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I was wondering if someone could check if I'm doing this correctly. I am given the stereographic projection map from $S^3\backslash \{(0,0,0,1)\}\rightarrow R^3$ $u_i=\frac{x_i}{1-x_4}$ and asked to express the metric $\sum_{i=1}^4 dx_i^2$ in the form $f(u_1, u_2, u_3) (du_1^2+du_2^2+du_3^2)$. The work that I've done so far is below, but I dont seem to be getting the correct form. I was hoping someone could helpj me to see where I'm going wrong:

Let the variable $j$ range over the values $1,2,3$ and $i$ over the values $1,2,3,4$ so that \begin{equation} \sum x_i^2=1 \label{1} \end{equation} Given that \begin{equation} u_j=\frac{x_j}{1-x_4} \label{2} \end{equation} we find, upon combining (\ref{1}) and (\ref{2}), that $$\sum u_j^2=\frac{\sum x_i^2 -x_4^2}{(1-x_4)^2}=\frac{1+x_4}{1-x_4} $$ which implies \begin{equation} x_4=\frac{\sum u_j^2-1}{\sum u_j^2+1} =\frac{\sum u_j^2+1-2}{\sum u_j^2+1}=1-\frac{2}{\sum u_j^2+1} \label{3} \end{equation} Hence

\begin{align*} dx_4&=\sum \frac{\partial x_4}{\partial u_j }du_j\\ &=\frac{\sum 4u_jdu_j}{\left(\sum u_j^2+1\right)^2} \end{align*}

hence

\begin{align} dx_4^2&=\left(\frac{4\sum u_jdu_j}{\sum u_j^2+1}\right)^2\\ %&=\frac{16 \sum u_j^2du_j^2+32e_2(u_jdu_j)}{\left(\sum u_j^2+1 \right)^2} \label{5}\\ \end{align} Next note that (\ref{2}) gives \begin{align*} dx_j&=\frac{\partial x_j}{\partial u_j}du_j\\ &=(1-x_4)du_j \end{align*} so that \begin{align} dx_j^2&=(1-x_4)^2du_j^2 \label{4} \end{align} and, combining (\ref{3}) and (\ref{4}), we obtain \begin{align*} dx_j^2=\frac{4u_jdu_j^2}{\left( \sum u_j^2+1 \right)^2} \end{align*} finally, (\ref{5}) and (\ref{4}) imply \begin{align*} \sum dx_i^2 &=\sum dx_j^2+dx_4^2\\ &=\frac{4\sum u_jdu_j^2+\left( 4\sum u_jdu_j \right)^2}{\left( \sum u_j^2+1 \right)^2}\\ &=\frac{4\sum u_j du_j^2+16\sum u_jdu_j\sum u_j du_j}{\left(\sum u_j^2+1\right)^2}\\ &=\frac{4\sum u_j du_j^2+16\left( \sum u_j^2du_j^2 +2e_2(u_j du_j) \right)}{\left(\sum u_j^2+1\right)^2}\\ %&= \frac{\sum 4u_jdu_j+16 \sum {u_j}^2du_j^2+32e_2(u_jdu_j)}{\left( \sum u_j^2+1 \right)^2}\\ %&= \frac{\sum 4u_jdu_j\left(1+4 \sum {u_j}^2du_j^2 \right)+32e_2(u_jdu_j)}{\left( \sum u_j^2+1 \right)^2} \end{align*} where $e_2(\alpha_j)$ denotes the elementary symmetric polynomial consisting of all products of distinct pairs of the symbols $\alpha_i$ (e.g. $e_2(\alpha_1, \alpha_2, \alpha_3)=\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3$)

This last expression is nasty and seems incorrect. Any help would be greatly appreciated. Thanks!

user140776
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  • When you turn to the computation of $dx_j$ for $j\ne4$, note that $x_j=(1-x_4)u_j$ hence $$dx_j=(1-x_4)du_j-u_jdx_4$$ instead of $dx_j=(1-x_4)du_j$, hence $$dx^2=\sum_{j=1}^4dx_j^2$$ is actually $$dx^2=dx_4^2+(1-x_4)^2\sum_{j=1}^3du_j^2+\sum_{j=1}^3u_j^2dx_4^2-2(1-x_4)\sum_{j=1}^3u_jdu_jdx_4$$ Now use $$dx_4=(1-x_4)^2dv$$ with the shorthand $$dv=\sum_{j=1}^3u_jdu_j$$ and the identity already in your post saying that $$\sum_{j=1}^3u_j^2=\sum_{j=1}^3\frac{x_j^2}{(1-x_4)^2}=\frac{1-x_4^2}{(1-x_4)^2}=\frac{1+x_4}{1-x_4}$$ to deduce ... – Did Dec 14 '16 at 10:25
  • $$dx^2=(1-x_4)^4dv^2+(1-x_4)^2\sum_{j=1}^3du_j^2+(1-x_4^2)(1-x_4)^2dv^2-2(1-x_4)^3dv^2$$ which simplifies into $$dx^2=(1-x_4)^2\sum_{j=1}^3du_j^2=\ldots$$ – Did Dec 14 '16 at 10:25

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