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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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$(g_{ij})$ cannot be that matrix: it is a $2\times 2$ matrix! – Mariano Suárez-Alvarez Oct 4 '11 at 5:19
I essentially used the wikipedia article under "induced metric" to compute the metric. Perhaps I am not understanding something here? – kobebryant Oct 4 '11 at 5:50
Indeed, you are not understanding something: the matrix you are trying to compute is the matrix of coefficients of an inner product defined on the tangent spaces to the spheres, which are of dimension two. You should try a more systematic textbook, really. – Mariano Suárez-Alvarez Oct 4 '11 at 5:52
@kobebryant: I think you have your matrices the wrong way around. $(DF)$ should be a $4 \times 2$ matrix, so $(DF)^T (DF)$ should be a $2 \times 2$ matrix. – Zhen Lin Oct 4 '11 at 8:56
In that case, I presume that $z$'s are replaced by $\sqrt{1-x^{2}-y^{2}}$ to get a 4x2 matrix for $DF$? – kobebryant Oct 4 '11 at 15:23

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 $$ g_{ij}(0,0) = D \varphi ^T \cdot D \varphi|_{(0,0)} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$

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