I am learning alone and I have just start studying about manifolds and below a question. I want to be sure that my answer and so my understanding is correct.
Question:
Let $ \lambda \in \mathbb{R}$ and let $S = \left \{ (x_1,x_2,x_3) \in \mathbb{R}^3 : x_1^2+x_2^2-x_3^2 = \lambda \right \} $. Find the $\lambda$ for which $S$ is a submanifold of $\mathbb{R}^3 $
Answer:
1- According to the definition I have learned, in order for $S$ to be a manifold of $ \mathbb{R}^3 $ of dimension $p \in \left \{ 0;1;2;3 \right \} $, $S$ must verify that:
$\forall a \in S $ it exists at least one neighborhood $U_a$ of $a$ and at least one mapping $ f \in C^1(U_a; \mathbb{R}^{3-p} )$ with a differential $D\left \{ f(x) \right \}$ (so the mapping $f$ is differentiable ) s.t.$rank ( D\left \{ f(x) \right \} ) = 3 - p$ and $S \cap U_a = \left \{ f = 0 \right \} $. With $S \cap U_a = \left \{ f = 0 \right \} $ meaning that all the points $x' \in \mathbb{R}^3 $ that are in $ S $ and in the neighborhood $U_a$ of $a$ verify $f(x')=0$
2- $\forall a \in S$ we begin by taking the following neighborhood $U_a = \mathbb{R}^{3} $. In consequence $\forall a \in S \Rightarrow S \cap U_a = S \cap \mathbb{R}^{3} =S $. Hence it cames naturally that $ \forall a \in S \cap U_a \Rightarrow \left \{ f = 0 \right \}$.
3- Let prove that such a mapping $f$ exists.
We begin to build it by defining $ f (x_1 ; x_2 ; x_3) = x_1^2+x_2^2-x_3^2 - \lambda $ which is a polynomial function those it is in $C^{\infty}$.
Hence the differential is directly obtained via the gradient $D\left \{ f(x_1 ; x_2 ; x_3) \right \} = \vec{\bigtriangledown} \left \{ f (x_1 ; x_2 ; x_3) \right \} = \begin{pmatrix}
2x_1
\\ 2x_2
\\
-2x_3
\end{pmatrix} $
Note that the gradient here is a $ 3 \times 1 $ matrix so the definition of the rank here will be the same as the one for other matrix. That means the number of line linearly independent.
4-Case $\lambda \neq 0$
As we explained in "2-" $ \forall a \in S \cap U_a = S $ they verify $x_1^2+x_2^2-x_3^2 - \lambda = 0 \Rightarrow x_1^2+x_2^2-x_3^2 = \lambda $. And now if $\lambda \neq 0$ it means that at least one of the $x_1,x_2,x_3$ is different from $0$.
If only one $x_1,x_2,x_3$ is different from $0 \Rightarrow rank(\vec{\bigtriangledown} \left \{ f (x_1 ; x_2 ; x_3) \right \}) = 1 $
If two or three elements of $x_1,x_2,x_3$ are strictly different from $0$ we have that two of the three line of the gradiant vector are linearly dependent of a third one, a.k.a, two of the lines can be obtained by multiplying the element of the third line by a scalar $\Rightarrow rank(\vec{\bigtriangledown} \left \{ f (x_1 ; x_2 ; x_3) \right \}) = 1 $
In conclusion in such a case $S$ is a manifold of dimension $2$ of $\mathbb{R}^3$
5-Case $\lambda = 0$
In such a case we have that $ S = \left \{ (x_1,x_2,x_3) \in \mathbb{R}^3 : \pm \sqrt{x_1^2+x_2^2}= x_3^2 \right \} $. In other words $S$ is a cone.
If we focus on this cone without the point $(0;0;0)$, in other words $S - (0;0;0)$, will be a manifold of dimension $2$ with the same reasoning as above.
Now let suppose that $S$ ( so with the point $(0;0;0)$ ) is still a manifold of $\mathbb{R}^3$ of dim $ = 2$. It will mean that $T_0 S = $ the tangent plane contening the set of all vectors tangent to $S$ in $(0;0;0)$ = $ker( \vec{\bigtriangledown} \left \{ f (0;0;0) \right \}) ) $.
But obviously $dim (ker( \vec{\bigtriangledown} \left \{ f (0;0;0) \right \}) ) = ker (\begin{pmatrix}
0
\\ 0
\\
0
\end{pmatrix}) = 3$ as $\vec{\bigtriangledown} f : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ and all vectors $v$ in $\mathbb{R}^3$ satisfy $ (\begin{pmatrix}
0
\\ 0
\\
0
\end{pmatrix}) v = (\begin{pmatrix}
0
\\ 0
\\
0
\end{pmatrix}) $
Is it correct?
I doubt mostly on the end of my part "4-".
I will read with pleasure any comment or precision, but just take into account that I am studying (for the moment at least) manifold in $ \mathbb{R}^n$
Thank for your help.
If you define $ F: \mathbb R^3 \to \mathbb R$ as the map $F(x_1, x_2,x_3) = x_1^2 +x^2_2 - x_3^3$ then $S$ can be viewed as $S=F^{-1}(\lambda)$. By a standard result in Differential Geometry, $S$ is an embedded submanifold if and only if $\lambda$ is a regular value for $F$.
Being a regular value means that, for any $x \in F^{-1}(\lambda)$, $dF_x$ is a surjective linear map. As you correctly pointed out, the only point where the differential has not full rank is $x = 0$, i.e. the origin. So you just have to assure that the origin does not lie in $S$; but it is clear that $0 \in F^{-1}(\lambda)$ if and only if $\lambda = 0$. This proves that $S$ is an embedded submanifold (so itself a manifold) if and only if $\lambda \neq 0$.