# Implicit Function Theorem on Surfaces.

(a) State the Implicit Function Theorem in the most general way that you know

(b) Let $$\Sigma$$ the set of $$2 \times 2$$ matrices with determinant zero. Show that if $$0 \neq M \in \Sigma$$, then there is a neighborhood $$V \ni M$$ such that $$\Sigma \cap V$$ is a parameterized surfaces.

(c) Write explicitly a parameterization for $$\Sigma \cap V$$ with $$M = \left(\begin{array}{cc}1 & 1 \\ 1 & 1\end{array} \right)$$ and $$V \ni M$$ of your choice.

Attempt.

(a) Let $$F: \mathbb{R}^{n} \times \mathbb{R}^{m} \to \mathbb{R}^{m}$$ be a $$C^{1}$$ function. Suppose that $$F(a,b) = 0$$ and $$\displaystyle \det\left(\frac{\partial}{\partial y}F(a,b)\right) \neq 0$$. So there is $$a \in A$$ and $$b \in B$$ open sets and a $$C^{1}$$ function $$f: A \to B$$ such that

• $$b = f(a)$$,
• $$F(x,f(x)) = 0$$ for each $$x \in A$$

and $$f$$ is a

• $$\frac{\partial}{\partial x_{i}}f(x,y) = -\left(\frac{\partial}{\partial y}F(a,b)\right)^{-1}\left(\frac{\partial}{\partial x_{i}}F(a,b)\right)$$.

(b) Consider the function $$F: (\mathbb{R}^{3}\times \mathbb{R}) \to \mathbb{R}$$ given by $$F(X) = \det(X) = a_{11}a_{22} - a_{12}a_{21}$$ for $$X = \left(\begin{array}{cc}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right).$$

We have $$F \in C^{1}$$, $$F(M) = F(m_{11},m_{12},m_{21},m_{22}) = 0$$ and $$\det\left(\frac{\partial}{\partial a_{22}}F(M)\right)^{-1} \neq 0$$. By the Implicit Function Theorem, $$F^{-1}(0)\cap(V \times U) = (\Sigma\cap V)\times(\Sigma\cap U)$$ is the graph of a continuously differentiable function $$f: M \in V \to F(M) \in U$$. But the graph of a $$C^{1}$$ function is a surface. Therefore, $$\Sigma \cap V$$ is a parameterized surface.

(c) I can write $$f$$ as $$\displaystyle a_{22} \mapsto \frac{a_{12}a_{21}}{a_{11}}$$. But I dont know how to choice $$V$$ and find a parameterization.

## 1 Answer

(a) I believe you are missing some subtleties. I'd recommend Krantz' book

(b) We have $$\nabla F(X) = (a_{22},a_{21},a_{12},a_{11})$$. This ensures that $$\nabla F(M)\neq 0$$ for any $$M\neq 0$$, Hence you may apply the Implicit Function Theorem. In your attempt you covered just the case $$M=(m_{ij})$$ with $$m_{11}\neq 0$$. However, the other cases follow analogously.

(c) you have that $$\displaystyle \frac{\partial F}{\partial a_{11}}(M) = m_{22}=1 \neq 0$$. Then you can choose the parameterization: $$(a_{12}, a_{21}, a_{22}) \mapsto \begin{pmatrix} \frac{a_{12}a_{21}}{a_{22}} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$$ And you can take $$V= \{X \mid a_{22} \neq 0\}$$. Indeed, $$V = \{(x,y,z) \in \mathbb{R}^3 \mid z\neq 0\}\times \mathbb{R}$$ and $$\Sigma \cap V$$ is the graph of $$\displaystyle f\colon \{(x,y,z) \in \mathbb{R}^3 \mid z\neq 0\} \to \mathbb{R}$$ given by $$\displaystyle f(x,y,z) = \frac{xy}{z}$$.

• Thank you, its very helpful! I will check the reference. – Lucas Corrêa Feb 21 at 3:54