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\title{MATH 208 - HW \# 6 - Corrections}
\author{Nathan Marianovsky \\ October 27, 2017}

\begin{document}

\maketitle

\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}

\begin{solution}
Begin by considering the map $f: GL(3,\mathbb{R}) \rightarrow SO(3,\mathbb{R})$ given by $f(A) = AA^T$. We can explicitly compute the derivative as:
\begin{equation*}
\begin{split}
f'_A(h) &= \frac{\text{d}}{\text{d}\epsilon} \Big|_{\epsilon=0} f(A+\epsilon h) \\
&= \frac{\text{d}}{\text{d}\epsilon} \Big|_{\epsilon=0} (A+\epsilon h)(A+\epsilon h)^T \\
&= \frac{\text{d}}{\text{d}\epsilon} \Big|_{\epsilon=0} (AA^T + \epsilon(Ah^T + hA^T) + \epsilon^2 hh^T) \\
&= \Big( (Ah^T + hA^T) + 2\epsilon hh^T \Big) \Big|_{\epsilon=0} \\
&= Ah^T + hA^T
\end{split}
\end{equation*}
If we restrict $A \in SO(3,\mathbb{R})$, then $f(A) = I$ and consequently $Ah^T + hA^T = 0$. Furthermore, if $A = I$, then we actually end up with $h^T = -h$ implying $h \in \mathfrak{so}(3,\mathbb{R})$. Thus, $3 \times 3$ skew-symmetric matrices span the tangent space for $SO(3,\mathbb{R})$ (naturally dimensions also match up). Knowing this information tells us that a vector field will consist of component functions tagged to a direction given by a skew-symmetric matrix. In fact, $E_3 \in \mathfrak{so}(3,\mathbb{R})$ which tells us $X(g)$ is a vector field. To see that is smooth, we just observe that the action of the coordinate function is matrix multiplication which has already been proven to be smooth. 
\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. First we prove the Equivariant Rank Theorem: Let $M$ and $N$ be smooth manifolds and $G$ a Lie group. Suppose $F:M \rightarrow N$ is a smooth map that is equivariant with respect to a transitive smooth $G$-action on $N$ and $M$. Then $F$ has constant rank.  
\begin{itemize}

\item
Pick an arbitrary point $p \in M$ and let $\phi$ and $\psi$ denote the $G$-actions on $M$ and $N$ respectively. Due to the transitivity of $G$ we can always find a $g \in G$ s.t. $\phi_g(p) = q$ and $\psi_g(F(p)) = F(q)$, which combined with $\psi_g \circ F = F \circ \phi_g$ provides the commutative diagram:
\begin{equation*}
\begin{split}
&T_pM \xrightarrow{\text{d}F_p} T_{F(p)}N \\
\text{d}\phi_g &\big\downarrow \vin \hspace{1.85cm} \win \big\downarrow \text{d}\psi_g \\
&T_qM \xrightarrow[\text{d}F_q]{} T_{F(q)}N
\end{split}
\end{equation*}
The vertical lines are isomorphisms thereby giving that $\text{d}F_p$ and $\text{d}F_q$ have the same rank. Since $p$ and $q$ are arbitrary, $F$ has constant rank. 


\end{itemize}
This applies directly to our scenario because if we take $G = SO(3,\mathbb{R})$, $M = SO(3,\mathbb{R})$, and $N = S^2$ we know that there is always a rotation taking $\overrightarrow{\pi(R)} \in S^2$ to $\overrightarrow{\pi(P)} \in S^2$. Furthermore, there is always a rotation that can take any $R \in SO(3,\mathbb{R})$ to $P \in SO(3,\mathbb{R})$. Thus, $\pi: SO(3,\mathbb{R}) \rightarrow S^2$ has constant rank. Now we want to show that a smooth surjective map of constant rank is submersion. 
\begin{itemize}

\item
Assume we have $F: M \rightarrow N$ smooth and constant rank. Suppose $F$ is not a submersion, implying that \newline $\text{rank}(F) = k < n = \dim(N)$. The rank theorem lets us know that at each point there is a smooth coordinate neighborhood in which $F$ obtains the representation:
\begin{equation*}
F(x^1,\dots,x^k,x^{k+1},\dots,x^m) = (x^1,\dots,x^k,0,\dots,0)
\end{equation*}
For any open cover we can pick countably many smooth charts $\{(U_i,\phi_i)\}$ for $M$ and corresponding charts $\{(V_i,\psi_i)\}$ for $N$ s.t. the sets $\{U_i\}$ cover $M$ and $F$ maps $U_i$ into $V_i$. Now using the fact that any $k<n$ dimensional space will be a measure zero set in $\mathbb{R}^n$ when using the Lebesgue measure implies $m(F(U_i)) = 0$. Consequently:
\begin{equation*}
m\Big( \bigcup F(U_i) \Big) \leq \sum m(F(U_i)) = 0
\end{equation*}
telling us that $F(M) \subset N$ is a measure zero set. Therefore, $F$ cannot be surjective. We have proven that \newline "not submersion $\implies$ not surjective" which is logically equivalent to "surjective $\implies$ submersion" for a constant rank map.

\end{itemize}
The only thing left is to convince ourselves of the surjectivity of $F: SO(3,\mathbb{R}) \rightarrow S^2$. The Gram-Schmidt process, given the set $\{v_1,v_2,v_3\}$, spits out the vectors $\{e_1,e_2,e_3\}$ where:
\begin{equation*}
\begin{split}
u_1 = v_1 \hspace{.3cm} &\text{and} \hspace{.3cm} e_1 = \frac{u_1}{\|u_1\|} \\
u_2 = v_2 - \text{proj}_{u_1}(v_2) \hspace{.3cm} &\text{and} \hspace{.3cm} e_2 = \frac{u_2}{\|u_2\|} \\
u_3 = v_3 - \text{proj}_{u_1}(v_3) - \text{proj}_{u_2}(v_3) \hspace{.3cm} &\text{and} \hspace{.3cm} e_3 = \frac{u_3}{\|u_3\|} \\
\end{split}
\end{equation*}
where the new set consists of orthonormal vectors that could be identified as the columns or rows of a rotation matrix. To achieve surjectivity we take $v_1 = (a,b,c) \in S^2$ and note that $e_1 = v_1$ where this can be the last column specifically since order will not matter. So the rest of the task rests on finding two non-zero vectors in $\mathbb{R}^3$, not necessarily normalized, that are linearly independent of $v_1$. Since this is always possible we can take our list of vectors, input them into the Gram Schmidt process, and have the process spit out the necessary vectors to write a rotational matrix associated to our point on $S^2$. Finally, surjectivity combined with the fact that $F$ is constant rank tells us $F$ is a submersion. Note that all of this logic can be extended to say $F: SO(n+1,\mathbb{R}) \rightarrow S^n$ is a submersion. 
\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}
To determine such a transformation we just need to look at three points. First we have the extremes:
\begin{equation*}
\mathcal{M}\begin{pmatrix}
0 \\ 1
\end{pmatrix} = \begin{pmatrix}
k_1 \\ 0
\end{pmatrix} \hspace{.3cm} \text{and} \hspace{.3cm} \mathcal{M}\begin{pmatrix}
1 \\ 0
\end{pmatrix} = \begin{pmatrix}
0 \\ k_2
\end{pmatrix}
\end{equation*}
which forces:
\begin{equation*}
\mathcal{M} = \begin{pmatrix}
0 & k_1 \\ k_2 & 0
\end{pmatrix}
\end{equation*}
Since we are looking to invert over the line $y=x$, we also need:
\begin{equation*}
\mathcal{M}\begin{pmatrix}
1 \\ 1
\end{pmatrix} = \begin{pmatrix}
m \\ m
\end{pmatrix}
\end{equation*}
which actually forces $k_1=k_2$. Furthermore, since any scalar multiple will do, the nicest choice is:
\begin{equation*}
\mathcal{M} = \begin{pmatrix}
0 & 1 \\ 1 & 0
\end{pmatrix}
\end{equation*}
\end{solution}

\end{document}