Prove that the unit circle centred at the origin is a submanifold of $R^2$.

Question

I am having trouble applying the definition of submanifold. In this question, how do I select an atlas for $R^2$ and then how do I choose an atlas for $S^1$?

This is the definition I am using:

Answer 1

Let $B$ be the ball centered at $0$ with radius $1/2$ in $\mathbb R^{n+1}$. $p\in \mathbb S^n$ be fixed. let $\{e_1, \cdots, e_n\}$ be an orthonormal basis of $p^\perp: \{ v\in \mathbb R^{n+1} : v\cdot p = 0\}$. Then write $(x_1, \cdots, x_n, x_{n+1}) = (x, x_{n+1})$ and define $ F: B \to \mathbb R^{n+1}$ by $$F(x, x_{n+1}) = x_1 e_1 + x_2 e_2 + \cdots + x_n e_n+\left(\sqrt{1- \|x\|^2} + x_{n+1}\right)p$$

Note that if $p = (0,0,\cdots, 0, 1)$, then this map is just $$\left(x, \sqrt{1- \|x\|^2} + x_{n+1}\right).$$

The differential of $F$ is given by (with respect to the basis $\{e_1, \cdots, e_n, p\}$ in the image)

$$DF = I_{n+1} - \frac{1}{\sqrt{1- \|x\|^2}}\left(x_1 + \cdots +x_n\right) p$$

In particular, $DF_{(0,0)}$ is invertible and so by the inverse function theorem, there is $U \subset \mathbb R^{n+1}$ containing $p = F(0,0)$ and a diffeomorphism $\phi : U \to U' \subset B$ so that $\phi^{-1} = F|_{U'}$. From the construction. we see that $F(B) \cap \mathbb S^n = \{(x, 0) \in B\}$. This implies $\phi(U \cap \mathbb S^n) = U' \cap \mathbb R^n \times \{0\}$. As $p$ is arbitrary, we've checked that $\mathbb S^n$ is a submanifold of $\mathbb R^{n+1}$.