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I have the following problem:

Let $$M_{a,b}=\{(x,y,z): x^1+y^2=z^2,\,\, z=ax+b\}$$ Investigate for which (a,b) this is a submanifold.

I thought that we could look at the following function $$F(x,y,z)=(x^2+y^2-z^2,\,\,\,z-ax-b)$$ where $F:\mathbb{R}^3\rightarrow \mathbb{R}^2$. But with this function I only get that $a\neq 0$ but nothing about b. So I think that we need another approach, but I can't imagine which one. Could someone give me a hint?

Thank you.

We have the following definition for submanifolds Def: $M\subset \mathbb{R}^n$ is submanifold of dimension k if for all $a\in M$ one of the following three equivalent points holds:

  1. For each $a\in M$ there exists an open neighbourhood $U\subset \mathbb{R}^n$ and a diffeomorphismen $\phi:U\rightarrow V$ where V is open in $\mathbb{R}^n$ such that $$\phi(U\cap M)=V\cap(\mathbb{R}^k\times \{0\})$$
  2. For each $a\in M$ there exists an open Neighbourhood $U\subset \mathbb{R}^n$ and a submersion $F:U\rightarrow \mathbb{r}^{n-k}$ at a such that $$U\cap M=U\cap\{F=0\}$$
  3. For each $a\in M$ there exists an open $O\subset M$ and and open $O'\subset \mathbb{R}^k$ and a homeomorphism $\phi:O'\rightarrow O$ such that $\phi$ is a immersion at $\phi^{-1}(a)$ and $O=\phi^{-1}(O)$
user123234
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  • Why $a \ne 0$? For $a = b = 0$, this is just the set ${ (x, x^2, z): x, z \in \mathbb{R}}$. – ViktorStein Nov 14 '21 at 21:58
  • ah yes now I see it, so then I don't know how to procede. Could you give me a hint, do I need to use another definition instead of the second one? – user123234 Nov 14 '21 at 22:04

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