Coordinate system $(\phi, U, x_1, x_2)$ around $p \in S^1$ such that $S^1 \cap U = \{(x_1, x_2): x_2 = 0\}$.

Question

Consider $S^1$ as a subspace $\Phi^{-1}(1) \subseteq \mathbb{R}^2$, where $\Phi(x_1, x_2) = x_1^2 + x_2^2$. Using the Implicit Function Theorem, show that for all points $p \in S^1$, there is a coordinate system $(\phi, U, x_1, x_2)$ around $p$ such that $S^1 \cap U = \{(x_1, x_2): x_2 = 0\}$.

I think that I have to show the existence of a chart $(\phi, U)$ around $p\in U \subseteq \mathbb{R}^2$ such that $\pi_2\circ \phi(u, v) = x_2(u, v) = 0$ precisely when $(u, v) \in S^1 \cap U$, i.e. $\Phi(u, v) = 1$. However, I am not sure how to involve the Implicit Function Theorem here.

Answer 1

We consider the continously differentiable function $$\Phi : \mathbb R^2 \to \mathbb R, f(x_1,x_2) = x_1^2+x_2^2 $$ with $S^1 = \Phi^{-1}(1)$.

The Jacobian matrix of $\Phi$ at $ p = (p_1,p_2)$ is $$J\Phi(p) = \left(\frac{\partial \Phi}{\partial x_1}(p),\frac{\partial \Phi}{\partial x_2}(p) \right)= (2p_1,2p_2) .$$ The implicit function theorem says that if $p \in \Phi^{-1}(1)$ and the matrix $\left(\frac{\partial \Phi}{\partial x_2}(p)\right)$ is invertible (which means that $\frac{\partial \Phi}{\partial x_2}(p) = 2p_2 \ne 0$, i.e. $p_2 \ne 0$), then there exist open neighborhoods $W$ of $p$ in $\mathbb R^2$ and $V$ of $p_1$ in $\mathbb R$ with the following properties:

1. For each $x_1 \in V$ there exists a unique $g(x_1) \in \mathbb R$ such that $(x_1,g(x_1)) \in \phi^{-1}(1) \cap W$.

2. The function $g : V \to \mathbb R$ is continuously differentiable.

Note that 1. implies $g(p_1) = p_2$. We have

$$ (x_1,g(x_1)) \in \Phi^{-1}(1) \cap W = S^1 \cap W \text{ for all } x_1 \in V . \tag{1}$$

Define $$\tilde U = V \times \mathbb R ,\\ \tilde \phi : \tilde U \to \tilde U, \tilde \phi(x_1,x_2) = (x_1,x_2 - g(x_1)) .$$ Clearly $\tilde \phi$ is a diffeomorphism (its inverse is given by $\psi(x_1,x_2) = (x_1,x_2 + g(x_1))$). The set $U = W \cap \tilde U$ is an open neigborhood of $p$ in $\mathbb R^2$, thus

$$\phi : U \stackrel{\tilde \phi}{\to} \tilde \phi(U)$$

is a chart around $p$. For $x = (x_1,x_2) \in U$ we have $$\pi_2(\phi(x_1,x_2)) = 0 \Longleftrightarrow x_2 = g(x_1).$$

Thus if $x \in U$ with $\pi_2(\phi(x)) = 0$, $(1)$ shows that $x \in S^1$, hence $x \in S^1 \cap U$.

For $x =(x_1,x_2) \in S^1 \cap U$ we have $\phi(x) = (x_1, x_1 - g(x_1))$ and since $x_1 \in V$ we get $\pi_2(\phi(x)) = 0$.

The case $p_2 = 0$ is not covered by the above arguments. But in that case we have $p_1 \ne 0$ and we can use the same approach by exchanging the coordinates (and getting a function $f$ such that $(f(x_2),x_2) \in S^1 \cap W$).