \\ 0 & 1 \end{pmatrix}. $$