Representing a Riemannian metric in $\mathbb R^3$ restricted to the upper half of $S^2$ Consider $M := \mathbb R^3$ as a smooth manifold with a Riemannian metric $g := \sum_{i=1}^3 dx^i\otimes dx^i$, where $(x^1, x^2, x^3)$ is the standard coordinate of $M$.  Let $N\subset M$ be a submanifold defined as $S^2 \cap H$, where $H = \{(x^1, x^2, x^3) \mid x^3 > 0\}$. Then $N$ has the coordinate $(x^1,x^2)$.
The problem is to represent the restriction of $g$ to the tangent bundle $TN$ w.r.t. the (global) frame $(dx^1, dx^2)$.
I tried to solve it in the following way: I introduced a coordinate neighborhood $(M \cap H , \psi)$ to $M$, where $\psi:M\cap H \rightarrow(0,\infty)\times(0,\pi)\times(0,\pi)$ is a diffeomorphism defined by $\psi^{-1}(r, \phi, \theta) = (r\cos\phi, r\sin\phi\cos\theta, r\sin\phi\sin\theta)$.  Then I write $g$ in terms of $dr$, $d\phi$ and $d\theta$. Since I get $g|_{TN^{\otimes 2}}$ by dropping the terms that involve $dr$ from $g$, I rewrite this in terms of $dx^1, dx^2$ by applying inverse mapping theorem to obtain the result.
While this seems to work in theory, it is a bit cumbersome.  I would be most grateful if you could help me solve this problem more easily.
 A: Let me elaborate on Avitus's comment. 


*

*Let $\iota : N \to \mathbb{R}^3$ be the canonical embedding. Then what you want to understand is actually $\iota^\ast g = \delta_{ij} \iota^\ast dx^i \otimes \iota^\ast dx^j$.

*As you observed, you actually have a diffeomorphism $\phi : N \to U := \{(y^1,y^2) \in \mathbb{R}^2 \mid (y^1)^2 + (y^2)^2 < 1\}$ given by $$\phi(x^1,x^2,x^3) := (x^1,x^2),$$ so that $T^\ast N$ admits the global frame $$\{\phi^\ast dy^1,\phi^\ast dy^2\}.$$ What you would like to do is express $\iota^\ast g$ in terms of $\phi^\ast dy^1$ and $\phi^\ast dy^2$, which involves expressing $\{\iota^\ast dx^1,\iota^\ast dx^2,\iota^\ast dx^3\}$ in terms of $\{\phi^\ast dy^1,\phi^\ast dy^2\}$.

*As Avitus observed, $\phi^{-1} : U \to N$ is given by
$$
\phi^{-1}(y^1,y^2) = \left(y^1,y^2,\sqrt{1-(y^1)^2-(y^2)^2}\right).
$$ You can readily check, then, that
$$
 (\iota \circ \phi^{-1})^\ast(dx^1) = dy^1, \quad (\iota \circ \phi^{-1})^\ast(dx^2) = dy^1,\\(\iota \circ \phi^{-1})^\ast(dx^3)
= -\frac{y^1}{\sqrt{1-(y^1)^2-(y^2)^2}} dy^1 -\frac{y^2}{\sqrt{1-(y^1)^2-(y^2)^2}} dy^2.
$$


Now, observe that by the basic properties of pullbacks, for any $\omega \in \Omega^1(\mathbb{R}^3)$,
$$
 \phi^\ast\left(\iota \circ \phi^{-1}\right)^\ast(\omega) = \left(\left(\iota \circ \phi^{-1}\right) \circ \phi\right)^\ast(\omega) = \iota^\ast(\omega).
$$
What does this imply, then, for $\iota^\ast dx^1$, $\iota^\ast dx^2$, $\iota^\ast dx^3$? What does this, in turn, imply for $\iota^\ast g$?
A: Here is a pedestrian approach:
Denote the coordinates in $N$ by $z=(z_1,z_2)$. Then we have a map
$$f:\quad N\to {\mathbb R}^3, \qquad (z_1,z_2)\mapsto\left(z_1,z_2,\sqrt{1-z_1^2-z_2^2}\right)\ .$$
One computes
$$df(z).e_1=f_{.1}(z)=\bigl(1,0,-z_1/\sqrt{1-z_1^2-z_2^2}\bigr),\quad 
df(z).e_2=f_{.2}(z)=\bigl(0,1,-z_2/\sqrt{1-z_1^2-z_2^2}\bigr)\ .\tag{1}$$
Now the fundamental form $g$ in ${\mathbb R}^3$ is just the scalar product. The matrix $\bigl[g^*_{ik}(z)\bigr]$ of the pullback $g^*$ on $N$ is therefore given by
$$\eqalign{g^*_{ik}(z)&=g^*(z).(e_i,e_k)= df(z).e_i\>\cdot\>df(z).e_k\cr &=f_{.i}(z)\cdot f_{.k}(z)\ .\cr}$$
Now plug in the expressions obtained in $(1)$ and obtain
$$\bigl[g^*_{ik}(z)\bigr]=\left[\matrix{{1-z_2^2\over 1-z_1^2-z_2^2} &{z_1z_2\over 1-z_1^2-z_2^2} \cr {z_1z_2\over 1-z_1^2-z_2^2}  & {1-z_1^2\over 1-z_1^2-z_2^2}  \cr}\right]\ .$$
