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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.

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  • $\begingroup$ 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? $\endgroup$ – James Aug 16 '18 at 13:37
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You have many important ideas right, but some pieces are also missing. For example, you said that the matrix has at least one non-zero element and then somehow jumped to the conclusion that this element must be on the fourth row. You need to argue why the fourth row has a non-zero element. I recommend writing the proof more carefully, explaining what happens and why. Instead of just claiming that "it follows from the implicit function theorem" (which it does), tell what the implicit function theorem exactly gives you and observe that it is what you wanted. Below is a sketchy proof that I hope to be helpful.

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 $i$.

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