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Let $F:S^{2}\rightarrow\mathbb{R}^{4}$ be the immersion defined as $(x^{2}-y^{2},xy,xz,yz)$ and consider the metric on $S^{2}$ induced by $F$. Find $g_{ij}(0,0)$ for the upper hemisphere parameterization $\phi(x,y)=(x,y,\sqrt{1-x^{2}-y^{2}})$

For the metric on $S^{2}$ induced by $F$, we can explicity determine $g_{ij}$ (this is just finding $(DF)^{T}(DF)$): $ g_{ij}=\left(\begin{array}{cccc} 4x^{2}+4y^{2} & 0 & 2xz & 0\\ 0 & y^{2}+z^{2} & yz & xz\\ 2xz & yz & x^{2}+z^{2} & xy\\ -2yz & xz & xy & z^{2}+y^{2}\end{array}\right)$

I am wondering for the parameterization $\phi$, would I just substitute $z$ by $\sqrt{1-x^{2}-y^{2}}$ and calculuate $g_{ij}(0,0)$?

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    In that case, I presume that $z$'s are replaced by $\sqrt{1-x^{2}-y^{2}}$ to get a 4x2 matrix for $DF$?2011-10-04

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As we see, everything is defined over $U:=\{(x,y)\in \mathbb{R}^2 | x^2 + y^2 < 1\}$, and we are asked to find $ g_{ij}(x,y)|_{(x,y)=(0,0)} $ For that it is necessary to figure out what is the immersion. The parametrization identifies the upper hemisphere with $U$, and we have really two maps $ f:U \rightarrow \mathbb{R}^3 : (x,y) \mapsto (x,y,z=\sqrt{1-x^2-y^2}) $ and $ F:\mathbb{R}^3 \rightarrow \mathbb{R}^4 : (x,y,z) \mapsto (x^{2}-y^{2},xy,xz,yz) $ Hence we have a map $\varphi = F \circ f$ explicitly given by $ \varphi: U \rightarrow \mathbb{R}^4 : (x,y) \mapsto (x^2 - y^2, xy, x \sqrt{1-x^2-y^2}, y \sqrt{1-x^2-y^2}) $ which is the actual immersion: $ \varphi_x = (2x,y,\sqrt{1-x^2-y^2} - \frac{x^2}{\sqrt{1-x^2-y^2}}, \frac{-xy}{\sqrt{1-x^2-y^2}})=|_{(0,0)}=(0,0,1,0) $ $ \varphi_y = (-2y,x, \frac{-xy}{\sqrt{1-x^2-y^2}} , \sqrt{1-x^2-y^2} - \frac{y^2}{\sqrt{1-x^2-y^2}})=|_{(0,0)}=(0,0,0,1) $ In particular, $D \varphi |_{(0,0)}= \begin{pmatrix} 0 & 0 \\ 0 & 0 \\ 1 & 0 \\ 0 & 1 \end{pmatrix} $ and therefore $ g_{ij}(0,0) = D \varphi ^T \cdot D \varphi|_{(0,0)} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $