The main Question: define paraboloid in $\mathbb{R}^3$ as $x^2+y^2-z^2=a$. Prove that when $a>0$, then it is a manifold. Moreover, when $a=0$ it is not a manifold.
Define $X =\{(x,y,z)|x^2+y^2-z^2=a\} $.
The first part when $a>0$, Consider these opens:
$$
\begin{equation}
\begin{cases}
X_1 &=\{(x,y,z) \in \mathbb{R}^3|z \gt 0\}\\[2ex]
X_2 &= \{(x,y,z) \in \mathbb{R}^3|z \lt 0\}\\[2ex]
X_3 &= \{(x,y,z) \in \mathbb{R}^3|x \gt 0\}\\[2ex]
X_4 &= \{(x,y,z) \in \mathbb{R}^3|x \lt 0\}\\[2ex]
X_5 &= \{(x,y,z) \in \mathbb{R}^3|y \gt 0\}\\[2ex]
X_6 &= \{(x,y,z) \in \mathbb{R}^3|y \lt 0\}\\[2ex]
\end{cases}
\end{equation}\tag{1}\label{eq1}$$
We should have $X = \cup_{i=1}^{6}X_i$, because there is $(0,0,0)$ is not in $X$.
Now we can make each of $X_i$ diffeomorphic to an open set in $\mathbb{R}^2$.
$$
\begin{equation}
\begin{cases}
U_1 &=\{(x,y) \in \mathbb{R}^2 |x^2+y^2 \gt a\} = \mathbb{R}^2 - \overline{B_a(0)}\\[2ex]
U_2 &= \{(z,y) \in \mathbb{R}^2|y^2 \lt z^2+a\}\\[2ex]
U_3 &= \{(z,x) \in \mathbb{R}^2|x^2 \lt z^2+a\}\\[2ex]
\end{cases}
\end{equation}\tag{2}\label{eq2}$$
Now we can define diffeomorphism like the following:
$$
\begin{equation}
\begin{cases}
\phi_1 : &U_1 &\rightarrow X_1\\
&(x,y) &\to (x,y, \sqrt{x^2+y^2-a})
\end{cases}
\end{equation}\tag{3}\label{eq3}
$$
$$
\begin{equation}
\begin{cases}
\phi_2 : &U_1 &\rightarrow X_2\\
&(x,y) &\to (x,y, -\sqrt{x^2+y^2-a})
\end{cases}
\end{equation}\tag{4}\label{eq4}$$
$$ \begin{equation} \begin{cases} \phi_3 : &U_2 &\rightarrow X_3\\ &(z,y) &\to (\sqrt{a+z^2-y^2},y,z) \end{cases} \end{equation}\tag{5}\label{eq5}$$ $$ \begin{equation} \begin{cases} \phi_4 : &U_2 &\rightarrow X_4\\ &(z,y) &\to (-\sqrt{a+z^2-y^2},y,z) \end{cases} \end{equation}\tag{6}\label{eq6}$$ $$ \begin{equation} \begin{cases} \phi_5 : &U_3 &\rightarrow X_5\\ &(z,x) &\to (x,\sqrt{a+z^2-x^2},z) \end{cases} \end{equation}\tag{7}\label{eq7}$$ $$ \begin{equation} \begin{cases} \phi_5 : &U_3 &\rightarrow X_6\\ &(z,x) &\to (x,-\sqrt{a+z^2-x^2},z) \end{cases} \end{equation}\tag{8}\label{eq8}$$ As a result, $X$ when $a>0$ is a $2-$dimension manifold!
The problem is the second part. I know that each neighborhood of $0$ in $X$ could not be homeomorphic to any $\mathbb{R}^n$ for all $n \ge 2$. This is because if we remove the point zero from the zero neighborhood $U$ in $X$, then the new space is not connected. The new space can be separated by $X_1 \cap U$ and $X_2 \cap U$. This means that the subspace $X$ when $a=0$ can not be a $k$-dimension manifold ($\forall k \ge 2$). As these spaces will remain connected if we remove one and even countable points of them.
Consequently, The only part that remains is that we should show that this space cannot be a $1-$dimension manifold. For this part, I was thinking about something like that considering neighborhood of zero $U$ that is diffeomorphic to $\mathbb{R}$. In the space $\mathbb{R}$, if we remove each point then we will have disconnected subspace. Even though in the $U$, if we remove any other points except zero, the space will remain path connected(so connected).
My Question:
- If I am wrong somewhere please let me know.
- I think using 6 different opens to prove that the space is $2-$dimension manifold is a lot. What is the least number of functions and opens that are needed for this task?
- For the second part of prove, that the space cannot be a $1-dimension$ manifold, Do you know any other proofs?
- This is a latex question. In the equations $3-8$, the cases are not aligned. How should I write this part in latex to not have this problem?