Hi am reading An introduction to smooth manifold by Loring and have 1 doubt in example 9.6. It is written that
In this chart, the set $U_p\cap S^2$ is defined by the vanishing of the first coordinate $f$.
My doubts are as follows:
- How do we know that the in this chart the set $U_p\cap S^2$ is defined by the vanishing of the first coordinate $f$? As you can see in the screenshot we assumed $\frac{\partial f}{\partial x}(p)\neq0$, which means that f is not a constant value (otherwise derivative of constant is $0$). So why are we saying it is defined by the vanishing of the first coordinate?
For reference i am attaching the screenshot where i have highlighted the part where this is mentioned.
Example 9.6 (The $2$-sphere in $\mathbb{R}^3$). The unit $2$-sphere
$$S^2=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2+z^2 =1\}$$
is the level set $g^{-1}(1)$ of level $1$ of the function $g(x,y,z)=x^2+y^2+z^2$. We will use the inverse function theorem to find adapted charts of $\mathbb{R}^3$ that cover $S^2$. As the proof will show, the process is easier for a zero set, mainly because a regular submanifold is defined locally as the zero set of coordinate functions. To express $S^2$ as a zero set, we rewrite its defining equation as
$$f(x,y,z)=x^2+y^2+z^2-1=0.$$
Then $S^2=f^{-1}(0)$.Since
$$\frac{\partial f}{\partial x}=2x,\quad\frac{\partial f}{\partial y}=2y,\quad\frac{\partial f}{\partial z}=2z,$$
the only critical point of $f$ is $(0,0,0)$, which does not lie on the sphere $S^2$. Thus, all points on the sphere are regular points of $f$ and $0$ is a regular value of $f$.Let $p$ be a point of $S^2$ at which $(\partial f/\partial x)(p)=2x(p)\ne0$. Then the Jacobian matrix of the map $(f,y,z):\mathbb{R}^3\longrightarrow\mathbb{R}^3$ is
$$\left[\begin{array}{ccc}
\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\\[2mm]
\frac{\partial y}{\partial x} & \frac{\partial y}{\partial y} & \frac{\partial y}{\partial z}\\[2mm]
\frac{\partial z}{\partial x} & \frac{\partial z}{\partial y} & \frac{\partial z}{\partial z}
\end{array}\right]=\left[\begin{array}{ccc}
\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right],$$
and the Jacobian determinant $\partial f/\partial x(p)$ isnonzero. By Corollary 6.27 of the inverse function theorem (Theorem 6.26), there is a neighborhood $U_p$ of $p$ in $\mathbb{R}^3$ such that $(U_p,f,y,z)$ is a chart in the atlas of $\mathbb{R}^3$. In this chart, the set $U_p\cap S^2$ is defined by the vanishing of the first coordinate $f$. Thus, $(U_p,f,y,z)$ is an adapted chart relative to $S^2$, and $(U_p\cap S^2,y,z)$ is a chart for $S^2$.
Best Answer
Tu considers the smooth map $F : \mathbb R^3 \to \mathbb R^3, F(x,y,z) = (f(x,y,z),y,z)$ and shows that on some $U_p$ it restricts to a chart $F \mid_{U_p} : U_p \to F(U_p)$ on $\mathbb R^3$. In this special case this means nothing else than that $F \mid_{U_p}$ is diffeomorphism between open subsets of $\mathbb R^3$. The three coordinate functions of $F$ are $F_1= f$, $F_2 = p_2$, $F_3 = p_3$, where $p_i : \mathbb R^3 \to \mathbb R,p_i(x_1,x_2,x_3) = x_i$, is the projection onto the $i$-th coordinate.
But now by definition of $f$ we have have $S^2 = f^{-1}(0)$, i.e. $S^2 = \{(x,y,z) \in \mathbb R^3 \mid f(x,y,z) = 0 \}$. Therefore $$U_p \cap S^2 = \{(x,y,z) \in U_p \mid f(x,y,z) = 0 \} = \{(x,y,z) \in U_p \mid F_1(x,y,z) = 0\} = \{(x,y,z) \in U_p \mid F(x,y,z) \in \{ 0 \} \times \mathbb R^2\} = (F \mid_{U_p})^{-1}(\{ 0 \} \times \mathbb R^2) .$$
This means that $U_p \cap S^2$ is the set of points $\xi \in U_p$ where the first coordinate of $F(\xi)$ vanishes, i.e. where $f(\xi)$ vanishes. This is what Tu expresses in the form
Note that this shows that (cf. Definition 9.1) $S^2 \subset \mathbb R^3$ is a regular submanifold of dimension $2$ because for each $p \in S^2$ we found the chart $(U_p,F \mid_{U_p})$ in the maximal atlas of $\mathbb R^3$ such that $U_p \cap S^2$ is defined by the vanishing of $n-2 = 1$ of the coordinate functions of $F \mid_{U_p}$.