In introduction I would like to say that I just begin to study (alone) what is are submanifold so I need your help in order to be sure that my understanding of those concept is correct.
Question:
Let $S = \left \{ (x;y;z) \in \mathbb{R} \times \mathbb{R}^{+*} \times \mathbb{R} : x^2 + z^2 = 1/y \right \}$ . Prove that $S$ is a submanifold and justify his dimension.
Answer:
1- First of all let chose $\forall a \in S$ the neigboorhood $U_a$ to be equal to $ \mathbb{R}^3 - (0;0;0)$
2- Now let define $f(x;y;z) = x^2 + z^2 - 1/y $. We can remark two things. First that $ f \in C^1(U_a; \mathbb{R})$ as the sum of function $C^1(U_a)$, and segondly that $ U_a \cap S = \left \{ f = 0 \right \} $
3- Now as $f \in C^1(U_a; \mathbb{R}) \Rightarrow D \left \{ f(x;y;z) \right \} = \vec{\bigtriangledown} \left \{ f(x;y;z) \right \} = \begin{pmatrix} 2x\\ \frac{1}{y^2}\\ 2z \end{pmatrix} \neq \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$ on $U_a$. It is obvious that the $rank(\vec{\bigtriangledown} \left \{ f(x;y;z) \right \}) = 1$. As for any line we choose all the $2$ other lines can be write as a linear combinaison of the first one (trivial).
4- In conclusion we can write that by theorem/definition $rank(\vec{\bigtriangledown} \left \{ f(x;y;z) \right \}) = 3-p=1 $ and thus that $S$ is a submanifold of order $2$ of $ \mathbb{R}^3$
Is this correct?
I have mostly a doubt on my part "3- ".
Thank you for your help.
According to https://math.stackexchange.com/users/1152406/federico-t it seems that my answer is correct.
Rem: The gradient is of $rank=1$ as for any line we choose all the $2$ other lines can be write as a linear combinaison of the first one (trivial to prove).