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}.

1 Comment »

  1. […] via Jacobian of the stereographic projection not at the North pole — Ngô Quốc Anh […]

    Pingback by Jacobian of the stereographic projection not at the North pole — Ngô Quốc Anh – Alabair's Weblog — January 1, 2019 @ 13:59


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