0
$\begingroup$

I will use the following notations for tuples: $\mathbb R^{n+m} \ni (x, y) = (x_1, \dots, x_n, y_1, \dots, y_m)$. Furthermore, let $\mathbf f : \Omega \subset \mathbb{R}^{n + m} \rightarrow \mathbb{R}^m$ be of class $\mathcal C^1$ on $\Omega$, and write $$\mathbf f'(x, y) = (\mathbf f_x'(x, y), \mathbf f_y'(x, y)) \in \mathbb{R}^{m \times (n + m)},$$ where $\mathbf f_x'(x, y) = (D_jf_i(x, y))_{1 \leq i \leq m, 1 \leq j \leq n} \in \mathbb{R}^{m \times n}$ and $\mathbf f_y'(x, y) = (D_{n + j}f_i(x, y))_{1 \leq i, j \leq m} \in \mathbb{R}^{m \times m}$.

Theorem (implicit function). Let $\mathbf f : \Omega \subset \mathbb{R}^{n + m} \rightarrow \mathbb{R}^m$ be of class $\mathcal C^1$ on $\Omega$ and let $(a, b) \in \Omega$ such that $\mathbf f(a, b) = 0$. If $\det \mathbf f_y'(a, b) \neq 0$, then there exists an open neighborhood $A$ of $a$ in $\mathbb{R}^n$, and an open neighborhood $B$ of $b$ in $\mathbb{R}^m$, with $A \times B \subset \Omega$, and a unique $\mathcal C^1$-function $\mathbf g : A \rightarrow B$ such that $\mathbf f(x, \mathbf g(x)) = 0$ for all $x \in A$.

Definition ($k$-dimensional submanifold). A set $M \subset \mathbb{R}^n$ is called a $k$-dimensional submanifold of $\mathbb{R}^n$ if for all $x_0 \in M$, there exists an open neighborhood $\Omega$ of $x_0 \in \mathbb{R}^n$ and a $\mathcal C^1$-function $\mathbf f : \Omega \subset \mathbb{R}^n \rightarrow \mathbb{R}^{n - k}$ such that $$M \cap \Omega = \mathbf f^{-1}(\{0\}) \hspace{10pt} \mbox{and} \hspace{10pt} \mbox{rank } D\mathbf f(x) = n - k \mbox{ for all } x \in M \cap \Omega.$$

What I want to prove is this:

Proposition. The following are equivalent:

(a) $M$ is a $k$-dimensional submanifold of $\mathbb{R}^n$;

(b) For each $x_0 \in M$, write $x_0 = (y_0, z_0)$ with $y_0 \in \mathbb{R}^k$, $z_0 \in \mathbb{R}^{n - k}$, there exists an open neighborhood $U$ of $y_0$ in $\mathbb{R}^k$, an open neighborhood $V$ of $z_0 \in \mathbb{R}^{n - k}$, and a $\mathcal C^1$-function $\mathbf g : U \rightarrow V$ with $\mathbf g(y_0) = z_0$ such that $$M\cap(U \times V) = \{(y_0, \mathbf g(y_0)) : y_0 \in U\}.$$

I can do the proof from (b) to (a):

Define $\Omega = U \times V$, $\mathbf f : \Omega \rightarrow \mathbb{R}^{n - k}$ by $$\mathbf f(y, z) = z - \mathbf g(y).$$ Then $\mathbf f$ is a $ \mathcal C^1$-function, $M \cap \Omega = M \cap (U \times V) = \{(y_0, \mathbf g(y_0)) : y_0 \in U\} = \mathbf f^{-1}(\{0\})$ and $$D\mathbf f(x_0) = D\mathbf f(y_0, z_0) = (-D\mathbf g(y_0), I_{n - k})$$ which is of rank $n - k$ obviously.

There is a little problem when I do the proof from (a) to (b):

By the definition of $k$-dimensional submanifold, for all $x_0 = (y_0, z_0) \in M$, there exists $\Omega$ of $x_0$, and a $\mathcal C^1$-function $\mathbf f : \Omega \rightarrow \mathbb{R}^{n - k}$ such that $$M \cap \Omega = \mathbf f^{-1}(\{0\}) \hspace{10pt} \mbox{and} \hspace{10pt} \mbox{rank } D\mathbf f(x) = n - k \mbox{ for all } x \in M \cap \Omega.$$ Since $x_0 \in M \cap \Omega$, $\mathbf f(x_0) = \mathbf f(y_0, z_0) = 0$ and since rank $D\mathbf f(x) = n - k$, $\det \mathbf f_z'(y_0, z_0) \neq 0$. Hence by the implicit function theorem, there exists an open neighborhood $U$ of $y_0$, an open neighborhood $V$ of $z_0$ and a $\mathcal C^1$-function $\mathbf g : U \rightarrow V$ such that $\mathbf f(y_0, \mathbf g(y_0)) = 0$ for all $y \in U$.

The problem is that although I know that $\mathbf f(x_0) = \mathbf f(y_0, z_0) = \mathbf f(y_0, \mathbf g(y_0)) = 0$, can I conclude $x_0 = (y_0, \mathbf g(y_0))$ and hence $z_0 = \mathbf g(y_0)$? Does $\mathbf g$ need to be injective?

$\endgroup$
0
$\begingroup$

Yes, you can conclude that. The $\mathcal C^1$-function $\mathbf g : U \to V$ you found is exactly the function you need to associate to each $y \in U$ that one and only one $z \in V$ such that $\mathbf f(y,z) = 0$. Since $y_0 \in U$ and $z_0 \in V$ by construction, this also works in the particular case of $(y_0,z_0)$, meaning that you can very well state $z_0 = \mathbf g(y_0)$. This concludes the proof that $M$ is locally the graph of a $\mathcal C^1$-function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.