Riemannian metric induced by immersion 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 $g_{ij}(0,0)$?
 A: 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
$$
g_{ij}(0,0) = D \varphi ^T \cdot D \varphi|_{(0,0)} =
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
$$
