# Ngô Quốc Anh

## October 9, 2017

### Jacobian of the stereographic projection not at the North pole

Filed under: Uncategorized — Ngô Quốc Anh @ 18:57

Denote $\pi_P : \mathbb S^n \to \mathbb R^n$ the stereographic projection performed with $P$ as the north pole to the equatorial plane of $\mathbb S^n$. Clearly when $P$ is the north pole $N$, i.e. $N = (0,...,0,1)$, then $\pi_N$ is the usual stereographic projection.

As we have already known that, for arbitrary $\xi \in \mathbb S^n$, the image of $\xi$ is

$\displaystyle \pi_P : \xi \mapsto x = P+\frac{\xi-P}{1-\xi \cdot P}.$

Here the point $x \in \mathbb R^n$ is being understood as a point in $\mathbb R^{n+1}$ by adding zero in the last coordinate. For the inverse map, it is not hard to see that

$\displaystyle \pi_P^{-1} : x \mapsto \xi =\frac{|x|^2-1}{|x|^2+1}P+\frac 2{|x|^2+1}x.$

The purpose of this entry is to compute the Jacobian of the, for example, $\pi_P^{-1}$ by comparing the ratio of volumes.

First pick two arbitrary points $x, y \in \mathbb R^n$ and denote $\xi = \pi_P^{-1}(x)$ and $\eta = \pi_P^{-1}(y)$. The Euclidean distance between $\xi$ and $\eta$ is

$\displaystyle |\xi -\eta|^2 = \sum_{i=1}^{n+1} |\xi_i - \eta_i|^2 =\sum_{i=1}^n |\xi_i - \eta_i|^2+|\xi_{n+1} - \eta_{n+1}|^2.$

Note that

$\displaystyle \sum_{i=1}^n |\xi_i - \eta_i|^2=\sum_{i=1}^n 4\left[ -\Big(\frac 1{1+|x|^2}-\frac 1{1+|y|^2} \Big)p_i+ \frac{x_i}{1+|x|^2}-\frac{y_i}{1+|y|^2}\right]^2=4\Big(\frac 1{1+|x|^2}-\frac 1{1+|y|^2} \Big)^2 \sum_{i=1}^n p_i^2 - 8\Big(\frac 1{1+|x|^2}-\frac 1{1+|y|^2} \Big)\sum_{i=1}^n p_i \Big( \frac{x_i}{1+|x|^2}-\frac{y_i}{1+|y|^2} \Big) + 4 \sum_{i=1}^n\Big( \frac{x_i}{1+|x|^2}-\frac{y_i}{1+|y|^2} \Big)^2.$

As already computed in this entry, we know that

$\displaystyle \sum_{i=1}^n 4\left( \frac{x_i}{1+|x|^2}-\frac{y_i}{1+|y|^2}\right)^2=\frac{4|x|^2}{(1+|x|^2)^2} + \frac{4|y|^2}{(1+|y|^2)^2} - \frac{8\langle x,y \rangle}{(1+|x|^2)(1+|y|^2)}.$

Therefore,

$\displaystyle \sum_{i=1}^n |\xi_i - \eta_i|^2=4\Big(\frac 1{1+|x|^2}-\frac 1{1+|y|^2} \Big)^2 \sum_{i=1}^n p_i^2 - 8\Big(\frac 1{1+|x|^2}-\frac 1{1+|y|^2} \Big) \Big( \frac{\langle p, x \rangle}{1+|x|^2}-\frac{\langle p, y\rangle}{1+|y|^2} \Big) + \frac{4|x|^2}{(1+|x|^2)^2} + \frac{4|y|^2}{(1+|y|^2)^2} - \frac{8\langle x,y \rangle}{(1+|x|^2)(1+|y|^2)}.$

Finally, we have

$\displaystyle |\xi_{n+1} - \eta_{n+1}|^2 = 4 \left( \frac 1{1+|x|^2} - \frac 1{1+|y|^2} \right)^2p_{n+1}^2.$

Thus,

$\displaystyle |\xi -\eta|^2 =4 \left( \frac 1{1+|x|^2} - \frac 1{1+|y|^2} \right)^2- 8\Big(\frac 1{1+|x|^2}-\frac 1{1+|y|^2} \Big) \Big( \frac{\langle p, x \rangle}{1+|x|^2}-\frac{\langle p, y\rangle}{1+|y|^2} \Big) + \frac{4|x|^2}{(1+|x|^2)^2} + \frac{4|y|^2}{(1+|y|^2)^2} - \frac{8\langle x,y \rangle}{(1+|x|^2)(1+|y|^2)}.$

thanks to $|P|=1$. Again, from this entry, we obtain

$\displaystyle |\xi -\eta|^2 =\frac 4{(1+|x|^2)(1+|y|^2)}|x-y|^2- 8\Big(\frac 1{1+|x|^2}-\frac 1{1+|y|^2} \Big) \Big( \frac{\langle p, x \rangle}{1+|x|^2}-\frac{\langle p, y\rangle}{1+|y|^2} \Big).$

In other words, we have just shown that the Euclidean distance has been scaled by

$\displaystyle \frac2{\big((1+|x|^2)(1+|y|^2)\big)^{1/2}} \left( 1+ 2 \frac {|x|^2-|y|^2}{|x-y|^2} \Big( \frac{\langle p, x \rangle}{1+|x|^2}-\frac{\langle p, y\rangle}{1+|y|^2} \Big)\right)^{1/2}.$

Clearly this scaling factor has limit as $y \to x$. To find the limit, it suffices to find the limit of

$\displaystyle\frac {|x|^2-|y|^2}{|x-y|^2} \Big( \frac{\langle p, x \rangle}{1+|x|^2}-\frac{\langle p, y\rangle}{1+|y|^2} \Big).$

We rewrite this as follows

$\displaystyle\frac{{|x| + |y|}}{{(1 + |x{|^2})(1 + |y{|^2})}}\left\langle {p,\frac{{|x| - |y|}}{{|x - y{|^2}}}[x(1 + |y{|^2}) - y(1 + |\vec x{|^2}]} \right\rangle.$

Hence, we have to find the limit of

$\displaystyle{\frac{{|x| - |y|}}{{|x - y{|^2}}}[x(1 + |y{|^2}) - y(1 + |\vec x{|^2}]}$

as $y \to x$. If we write $x=(x_1,...,x_n,0)$ and $y=(y_1,...,y_n,0)$, then by sending $y_i \to x_i$ step by step, it is not hard to verify that

$\displaystyle\frac{{|x| - |y|}}{{|x - y{|^2}}}[x(1 + |y{|^2}) - y(1 + |\vec x{|^2}] \to \frac{1}{|x|}( - 2{x_1}x_n^2,..., - 2{x_{n - 1}}x_n^2,(|x{|^2} - 2x_n^2 + 1){x_n})$

as $y \to x$. Hence,

$\displaystyle\lim_{y \to x} \frac{{|\xi - \eta |}}{{|x - y|}} = \frac{2}{{1 + |x{|^2}}}\sqrt {1 + {{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^2}\left\langle {p',{x_n}( - 2{x_n}x'',|x{|^2} - 2x_n^2 + 1)} \right\rangle } = : \Phi$

where $z''$ and $z'$ denote the first $n-1$ and $n$ coordinates of $z$, respectively.

Hence by taking $y \to x$ implies that tangent vectors at $x$ are scaled by a (conformal) factor $\Phi$. Therefore, the standard metric on $\mathbb S^n$ is expressed in terms of $\pi_P^{-1}$ by

$\displaystyle g_{ij}(x) = \Phi^2 \delta_{ij}.$

Thus the Jacobian of $\pi_P^{-1}: x \mapsto \xi$ at the point $x$ is $\Phi^n$.

Corollary. If $P$ is the North pole $N$, then $p'$ is the zero vector. Consequently,

$\displaystyle \Phi = \frac 2{1+|x|^2}.$

Application. As a consequence of the above finding, for the conformal transformation $\varphi_{N,t} : \mathbb S^n \to \mathbb S^n$ given by

$\displaystyle \varphi_{N,t} : x \mapsto \pi^{-1} \big( t \pi ( x) \big)$

the Jacobian of its inverse $\varphi_{N,t}^{-1}$ is

$\displaystyle \det (d\varphi_{N,t}^{-1})=\det d\big(\pi^{-1} (t^{-1} \cdot) \big) t^{-n}\det d \big(\pi ( \cdot ) \big).$

Clearly,

$\displaystyle \det (d\varphi_{N,t}^{-1})=\left( \frac 2{1+t^{-2}|x|^2} \right)^n t^{-n} \left( \frac 2{1+|x|^2} \right)^{-n}$

and

$\displaystyle \det (d\varphi_{N,t}(\pi^{-1} (x)))=\left( \frac 2{1+t^2|x|^2} \right)^n t^n \left( \frac 2{1+|x|^2} \right)^{-n}.$