Pullback metric on sphere

Question

I am learning differential geometry, and wanted to see a calculation for the round (induced) metric on the sphere $S^n$.

To do this, I wanted to consider the immersion $\iota:S^n \rightarrow \mathbb{R}^{n+1}$, and consider the pullback formula for an immersion $\phi: M \rightarrow N$ and a Riemannian metric $g$ on $N$ given at each $p \in M$ by $$\phi^{*}g(v, w) = \langle \mathrm{d}\phi_p(v), \mathrm{d}\phi_p(w) \rangle$$ which in this case becomes $$\iota^{*}g(v, w) = \langle \mathrm{d}\iota_p(v), \mathrm{d}\iota_p(w) \rangle$$ and where the metric tensor is given on $\mathbb{R}^{n+1}$ by $g_{ij} = \delta_{ij}$.

I would appreciate seeing this calculation in full, since I don't know how to compute with pullbacks, and cannot find a reference. Thank you!

Answer 1

Consider the immersion $\varphi:S^n\hookrightarrow\mathbb{R}^{n+1}$ where we have;

$$\varphi(\theta_1,\theta_2,\cdots\theta_{n})= \begin{bmatrix} x_1 \\ x_2 \\ \vdots\\ x_{n+1} \end{bmatrix}$$

Where our coordinate components, $x_i$, are defined as;

$$x_1=\cos\theta_1$$ $$x_2=\cos\theta_2\sin\theta_1$$ $$\vdots$$ $$x_k=\cos\theta_k\prod_{m=1}^{k-1}\sin\theta_m$$ $$\vdots$$ $$x_{n+1}=\prod_{m=1}^{n}\sin\theta_m$$

As you have noted, the pull back of our metric will be;

$$\varphi^*g(v,w)=g(\varphi_*(v),\varphi_*(w))$$

Which will give us;

$$\hat{g}_{ij}=g_{ab}\frac{\partial x_a}{\partial\theta_i}\frac{\partial x_b}{\partial\theta_j}$$

The metric, $g_{ab}$, on $\mathbb{R}^{n+1}$ is simply equal to $\delta_{ab}$. Therefore;

$$\hat{g}_{ij}=\frac{\partial x_a}{\partial\theta_i}\frac{\partial x_a}{\partial\theta_j}$$

The diagonal elements of our tensor should be fairly easy for you to calculate on your own. You will find that;

$$\hat{g}_{11}=\bigg(\frac{\partial x_a}{\partial \theta_1}\bigg)^2=1$$ $$\hat{g}_{ii}=\bigg(\frac{\partial x_a}{\partial \theta_i}\bigg)^2=\prod_{m=1}^{i-1}\sin^2\theta_m$$

The calculation for the off-diagonal elements is a bit more involved and requires you to notice the following. Consider the $n=2$ dimensional case of our immersion and notice that the off-diagonal elements of our metric would be;

$$\hat{g}_{12}=\frac{\partial x_a}{\partial\theta_1}\frac{\partial x_a}{\partial\theta_2}=0-(\cos\theta_2\cos\theta_1)(\sin\theta_2\sin\theta_1)+(\sin\theta_2\cos\theta_1)(\cos\theta_2\sin\theta_1)=0$$

Because the metric is symmetric we then have $\hat{g}_{12}=\hat{g}_{21}=0$. So all of our off-diagonal elements are equal to $0$. While the $n$-dimensional case is certainly messier, the calculations themselves are trivial and you will arrive at the conclusion that for $i\neq j$ we have;

$$\hat{g}_{ij}=0$$

Leaving us with our final expression for the induced metric on $S^n$;

$$\hat{g}=d\theta_1\otimes d\theta_1+\sum_{i=2}^{n}\bigg[\prod_{m=1}^{i-1}\sin^2\theta_m\bigg]d\theta_i\otimes d\theta_i$$

Answer 2

Note that to do this calculation, you need a parameterization of $S^n$, also known as a coordinate chart (or more than one if you prefer). I suggest doing this calculation first for $S^2$ by using the parameterization given by spherical coordinates. Another one is to parameterize the upper hemisphere as a graph of a function.

In general, "pullback" means the following: Suppose you have a map $f: M \rightarrow N$, coordinates $(x^1, \dots, x^m)$ on $M$, and coordinates $(y^1, \dots, y^n)$ on $N$.You can use $f$ to treat each coordinate $y^k$ as a function of the coordinates $x^1, \dots, x^m$. The pullback of a differential form on $N$, written in terms of $y^1, \dots, y^n$ and $dy^1, \dots, dy^n$, is simply the same differential form but treating each $dy^k$ as the exterior derivative of the function $y^k$ in terms of $dx^1, \dots, dx^m$.