Prove that $M=\{ x\in S^{n-1} \ | \ f(x)=0\}$ is an embedded submanifold of $S^{n-1}$.

Question

Let $n>1$. Define $f:\mathbb{R}^n \to \mathbb{R}$ by

$$f(x_1,…,x_n)=x_1^3+…+x_n^3$$

Prove that

$$M=\{ x\in S^{n-1} \ | \ f(x)=0\}$$

is an embedded submanifold of $S^{n-1}$. What is its dimension? $S^{n-1}$ is the unit sphere in $\mathbb{R^n}$.

My attempt:

We have $M=f^{-1}(0)$. The Jacobian matrix

$[df_*]=\begin{pmatrix} 3x_1^2 & … & 3x_n^2 \end{pmatrix}$

Since $x\in S^{n-1}$, we have $x_1^2+…+x_n^2=1$, hence $x_1,…, x_n$ cannot be all zeros and then $[df_*]$ has a full rank. As a result, $M=f^{-1}(0)$ is an embedded submanifold of dim $1$.

Is it correct? Thank you!

Answer 1

You want to check if $f^{-1}(0) \cap \mathbb S^{n-1}$ is an embedded submanifold of $\mathbb S^{n-1}$. Hence you need to check $d(f|_{\mathbb S^{n-1}}\ )$, instead of $df$, is non-zero in $M$.

Example where $df \neq 0$ is not sufficient: Let $$f(x_1, \cdots, x_n) = x_n - F(x_1, \cdots, x_{n-1})$$ (note that $\nabla f = (-\nabla F, 1)$ is always non-zero), where $F : \mathbb R^{n-1} \to \mathbb R$ is smooth, $F = 1$ when $\sqrt{x_1^2+\cdots + x_{n-1}^2} >3$ and $$ F(x_1, \cdots, x_{n-1}) = \sqrt{1-x_1^2 - \cdots - x_{n-1}^2}$$ when $\sqrt{x_1^2+\cdots + x_{n-1}^2} <1/2$. Then $f^{-1}(0)$ is just the graph of $F$, and $f^{-1}(0) \cap \mathbb S^{n-1}$ is not an embedded submanifold of $\mathbb S^{n-1}$ (it contains an open set of $\mathbb S^{n-1}$, but is not the whole $\mathbb S^{n-1}$).

Going back to your question. You need to check if $d(f|_{\mathbb S^{n-1}}\ )$ is non-zero. This is the same as checking that $df(x)$ is not parallel to $n(x)=x$, the normal vector to the sphere $\mathbb S^{n-1}$. This is indeed true: if $$df=\begin{pmatrix} 3x_1^2 & … & 3x_n^2 \end{pmatrix} = k\begin{pmatrix} x_1 & … & x_n \end{pmatrix}$$ one has $3x_i^2 = kx_i$ for all $i$. If $x_i$ is nonzero, $x_i = k/3$. That is, all non-zero terms must be the same. However $x\in f^{-1}(0)$ implies $x_1^3+ \cdots x_n^2 = 0$, so it is impossible. Thus all $x_i = 0$ which is again impossible since $x\in \mathbb S^{n-1}$. Thus $df$ is nowhere parallel to $n$, and thus $d(f|_{\mathbb S^{n-1}}\ )$ is non-zero in $M$.

Answer 2

It is not entirely correct. You use that regular level sets are embedded submanifolds, which is true. But the dimension of the level set is the dimension of the sphere minus 1. The level set has codimension 1.