20
$\begingroup$

The metric on the $n$-sphere is the metric induced from the ambient Euclidean metric. Find the metric, $d\Omega^2_n$, on the $n$-sphere and the volume form, $\Omega_{S_n}$ , of $S^n$ in terms of the stereographic coordinates on $U_N =S^n − (0, . . . , 0, 1)$.

The stereographic projection coorinates $u_j$ are given by $u_j=\frac{x_j}{1-x_{n+1}}$ for $j=1,\dots,n$.

We also have $x_j=\frac{2u_j}{1+\sum_{k=1}^nu_k^2}$ and $x_{n+1}=\frac{1-\sum_{k=1}^nu_k^2}{1+\sum_{k=1}^nu_k^2}$

I took differentials and put them into the expression for the Euclidean metric put things are getting messy so I'm not sure if it's right. I also have to find the components of the Levi-Civitta connection, curvature tensor, Ricci tensor.

$\endgroup$
1
  • 1
    $\begingroup$ Your approach is correct, and should be tractable. To manage the algebra, it may help to give a name (such as "$|u|^{2}$") to $\sum_{k} u_{k}^{2}$. Alternatively, you might work out the case $n = 1$ by hand, then observe that stereographic projection is equivariant with respect to the action of $SO(n)$ on the orthogonal complement of the $x_{n+1}$-axis. $\endgroup$ Mar 21, 2015 at 14:46

2 Answers 2

15
$\begingroup$

Let's find the metric on the sphere in stereographic projection. We are going to give the chart and its inverse for $S^n(R)$, the sphere centre at the origin and radius $R$. The chart $\varphi:S^n(R)\subset\mathbb{R}^{n+1} \to \mathbb{R}^n$ is given by \begin{equation} \varphi(\xi,\tau)=\left(\frac{R\xi}{R-\tau}\right), \end{equation} where $\xi=(\xi_1,\ldots,\xi_n)\in\mathbb{R}^n, \tau\in \mathbb{R}$, $|\cdot|$ is the euiclidean norm, and $\xi, \tau$ are such that $|\xi|^2 + |\tau|^2=R^2$.

Its inverse map is the parametrization $\varphi^{-1}:\mathbb{R}^n\to S^n(R)$ given by \begin{equation} \varphi^{-1}(u)=\left(\frac{2R^2u}{|u|^2+R^2},R\frac{|u|^2-R^2}{|u|^2+R^2}\right), \end{equation} with $u=(u_1,\ldots,u_n)\in\mathbb{R}^n$.

Let $V\in T_u\mathbb{R}^n$ a tangent vector at $u$. Then it can be written as \begin{equation} V = \sum_{k=1}^{n}V(u_k)\left.\frac{\partial}{\partial u_k}\right|_u \end{equation}

Now, since $d_u\varphi^{-1}:T_u\mathbb{R}^n\to S^n(R)$ we compute: \begin{equation} d_u\varphi^{-1}(V)= \sum_{k=1}^{n}V(u_k) d_u\varphi^{-1}\left(\left.\frac{\partial}{\partial u_k}\right|_u\right), \end{equation} since $d_u\varphi(\partial/\partial u_k|_u)\in T_pS^n\subset\mathbb{R}^{n+1}$ then, is linear combination of the $\{\partial/\partial \xi_j,\partial/\partial\tau\}$, we omit the $u$ subscript, we have: \begin{align*} d_u\varphi^{-1}\left( \frac{\partial}{\partial u_k} \right)&= \sum_{j=1}^n d_u\varphi^{-1}\left(\frac{\partial}{\partial u_k}\right)(\xi_j)\frac{\partial}{\partial \xi_j} + d_u\varphi^{-1}\left(\frac{\partial}{\partial u_k}\right)(\tau)\frac{\partial}{\partial\tau}\\ &= \sum_{j=1}^n \frac{\partial}{\partial u_k}(\xi_j\circ\varphi^{-1})\frac{\partial}{\partial \xi_j} + \frac{\partial}{\partial u_k}(\tau\circ\varphi^{-1})\frac{\partial}{\partial\tau},\\ \end{align*} now, from the above formula we have \begin{align*} d_u\varphi^{-1}(V)&= \sum_{k=1}^{n}V(u_k) \sum_{j=1}^n \frac{\partial}{\partial u_k}(\xi_j\circ\varphi^{-1})\frac{\partial}{\partial \xi_j} +\sum_{k=1}^{n}V(u_k) \frac{\partial}{\partial u_k}(\tau\circ\varphi^{-1})\frac{\partial}{\partial\tau}\\ &= \sum_{j=1}^{n}\sum_{k=1}^n V(u_k)\frac{\partial}{\partial u_k}(\xi_j\circ\varphi^{-1})\frac{\partial}{\partial \xi_j} +\sum_{k=1}^{n}V(u_k) \frac{\partial}{\partial u_k}(\tau\circ\varphi^{-1})\frac{\partial}{\partial\tau}\\ &= \sum_{j=1}^{n}V(\xi_j\circ\varphi^{-1})\frac{\partial}{\partial \xi_j} +V (\tau\circ\varphi^{-1})\frac{\partial}{\partial\tau}\\ \end{align*}

We are going to abuse on the notation and we'll write $V_k$ instead of $V(u_k)$ and also we write $V(\xi_j)$ instead of $V(\xi_j\circ \varphi^{-1})$. Then the tensor metric is $d\Omega_n^2 = \sum_{j=1}^n V(\xi_j)^2 + V(\tau)^2$. Then \begin{align*} V(\xi_j)&=\sum_{k=1}^n V_k\frac{\partial }{\partial u_k}\left(\frac{2R^2u_j}{|u|^2+R^2}\right)\\ &=\sum_{k=1}^n V_k\left(\frac{2R^2\delta_{jk}}{|u|^2+R^2}- \frac{4R^2u_ju_k}{(|u|^2+R^2)^2}\right)\\ &=\frac{2R^2V_j}{|u|^2+R^2}- \frac{4R^2u_j}{(|u|^2+R^2)^2}\sum_{k=1}^n V_k u_k \\ &=\frac{2R^2V_j}{|u|^2+R^2}- \frac{4R^2u_j}{(|u|^2+R^2)^2}\langle V, u\rangle \\ \end{align*} where we used that $\delta_{jk}$ equals $1$ if $j=k$ and vanished otherwise. In the same way we have \begin{align*} V(\tau)&=\sum_{k=1}^n V_k\frac{\partial }{\partial u_k}\left(R\frac{|u|^2-R^2}{|u|^2+R^2}\right)\\ &=\sum_{k=1}^n V_k\left(\frac{2Ru_{k}}{|u|^2+R^2}- \frac{2R(|u|^2-R^2)u_k}{(|u|^2+R^2)^2}\right)\\ &=\frac{2R}{|u|^2+R^2}\langle V,u\rangle- \frac{2R(|u|^2-R^2)}{(|u|^2+R^2)^2}\langle V,u\rangle\\ &=\frac{2R(|u|^2+R^2)-2R(|u|^2-R^2)}{(|u|^2+R^2)^2}\langle V,u\rangle\\ &=\frac{4R^3}{(|u|^2+R^2)^2}\langle V,u\rangle\\ \end{align*}

Now we are able to compute the metric: \begin{align*} d\Omega_n^2 &= \sum_{j=1}^n V(\xi_j)^2 + V(\tau)^2\\ &= \sum_{j=1}^n \left(\frac{2R^2V_j}{|u|^2+R^2}- \frac{4R^2u_j}{(|u|^2+R^2)^2}\langle V, u\rangle\right)^2+ \left(\frac{4R^3}{(|u|^2+R^2)^2}\langle V,u\rangle\right)^2\\ &=\sum_{j=1}^n\left\{\frac{4R^4V_j^2}{(|u|^2+R^2)^2}- \frac{16R^4V_ju_j}{(|u|^2+R^2)^3}\langle V,u\rangle+ \frac{16R^4u_j^2}{(|u|^2+R^2)^4}\langle V,u\rangle^2\right\}+ \frac{16R^6}{(|u|^2+R^2)^4}\langle V,u\rangle^2\\ &=\frac{4R^4|V|^2}{(|u|^2+R^2)^2}- \frac{16R^4}{(|u|^2+R^2)^3}\langle V,u\rangle^2+ \frac{16R^4|u|^2}{(|u|^2+R^2)^4}\langle V,u\rangle^2+ \frac{16R^6}{(|u|^2+R^2)^4}\langle V,u\rangle^2\\ &=\frac{4R^4|V|^2}{(|u|^2+R^2)^2} \end{align*}

The we conclude that in stereographic projection \begin{equation} d\Omega_n^2 (d\varphi^{-1}(V))= \frac{4R^4}{(|u|^2+R^2)^2}ds^2(V) \end{equation} where $ds^2$ stands for the stantdard metric in $\mathbb{R}^n$.

This calculation can be found in John M. Lee's Riemannian Manifolds. An introduction to Curvature Chap. 3.

$\endgroup$
2
  • $\begingroup$ For the last equality, the last 3 terms drop because $<V,u>=0$ correct? Why is this the case? $\endgroup$
    – Mr. Brown
    Dec 20, 2020 at 2:39
  • 1
    $\begingroup$ Hi, @ZackFox. We do not necessarily have $\langle V, u \rangle$ = 0. Just factorise $16R^4$ form the last two terms and it cancels with the second term. $\endgroup$
    – Daniel B.
    Feb 6, 2021 at 13:55
9
$\begingroup$

Let me present the calculation using a slightly different, in my opinion a bit clearer, notation:

  1. (Definition of stereographic coordinates) Let $\newcommand{\bs}[1]{\boldsymbol{#1}}\newcommand{\bfx}{\mathbf{x}}\newcommand{\bfu}{\mathbf{u}}(\bfx,x_{n+1})\in S^n$ be an element of the unit $n$-sphere, where $\bfx\in\mathbb R^n$, and define the stereographic coordinates via the map $\varphi:S^n\to\mathbb R^n$ such that $$\bfu\equiv\varphi(\bfx) \equiv \frac{\bfx}{1-x_{n+1}}.$$ Note that I'm assuming unit radius here. This map is a diffeomorphism on $S^n\setminus\{N\}$ (the sphere without the north pole), and its inverse $\phi:\mathbb R^n\to S^n$ defines the parametrisation: $$\phi(\bfu) = \left(\frac{2\bfu}{1+u^2}, \frac{1-u^2}{1+u^2}\right),$$ where $u^2\equiv\|\bfu\|^2\equiv\sum_{i=1}^n u_i^2$. The intuition is that if we draw a line from the north pole $N=(\mathbf 0,1)\in S^n$ towards the point $(\mathbf x,x_{n+1})\in S^n$, the intersection of this line with the $x_{n+1}=0$ plane is at $\varphi(\mathbf x)\in\mathbb{R}^n$.

  2. (Overall objective) We want to compute the metric on $\mathbb R^n$ that corresponds to the Euclidean flat metric on $S^n$. In other words, let $\bs\xi,\bs \xi'\in T_p S^n$ be tangent vectors to the sphere, for some $p\in S^n$, and denote with $\langle \bs\xi,\bs\xi'\rangle_p$ the corresponding inner product defined in the usual (Euclidean) way. We want to exploit $\phi$ to see what this inner product/metric corresponds to in the parameter space.

    More explicitly, this means to take $\bs\eta,\bs\eta'\in T_{\bfu}\mathbb R^n$ and compute their inner product using $$\langle\bs\eta,\bs\eta'\rangle_{\bfu} = \langle \mathrm d\phi(\bfu,\bs\eta), \mathrm d\phi(\bfu,\bs\eta')\rangle,$$ where the inner product in the RHS is here just the standard Euclidean product in the $\mathbb R^{n+1}$ space in which $S^n$ is embedded, and $\mathrm d\phi(\bfu,\bs\eta)\equiv\mathrm d\phi(\bfu)(\bs\eta)\equiv\partial_{\bs\eta}\phi(\bfu)$ denotes the differential of $\phi$ at the point $\bfu$ in the direction $\bs\eta$.

  3. (Calculation of differential of the parametrisation) Consider then two tangent vectors in the parameter space, $\bs\eta,\bs\eta'\in T_{\bfu}\mathbb R^n\simeq\mathbb R^n$, and observe that $\mathrm d\phi:T\mathbb R^n\simeq\mathbb R^n\times\mathbb R^n\to TS^n$, where the differential can be represented as $$\mathrm d\phi(\bfu):\mathbb R^n\to T_\bfu S^n: \bs\eta\mapsto \sum_i \eta_i \partial_i \phi(\bfu), \\ \partial_1\phi(\bfu) = \frac{1}{(1+u^2)^2} \left( 2(1+u^2)-(2u_1)^2, -4 u_1 u_2, ... , -4 u_1 u_n, -4u_1 \right), $$ and thus more generally, $$ \partial_i\phi(\bfu) = \frac{2}{1+u^2} \mathbf e_i - \frac{4 u_i}{(1+u^2)^2} (\bfu,1), \\ \mathrm d\phi(\bfu,\bs\eta) = \frac{2}{1+u^2} (\bs\eta, 0) - \frac{4\langle\bfu,\bs\eta\rangle}{(1+u^2)^2} (\bfu,1), $$ where $\langle\bfu,\bs\eta\rangle\equiv\sum_{i=1}^n u_i \eta_i$, and $\bs\eta\equiv\sum_{i=1}^n \eta_i \mathbf e_i$.

  4. (Final result) From the above, a brief calculation shows that only the first factor in the expression of $\mathrm d\phi(\bfu,\bs\eta)$ contributes, and the result is: $$\langle\bs\eta,\bs\eta'\rangle_{\bfu} = \frac{4}{(1+u^2)^2} \langle\bs\eta,\bs\eta'\rangle \equiv \frac{4}{(1+u^2)^2} \sum_{i=1}^n \eta_i \eta'_i.$$ The end result is thus that the metric remains Euclidean, up to a (position-dependent) multiplicative factor. We thus say that the metric is conformal. This result is also often expressed concisely saying that the metric is $$\frac{4}{(1+u^2)^2}\sum_{i=1}^n {\mathrm du}_i^2.$$

  5. (Shorter derivation) A less formal but more direct approach to obtain the same result is to just consider the relation between $\boldsymbol u$ and $\boldsymbol x$, and write $$dx_k = \frac{2(1+u^2)du_k - 2u_k d(u^2)}{(1+u^2)^2}, \\ dx_{n+1} = \frac{-(1+u^2)d(u^2)-(1-u^2) d(u^2)}{(1+u^2)^2} = -2\frac{d(u^2)}{(1+u^2)^2},$$ and thus $$dx_{n+1}^2 + \sum_k dx_k^2 = \frac{4 (d(u^2))^2 + 4 (1+u^2)^2\sum_k du_k^2-4(1+u^2)(d(u^2))^2+4u^2(d(u^2)^2)}{(1+u^2)^4} \\ = \frac{4}{(1+u^2)^2} \sum_k du_k^2.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .