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Let $f: \mathbb{R}^+ \times \mathbb{R} \rightarrow \mathbb{R}$ with $f \in C^1$ and the Jacobian of $f$ has full rank (1) for all $z \in \mathbb{R}^+ \times \mathbb{R}$.

Then $M=\{z=(z_1,z_2,z_3) \in \mathbb{R}^3|(z_1,z_2)\neq (0,0),f(\sqrt{z_1^2+z_2^2},z_3)=0\}$ is a 2-dimensional submanifold of $\mathbb{R}^3$.

I am stuck on this task, I have some approach but I couldn't quite finish my proof (and I am especially not sure if I did any mistake):

My idea was to write $M$ as the roots of a function $g=f*h$ with a Jacobian of maximal rank (it follows that M is a submanifold).

Let use take $h: \mathbb{R}^2\backslash\{0\} \times \mathbb{R} \rightarrow \mathbb{R}^+ \times \mathbb{R}$ with $h(z_1,z_2,z_3)=(\sqrt{z_1^2+z_2^2},z_3)$ we get that $h$ is continuous because both components are continuous and therefore the composition $g=f*h$ is continuous.

And therefore it follows that $$J(f*h)=\left( \begin{array}{ccc} \frac{z_1 \partial_1f\left(\sqrt{z_1^2+z_2^2},\text{z3}\right)} {\sqrt{z_1^2+z_2^2}} & \frac{z_2 \partial_1f\left(\sqrt{z_1^2+z_2^2},\text{z3}\right)} {\sqrt{z_1^2+z_2^2}} & \partial_2f\left(\sqrt{z_1^2+z_2^2},\text{z3}\right) \end{array} \right)$$

Now I think I am done if I can show that at least one entry of that matrix is always unequal to zero. Thank you for your help.

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Why not show that $h$ has full rank? – JSchlather Jun 24 '11 at 21:02
Thats easy but does it follow that $f*h$ has full rank? – Listing Jun 24 '11 at 21:41
up vote 1 down vote accepted

Just look at your $J(f\circ h)$. If for some point $p\in{\rm dom}(h)$ one has $\partial_2 f(h(p))=0$ then by assumption on $f$ we are sure that $\partial_1 f(h(p))\ne 0$; furthermore at least one of $z_1$, $z_2$ is $\ne0$, so the corresponding component of $J(f\circ h)$ is $\ne 0$. It follows that $J(f\circ h)(p)$ is $\ne0$ for all $p\in{\rm dom}(h)$. Assume now that you have a point $p$ with $f(h(p))=0$. Then by the implicit function theorem you can select a suitable coordinate, say $z_1$, such that in the neigbourhood of $p$ the equation $f(h(z))=0$ is equivalent to $z_1= g(z_2,z_3)$ for some $C^1$-function $g$.

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You can do this using regular values. I think the necessary machinary comes from the Regular Value Theorem but the idea is to show that 0 is a regular value of the image, that is f^{-1}(0) contains no critical points, but seeing as we're told that the jacobian has full rank (of 1) this must be true and hence M is a regular submanifold of codim 1.

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Thanks that gives the result very nicely. I am still thinking if there is a way to do it directly. – Listing Jun 24 '11 at 22:01

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