How can I show that the following set is not a submanifold?

Question

I have the following problem:

Show if the set $M_{a,0}=\{(x,y,z):x^2+y^2=z^2, z=ax\}$ is a submanifold for $1<a<0$ and $a>1$.

Geometrically I see that if $1<a<0$ then we have a submanifold and in the other case we don't but I somehow think that I have an error in my computations:

$a>1,b=0$ (I wanted to use definition 2 at the end of the post) Here we have $$M_{a,0}=\{(x,y,z): x^2+y^2=z^2, z=ax\}$$ Then I thought I would define $$F(x,y,z)=(x^2+y^2-z^2,z-ax)$$ then $F^{-1}(\{0\})=M_{a,0}$. Then I calculated the differential $$DF=\begin{pmatrix} 2x&2y&-2z // -a&0&1\end{pmatrix}$$. Now I have assumed that F is a subversion, then for all $(x,y,z)$ there should be $l$ so that $(2x,2y,-2z)\cdot l=(-a,0,1)$. But if we now choose $(x,y,z)=(0,0,0)\in M_{a,0}$ then this does not work, so this is a contradiction and $M_{a,0}$ is not a submanifold.

Now I can do the same with $1>a>0$ and that doesn't work, because for $1>a>0$ you have a submanifold, so I think my proof above is wrong, but I don't see why, or how I can still use my property of a to make it right.

Thanks a lot for your help.

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 equivalence points holds:

1. for every $a\in M$ there is an open neighbourhood $U\subset \mathbb{R}^n$ and a diffeomorphism $\phi:U\rightarrow V$ where V is open in $\mathbb{R}^n$ such that $$\phi(U\cap M)=V\cap(\mathbb{R}^k\times \{0\})$$.
2. for every $a\in M$ there is 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 every $a\in M$ there is an open $O\subset M$ and an open $O'\subset \mathbb{R}^k$ and a homeomorphism $\phi:O'\rightarrow O$ such that $\phi$ is an immersion in $\phi^{-1}(a)$ and $O=\phi^{-1}(O)$.

Answer 1

…then this does not work, so this is a contradiction and $M_{a,0}$ is not a submanifold.

This implication is wrong. It’s true that you have a contradiction, but this doesn’t imply that the set is not a manifold. Maybe it’s just a matter of a poor choice of the function $F$.

Consider for instance the submanifold $X=\{(x,y)\in\mathbb R^2\;|\;y=0\}\subset \mathbb R^2$. You can choose $F(x,y)=y$, which gives a submersion, but also $F(x,y)=y^2$, which doesn‘t. Both functions are such that $F^{-1}(0)=X$.