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

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

\newcommand{\hookuparrow}{\mathrel{\rotatebox[origin=c]{90}{$\hookrightarrow$}}}

\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 \# 6}
\author{Nathan Marianovsky \\ October 23, 2017}

\begin{document}

\maketitle

\begin{problem}
Look at all lines in $\mathbb{R}^2$ and note that it forms a manifold. What is its dimension? Provide explicit charts.
\end{problem}

\vspace{.5cm}

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

\item
We note that the dimension of $\mathbb{RP}^1$ came out to be one since it only required information about the slope to identify a unique line in the space. Building off of this we can say that our manifold, $M$, is characterized as $M \cong \mathbb{RP}^1 \times \mathbb{R}$ because it needs the information about the slope from $\mathbb{RP}^1$ and another value for the intercept of the line. Therefore the dimension of $M$ must be two. 

\item
For $\mathbb{RP}^1$ it took two charts to cover the space since one line would always be missed by any chart. Using this as motivation we can take the two lines $x = 0$ and $y = 0$ and notice that it takes two charts to cover the space. 
\begin{itemize}

\item
To obtain all non-vertical lines we define the chart $f: U \subseteq M \rightarrow \mathbb{R}^2$ given by:
\begin{equation*}
f: (m,c) \rightarrow \{(x,y) \in \mathbb{R}^2 \hspace{.1cm} | \hspace{.1cm} y = mx + b\}
\end{equation*}

\item
To obtain all non-horizontal lines we define the chart $g: V \subseteq M \rightarrow \mathbb{R}^2$ given by:
\begin{equation*}
g: (m,c) \rightarrow \{(x,y) \in \mathbb{R}^2 \hspace{.1cm} | \hspace{.1cm} x = my + b\}
\end{equation*}

\end{itemize}
Intuition tells us that the overlap map will consist of all lines having non-zero/non-infinite slope. To see this explicitly we have:
\begin{equation*}
g \circ f^{-1}: \mathbb{R}^2 \rightarrow \mathbb{R}^2 \hspace{.3cm} \text{given by} \hspace{.3cm} \{(x,y) \in \mathbb{R}^2 \hspace{.1cm} | \hspace{.1cm} y = mx + b\} \rightarrow \Bigg\{(x,y) \in \mathbb{R}^2 \hspace{.1cm} \Bigg| \hspace{.1cm} x = \frac{1}{m}y - \frac{b}{m} \hspace{.1cm} \text{and} \hspace{.1cm} m \neq 0\Bigg\}
\end{equation*}

\item
It is also interesting to note that $M \cong S^1 \times \mathbb{R}$. The reason behind this is that $\mathbb{RP}^1 \cong S^1$.  To show this explicitly recall the stereographic atlas $\{(U_N,\phi_N),(U_S,\phi_S)\}$ for $U_N = S^1 \setminus \{(0,1)\}$ and $U_S = S^1 \setminus \{(0,-1)\}$ where the maps are given by:
\begin{equation*}
\phi_N: (x,y) \rightarrow \frac{x}{1-y} \hspace{.3cm} \text{and} \hspace{.3cm} \phi_S: (x,y) \rightarrow \frac{x}{1+y}
\end{equation*}
For $\mathbb{RP}^1$ we use the atlas $\{(U_1,\phi_1),(U_2,\phi_2)\}$ where $U_1 = U_2 = \mathbb{R} \setminus \{0\}$. The maps are given by:
\begin{equation*}
\phi_1: \{(x,y) \in \mathbb{R}^2 \hspace{.1cm} | \hspace{.1cm} y = mx\} \rightarrow \frac{y}{x} \hspace{.3cm} \text{and} \hspace{.3cm} \phi_2: \{(x,y) \in \mathbb{R}^2 \hspace{.1cm} | \hspace{.1cm} x = my\} \rightarrow \frac{x}{y}
\end{equation*}
Now we define the mapping:
\begin{equation*}
\psi: \mathbb{RP}^1 \rightarrow S^1 \hspace{.3cm} \text{given by} \hspace{.3cm} \psi(\mathit{l}) = \begin{cases}
(\phi_N^{-1} \circ \phi_1)(l) & \text{if} \hspace{.3cm} \mathit{l} \in U_1 \\
(\phi_S^{-1} \circ \phi_2)(l) & \text{if} \hspace{.3cm} \mathit{l} \in U_2
\end{cases}
\end{equation*}
which is a diffeomorphism between $\mathbb{RP}^1$ and $S^1$. 

\end{itemize}
\end{solution}

\vspace{1cm}
\begin{problem}
Prove that multiplication $(A,B) \rightarrow AB$ is a smooth map $SO(3,\mathbb{R}) \times SO(3,\mathbb{R}) \rightarrow SO(3,\mathbb{R})$. 
\end{problem}

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

\item
First we notice that $SO(3,\mathbb{R}) \times SO(3,\mathbb{R}) \hookrightarrow gl(3,\mathbb{R}) \times gl(3,\mathbb{R})$ via the identity mapping. Furthermore, the identity mapping is continuous and injective telling us that $SO(3,\mathbb{R}) \times SO(3,\mathbb{R})$ is an embedded manifold in \newline $gl(3,\mathbb{R}) \times gl(3,\mathbb{R})$. 

\item
Take $A,B \in SO(3,\mathbb{R})$ and notice $\det(AB) = \det(A)\det(B) = 1$ and $(AB)^{-1} = B^{-1}A^{-1} = B^TA^T = (AB)^T$. This tells us that $\mathfrak{Im}(f) \subseteq SO(3,\mathbb{R})$ where $f$ represents the matrix multiplication. To show inclusion in the other direction we consider the surjectivity of the map. For any $R \in SO(3,\mathbb{R})$ we can always find the pair \newline $(R,I) \in SO(3,\mathbb{R}) \times SO(3,\mathbb{R})$ s.t. $RI = R$ where $I$ is the identity matrix. Thus, $\mathfrak{Im}(f) = SO(3,\mathbb{R})$. 

\newpage
\item
Now we show that $f: gl(3,\mathbb{R}) \times gl(3,\mathbb{R}) \rightarrow gl(3,\mathbb{R})$ is a smooth map from which it will follow that it is also smooth on the multiplication of rotation matrices, an embedded manifold. Explicitly we have:
\begin{equation*}
AB = \begin{pmatrix}
a_1 & a_2 & a_3 \\
a_4 & a_5 & a_6 \\
a_7 & a_8 & a_9
\end{pmatrix} \begin{pmatrix}
b_1 & b_4 & b_7 \\
b_2 & b_5 & b_8 \\
b_3 & b_6 & b_9
\end{pmatrix} = \begin{pmatrix}
\leftarrow & \overrightarrow{v_1} & \rightarrow \\
\leftarrow & \overrightarrow{v_2} & \rightarrow \\
\leftarrow & \overrightarrow{v_3} & \rightarrow 
\end{pmatrix} \begin{pmatrix}
\uparrow & \uparrow & \uparrow \\
\overrightarrow{w_1} & \overrightarrow{w_2} & \overrightarrow{w_3} \\
\downarrow & \downarrow & \downarrow
\end{pmatrix} = \begin{pmatrix}
\langle v_1,w_1 \rangle & \langle v_1,w_2 \rangle & \langle v_1,w_3 \rangle \\
\langle v_2,w_1 \rangle & \langle v_2,w_2 \rangle & \langle v_2,w_3 \rangle \\
\langle v_3,w_1 \rangle & \langle v_3,w_2 \rangle & \langle v_3,w_3 \rangle
\end{pmatrix}
\end{equation*}
So using the charts $(W_1,\psi_1)$ and $(W_2,\psi_2)$ where:
\begin{equation*}
\begin{split}
\psi_1: W_1 \rightarrow \mathbb{R}^{18} \hspace{.3cm} &\text{given by} \hspace{.3cm} \psi_1(A,B) = (a_1,\dots,a_9,b_1,\dots,b_9) \\
\psi_2: W_2 \rightarrow \mathbb{R}^9 \hspace{.3cm} &\text{given by} \hspace{.3cm} \psi_2(C) = (c_1,\dots,c_9)
\end{split}
\end{equation*}
gives us a map from $\zeta = \psi_2 \circ f \circ \psi_1^{-1}: \mathbb{R}^{18} \rightarrow \mathbb{R}^9$:
\begin{equation*}
\zeta(a_1,\dots,a_9,b_1,\dots,b_9) = \begin{pmatrix}
\langle v_1,w_1 \rangle \\
\langle v_1,w_2 \rangle \\
\langle v_1,w_3 \rangle \\
\langle v_2,w_1 \rangle \\
\langle v_2,w_2 \rangle \\
\langle v_2,w_3 \rangle \\
\langle v_3,w_1 \rangle \\
\langle v_3,w_2 \rangle \\
\langle v_3,w_3 \rangle
\end{pmatrix}
\end{equation*}
Since each component of $\zeta$ is smooth, then so is $\zeta$. 

\end{itemize}
\end{solution}

\vspace{1cm}
\begin{problem}
Prove that inversion $A \rightarrow A^{-1}$ is a smooth map $SO(3,\mathbb{R}) \rightarrow SO(3,\mathbb{R})$. 
\end{problem}

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

\item
First we notice that $SO(3,\mathbb{R}) \hookrightarrow GL(3,\mathbb{R})$ via the identity mapping. Furthermore, the identity mapping is continuous and injective telling us that $SO(3,\mathbb{R})$ is an embedded manifold in $GL(3,\mathbb{R})$.

\item
Take $A \in SO(3,\mathbb{R})$ and notice $\det(A^{-1}) = \frac{1}{\det(A)} = 1$ and $(A^{-1})^T = (A^T)^{-1} = (A^{-1})^{-1}$. This tells us that $\mathfrak{Im}(g) \subseteq SO(3,\mathbb{R})$ where $g$ represents the matrix inversion. To show inclusion in the other direction we consider the surjectivity of the map. For any $R \in SO(3,\mathbb{R})$ we can always find a unique inverse $R^{-1} \in GL(3,\mathbb{R})$ s.t. $RR^{-1} = R^{-1}R = I$. Furthermore, if $R$ is rotation, then so is the inverse. Thus, $\mathfrak{Im}(f) = SO(3,\mathbb{R})$. 

\item
Now we show that $g: GL(3,\mathbb{R}) \rightarrow GL(3,\mathbb{R})$ is a smooth map from which it will follow that it is also smooth on rotation matrices, an embedded manifold. Explicitly we have:
\begin{equation*}
A^{-1} = \frac{1}{\det(A)} \begin{pmatrix}
M_{11} & -M_{21} & M_{31} \\
-M_{12} & M_{22} & -M_{32} \\
M_{13} & -M_{23} & M_{33}
\end{pmatrix}
\end{equation*}
where $M_{ij}$ represents the determinant of the $2 \times 2$ matrix that results from deleting row $i$ and column $j$ of $A$. This tells us that matrix inversion can be thought of as a map from $\mathbb{R}^9$ to $\mathbb{R}^9$ that sends $(a_1,\dots,a_9)$ to rational functions of those inputs. Since the determinant is non-zero, we can say that each rational function is smooth implying that $g$ is a smooth map.

\end{itemize}
\end{solution}

\vspace{1cm}
\begin{problem}
Let:
\begin{equation*}
E_3 = \begin{pmatrix}
0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\end{equation*}
Prove that $X(g) = gE_3$ defines a smooth vector field on $SO(3,\mathbb{R})$. 
\end{problem}

\newpage
\begin{solution}
For an arbitrary $g \in SO(3,\mathbb{R})$ we can see the action of the derivation directly:
\begin{equation*}
X(g) = gE_3 = \begin{pmatrix}
g_1 & g_2 & g_3 \\
g_4 & g_5 & g_6 \\
g_7 & g_8 & g_9
\end{pmatrix} \begin{pmatrix}
0 & -1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix} = \begin{pmatrix}
g_2 & -g_1 & 0 \\
g_5 & -g_4 & 0 \\
g_8 & -g_7 & 0
\end{pmatrix}
\end{equation*}
Therefore, the corresponding vector field has to take the form $V(x,y) = y\frac{\partial}{\partial x} - x\frac{\partial}{\partial y}$. The correspondence comes from observing the behavior of the column vectors. Furthermore, since each component function is smooth, so is the vector field.
\end{solution}

\vspace{1cm}
\begin{problem}
Prove that the map $\pi: SO(3,\mathbb{R}) \rightarrow S^2$ given by $\pi(g) = ge_3$ is a submersion. 
\end{problem}

\begin{solution}
Notice what this map provides explicitly:
\begin{equation*}
\pi(g) = \begin{pmatrix}
g_1 & g_2 & g_3 \\
g_4 & g_5 & g_6 \\
g_7 & g_8 & g_9
\end{pmatrix} \begin{pmatrix}
0 \\ 0 \\ 1
\end{pmatrix} = \begin{pmatrix}
g_3 \\ g_6 \\ g_9
\end{pmatrix}
\end{equation*}
Using the fact that the column vectors of any rotation matrix make up an orthonormal set implies $g_3^2 + g_6^2 + g_9^2 = 1$. Thus, $\pi(g) \in S^2$. To show that this map is a submersion means we have to show the Jacobian is surjective. To accomplish such a task we need to consider the lifting:
\begin{equation*}
\begin{split}
GL(3,\mathbb{R}) &\xrightarrow{\tilde{\pi}} (\mathbb{R}^3)^\times \\
\kappa \big\uparrow \hspace{1cm} & \hspace{.6cm} \big\uparrow \alpha \\
SO(3,\mathbb{R}) & \xrightarrow{\pi} S^2 
\end{split}
\end{equation*}
where $\kappa(R) = R$, $\tilde{\pi}(A) = Ae_3$, and $\alpha(\overrightarrow{p}) = \overrightarrow{p}$. Since $SO(3,\mathbb{R}) \hookrightarrow GL(3,\mathbb{R})$ and $S^2 \hookrightarrow (\mathbb{R}^3)^\times$ via the identity maps, they are embedded manifolds. Furthermore, we can see that:
\begin{equation*}
J(\tilde{\pi}) = \begin{pmatrix}
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{pmatrix}
\end{equation*}
which has full rank showing that $\tilde{\pi}$ is a submersion of $GL(3,\mathbb{R})$ into $(\mathbb{R}^3)^\times$. Therefore, $\pi$ is also a submersion since it inherits the behavior from $\tilde{\pi}$. 
\end{solution}

\vspace{1cm}
\begin{problem}
As one of the HWs the coordinate transition map $x \rightarrow \frac{1}{x}$ came up for switching from one chart to another in the standard affine charts of $\mathbb{RP}^1$. Instead of viewing this map passively as a change of coordinates, we can look at it actively, as a transformation of $F: \mathbb{RP}^1 \rightarrow \mathbb{RP}^1$, defined within the single affine chart $Y \neq 0$. Find the linear transformation $L: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ which induces the map $f(x) = \frac{1}{x}$ in this standard affine chart.
\end{problem}

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

\item
From the viewpoint of $X = 1$ we have to satisfy the mappings $(1,m) \rightarrow \Big( \frac{1}{m},1 \Big)$ and $(1,0) \rightarrow (0,1)$. The linear transformation must therefore take the form:
\begin{equation*}
L = \begin{pmatrix}
0 & \frac{1}{m^2} \\
1 & 0
\end{pmatrix}
\end{equation*}
which misses the infinite point $(X = 0)$.

\item
From the viewpoint of $Y = 1$ we have to satisfy the mappings $\Big( \frac{1}{m},1 \Big) \rightarrow (1,m)$ and $(0,1) \rightarrow (1,0)$. The linear transformation must therefore take the form:
\begin{equation*}
L = \begin{pmatrix}
0 & 1 \\
m^2 & 0
\end{pmatrix}
\end{equation*}
which misses the infinite point $(Y = 0)$.

\end{itemize}
\end{solution}

\end{document}