# Implicit function theorem hyperplane

Let $Q=\{(x_1,x_2,x_3,x_4):x_4=0\}$ and let $\phi:\mathbb{R}^4\to\mathbb{R}^4$ be a $C^1$ diffeomorphism. Prove that for every $x\in\phi(Q)$, there is an $\epsilon$ such that $B_{\epsilon}(x)\cap\phi(Q)$ is the graph of a $C^1$ function having one of the following forms

$x_1=f(x_2,x_3,x_4)$, $x_2=f(x_1,x_3,x_4)$, $x_3=f(x_1,x_2,x_4)$ or $x_4=f(x_1,x_2,x_3)$.

My attempt: Let $g=\phi^{-1}$, let $\pi_j:\mathbb{R}^4\to\mathbb{R}$ be the projection onto the jth coordinate, and $x\in\phi(Q)$. Note $\pi_4(g(x))=0$. Since $g$ is a diffeomorphism, $Dg(x)\not=0$ for every $x$. Since $Dg(x)$ is a 4x4 matrix, we know at least one of these entries is nonzero. Suppose entry $ij$ is nonzero. We want to show that $\frac{\partial}{\partial x_i}\pi_j(f(x))\not=0$, and then the result will follow from the implicit function theorem. $\frac{\partial}{\partial x_4}\pi_j(f(x))=(Df(x))_{ij}\not=0$.

Is this the right idea? Also, I'm not really seeing where the assumption that everything in $Q$ has zero for its fourth coordinate is being used, so I suspect this isn't completely right.

• Your $Q$ is basically a $\mathbb R^3$ sitting in $\mathbb R^4$, therefore $\phi(Q)$ likewise be a three-dimensional, hence locally be diffeomorphic to $\mathbb R^3$. Therefore if you take sufficient local coordinates, $\phi(Q)$ will depend on three coordinates only and you can solve for the fourth one implicitly. I know this is very sketchy but does it help you? – James Aug 16 '18 at 13:37

We want to study the surface $\phi(G)$ locally (near a point). We have that $x\in\phi(G)$ if and only if $g(x)\in G$, which can be rewritten as $\pi_4(g(x))=0$. I denote $g=\phi^{-1}$ as you did.
Since the surface is defined in such a way, we should use the implicit function theorem for the function $F=\pi_4\circ g$. The gradient of $F$ is the fourth row of the derivative matrix (Jacobian matrix) of $g$. Since $g$ is a diffeomorphism, the said matrix is invertible, so the row is non-zero. Therefore $\nabla F$ is never zero.
Now consider any point $x_0\in\phi(Q)$. (I find it cleanest to decorate the base point with an index or something, so that $x$ is always a free variable.) Since $\nabla F(x_0)\neq$, there is $i\in\{1,2,3,4\}$ so that $\partial_iF(x_0)\neq0$. For simplicity, let us assume that $i=1$; the other cases are similar, but it is notationally simplest to keep $i$ fixed.
Let us write $x=(x_1,x_2,x_3,x_4)$ as $x=(y,x)$, where $y=x_1$ and $z=(x_2,x_3,x_4)$. We write $x_0=(y_0,z_0)$. We can consider $F$ as a function $F(y,z)$. Now the derivative of $F$ with respect to $y$ is $\partial_1F(x_0)$, which is non-zero. Therefore by the implicit function theorem there is a function $f$ of $z$ defined near $z_0$ so that $F(y,z)=0 \iff y=f(z)$ near $x_0$. (The exact formulation depends on the variant of the implicit function theorem you have.) This means that $x_1=f(x_2,x_3,x_4)$. The ther options correspond to different values of $i$.