Ngô Quốc Anh

August 9, 2011

Stereographic projection, 5

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 14:39

I put here several common formulas that are needed when we work on stereographic projection. First we try to calculate $\displaystyle\Delta \left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{n - 2}}} \right).$

We do step by step.

Step 1. A direct computation leads us to $\displaystyle\begin{gathered} {\partial _i}\left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{n - 2}}} \right) = (n - 2){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{n - 3}}{\partial _i}\left( {\frac{2}{{1 + |x{|^2}}}} \right) \hfill \\ \qquad= (n - 2){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{n - 3}}\frac{{ - 4{x_i}}}{{{{(1 + |x{|^2})}^2}}}. \hfill \\ \end{gathered}$

Therefore, $\displaystyle\begin{gathered} \partial _i^2\left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{n - 2}}} \right) = {\partial _i}\left[ {(n - 2){{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{n - 3}}\frac{{ - 4{x_i}}}{{{{(1 + |x{|^2})}^2}}}} \right] \hfill \\ \qquad= - 4(n - 2){\partial _i}\left[ {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{n - 3}}\frac{{{x_i}}}{{{{(1 + |x{|^2})}^2}}}} \right] \hfill \\ \qquad= - 4(n - 2){\partial _i}\left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{n - 3}}} \right)\frac{{{x_i}}}{{{{(1 + |x{|^2})}^2}}} - 4(n - 2){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{n - 3}}{\partial _i}\left( {\frac{{{x_i}}}{{{{(1 + |x{|^2})}^2}}}} \right). \hfill \\ \end{gathered}$

Observe that $\displaystyle {\partial _i}\left( {\frac{{{x_i}}}{{{{(1 + |x{|^2})}^2}}}} \right) = \frac{{{{(1 + |x{|^2})}^2} - 4x_i^2(1 + |x{|^2})}}{{{{(1 + |x{|^2})}^4}}} = \frac{1}{{{{(1 + |x{|^2})}^2}}} - \frac{{4x_i^2}}{{{{(1 + |x{|^2})}^3}}}.$

Thus, $\displaystyle\begin{gathered} \partial _i^2\left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{n - 2}}} \right) = 16(n - 2)(n - 3){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{n - 4}}\frac{{x_i^2}}{{{{(1 + |x{|^2})}^4}}} \hfill \\ \qquad- 4(n - 2){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{n - 3}}\frac{1}{{{{(1 + |x{|^2})}^2}}} + 16(n - 2){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{n - 3}}\frac{{x_i^2}}{{{{(1 + |x{|^2})}^3}}} \hfill \\ \qquad= (n - 2)(n - 3)x_i^2{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^n} - (n - 2){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{n - 1}} + 2(n - 2)x_i^2{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^n} \hfill \\ \qquad= (n - 2)(n - 1)x_i^2{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^n} - (n - 2){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{n - 1}}. \hfill \\ \end{gathered}$

Step 2. Summarize all to get $\displaystyle\begin{gathered} \Delta \left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{n - 2}}} \right) = (n - 2)(n - 1)|x{|^2}{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^n} - (n - 2)n{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{n - 1}} \hfill \\ \qquad= (n - 2){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^n}\left( {(n - 1)|x{|^2} - \frac{n}{2}(1 + |x{|^2})} \right) \hfill \\ \qquad= (n - 2){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^n}\left( {\left( {\frac{n}{2} - 1} \right)|x{|^2} - \frac{n}{2}} \right). \hfill \\ \end{gathered}$

Now we try to calculate $\displaystyle\nabla \left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{\frac{{n - 2}}{2}}}} \right).$

We have $\displaystyle\begin{gathered} {\partial _i}\left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{\frac{{n - 2}}{2}}}} \right) = \frac{{n - 2}}{2}{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{\frac{{n - 4}}{2}}}{\partial _i}\left( {\frac{2}{{1 + |x{|^2}}}} \right) \hfill \\ \qquad= \frac{{n - 2}}{2}{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{\frac{{n - 4}}{2}}}\frac{{ - 4{x_i}}}{{{{(1 + |x{|^2})}^2}}}. \hfill \\ \end{gathered}$

In other words, $\displaystyle\nabla \left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{\frac{{n - 2}}{2}}}} \right) = - \frac{{n - 2}}{2}{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{\frac{n}{2}}}x.$

Consequently, $\displaystyle {\left| {\nabla \left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{\frac{{n - 2}}{2}}}} \right)} \right|^2} = {\left( {\frac{{n - 2}}{2}} \right)^2}{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^n}|x{|^2}.$

Interestingly $\displaystyle\begin{gathered} {\left| {\nabla \left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{\frac{{n - 2}}{2}}}} \right)} \right|^2} - \frac{1}{2}\Delta \left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{n - 2}}} \right) \hfill \\ \qquad= {\left( {\frac{{n - 2}}{2}} \right)^2}{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^n}|x{|^2} - \frac{{n - 2}}{2}{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^n}\left( {\left( {\frac{n}{2} - 1} \right)|x{|^2} - \frac{n}{2}} \right) \hfill \\ \qquad= \frac{{n - 2}}{2}{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^n}\left( {\frac{{n - 2}}{2}|x{|^2} - \left( {\frac{n}{2} - 1} \right)|x{|^2} + \frac{n}{2}} \right) \hfill \\ \qquad= \frac{{n(n - 2)}}{2}{\left( {\frac{2}{{1 + |x{|^2}}}} \right)^n}. \hfill \\ \end{gathered}$