\documentclass[11pt, letterpaper, onecolumn, conference, final]{IEEEtran}

\usepackage[margin = .5in]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{graphicx}
\usepackage{subfigure}
\usepackage[utf8]{inputenc}
\usepackage{amsthm}

\newfam\hebfam
\font\tmp=rcjhbltx at10pt \textfont\hebfam=\tmp
\font\tmp=rcjhbltx at7pt  \scriptfont\hebfam=\tmp
\font\tmp=rcjhbltx at5pt  \scriptscriptfont\hebfam=\tmp

\edef\declfam{\ifcase\hebfam 
     0\or1\or2\or3\or4\or5\or6\or7\or8\or9\or A\or B\or C\or D\or E\or F\fi}

\mathchardef\aleph  = "0\declfam 27

\DeclareMathOperator*{\argmin}{arg\,min}

\theoremstyle{definition}
\newtheorem{problem}{Problem}
\newtheorem{solution}{Solution}

\title{MATH 208 - HW \# 2}
\author{Nathan Marianovsky \\ October 9, 2017}

\begin{document}

\maketitle

\begin{problem}
Prove that the cone, $x^2+y^2=z^2$ for $z \geq 0$, is a topological manifold, but is not a smooth embedded manifold in Euclidean 3-space.
\end{problem}

\vspace{.5cm}

\begin{solution}
\hspace{.1cm}
\begin{itemize}

\item
To show that the cone, $\mathcal{C} = \{(x,y,z) \in \mathbb{R}^3 \hspace{.1cm} | \hspace{.1cm} x^2+y^2=z^2 \hspace{.1cm} \text{and} \hspace{.1cm} z \geq 0\}$, is a topological manifold we need to find a homeomorphism between $\mathbb{R}^n$ and $\mathcal{C}$ for some fixed $n$. The most natural choice is projection onto the $xy$-plane ($n = 2$) given by:
\begin{equation*}
f(x,y,z) = (x,y) \hspace{.3cm} \text{and} \hspace{.3cm} f^{-1}(x,y) = (x,y,\sqrt{x^2+y^2})
\end{equation*}
We check that $f$ is a homeomorphism directly:
\begin{itemize}

\item
If $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$, then $(x_1,y_1) = (x_2,y_2)$ implying $x_1 = x_2$ and $y_1 = y_2$. Thus, $f$ is injective. 

\item
For any $\overrightarrow{q} = (x,y) \in \mathbb{R}^2$ we can always find a unique point $\overrightarrow{p} =  (x,y,\sqrt{x^2+y^2}) \in \mathcal{C}$ s.t. $f(\overrightarrow{p}) = \overrightarrow{q}$. Thus, $f$ is surjective.

\item
Now using the fact that projection maps are continuous we know that $f$ is continuous.

\item
For $f^{-1}$ we need to consider open sets in the topology endowed on the cone as a subspace of $\mathbb{R}^3$. Any such open set takes the form $U = \mathcal{C} \bigcap B_r(\overrightarrow{p})$ where $B_r(\overrightarrow{p})$ is an open ball in $\mathbb{R}^3$. Therefore, the open sets in this topology correspond exactly to the level curves of the cone. So taking any level curve $U \subseteq \mathcal{C}$ provides $f^{-1}(U) = \mathcal{B}_\rho(0)$ for $\mathcal{B}_\rho(0) \subseteq \mathbb{R}^2$ open and $\rho = \sqrt{x^2+y^2}$.

\end{itemize}

\vspace{1cm}
\item
To see that $\mathcal{C}$ is not a smooth manifold we need to show that there is weird behavior occuring in the derivative at the origin. The gradient corresponding to our surface takes the form:
\begin{equation*}
\renewcommand*{\arraystretch}{1.5}
\nabla f = \begin{pmatrix}
\frac{x}{\sqrt{x^2+y^2}} \\
\frac{y}{\sqrt{x^2+y^2}} \\
-1
\end{pmatrix}
\end{equation*}
To show that the gradient does not behave nicely at the origin we observe the value it decides to take on depending on the direction from which we approach. By letting $y = 0$ we arrive at:
\begin{equation*}
\nabla f = \begin{pmatrix}
\text{sgn}(x) \\
0 \\
-1
\end{pmatrix}
\end{equation*}
Knowing that $\text{sgn}(x) \rightarrow -1$ if $x \rightarrow 0^-$ and $\text{sgn}(x) \rightarrow 1$ if $x \rightarrow 0^+$ we arrive at the conclusion that $\nabla f$ cannot be defined at the origin. Consequently, there does not exist a diffeomorphism between $\mathcal{C}$ and $\mathbb{R}^2$ implying $\mathcal{C}$ is not a smooth manifold. 

\end{itemize}
\end{solution}

\newpage
\begin{problem}
Show that $u = xy$ and $v = y$ is not a good change of coordinates near the origin, while on the other hand $u = (x+.005)(y+.001)$ and $v = y$ is a good change of coordinates near the origin. 
\end{problem}

\vspace{.5cm}
\begin{solution}
\hspace{.1cm}
\begin{itemize}

\item
The first change of coordinates can be characterized as:
\begin{equation*}
f: \mathbb{R}^2 \rightarrow \mathbb{R}^2 \hspace{.3cm} \text{given by} \hspace{.3cm} f(x,y) = \begin{pmatrix} xy \\ y \end{pmatrix}
\end{equation*}
Where the Jacobian takes the form:
\begin{equation*}
J(f) = \begin{pmatrix}
y & x \\
0 & 1
\end{pmatrix}
\end{equation*}
As a result, $\det(J(f)) = y$. Therefore, the Jacobian is not of full rank along the $x$-axis. So for any neighborhood of the origin, $\mathcal{B}_r(0)$, we will always have the Jacobian non-invertible. By the Inverse Function Theorem we can say that the change of coordinates cannot be inverted at the origin.

\vspace{1cm}
\item
The second change of coordinates can be characterized as:
\begin{equation*}
g: \mathbb{R}^2 \rightarrow \mathbb{R}^2 \hspace{.3cm} \text{given by} \hspace{.3cm} g(x,y) = \begin{pmatrix} (x+.005)(y+.001) \\ y \end{pmatrix}
\end{equation*}
Where the Jacobian takes the form:
\begin{equation*}
J(g) = \begin{pmatrix}
y+.001 & x+.005 \\
0 & 1
\end{pmatrix}
\end{equation*}
As a result, $\det(J(g)) = y+.001$. Therefore, the Jacobian is not of full rank along a copy of the $x$-axis at height $y=-.001$. By the Inverse Function Theorem we can say that $g^{-1}$ exists in a neighborhood of the origin that does contain any of the bad points mentioned right before. For example you can consider $\mathcal{B}_{.0005}(0)$ to be such a neighborhood.

\end{itemize}
\end{solution}

\end{document}