Proposition

For each $n \in \mathbb{N}$, $S^n = \{ x \in \mathbb{R}^{n + 1} \mid \abs{x} = 1 \}$ is an $n$-manifold.

Solution

Let $n \in \mathbb{N}$ be given. We will show that $S^n$ is an $n$-manifold. Since $S^n$ is a subspace of $\mathbb{R}^{n + 1}$, it is Hausdorff and second countable. We will show that $S^n$ is locally Euclidean. Let $a = (a_1, \cdots, a_{n + 1}) \in S^n$. Since $a_1^2 + \cdots + a_{n + 1}^2 = 1$, there must exists an $i$ such that $a_i \ne 0$.

First, we will assume $a_i > 0$.

Let $U = \{ (x_1, \cdots, x_{n + 1}) \in \mathbb{R}^{n + 1} \mid x_i > 0 \}$. Then $U$ is an open set subset of $\mathbb{R}^{n + 1}$. Therefore, $U \cap S^n$ is an open subset of $S^n$. Since $a \in U \cap S^n$, $U \cap S^n$ is a neighborhood of $a$ in $S^n$.

Let $V = \{ x \in \mathbb{R}^n \mid \abs{x} < 1 \} $. Let $f(x_1, \cdots, x_{n - 1}) = (x_1, \cdots, x_{i - 1}, \sqrt{1 - x_1^2 - \cdots - x_{n - 1}^2}, x_i, \cdots, x_{n - 1})$. Then $f$ maps $V$ into $U \cap S^n$. We claim that this is a homeomorphism.

• Well defined?
  • $x_1^2 + \cdots + x_{i - 1}^2 + (1 - x_1^2 - \cdots - x_{n - 1}^2) + x_i^2 + \cdots + x_{n - 1}^2 = 1$.
  • Since $\sqrt{1 - x_1^2 - \cdots - x_{n - 1}^2} > 0$, $f$ indeed maps $V$ into $U \cap S^n$.
• Continuous?
  • Each component function is continuous, so it is continuous.
• Injective?
  • Suppose $f(x_1, \cdots, x_{n - 1}) = f(y_1, \cdots, y_{n - 1})$. Then $x_1 = y_1, x_2 = y_2, \cdots, x_{n - 1} = y_{n - 1}$.
• Surjective?
  • Let $x_1, \cdots, x_{n + 1} \in U \cap S^n$. Then $x_1^2 + \cdots + x_{n + 1}^2 = 1$. Moreover, $x_i > 0$. Therefore, $x_i = \sqrt{x_1^2 + \cdots + x_{i - 1}^2 + x_{i + 1}^2 + \cdots + x_{n + 1}^2}$.
• Continuous inverse?
  • The inverse function is $(\pi_1, \pi_2, \cdots, \pi_{i - 1}, \pi_{i + 1}, \cdots, \pi_{n + 1})$ where each $\pi_k$ is the projection of the $k$-th coordinate. Since each coordinate function is continuous, the inverse is continuous.

Thus $f$ is indeed a homeomorphism, so $V$ is homeomorphic to $U \cap S^n$. This means that we found a neighborhood of $a$ that is homemorphic to an open subset of $\mathbb{R}^n$. $a$ was chosen arbitrarily, so this means $S^n$ is locally Euclidean. Thus $S^n$ is an $n$-manifold.

The case for $a_i < 0$ is similar.