% file: DGDT_dump.tex
% Differential Geometry, Differential Topology, in unconventional ``grande'' format; fitting a widescreen format
%
% github        : ernestyalumni
% linkedin      : ernestyalumni
% wordpress.com : ernestyalumni
%
% This code is open-source, governed by the Creative Common license.  Use of this code is governed by the Caltech Honor Code: ``No member of the Caltech community shall take unfair advantage of any other member of the Caltech community.''
%

\documentclass[10pt]{amsart}
\pdfoutput=1
%\usepackage{mathtools,amssymb,lipsum,caption}
\usepackage{mathtools,amssymb,caption}

\usepackage{graphicx}
\usepackage{hyperref}
\usepackage[utf8]{inputenc}
\usepackage{listings}
\usepackage[table]{xcolor}
\usepackage{pdfpages}
%\usepackage[version=3]{mhchem}
%\usepackage{mhchem}

\usepackage{tikz}
\usetikzlibrary{matrix,arrows}

\usepackage{multicol}

\hypersetup{colorlinks=true,citecolor=[rgb]{0,0.4,0}}

\oddsidemargin=15pt
\evensidemargin=5pt
\hoffset-45pt
\voffset-55pt
\topmargin=-4pt
\headsep=5pt
\textwidth=1120pt
\textheight=595pt
\paperwidth=1200pt
\paperheight=700pt
\footskip=40pt

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}{Corollary}
%\newtheorem*{main}{Main Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{proposition}{Proposition}

\newtheorem{definition}{Definition}
\newtheorem{remark}{Remark}

\newenvironment{claim}[1]{\par\noindent\underline{Claim:}\space#1}{}
\newenvironment{claimproof}[1]{\par\noindent\underline{Proof:}\space#1}{\hfill $\blacksquare$}

%This defines a new command \questionhead which takes one argument and
%prints out Question #. with some space.
\newcommand{\questionhead}[1]
  {\bigskip\bigskip
   \noindent{\small\bf Question #1.}
   \bigskip}

\newcommand{\problemhead}[1]
  {
   \noindent{\small\bf Problem #1.}
   }

\newcommand{\exercisehead}[1]
  { \smallskip
   \noindent{\small\bf Exercise #1.}
  }

\newcommand{\solutionhead}[1]
  {
   \noindent{\small\bf Solution #1.}
   }

\title{The Differential Geometry Differential Topology Dump}
\author{Ernest Yeung \href{mailto:ernestyalumni@gmail.com}{ernestyalumni@gmail.com}}
\date{28 juillet 2016}
\keywords{Differential Geometry, Differential Topology}
\begin{document}

\definecolor{darkgreen}{rgb}{0,0.4,0}
\lstset{language=Python,
 frame=bottomline,
 basicstyle=\scriptsize,
 identifierstyle=\color{blue},
 keywordstyle=\bfseries,
 commentstyle=\color{darkgreen},
 stringstyle=\color{red},
 }
%\lstlistoflistings

\maketitle

\begin{multicols*}{2}

\setcounter{tocdepth}{1}
\tableofcontents

\begin{abstract}
Everything about Differential Geometry, Differential Topology

\end{abstract}

\part{Combinatorics, Probability Theory}

\begin{theorem}[4.2. of Feller (1968)  \cite{Fell1968}]
	Let $r_1,\dots r_k \in \mathbb{Z}$, s.t. $r_1 + r_2 + \dots + r_k = n$;, $r_i \geq 0$.

Let
\begin{equation}
\frac{ N! }{r_1! r_2! \dots r_k!} =
\end{equation}
number of ways in which $n$ elemnts can be divided into $k$ ordered parts (partitioned into $k$ subpopulations).  cf. Eq. (4.7) of Feller (1968)  \cite{Fell1968}.

Note that the order of the subpopulations is essential in the sense that ($r_1 = 2,r_2 =3$) and ($r_1=3,r_2=2$) represent different partitions.  However, no attention is paid to the order within the groups.

\end{theorem}
\begin{proof}
\begin{equation}
\binom{n}{r_1} \binom{n-r_1}{r_2} \binom{n-r_1-r_2}{r_3} \dots \binom{n-r_1-\dots - r_{k-2} }{ r_{k-1} } = \frac{n!}{ r_1! r_2! \dots r_k! }
\end{equation}
i.e. in order to effect the desired partition, we have to select $r_1$ elementsout of $n$, remaining $n-r_1$ elements select a second group of size $r_2$, etc.  After forming the $(k-1)$st group there remains $n-r_1 -r_2 - \dots - r_{k-1} = r_k$ elements, and these form the last group.
\end{proof}

cf. pp. 37 of Feller (1968)  \cite{Fell1968}
Examples.  (g) Bridge.  32 cards are partitioned into 4 equal groups $\to 52!/(13!)^4$.

Probability each player has an ace (?).

The 4 aces can be ordered in $4! = 24$ ways, each order presents 1 possibility of giving 1 ace to each player.  \\
Remaining 48 cards distributed $(48!)/( 12!)^4$ ways.

\[
\to p = 24 \frac{ 48!}{ (12!)^4 } / \frac{ 52!}{ (13!)^4}
\]

(h) A throw of 12 dice $\to  6^{12}$ different outcomes total.  Event each face appears twice can occur in as many ways as 12 dice can be arranged in 6 groups of 2 each.
\[
\frac{12!}{(2!)^6} / \frac{ 52!}{ (13!)^4}
\]

\subsubsection{Application to Occupancy Problems; binomial coefficients}

cf. Sec. 5 Application to Occupancy Problems of Feller (1968)  \cite{Fell1968}.

Consider randomly placing $r$ balls intos $n$ cells.

Let $r_k = $ occupancy number $=$ number of balls in $k$th cell.

Every $n$-tuple of integers satisfying $r_1 + r_2 + \dots + r_n = r$;  $r_k \geq 0$.  describes a possible configuration of occupancy numbers.

With indistinguishable balls 2 distributions are distinguishable only if the corresponding $n$-tuples ($r_1,\dots r_n$) are not identical.

\begin{enumerate}
\item[(i)] number of distinguishable distributions is
\begin{equation}
A_{r,n} = \binom{n+r-1}{r} = \binom{n+r-1}{n-1}
\end{equation} cf. Eq. (5.2) of Feller (1968)  \cite{Fell1968}
\item[(ii)]  number of distinguishable distributions in which no cell remains empty is $\binom{r-1}{n-1}$.
\end{enumerate}

\begin{proof}
Represent balls by stars, indicate $n$ cells by $n$ spaces between $n+1$ bars.  e.g. $\begin{aligned} & \quad \\
& r= 8 \text{ balls } \\
& n = 6 \text{ cells } \end{aligned}$.

\[
\begin{aligned}
	& 3 	  & 1   & 0 & 0 & 0 & 4 \\
	& | * * * & | * & | & |   & | & | **** |
\end{aligned}
\]

Such a symbol necessarily starts and ends with a bar, but remaining $n-1$ bars and $r$ starts appear in an arbitrary order.  In this way, it becomes apparent that the number of distinguishable distributions equals the number of ways of selecting.

$r$ places out of $n+r-1$, $\frac{ (n+r-1)! }{ (n-1)! r!} = \binom{n-1+r}{r} $

\[
\begin{aligned}
	& | | | | \dots | | \qquad \, n+1 \text{ bars } \\
	& * * * \dots * * \qquad \, r \text{ stars leave } r-1 \text{ spaces }
\end{aligned}
\]

Condition that no cell be empty imposes the restriction that no 2 bars be adjacent.  $r$ stars leave $r-1$ spaces of which $n-1$ are to be occupied by bars.  Thus $\binom{r-1}{n-1}$ choices.

\end{proof}

Probability to obtain given occupancy numbers $r_1,\dots r_n = \frac{r!}{ r_1! r_2! \dots r_n! } / n^r$, with $ \begin{aligned} & \quad \\
	& r \text{ balls } \\
& n \text{ cells } \end{aligned}$  given by Thm. 4.2. of Feller (1968)  \cite{Fell1968}, which is the Maxwell-Boltzmann distribution.

\begin{enumerate}
\item[(a)] Bose-Einstein and Fermi-Dirac statistics.  Consider $r$ indistinguishable particles, $n$ cells, each particle assigned to 1 cell.

State of the system - random distribution of $r$ particles in $n$ cells.

If $n$ cells distinguishable, $n^r$ arrangements equiprobable $\to $ Maxwell-Boltzmann statistics.

Bose-Einstein statistics: only distinguishable arrangements are considered, and each assigned probability $\frac{1}{A_{r,n}}$
\begin{equation}
A_{r,n} = \binom{n+r-1}{r} = \binom{n-1+r}{n-1}
\end{equation}
cf. Eq. 5.2 of Feller (1968)  \cite{Fell1968}

Fermi-Dirac statistics.
\begin{enumerate}
\item[(1)] impossible for 2 or more particles to be in the same cell.  $\to r \leq n$.
\item[(2)] all distinguishable arrangements satisfying the first condition have equal probabilities.

$\to$ an arrangement is completely described by stating which of the $n$ cells contain a particle

$r$ particles $\to \binom{n}{r}$ ways $r$ cells chosen.

Fermi-Dirac statistics, there are $\binom{n}{r}$ possible arrangements, prob. $1/\binom{n}{r}$.
\end{enumerate}
\end{enumerate}
pp. 39.  Feller (1968)  \cite{Fell1968}.  Consider cells themselves indistinguishable!  Disregard order among occupancy numbers.

cf. Feller (1968)  \cite{Fell1968}

\part{Linear Algebra Review}

cf. \emph{Change of Basis}, of Appendix B of  John Lee (2012) \cite{JLee2012}.

\exercisehead{B.22} Suppose $V,W, X$ finite-dim. vector spaces \\
$S:V\to W$, \, $T:W \to X$

\begin{enumerate}
	\item[(a)] $\text{rank}S \leq \text{dim}V$ \quad \, with $\text{rank}S = \text{dim}V$ iff $S$ injective
	\item[(b)] $\text{rank}S \leq \text{dim}W$ \quad \, with $\text{rank}S = \text{dim}W$ iff $S$ surjective
	\item[(c)] if $\text{dim}V = \text{dim}W$ and $S$ either injective or surjective, then $S$ isomorphism
	\item[(d)] $\text{rank}TS \leq \text{rank}S$ \quad \, $\text{rank}TS = \text{rank}S$ iff $\text{im}S \bigcap \text{ker}T = 0$
	\item[(e)] $\text{rank}TS \leq \text{rank}T$ \quad \, $\text{rank}TS = \text{rank}T$ iff $\text{im}S + \text{ker}T = W$
	\item[(f)] if $S$ isomorphism, then $\text{rank}TS = \text{rank}T$
	\item[(g)] if $T$ isomorphism, then $\text{rank}TS = \text{rank}S$
\end{enumerate}

EY : Exercise B.22(d) is useful for showing the chart and atlas of a Grassmannian manifold, found in the More examples, for smooth manifolds.

\begin{proof}
	\begin{enumerate}
		\item[(a)] Recall the \textbf{rank-nullity theorem}:
		\begin{theorem}[Rank-Nullity Theorem]
\begin{equation}
\text{dim}(\text{im}(S)) + \text{dim}(\text{ker}(S)) = \text{dim}V
\end{equation}
			\end{theorem}
		Now
		\[
		\begin{gathered}
		\text{rank}(S) + \text{dim}(\text{ker}(S)) \equiv \text{dim}(\text{im}(S)) + \text{dim}(\text{ker}(S)) = \text{dim}V  \\
\Longrightarrow \boxed{ 	\text{rank}(S) \leq \text{dim}V  }
\end{gathered}
		\]

		If $\text{rank}(S) = \text{dim}{V}$, \\
		then by rank-nullity theorem, $\text{dim}(\text{ker}(S)) = 0$, implying that $\text{ker}S = \lbrace 0 \rbrace$.  \\
		Suppose $v_1, v_2 \in V$ and that $S(v_1) = S(v_2) $.  By linearity of $S$, $S(v_1) - S(v_2) = S(v_1-v_2) = 0$, which implies, since $\text{ker}S = \lbrace 0 \rbrace$, that $v_1 - v_2 = 0$.  \\
		$\Longrightarrow v_1 = v_2$.  Then by definition of injectivity, $S$ injective.

		If $S$ injective, then $S(v)=0$ implies $v=0$.  Then $\text{ker}S = \lbrace 0 \rbrace$.  Then by rank-nullity theorem, $\text{rank}(S) = \text{dim}{V}$.

		\item[(b)] 		$\forall \, w \in \text{im}(S)$, $w\in W$.  Clearly $\text{rank}S \leq \text{dim}W$.

		If $S$ surjective, $\text{im}(S) = W$.  Then $\text{dim}(\text{im}(S)) = \text{rank}S = \text{dim}W$.  \\

		If $\text{rank}S = \text{dim}W = m$, then $\text{im}(S)$ has basis $\lbrace y_i \rbrace_{i=1}^m$, $y_i \in \text{im}(S)$, so $\exists \, x_i \in V$, $i=1\dots m$ s.t. $S(x_i) = y_i$, with $\lbrace S(x_i) \rbrace_{i=1}^m $ linearly independent.

		Since $\lbrace S(x_i) \rbrace_{i=1}^m$ linearly independent and $\text{dim}W = m$, $\lbrace S(x_i) \rbrace_{i=1}^m$ basis for $W$.   \\
		$\forall \, w \in W$, $w=\sum_{i=1}^m w^i S(x_i) = S(\sum_{i=1}^m w^i x_i)$.  $\sum_{i=1}^m w^i x_i \in V$.  $S$ surjective.

		\item[(c)]
		\item[(d)] Now
		\[
		\begin{aligned}
		& \text{dim}V = \text{rank}TS + \text{nullity}TS \\
		&  \text{dim}V = \text{rank}S + \text{nullity}S
		\end{aligned}
		\]
		$\text{ker}S \subseteq \text{ker}TS$, clearly, so $\text{nullity}S \leq \text{nullity}TS$
		\[
		\Longrightarrow \boxed{ \text{rank}TS \leq \text{rank}S }
		\]

		If $\text{rank}TS = \text{rank}S$, \\
		\phantom{ \quad } then $\text{nullity}S = \text{nullity}TS$ \\
		\phantom{ \, } Suppose $w \in \text{Im}S \bigcap \text{ker}T$, $w \neq 0$ \\
		\phantom{ \quad } Then $\exists \,  v\in S$, s.t. $w = S(v)$ and $T(w)=0$ \\
		\phantom{ \quad \, } Then $T(w) = TS(v) =0$.  So $v\in \text{ker}TS$ \\
		\phantom{ \quad \quad \, } $v\notin \text{ker}S$ since $w = S(v) \neq 0$ \\
		\phantom{ \quad \quad \, } This implies $\text{nullity}TS > \text{nullity}S$.  Contradiction. \\
		$\Longrightarrow \text{Im}S \bigcap \text{ker}T =0$ \\

		If $\text{Im}S \bigcap \text{ker}T =0$, \\
		\phantom{ \quad } Consider $v \in \text{ker}TS$.  Then $TS(v)=0$.  \\
		\phantom{ \quad  Consider $v \in \text{ker}TS$}.  Then $S(v)  \in \text{ker}T$ \\
		\phantom{ \quad  } $S(v) =0$; otherwise, $S(v) \in \text{Im}S$, contradicting given $\text{Im}S \bigcap \text{ker}T =0$ \\
		\phantom{ \quad \quad } $v\in \text{ker}S$ \\

		$\text{ker}TS \subseteq \text{ker}S$\\
		$\Longrightarrow \text{ker}TS = \text{ker}S$ \\
		So $\text{nullity}TS = \text{nullity}S$  \\
		$\Longrightarrow \text{rank}TS = \text{rank}S$

		\item[(e)]
		\item[(f)]
		\item[(g)]
	\end{enumerate}
\end{proof}

\part{Manifolds}

\section{Inverse Function Theorem}

Shastri (2011) had a thorough and lucid and explicit explanation of the Inverse Function Theorem \cite{AShastri2011}.  I will recap it here.  The following is also a blend of Wienhard's Handout 4 \url{https://web.math.princeton.edu/~wienhard/teaching/M327/handout4.pdf}

\begin{definition}
  Let $(X,a)$ metric space.

\textbf{contraction} $\phi:X \to X$ if $\exists \, $ constant $0<c<1$ s.t. $\forall \, x,y \in X$
\[
d(\phi(x),\phi(y)) \leq cd(x,y)
\]
\end{definition}

\begin{theorem}[Contraction Mapping Principle]
  Let $(X,d)$ complete metric space.  \\
Then $\forall \, $ contraction $\phi:X\to X$, $\exists \, ! y\in X$ s.t. $\phi(y) = y$, $y$ \emph{fixed pt.}
\end{theorem}

\begin{proof}
  Recall def. of complete metric space $X$, $X$ metric space s.t. $\forall \, $ Cauchy sequence in $X$ is convergent in $X$ (i.e. has limit in $X$).

$\forall \, x_0 \in X$,
Define $\begin{aligned} & \quad \\
  & x_1 = \phi(x_0) \\
  & x_2 = \phi(x_1) \\
  & \vdots \\
  & x_j = \phi(x_{j-1}) \\
  & \vdots \\
  & x_n = \phi(x_{n-1})
\end{aligned}$

\[
\begin{gathered}
  d(x_{n+1},x_n) = d(\phi(x_n),\phi(x_{n-1})) \leq c d(x_n,x_{n-1}) \leq \dots \leq c^nd(x_1,x_0)
\end{gathered}
\]
for some $0< c<1$.

\[
d(x_m,x_n) \leq d(x_n,x_{n-1}) + d(x_{n-1},x_m) \leq d(x_n,x_{n-1}) + d(x_{n-1},x_{n-2}) + \dots + d(x_{m+1},x_m) \leq \sum_{k=n-1}^m c^k d(x_1,x_0)
\]
Thus, $\forall \, \epsilon >0$, $\exists \, n_0 >0$, ($n_0$ large enough) s.t. $\forall \, m ,n\in \mathbb{N}$ s.t. $n_0 < n <m$,
\[
d(x_m,x_n) \leq \sum_{k=n-1}^m c^k d(x_1,x_0) < \epsilon d(x_1,x_0)
\]
Thus, $\lbrace x_n \rbrace$ Cauchy sequence.  Since $X$ complete, $\exists \, $ limit pt. $y \in X$ of $\lbrace x_n \rbrace$.
\[
\phi(y) = \phi(\lim_n x_n) = \lim_n \phi(x_n) = \lim_n x_{n+1} = y
\]
Since by def. of $y$ limit pt. of $\lbrace x_n \rbrace$, $\forall \, \epsilon >0$, then $\lbrace n | |x_n -y|\leq \epsilon, \, n \in \mathbb{N}\rbrace$ is infinite.

Consider $\delta > \mathbb{N}$.  Consider $\lbrace n | |x_n-y| \leq \delta, n \in \mathbb{N}\rbrace$

$\exists \, N_{\delta} \in \mathbb{N}$ s.t. $\forall \, n > N_{\delta}$, $|x_n-y|< \delta$; otherwise, $\forall \, N_{\delta}$, $\exists \, n > N_{\delta}$ s.t. $|x_n - y| \geq \delta$.  Then $\lbrace n | |x_n -y| \leq \delta , n \in \mathbb{N} \rbrace$ finite.  Contradiction.

$\phi$ cont. so by def. $\forall \, \epsilon >0$, $\exists \, \delta >0$ s.t. if $|x_n -y| < \delta$, then $|\phi(x_n) - \phi(y) | < \epsilon$.

Pick $N_{\delta}$ s.t. $\forall \, n > N_{\delta}$, $|x_n-y| < \delta$, and so $|\phi(x_n) - \phi(y)|< \epsilon$. There are infinitely many $\phi(x_n)$'s that satisfy this, and so $\phi(y)$ is a limit pt.

If $\exists \, y_1,y_2 \in X$ s.t. $\begin{aligned} & \quad \\
  & \phi(y_1) = y_1 \\
  & \phi(y_2) = y_2 \end{aligned}$, then
\[
d(y_1,y_2) = d(\phi(y_1), \phi(y_2)) \leq c d(y_1,y_2) \text{ with } c <1
\]
so $c=1$
\end{proof}

\begin{theorem}[Inverse Function Theorem]
  Suppose open $U \subset \mathbb{R}^n$, let $C^1 \, f: U \to \mathbb{R}^n$, $x_0 \in U$ s.t. $Df(x_0)$ invertible.

%  Let open $E \subset \mathbb{R}^n$, $0 \subset E$, let $f \in \mathcal{C}^1(E,\mathbb{R}^n)$ s.t. $Df(0)$ invertible.

Then $\exists \,$ open neighborhoods $V\ni x_0$, $W \ni f(x_0)$ s.t. $V\subseteq U$ and $W\subseteq \mathbb{R}^n$, respectively, and s.t.

\begin{enumerate}
\item[(i)] $f: V\to W$ bijection
\item[(ii)] $g = f^{-1}:V \to U$ differentiable, i.e. $g = f^{-1}:W\to V$ is $C^1$
\item[(iii)] $D(f^{-1}) $ cont. on $W$.
\item[(iv)] $Dg(y) = (Df(g(y)))^{-1}$ \, $\forall \, y \in W$
\end{enumerate}
Also, notice that $f(g(y)) = y \, \forall \, y \in W$.
\end{theorem}

\begin{proof}
%  Let $A = Df(0)$, consider $\widehat{f} = A^{-1}\circ f$.  Then $\widehat{f} \in (U;\mathbb{R}^n)$.  $D(\widehat{f})(0)=1$ since $D(\widehat{f})(0) = D(A^{-1} \circ f)(0)=A^{-1}Df(0)=1$.

Consider $\widetilde{f}(x) = (Df(x_0))^{-1}(f(x+x_0) - f(x_0))$.  Then \\
\phantom{Consider} $\widetilde{f}(0) = 0$ and
\[
\begin{aligned}
  &  D\widetilde{f}= (Df(x_0))^{-1}(Df(x+x_0) -0)  \\
  & D\widetilde{f}(0) = (Df(x_0))^{-1}Df(x_0)=1
\end{aligned}
\]
So let $\widetilde{f}\to f$ (notation) and so assume, without loss of generality, that $U\ni 0$, $f(0)=0$, $Df(0)=1$

Choose $0 < \epsilon \leq \frac{1}{2}$.  Let $0< \delta <1$ s.t. open ball $V = B_{\delta}(0) \subseteq U$, and $\| Df(x)-1\| < \epsilon$.  $\forall \, x \in U$, since $Df$ cont. at $0$.

Let $W=f(V)$.

$\forall \, y \in W$, define $\begin{aligned} & \quad \\
  & \phi_y : V \to \mathbb{R}^n \\
  & \phi_y(x) = x + (y-f(x))\end{aligned}$

\[
\begin{aligned}
  & D(\phi_y)(x) = 1 + - Df(x) \quad \, \forall \, x \in V \\
  & \| D(\phi_y)(x) \| = \| 1 - Df(x) \| \leq \epsilon <1
\end{aligned}
\]

$\forall \, x_1 ,x_2 \in V$, by mean value Thm. (not the equality that is only valid in 1-dim., but the inequality, that's valid for $\mathbb{R}^d$,
\[
\| \phi_y(x_1) - \phi_y(x_2) \| \leq \| D(\phi_y)(x') \| \| x_1 - x_2 \|
\]
for some $x' = cx_2 + (1-c)x_1$, $c\in [0,1]$.  $V$ only needed to be convex set.
\[
\Longrightarrow \| \phi_y(x_1) - \phi_y(x_2) \| \leq \epsilon \| x_1 - x_2 \|
\]
Then $\phi_y$ contraction mapping.

Suppose $f(x_1) = f(x_2)=y$, $x_1,x_2 \in V$.
\[
\begin{gathered}
\begin{aligned}
  & \phi_y(x_1) =x_1 \\
  & \phi_y(x_2) =x_2
\end{aligned} \\
\| \phi_y(x_1) - \phi_y(x_2) \| = \| x_1 - x_2 \| \leq \epsilon \| x_1 - x_2 \| \quad \, \forall \, \epsilon > 0 \Longrightarrow x_1 = x_2
\end{gathered}
\]
$\Longrightarrow \left. f\right|_U$ injective.

$W=f(V)$, so $f:V\to W$ surjective.  $f$ bijective.

Fix $y_0 \in W$, $y_0 = f(x_0)$, $x_0 \in V$.  \\
Let $r>0$ s.t. $B_r(x_0) \subset V$.  \\
Consider $B_{r\epsilon}(y_0)$.  If $y\in B_{r\epsilon}(y_0)$.
\[
\begin{gathered}
  r\epsilon > \| y-y_0 \| = \| y - f(x_0) \| = \| \phi_y(x_0) - x_0 \| \text{ with } \\
  \phi_y(x) = x + (y-f(x))
\end{gathered}
\]

If $x\in B_r(x_0)$,
\[
\| \phi_y(x) -x_0 \| \leq \| \phi_y(x) - \phi_y(x_0) \|  + \| \phi_y(x_0) - x_0 \| \leq \epsilon \| x-x_0 \| + r\epsilon < 2 r\epsilon = r
\]

%Consider $B_{r\epsilon }(y_0)$.

Thus $\phi(B_r(x_0)) = B_r(x_0)$.

By contraction mapping principle, $\exists \, a \in B_r(x_0)$, s.t. $\phi_y(a)=a$.  Then $\phi_y(a) = a+ (y-f(a)) = a \Longrightarrow f(a) =y$.

$y\in f(V) = W$.

So $B_{r\epsilon}(y_0) \subset W$.  $W$ open.

Let $\text{Mat}(n,n) \equiv $ space of all $n\times n$ matrices; $\text{Mat}(n,n)  = \mathbb{R}^{n^2}$.

\end{proof}

There is a proof of the implicit function theorem and its various forms in Shastri (2011) \cite{AShastri2011}, but I found Wienhard's Handout 4 for Math 327 to be clearer.\footnote{\url{https://web.math.princeton.edu/~wienhard/teaching/M327/handout4.pdf}}

\begin{theorem}[Implicit Function Theorem]
Let open $U \subset \mathbb{R}^{m+n} \equiv \mathbb{R}^m \times \mathbb{R}^n$  \\
\phantom{Let} $C^1 \, f:U \to \mathbb{R}^n $ \\
\phantom{Let} $(a,b) \in U$ s.t. $f(a,b) = 0$ and $\left. D_y f\right|_{(a,b)}$ invertible.

Then $\exists \, $ open $V \ni (a,b)$, $V \subset U$ \\
\phantom{Then} $\exists \, $ open neighborhood $W \ni a$, $W \subseteq \mathbb{R}^m$ \\
\phantom{Then} $\exists \, !$ \, $C^1 \, g:W \to \mathbb{R}^n$ s.t.
\[
\lbrace (x,y) \in V | f(x,y) =0 \rbrace = \lbrace (x,g(x)) | x \in W \rbrace
\]
Moreover,
\[
dg_x = - \left. (d_yf)^{-1} \right|_{(x,g(x))} \left. d_x f\right|_{(x,g(x))}
\]
and $g$ smooth if $f$.
\end{theorem}

\begin{proof}
  Define $\begin{aligned} & \quad \\
    & F: U \to \mathbb{R}^{m+n}   \\
    & F(x,y) = (x,f(x,y)) \end{aligned}$

Then $F(a,b) = (a,0)$ (given), and
\[
DF = \left[ \begin{matrix} 1 & \\
    \frac{ \partial f^i(x,y)}{ \partial x^j} & \frac{ \partial f^i(x,y) }{ \partial y^j } \end{matrix} \right] \equiv \left[ \begin{matrix} 1 & \\
    D_xf & D_yf \end{matrix} \right]
\]
$DF(a,b)$ invertible.

By inverse function theorem, since $DF(a,b)$ invertible at pt. $(a,b)$, \\
$\exists \, $ open neighborhoods $\begin{aligned} & \quad \\
  & V \ni (a,b) \subseteq \mathbb{R}^m \times \mathbb{R}^n \\
  & \widetilde{W} \ni (a,0) \subseteq \mathbb{R}^m \times \mathbb{R}^n \end{aligned}$ s.t. $F$ diffeomorphism with $F^{-1}: \widetilde{W} \to V$.

Set $W = \lbrace x \in \mathbb{R}^m | (x,0) \in \widetilde{W}\rbrace$.  Then $\pi_1(\widetilde{W}) =W$ open in $\mathbb{R}^m$.

Define $g:W\to \mathbb{R}^n$,
\[
\begin{aligned}
  & g(x) = \pi_2 \circ F^{-1}(x,0) \text{ or } \\
  & F^{-1}(x,0) = (h(x),g(x))
\end{aligned}
\]

Now $FF^{-1}(x,0) = (x,0) = (h(x), f(h(x),g(x)) )$ so $h(x)=x \, \forall \, x \in W$, $0 = f(x,g(x))$.

Then
\[
\lbrace (x,y) \in V | f(x,y) = 0 \rbrace = \lbrace (x,y) \in V | F(x,y) = (x,0) \rbrace = \lbrace (x,g(x)) | x \in W, 0 = f(x,g(x)) \rbrace
\]
Since $\pi$ smooth and $F^{-1}$ is $C^1$, $g$ is $C^1$.

To reiterate, $f(x,g(x)) =0$ on $W$.

Using chain rule while differentiating $f(x,g(x))=0$,
\[
\begin{gathered}
\partial_{x^j} f(x,g(x)) = \frac{ \partial f(x,g(x)) }{ \partial x^k} \frac{ \partial x^k}{ \partial x^j}+ \frac{ \partial f(x,g(x))}{ \partial y^k}\frac{ \partial g^k(x)}{\partial x^j} = \left. D_x f \right|_{(x,g(x))} + \left. (D_yf) \right|_{(x,g(x))} \cdot (Dg)_x = 0 \text{ or }  \\
(Dg)_x = -\left. (D_yf) \right|_{x,g(x)} \left. D_xf \right|_{(x,g(x))}
\end{gathered}
\]

\end{proof}

\section{Immersions}

\begin{definition}[Immersion]
  smooth $f:M \to N$, s.t. $Df(p) : T_pM \to T_{f(p)}N$ injective.  Then $f$ \textbf{immersion} at $p$.
\end{definition}

Absil, Mahony, and Sepulchre \cite{AMS2008} pointed out that another definition for a \emph{immersion} can utilize the theorem that $\text{rank}$ of $Df \equiv DF = \text{dim} T_pM$.  Indeed, recall these facts from linear algebra:

for $T:V \to W$,  \\
It's always true that $\begin{aligned} & \quad \\
& \text{rank}T \leq V \\
& \text{rank}T \leq W \end{aligned}$, and \\

$\text{rank}T = \text{dim}V$ iff $T$ injective.  \\
$\text{rank}T = \text{dim}W$ iff $T$ surjective.  \\

\[
\begin{gathered}
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=2.8em, column sep=4.8em, minimum width=2.2em]
{
	T_xM & T_{F(x)}N = T_yN \\
	x \in M & y=F(x) \in N  \\
	M & N \\
};
\path[->]
(m-1-1) edge node [above] {$DF(x)$} (m-1-2)
(m-2-1) edge node [auto] {$$} (m-1-1)
(m-2-2) edge node [auto] {$$} (m-1-2)
(m-3-1) edge node [auto] {$F$} (m-3-2)
;
\path[|->]
(m-2-1) edge node [auto] {$F$} (m-2-2)
;
\end{tikzpicture}   \\
\end{gathered}
\]

Now
\[
\begin{aligned}
& \text{dim}T_xM = \text{dim}M \\
& \text{dim}T_{F(x)}N = \text{dim}N
\end{aligned}
\]
And
\[
\text{rank}(DF(x)) \equiv \text{rank of $F$ }
\]
I know that the notation above is confusing, but this is what all Differential Geometry books apparently mean when they say "rank of $F$".

Now
\[
\text{rank}(DF(x)) = \text{dim}(\text{im}(DF(x)))  = \text{dim}T_xM \text{ iff } DF(x) \text{ injective }
\]

If $\forall \, x \in M$, this is the case, then $F$ an \textbf{ immersion }.

Apply the rank-nullity theorem in this case:

\[
\begin{gathered}
	\text{rank}(DF(x)) + \text{dim}\text{ker}(DF(x)) = \text{dim}T_xM = \text{dim}M \\
	\Longrightarrow \text{rank}(DF(x)) = \text{dim}M \leq \text{dim}T_{F(x)}N = \text{dim}N \text{ or } \text{dim}M \leq \text{dim}N
\end{gathered}
\]

Now

\[
\text{rank}(DF(x)) = \text{dim}T_{F(x)}N \text{ iff } DF(x) \text{ surjective }
\]

If $\forall \, x \in M$, this is the case, then $F$ an \textbf{ submersion }.

\[
 \text{rank}(DF(x)) = \text{dim}T_{F(x)}N = \text{dim}N  \leq \text{dim}M
\]

Shastri (2011) has this as the ``Injective Form of Implicit Function Theorem'', Thm. 1.4.5, pp. 23 and Guillemin and Pollack (2010) has this as the ``Local Immersion Theorem'' on pp. 15, Section 3 ``The Inverse Function Theorem and Immersions'' \cite{VGuilleminAPollack2010}.

\begin{theorem}[Local immersion Theorem i.e. Injective Form of Implicit Function Theorem]
  Suppose $f:M\to N$ immersion at $p$, $q=f(p)$.

Then $\exists \, $ local coordinates around $p,q$, $x,y$, respectively s.t. $f(x_1\dots x_m) = (x_1 \dots x_m,0 \dots 0)$.

\end{theorem}

\begin{proof}
  Choose local parametrizations
\[
\begin{tikzpicture}
  \matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
  {
    U \subseteq M & N \supseteq V \\
    \phi(U) & \psi(V) \\
};
  \path[->]
  (m-1-1) edge node [above] {$f $} (m-1-2)
          edge node [auto] {$\phi$} (m-2-1)
  (m-1-2) edge node [auto]  {$\psi$} (m-2-2)
  (m-2-1) edge node [auto] {$f$} (m-2-2)
  ;
\end{tikzpicture}
\quad \quad \, \begin{aligned} & \phi(p) = x \\
  & \psi(q) = y \end{aligned}
\]
$D(\psi f\varphi^{-1}) \equiv Df$.  $Df(p)$ injective (given $f$ immersion).  $Df(p) \in \text{Mat}(n,m)$

By change of basis in $\mathbb{R}^n$, assume $Df(p) = \left( \begin{matrix} I_m \\ 0 \end{matrix} \right)$.

Now define $\begin{aligned} & \quad \\
  & G : \phi(U) \times \mathbb{R}^{n-m} \to \mathbb{R}^n \\
  & G(x,z) = f(x) + (0,z) \end{aligned}$

Thus, $DG(x,z) =1$ and for open $\phi(U) \times U_2$, $ G(\phi(U)\times U_2)$ open.

By inverse function theorem, $G$ local diffeomorphism of $\mathbb{R}^n$, at $0$.

Now $f = G\circ \mathfrak{i}$, where $\mathfrak{i}$ is canonical immersion.
\[
\begin{gathered}
  G(x,0) = f(x) \\
  \Longrightarrow G^{-1}G(x,0) = (x,0) = G^{-1}f(x)
\end{gathered}
\]

Use $\psi \circ G$ as the local parametrization of $N$ around pt. $q$.  Shrink $U,V$ so that

\begin{tikzpicture}
  \matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
  {
    U \subseteq M & N \supseteq V \\
    \phi(U) & \psi\circ G(V) \\
};
  \path[->]
  (m-1-1) edge node [above] {$f $} (m-1-2)
          edge node [auto] {$\phi$} (m-2-1)
  (m-1-2) edge node [auto]  {$\psi\circ G$} (m-2-2)
  (m-2-1) edge node [auto] {$\mathfrak{i}$} (m-2-2)
  ;
\end{tikzpicture}

\end{proof}

\begin{theorem}[Implicit Function Thm.]
  Let open subset $U\subseteq \mathbb{R}^n \times \mathbb{R}^d$, $(x,y) = (x^1 \dots x^n, y^1 \dots y^k) $ on $U$.  \\
  Suppose smooth $\Phi:U\to \mathbb{R}^k$, $(a,b) \in U$, $c=\Phi(a,b)$

  If $k\times k$ matrix $\frac{ \partial \Phi^i}{ \partial y^j}(a,b)$ nonsingular, then $\exists $ neighborhoods $\begin{aligned} & \quad \\
    & V_0 \subseteq \mathbb{R}^n \text{ of $a$ } \\
    & W_0 \subseteq \mathbb{R}^k \text{ of $b$ } \end{aligned}$ and smooth $F:V_0 \to W_0$ s.t.

  $\Phi^{-1}(c) \bigcap (V_0\times W_0)$ is graph of $F$, i.e. \\
  $\Phi(x,y) =c$ for $(x,y) \in V_0\times W_0$ iff $y=F(x)$.
  \end{theorem}

\section{Submersions}
cf. pp. 20, Sec. 4 "Submersions", Ch. 1 of Guillemin and Pollack (2010) \cite{VGuilleminAPollack2010}.

Consider $X,Y\in \text{\textbf{Man}}$, s.t. $\text{dim}X \geq \text{dim}Y$.

\begin{definition}[submersion] If $f:X\to Y$, \\
if $Df_x \equiv df_x$ is \emph{surjective}, $f\equiv $ \textbf{submersion} at $x$.
\end{definition}
Recall that,
\[
\begin{gathered}
	Df_x:T_xX \to T_{f(x)}Y \\
	\text{dim}T_xX \geq \text{dim}T_{f(x)}Y
\end{gathered}
\]
\[
\begin{gathered}
\text{rank}Df_x \leq \text{dim}T_{f(x)}Y, \text{ in general, while } \\
\text{rank}Df_x = \text{dim}T_{f(x)}Y \text{ iff } Df_x \text{ surjective }
\end{gathered}
\]

Canonical submersion is standard projection: \\
If $\begin{gathered} \quad \\
\text{dim}X = k \\
\text{dim}Y = l \end{gathered}$, $k\geq l$,
\[
(a_1 \dots a_k ) \mapsto (a_1 \dots a_l)
\]

\begin{theorem}[Local Submersion Theorem]
	Suppose $f:X\to Y$ submersion at $x$, and $y = f(x)$,
	Then $\exists \, $ local coordinates around $x$, $y$ s.t.
	\[
	f(x_1\dots x_k) = (x_1 \dots x_l)
	\]
	i.e. $f$ locally equivalent to canonical submersion near $x$
\end{theorem}
\begin{proof}
I'll have a side-by-side comparison of my notation and the 1 used in Guillemin and Pollack (2010) \cite{VGuilleminAPollack2010} where I can.

For charts $(U,\phi), (V,\psi)$ for $X,Y$, respectively, $y=f(x)$ for $x\in X$,
\[
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	U \subseteq X & Y \supseteq V \\
	\mathbb{R}^k &  \mathbb{R}^l \\
};
\path[->]
(m-1-1) edge node [above] {$f $} (m-1-2)
edge node [auto] {$\phi$} (m-2-1)
(m-1-2) edge node [auto]  {$\psi\circ G$} (m-2-2)
(m-2-1) edge node [auto] {$\mathfrak{i}$} (m-2-2)
;
\end{tikzpicture} \quad \quad \,
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	 x & f(x)=y \\
	\phi(x)=(a^1\dots a^k) &  g(\phi(x))=g(a^1\dots a^k)=\psi(y) \\
};
\path[|->]
(m-1-1) edge node [above] {$f $} (m-1-2)
edge node [auto] {$\phi$} (m-2-1)
(m-1-2) edge node [auto]  {$\psi$} (m-2-2)
(m-2-1) edge node [auto] {$g$} (m-2-2)
;
\end{tikzpicture}
\]
$Dg_x$ surjective, so assume it's a $l\times k$ matrix $\left[ \begin{matrix} \mathbf{1}_l & 0 \end{matrix} \right]$.

Define
\begin{equation}
\begin{aligned}
& G:U \subset \mathbb{R}^k \to \mathbb{R}^k  \\
& G(a)\equiv G(a^1\dots a^k) := (g(a), a_{l+1}, \dots , a_k)
\end{aligned}
\end{equation}
Now
\begin{equation}
DG(a)  = \left[ \begin{matrix} \mathbf{1}_l & 0 \\ & \mathbf{1}_{k-l} \end{matrix} \right] = \mathbf{1}_k
\end{equation}
so $G$ local diffeomorphism (at $0$).

So $\exists \, $ $G^{-1}$ as local diffeomorphism of some $U'$ of $a$ into $U\subset \mathbb{R}^k$.

By construction,
\begin{equation}
g=\mathbb{P}_l \circ G
\end{equation}
where $\mathbb{P}_l$ is the \emph{canonical submersion}, the projection operator onto $\mathbb{R}^l$.
\[
g\circ G^{-1} = \mathbb{P}_l
\]
(since $G$ diffeomorphism)

\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	U\subseteq X &  V\subseteq Y \\
	\mathbb{R}^k &  \mathbb{R}^l \\
};
\path[->]
(m-2-1) edge node [auto] {$\phi^{-1}\circ G^{-1}  $} (m-1-1)
edge node [auto] {$\mathbb{P}_l$} (m-2-2)
(m-1-1) edge node [auto]  {$f$} (m-1-2)
(m-2-2) edge node [auto] {$\psi^{-1}$} (m-1-2)
;
\end{tikzpicture}
for \\
$\Longrightarrow $ \begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	\phi^{-1}\circ G^{-1}(a)\equiv \phi^{-1}\circ G^{-1}(a^1\dots a^k)=x & f(x)=y=\psi^{-1}(a^1\dots a^l) \\
	(a^1\dots a^k) &  (a^1\dots a^l) \\
};
\path[|->]
(m-2-1) edge node [auto] {$\phi^{-1}\circ G^{-1}  $} (m-1-1)
edge node [auto] {$\mathbb{P}_l$} (m-2-2)
(m-1-1) edge node [auto]  {$f$} (m-1-2)
(m-2-2) edge node [auto] {$\psi^{-1}$} (m-1-2)
;
\end{tikzpicture}

	\end{proof}
"An obvious corollary worth noting is that if $f$ is a submersion at $x$, then it is actually a submersion in a whole neighborhood of $x$." Guillemin and Pollack (2010) \cite{VGuilleminAPollack2010}

Suppose $f$ submersion at $x\in f^{-1}(y)$.

By local submersion theorem
\[
f(x_1\dots x_k)=(x_1 \dots x_l)
\]
Choose $y=(0, \dots , 0)$.

Then, near $x$, $f^{-1}(y) = \lbrace (0, \dots 0 , x_{l+1} \dots x_k)\rbrace$ i.e. let $V\ni x$ neighborhood of $x$, define $(x_1 \dots x_k)$ on $V$.

Then $f^{-1}(y) \bigcap V = \lbrace (0\dots 0 , x_{l+1} , \dots x_k) | x_1 = 0 , \dots x_l = 0\rbrace$.

Thus $x_{l+1}, \dots x_k$ form a coordinate system on open set $f^{-1}(y) \bigcap V \subseteq f^{-1}(y)$.

Indeed,
\[
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	U \subseteq X &  V\subseteq Y \\
	\mathbb{R}^k &  \mathbb{R}^l \\
};
\path[->]
(m-1-1) edge node [auto] {$ f  $} (m-1-2)
edge node [auto] {$ \phi $} (m-2-1)
(m-2-1) edge node [auto]  {$\mathbb{P}_l$} (m-2-2)
(m-1-2) edge node [auto] {$\psi$} (m-2-2)
;
\end{tikzpicture} \qquad \, \begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	x &  f(x)=y \\
	\phi(x)=(x^1\dots x^k) &  (x^1\dots x^l) \\
};
\path[|->]
(m-1-1) edge node [auto] {$ f  $} (m-1-2)
edge node [auto] {$ \phi $} (m-2-1)
(m-2-1) edge node [auto]  {$\mathbb{P}_l$} (m-2-2)
(m-1-2) edge node [auto] {$\psi$} (m-2-2)
;
\end{tikzpicture}
\]
and now
\[
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	f^{-1}(y) &  y \\
	\lbrace (0, \dots 0,x^1\dots x^k) \rbrace &  (0\dots 0) \\
};
\path[|->]
(m-1-2) edge node [auto] {$ f^{-1}  $} (m-1-1)
edge node [auto] {$ \psi $} (m-2-2)
(m-2-2) edge node [auto]  {$\mathbb{P}_l^{-1}$} (m-2-1)
(m-2-1) edge node [auto] {$\phi^{-1}$} (m-1-1)
;
\end{tikzpicture}
\]

\begin{definition}[regular value]
	For smooth $f:X\to Y$, $X,Y \in \text{\textbf{Man}}$, \\
	$y\in Y$ is a \textbf{regular value} for $f$ if $Df_x:T_xX \to T_y Y$ surjective $\forall \, x$ s.t. $f(x)=y$.

	$y\in Y$ \textbf{critical value} if $y$ not a regular value of $f$.
\end{definition}

Absil, Mahony, and Sepulchre \cite{AMS2008} pointed out that another definition for a \emph{regular value} can utilize the theorem that $\text{rank}$ of $Df \equiv DF = \text{dim} T_pN = \text{dim} N$, iff $DF(p)$ surjective, for $p\in M$, $F:M\to N$.  Then

\textbf{regular value} $y \in N$, of $F$, if rank of $F \equiv \text{rank}(DF(x)) = \text{dim}N$, $\forall \, x \in F^{-1}(y)$, for $F:M\to N$.

\begin{theorem}[Preimage theorem]
If $y$ regular value of $f:X\to Y$, \\
$f^{-1}(y)$ is a submanifold of $X$, with $\text{dim}f^{-1}(y)=\text{dim}X - \text{dim}Y$
\end{theorem}
\begin{proof}
Given $y$ is a regular value of $f:X\to Y$,  \\
$\forall \, x \in f^{-1}(y)$, $Df_x:T_xX \to T_yY$ is surjective.  By local submersion theorem,
\[
f(x^1 \dots x^k) = (x^1 \dots x^l)=y
\]
Since $x\in f^{-1}(y)$, $(x^1\dots x^k)=(y^1 \dots y^l,x^{l+1}\dots x^k)$.

For this chart for $(U,\varphi)$, $U\ni x$, consider $(U\cap f^{-1}(y),\psi)$ with $\psi(x) = (x^{l+1}\dots x^k) \quad \, \forall \, x\in U\cap f^{-1}(y)$.

$\forall \, f^{-1}(y)$ submanifold with $\text{dim}f^{-1}(y) = k-l = \text{dim}X-\text{dim}Y$.
	\end{proof}

\emph{Examples for emphasis}

If $\text{dim}X > \text{dim}Y$, \\
\phantom{\qquad \, } if $y\in Y$, regular value of $f:X\to Y$, \\
\phantom{\qquad \, \qquad \, } $f$ submersion, $\forall \, x \in f^{-1}(y)$ \\
If $\text{dim}X = \text{dim}Y$, \\
\phantom{\qquad \, } $f$ local diffeomorphism $\forall \, x\in f^{-1}(y)$ \\
If $\text{dim}X < \text{dim}Y$, $\forall \, y\in f(X)$ is a critical value.

\textbf{Example: $O(n)$ as a submanifold of $\text{Mat}(n,n)$}

Given $\text{Mat}(n,n)\equiv M(n) = \lbrace n \times n \text{ matrices } \rbrace$ is a manifold; in fact $\text{Mat}(n,n) \cong \mathbb{R}^{n^2}$, \\
Consider $O(n) = \lbrace A \in \text{Mat}(n,n) | AA^T = 1\rbrace$.
\begin{equation}
AA^T \in \text{Sym}(n) \equiv S(n) = \lbrace S\in \text{Mat}(n,n) | S^T = S \rbrace = \lbrace \text{ symmetric $n\times n$ matrices } \rbrace
\end{equation}
$\text{Sym}(n)$ submanifold of $\text{Mat}(n,n)$, $\text{Sym}(n)$ diffeomorphic to $\mathbb{R}^k$ (i.e. $\text{Sym}(n) \cong \mathbb{R}^k$), $k:= \frac{n (n+1)}{2}$.

\[
\begin{aligned}
	& f:\text{Mat}(n,n) \to \text{Sym}(n) \\
	& f(A) = AA^T
\end{aligned}
\]
Notice $f$ is smooth,
\[
\begin{gathered}
f^{-1}(1) = O(n) \\
Df_A(B) = \lim_{s\to 0} \frac{ f(A+sB) - f(A) }{s} = \lim_{s\to 0} \frac{(A+sB)(A^T + sB^T)- AA^T}{s} = AB^T +BA^T
\end{gathered}
\]
If $Df_A : T_A\text{Mat}(n,n) \to T_{f(A)}\text{Sym}(n)$ surjective when $A\in f^{-1}(1) = O(n)$ (???).

\begin{proposition} If smooth $g_1\dots g_l \in C^{\infty}(X)$ on $X$ are independent $\forall \, x\in X$, s.t. $g_i(x)=0$, $\forall \, i = 1\dots l$, \\
	then $Z=\lbrace x\in X | g_1(x) = \dots = g_l(x)=0 \rbrace = $ set of "common zeros" is a \emph{submanifold} of $X$ s.t. $\text{dim}Z = \text{dim}X- l$.

	Take \emph{note} that $g_1 \dots g_l$ are independent at $x$ means, really, that $D(g_1)_x \dots D(g_l)_x$ are linearly independent on $T_xX$.
\end{proposition}
\begin{proof}
Suppose smooth $g_1 \dots g_l \in C^{\infty}(X)$ on manifold $X$ s.t. $\text{dim}X = k\geq l$.

Consider $g=(g_1\dots g_l):X \to \mathbb{R}^l$, $Z\equiv g^{-1}(0)$.

Since $\forall \, g_i$ smooth, $D(g_i)_x:T_xX \to \mathbb{R}$ linear.

Now for
\[
Dg_x = (D(g_1)_x \dots D(g_l)_x):T_xX \to \mathbb{R}^l
\]
By rank-nullity theorem (linear algebra), $Dg_x$ surjective iff $\text{rank}Dg_x = l$ i.e. $l$ functionals $D(g_1)_x \dots D(g_l)_x$ are linearly independent on $T_xX$.

"We express this condition by saying the $l$ functions $g_1\dots g_l$ are independent at $x$."  (Guillemin and Pollack (2010) \cite{VGuilleminAPollack2010})

	\end{proof}

\section{Submanifolds; immersed submanifold, embedded submanifolds, regular submanifolds}

Recall immersion:  \\
$F:M \to N$ immersion iff $DF$ injective iff $\text{rank}DF = \text{dim}M$.

Consider manifolds $M\subseteq N$.  \\
Consider inclusion map $\begin{aligned} & \quad \\
	& i:M\to N \\
	& i: x \mapsto x \end{aligned}$.

	If $i$ immersion, $Di(x) = \frac{\partial y^i}{ \partial x^j} = \delta_j^{\  \  i}$ if $y^i = x^i$, $\forall \, i =1,\dots \text{dim}M$.

\begin{definition}[immersed submanifold]
	\textbf{immersed submanifold} $M \subseteq N$ if inclusion $i:M\to N$ is an immersion.
\end{definition}
cf. 3.3 Embedded Submanifolds of Absil, Mahony, and Sepulchre \cite{AMS2008}, also Ch. 5 Submanifolds, pp. 108, \textbf{Immersed Submanifolds} of  John Lee (2012) \cite{JLee2012}.

Immersed submanifolds often arise as images of immersions.

\begin{proposition}[Images of Immersions as submanifolds]
	Suppose smooth manifold $M$, \\
	\phantom{Suppose } smooth manifold with or without boundaries $N$, \\
	injective, smooth immersion $F:M\to N$ ($F$ injective itself, not just immersion)

Let $S=F(M)$.

Then $S$ has unique topology and smooth structure of smooth submanifolds of $N$ s.t. $F:M\to S$ diffeomorphism.

\end{proposition}
cf. Prop. 5.18 of John Lee (2012) \cite{JLee2012}.

\begin{proof}
Define topology of $S$: set $U\subseteq S$ open iff $F^{-1}(U) \subseteq M$ open ($F^{-1}(U\cap V) = F^{-1}(U) \cap F^{-1}(V), F^{-1}(U\cup V) =F^{-1}(U) \cup F^{-1}(V)$).  \\
Define smooth structure of $S$: $\lbrace F(U), \varphi \circ F^{-1}  | (U,\varphi) \in \text{atlas for $M$, i.e. $(U,\varphi)$ any smooth chart of $M$}\rbrace$.

"smooth compatibility condition":
\[
(\varphi_2\circ F^{-1}) (\varphi_i F^{-1})^{-1} = \varphi_2 \circ F^{-1}F\varphi_1^{-1} = \varphi_2 \varphi_1^{-1}
\]
since $\varphi_2\varphi_1^{-1}$ diffeomorphism ($\varphi_2 \varphi_1^{-1}$ bijection and it and inverse is differentiable)

$F$ diffeomorphism onto $F(M)$.

	and these are the only topology and smooth structure on $S$ with this property:
	\[
	S \xrightarrow{F^{-1}} M \xrightarrow{F} N \qquad \, = \qquad \, S \hookrightarrow M
	\]
	and $F^{-1}$ diffeomorphism, $F$ smooth immersion, so $i : S \to M$ smooth immersion.
	\end{proof}

Jeffrey Lee (2009) \cite{JLee2009}

\section{Tensors}

I'll go through Ch.7 \emph{Tensors} of Jeffrey Lee (2009) \cite{JLee2009}.

\begin{definition}[7.1\cite{JLee2009}] Let $V,W$ be modules over commutative ring $R$, with unity.

Then, algebraic $W$-valued tensor on $V$ is multilinear map.
\begin{equation}
	\tau: V_1 \times V_2 \times \dots \times V_m \to W
\end{equation}
where $V_i = \lbrace V,V^* \rbrace$ \quad \, $ \forall \, i=1,2,\dots m$.

If for $r,s$ s.t. $r+s =m$, there are $r$ \, $V_i = V^*$, $s \, V_i = V$, tensor is $r$-contravariant, $s$-covariant; also say tensor of total type $\binom{r}{s}$.
\end{definition}

EY : 20170404 Note that
\[
\begin{aligned}
	& ( \tau_{\beta}^{i\alpha} \frac{ \partial }{ \partial x^i } \text{ or } \tau_{\beta}^{i\alpha} e_i )(\omega_j dx^j \text{ or } \omega_je^j  \in V^*) \\
	& ( \tau^{\beta}_{i\alpha} dx^i \text{ or } \tau^{\beta}_{i\alpha} e^i )( X^j \frac{ \partial }{ \partial x^j} \text{ or } X^j e_j  \in V)
\end{aligned}
\]

$\exists \,$ natural map $\begin{aligned} & \quad \\
	& V\to V^{**} \\
& v \mapsto \widetilde{v} \end{aligned}$,  $\begin{aligned} & \quad \\
	& \widetilde{v} : \alpha \mapsto \alpha(v) \end{aligned}$.  If this map is an isomorphism, $V$ is \textbf{reflexive} module, and identify $V$ with $V^{**}$.

\exercisehead{7.5} Given vector bundle $\pi: E \to M$, open $U\subset M$, consider sections of $\pi$ on $U$, i.e. cont. $s:U\to E$, where $(\pi\circ s)(u)=u$, \, $\forall \, u \in U$.

Consider $E^* \ni \omega =\omega_i e^i$.

$\forall \, s\in \Gamma(E)$, $\omega(s) = \omega_i(s(x))^i$, \, $\forall \, x \in U\subset M$.  So define $\widetilde{s}: \omega,x\mapsto \omega(s(x))$, \, $\forall \, x \in U$.

If $\widetilde{s} =0$, $\widetilde{s}(\omega,x) = \omega(s(x)) =0$ \quad \, $\forall \, \omega \in E^*$, $\forall \, x\in U$, and so $s=0$.  (Let $\omega_i = \delta_{iJ}$ for some $J$, and so $s^J(x) =0$ \quad \, $\forall \, J$).

$s=0$.  So $\text{ker}(s\mapsto \widetilde{s}) = \lbrace 0 \rbrace$ (so condition for injectivity is fulfilled).

Since $\widetilde{s}:\omega,x\mapsto \omega(s(x))$, $\forall \, \omega \in E^*$, $\forall \, x \in U$, $s\mapsto \widetilde{s}$ is surjective.

$s\mapsto \widetilde{s}$ is an isomorphism so $\Gamma(E)$ is a \emph{reflexive} module.

\begin{proposition}
For $R$ a ring (special case), $\exists \, $ module homomorphism:  \\

tensor product space $\to $ tensor, as a multilinear map, i.e. $\exists$ \,
\begin{equation}
\begin{aligned}
&	\left( \otimes_{i=1}^r V \right) \otimes \left( \otimes_{j=1}^s V^* \right) \to T^r_{ \, \, s}(V;R)   \\
& u_1 \otimes \dots \otimes u_r \otimes \beta^1 \otimes \dots \otimes \beta^s \in \left( \otimes^r V \right) \otimes \left( \otimes^s V^* \right) \mapsto (\alpha^1 \dots \alpha^r, v_1 \dots v_s) \mapsto \alpha^1(u_1) \dots \alpha^r(u_r) \beta^1(v_1) \dots \beta^s(v_s)
\end{aligned}
\end{equation}
\end{proposition}

Indeed, consider
\[
(\alpha^1 \dots \alpha^r, v_1 \dots v_s) \in \underbrace{V^* \times \dots \times V^* }_{r} \times \underbrace{ V\times \dots \times V}_{s} \mapsto \alpha^1(u_1) \dots \alpha^r(u_r) \beta^1(v_1) \dots \beta^s(v_s)
\]
and so for
\[
\begin{aligned}
	& \alpha^i = \alpha^i_{\mu} e^{\mu} , \, & \, i =1,2, \dots r, \, & \, \mu = 1,2, \dots \text{dim}V^* \\
	& v_i = v_i^{\mu} e_{\mu} , \, & \, i = 1,2, \dots s, \, & \, \mu = 1, 2, \dots \text{dim}V
\end{aligned} \qquad \, \begin{aligned}
& \alpha^i(u_i) = \alpha^i_{\mu} u^{\mu}_i \\
 & \beta^i(v_i) = \beta^i_{\mu} v^{\mu}_i
\end{aligned}
\]
So that
\[
\begin{gathered}
	\alpha^1(u_1) \dots \alpha^r(u_r) \beta^1(v_1) \dots \beta^s(v_s) = \alpha^1_{\alpha_1}u^{\alpha_1}_1 \dots \alpha^r_{\alpha_r} u^{\alpha_r}_r \beta^1_{\mu_1} v^{\mu_1}_1 \dots \beta^s_{\mu_s} v^{\mu_s}_s = \\
	= (u^{\alpha_1}_1 \dots u_r^{\alpha_r} \beta^1_{\mu_1} \dots \beta^s_{\mu_s})(\alpha^1_{\alpha_1} \dots \alpha^r_{\alpha_r} v_1^{\mu_1} \dots v_s^{\mu_s} )
\end{gathered}
\]
Identify $u_1 \otimes \dots \otimes u_r \otimes \beta^1 \otimes \dots \otimes \beta^s$ with this multiplinear map.

\begin{proposition}
If $V$ is finite-dim. vector space, or if $V=\Gamma(E)$, for vector bundle $E\to M$, map
\begin{equation}
\left( \otimes_{i=1}^r V \right)  \otimes \left( \otimes_{j=1}^s V^* \right) \to T^r_{ \, \, s}(V;R)
\end{equation}
is an isomorphism.
\end{proposition}

\begin{definition}
tensor that can be written as
\begin{equation}
u_1\otimes \dots \otimes u_r \otimes \beta^1 \otimes \dots \otimes \beta^s \equiv u_1\otimes \dots \otimes \beta^s
\end{equation}
is \textbf{simple} or \textbf{decomposable}.
\end{definition}
Now well that not \emph{all} tensors are simple.

\begin{definition}[7.7\cite{JLee2009}, tensor product]
$\forall \, S\in T^{r_1}_{ \,\, s_1}(V)$, $\forall \, T \in T^{r_2}_{ \,\, s_2}(V)$, \\
define tensor product
\begin{equation}
\begin{gathered}
S\otimes T\in T^{r_1+r_2}_{ \, \, \, s_1+s_2}(V) \\
	S\otimes T( \theta^1\dots \theta^{r_1 + r_2}, v_1 \dots v_{s_1+s_2}) := S(\theta^1\dots \theta^{r_1}, v_1\dots v_{s_1})T(\theta^{r_1+1}\dots \theta^{r_1+r_2}, v_{s_1+1}\dots v_{s_1 + s_2} )
\end{gathered}
\end{equation}
\end{definition}

\begin{proposition}[7.8\cite{JLee2009}]
\end{proposition}

\[
\begin{gathered}
	\tau^{ i_1 \dots i_r }_{ \phantom{i_1 \dots i_r} j_1 \dots j_s} e_{i_1} \otimes \dots \otimes e_{i_r} \otimes e^{j_1}\otimes \dots \otimes e^{j_s} = \tau(e^{i_1} \dots e^{i_r}, e_{j_1} \dots e_{j_s} )e_{i_1} \otimes \dots \otimes e_{i_r} \otimes e^{j_1} \otimes \dots \otimes e^{j_s} = \tau
\end{gathered}
\]
So $\lbrace e_{i_1}\otimes \dots \otimes e_{i_r} \otimes e^{j_1} \otimes \dots \otimes e^{j_s} | i_1 \dots i_{r}, j_1\dots j_s \in 1 \dots n \rbrace$ spans $T^r_{\,\, s}(V;R)$

\exercisehead{7.11}  Let basis for $V$ \, $e_1 \dots e_n$, corresponding dual basis for $V^*$ \,  $e^1 \dots e^n$ \\
Let basis for $V$ \, $\overline{e}_1 \dots \overline{e}_n$, corresponding dual basis for $V^*$ \,  $\overline{e}^1 \dots \overline{e}^n$ \\
s.t.
\[
\begin{aligned}
	& \overline{e}_i = C^k_{ \,\, i} e_k \\
	& \overline{e}^i = (C^{-1})^i_{ \, \, k} e^k
\end{aligned}
\]
EY:20170404, keep in mind that

\[
\begin{aligned}
	&	Ax = e_i A^i_{ \, \, k} e^k(x^j e_j) = e_i A^i_{ \,\,j} x^j = A^i_{ \, \, j} x^j e_i \\
 & Ae_j = e_k A^k_{ \, \, i} e^i (e_j) = A^k_{ \,\, j} e_k = \overline{e}_j
\end{aligned}
\]
\[
\begin{gathered}
\overline{\tau}^i_{ \,\, jk} \overline{e}_i \otimes \overline{e}^j \otimes \overline{e}^k = \overline{\tau}^i_{ \, \, jk} C^l_{ \, \, i} e_l (C^{-1})^j_{ \,\, m} e^m(C^{-1})^k_{ \, \, n} e^n = \overline{\tau}^i_{ \, \, jk} C^l_{ \, \, i } (C^{-1})^j_{ \,\, m} (C^{-1})^k_{ \,\, n} = \tau^l_{ \,\, mn}   \\
\overline{\tau}^i_{ \,\, jk} = C^c_{ \,\, k} C^b_{ \,\, j} (C^{-1})^i_{ \,\, a} \tau^a_{\,\, bc}
\end{gathered}
\]

On Remark 7.13 of Jeffrey Lee (2009) \cite{JLee2009}:  first, egregious typo for $L(V,V)$; it shoudl be $L(V,W)$.  Onward, \\
for $L(V,W)$, \\
consider $W\otimes V^* \ni w\otimes \alpha$ s.t.
\[
(w\otimes \alpha)(v) = \alpha(v)w\in W, \, \forall \, v\in V, \text{ so } w\otimes \alpha \in L(V,W)
\]

Now consider (category of) left $R$-module,
\begin{equation}
{\,}_R\textbf{Mod} \ni {\,}_{\text{Mat}_{\mathbb{K}}(N,M) } \mathbb{K}^N
\end{equation}
where
\[
\begin{aligned}
	& V=\mathbb{K}^N \\
&	  W = \mathbb{K}^M
\end{aligned}
\]
For $A\in \text{Mat}_{\mathbb{K}}(N,M)$, $x\in \mathbb{K}^N$,
\[
e_i A^i_{ \,\ , \mu} e^{\mu}(x^{\nu} e_{\nu}) = Ax = e_iA^i_{\,\, \mu} x^{\mu} , \quad \, i=1,2,\dots M, \, \mu = 1,2, \dots N
\]
\[
A\in \text{Mat}_{\mathbb{K}}(N,M) \cong W\otimes V^* \cong L(V,W)
\]
Consider
\[
\begin{aligned}
	& \alpha \in (\mathbb{K}^N)^* = V^* \\
	& w\in \mathbb{K}^M = W
\end{aligned} \qquad \, \begin{aligned}
	& \alpha = \alpha_{\mu} e^{\mu} \\
	& w=w^ie_i
\end{aligned}
\]
\[
\alpha \otimes w = w \otimes \alpha = w^i\alpha_{\mu} e_i \otimes e^{\mu}
\]
(remember, isomoprhism between $\text{Mat}_{\mathbb{K}}(N,M)$ and $W\otimes V^*$ guaranteed, if $V,W$ are free $R$-modules, $R=\mathbb{K}$).

Let $V,W$ be left $R$-modules, i.e. $V,W \in {\,}_R\text{\textbf{Mod}}$.
\[
V^* \in \text{\textbf{Mod}}_R
\]
For $V^*\otimes W \in \text{\textbf{Mod}}_R\otimes {\,}_R\text{\textbf{Mod}}$
\[
\alpha \in V^*, w\in W
\]
\[
(\alpha \otimes w)(v) = \alpha(v)w, \text{ for } v\in V \in {\,}_R\text{\textbf{Mod}}
\]
But $(w\otimes \alpha)(v) = w\alpha(v)$.

Note $\alpha(v) \in R$.

Let $V,W$ be right $R$-modules, i.e. $V,W \in \text{\textbf{Mod}}_R$.
\[
V^* \in {\,}_R\text{\textbf{Mod}}
\]
For $W\otimes V^* \in \text{\textbf{Mod}}_R \otimes {\,}_R\text{\textbf{Mod}}$.
\[
\alpha \in V^*, \, w\in W
\]
\[
(v)(w\otimes \alpha) = w\alpha(v), \text{ with } \alpha(v)\in R, \, v\in V
\]
So $W\otimes V^* \cong L(V,W)$, for $V,W\in \text{\textbf{Mod}}_R$

\begin{definition}[7.20\cite{JLee2009}, \textbf{contraction}]
Let $(e_1,\dots e_n)$ basis for $V$, $(e^1\dots e^n)$ dual basis.

If $\tau \in T^r_{ \,\, s}(V)$, then for $k\leq r$, $l\leq s$, define
\begin{equation}
\begin{gathered}
	C^k_l \tau \in T^{r-1}_{ \, \, s-1}(V) \\
 C^k_l\tau(\theta^1 \dots \theta^{r-1}, w_1\dots w_{s-1}) := \\
	\sum_{a=1}^n \tau(\theta^1 \dots \underbrace{e^a}_{\text{$k$th position} } \dots \theta^{r-1}, w_1 \dots \underbrace{e_a}_{\text{$i$th position}} \dots w_{s-1}  )
\end{gathered}
\end{equation}
$C^k_l$ is called \textbf{contraction}, for some single $1\leq k \leq r$, some single $1\leq l \leq s$,
\[
C^k_l: T^r_s(V) \to T^{r-1}_{s-1}(V)
\]
s.t.
\[
(C^k_l\tau)^{i_1\dots \widehat{i}_k\dots i_r }_{ \phantom{i_1\dots \widehat{i}_k\dots i_r} j_1\dots \widehat{j}_l \dots j_s} := \tau^{i_1\dots a \dots i_r}_{ \phantom{i_1\dots a \dots i_r} j_1 \dots a \dots j_s }
\]
\end{definition}

Universal mapping properties can be invoked to give a basis free definition of contraction (EY : 20170405???).

IN general,
\[
\forall \, v_1 \dots v_s \in V, \forall \, \alpha^1 \dots \alpha^r \in V^*
\]
so that
\[
\begin{aligned}
	& v_j = v_j^{\mu} e_{\mu} \\
	& \alpha^i = \alpha^i_{\mu} e^{\mu}
\end{aligned} \quad \, \begin{aligned}
	& j=1\dots s, \quad \, \mu = 1,\dots \text{dim}V \\
	& i=1\dots r, \quad \, \mu = 1\dots \text{dim}V^*
\end{aligned}
\]
then $\forall \, \tau \in T^r_{ \,\, s} (V)$,
\[
\begin{gathered}
	\tau(\alpha^1\dots \alpha^r,v_1\dots v_s) = \tau( \alpha^1_{\mu_1} e^{\mu_1} \dots \alpha^r_{\mu_r} e^{\mu_r} , v_1^{\nu_1} e_{\nu_1} \dots v_s^{\nu_s}e_{\nu_s} ) = \\
	= \alpha^1_{\mu_1} \dots \alpha^r_{\mu_r} v_1^{\nu_1} \dots v_s^{\nu_s} \tau(e^{\mu_1}\dots e^{\mu_r} , e_{\nu_1} \dots e_{\nu_s} ) = \alpha^1_{\mu_1} \dots  \alpha_{\mu_r}^r v_1^{\nu_1} \dots v_s^{\nu_s} \tau^{\mu_1 \dots \mu_r}_{ \phantom{\mu_1 \dots \mu_r} \nu_1\dots \nu_s}
\end{gathered}
\]
which is equivalent to
\[
\begin{tikzpicture}
  \matrix (m) [matrix of math nodes, row sep=7.8em, column sep=12.8em, minimum width=5.2em]
  {
  \tau \in T^r_{\,\,s}(V)  &  \alpha^1 \otimes \dots \otimes \alpha^r \otimes v_1\otimes \dots \otimes v_s \otimes \tau  \\
& \tau(\alpha^1\dots \alpha^r,v_1\dots v_s)  \in R \\
};
  \path[|->]
  (m-1-1) edge node [above] {$ \alpha^1\otimes \dots \otimes \alpha^r \otimes v_1 \otimes \dots \otimes v_s \otimes $} (m-1-2)
  (m-1-2) edge node [left] {$C^1_{ s+1} C^2_{s+2}\dots C^r_{r+s} C^r_1 C_2^{r+1}\dots C_s^{r+s} $} (m-2-2)
  ;
\end{tikzpicture}
\]
where I've tried to express the right-$R$-module, "right action" on $\alpha^1 \otimes \dots \otimes \alpha^r \otimes v_1\otimes \dots \otimes v_s \in V^*\otimes \dots \otimes V$.

Conlon (2008) \cite{Conl2008}

\part{Pr\'{a}staro}

Pr\'{a}staro (1996) \cite{Pras1996}

\subsubsection{Affine Spaces}

cf. Sec. 1.2 - \emph{Affine Spaces} of Pr\'{a}staro (1996) \cite{Pras1996}

\begin{definition}[affine space]
  \begin{equation}
\begin{gathered}
    \text{ affine space \qquad \, } (M, \mathbf{M}, \alpha )  \\
    \text{ with } \\
    \begin{aligned}
      & M \equiv \text{ set (set of pts.) }  \\
      & \mathbf{M} \equiv \text{ vector space (space of free vectors) } \\
      & \alpha \equiv \mathbf{M} \times M \to M \equiv \text{ translation operator } \\
      & \alpha : (v,p ) \mapsto p' \equiv p + v
      \end{aligned}
\end{gathered}
  \end{equation}
  Note: $\alpha$ is a \textbf{transitive} action and without fixed pts. (free).

  i.e. $\forall \, p \in M$,
  \end{definition}

$\forall \, $ pt. $O \in M$, $\alpha:(v,O) \mapsto O' \equiv O + v$, $\alpha (\cdot , O) \equiv \alpha_O \equiv \alpha(O)$.  $\alpha_O(v) = O' = O + \mathbf{v}$ \qquad \, $\forall \, O' \in M$, $\exists \, \mathbf{v} \in \mathbf{M}$ s.t. $O' = O + \mathbf{v}$ \\
$\Longrightarrow M \equiv \mathbf{M}$.

$\forall \, (O, \lbrace e_i \rbrace)_{1 \leq i \leq n }$, where $\lbrace e_i \rbrace$ basis of $\mathbf{M}$, $M \equiv \mathbf{M} = \mathbb{R}^n$ so isomorphism $M \simeq \mathbb{R}^n$ \\
\begin{definition}
  $(O, \lbrace e_i \rbrace) \equiv $ affine frame.

  $\forall \, $ affine frame $(O,\lbrace e_i \rbrace)$, $\exists \, $ coordinate system $x^{\alpha} : M \to \mathbb{R}$, \\
  where $x^{\alpha}(p)$ is $\alpha$th component, in basis $\lbrace e_i \rbrace$, of vector $p-O$
  \end{definition}

\begin{theorem}[1.4 Pr\'{a}staro (1996) \cite{Pras1996}]
  Let $(x^{\alpha}), (\overline{a}^{\alpha})$ 2 coordinate systems correspond to affine frames $(O, \lbrace e_i \rbrace)$, $( \overline{O}, \lbrace \overline{e}_i \rbrace )$, respectively.
\begin{equation}
  \overline{x}^{\alpha} = A^{\alpha}_{ \, \, \beta} x^{\beta} + y^{\alpha}
\end{equation}
where
\[
y^{\alpha} \in \mathbb{R}^n, \qquad \, A^{\alpha}_{ \, \, \beta} \in GL(n; \mathbb{R})
\]
\end{theorem}

\begin{definition}[1.10 Pr\'{a}staro (1996) \cite{Pras1996}]
  \begin{equation}
    A(n) \equiv Gl(n,\mathbb{R}) \times \mathbb{R}^n
  \end{equation}
  affine group of dim. $n$
  \end{definition}

\begin{theorem}[1.5] symmetry group of $n$-dim. affine space, called affine group $A(M)$ of $M$.  $\exists \, $ isomoprhism,
  \begin{equation}
    A(M) \simeq A(n), \qquad \, f\mapsto (f^{\alpha}_{ \, \, \beta} , y^{\alpha}) \, ; \qquad \, f^{\alpha} \equiv x^{\alpha} \circ f = f^{\alpha}_{ \, \, \beta} x^{\beta} + y^{\alpha}
  \end{equation}
cf. Eq. 1.4 Pr\'{a}staro (1996) \cite{Pras1996}
  \end{theorem}

\part{Holonomy}

\begin{definition}[Conlon, 10.1.2] If $X,Y\in \mathfrak{X}(M)$, $M\subset \mathbb{R}^m$, \textbf{Levi-Civita connection} on $M\subset \mathbb{R}^m$
	\begin{equation}
	\begin{aligned}
	& \nabla : \mathfrak{X}(M) : \mathfrak{X}(M) \to \mathfrak{X}(M) \\
	\nabla_XY := p(D_XY)
	\end{aligned}
	\end{equation}
	with
	\[
	D_XY := \sum_{j=1}^m X(Y^j) \frac{ \partial }{ \partial x^j} = \sum_{i,j=1}^m X^i \frac{ \partial Y^j}{ \partial x^i} \frac{ \partial }{ \partial x^j} \qquad \,  \begin{aligned} & \quad \\
		& \forall \, X=\sum_{i=1}^m X^i \frac{ \partial }{ \partial x^i},  \\
		& \forall \, Y=\sum_{i=1}^m Y^i \frac{\partial }{ \partial x^i } \end{aligned}
	\]
\end{definition}

\[
\begin{aligned}
	& \nabla_{fX}Y = f(D_{fX}Y) = p(fD_XY) = fpD_XY = f\nabla_XY \\
	&  \nabla_X fY = p(D_XfY) = p \left( \sum_{i,j=1}^m \left( X^i f\frac{ \partial Y^j}{ dx^i } + X^i Y^j \frac{ \partial f}{ \partial x^i} \right) \frac{ \partial }{ \partial x^j} \right) = f\nabla_X Y + p \sum_{j=1}^m X(f) Y^j \frac{ \partial }{ \partial x^j} = f\nabla_XY + X(f) p(Y)
\end{aligned}
\]

\begin{definition}[Conlon, 10.1.4; Christoffel symbols]
\begin{equation}
\begin{gathered}
	\nabla_{\frac{ \partial }{ \partial x^i} \frac{ \partial }{ \partial x^j} = \Gamma^k_{ij} \frac{ \partial }{ \partial x^k} \qquad \, \text{ (Conlon's notation) }  }   \\
	\nabla_{\frac{ \partial }{ \partial x^j} \frac{ \partial }{ \partial x^i} = \Gamma^k_{ij} \frac{ \partial }{ \partial x^k} \qquad \, \text{ (F. Schuller's notation) }  }
\end{gathered}
\end{equation}
\end{definition}

\begin{definition}[torsion]
\begin{equation}
\begin{aligned}
	& T:\mathfrak{X}(M) \in \mathfrak{X}(M) \to \mathfrak{X}(M) \\
	& T(X,Y) = \nabla_XY - \nabla_Y X - [X,Y]
\end{aligned}
\end{equation}
If $T=0$, $\nabla$ torsion-free or symmetric.
\end{definition}
\[
\begin{gathered}
T(fX,Y) = f\nabla_XY - (f\nabla_Y X + Y(f)X) - \lbrace (fXY - (Y(f)X + fYX) \rbrace = fT(X,Y) \\
T(X,fY) = f\nabla_XY + X(f)Y - f\nabla_Y X   - \lbrace ( ( X(f) Y +fXY) -  fYX \rbrace = fT(X,Y)
\end{gathered}
\]
Thus, $T(X,Y)$ $C^{\infty}(M)$-bilinear.

$T\in \tau_1^2 (M)$.

$T(v,w) \in T_xM$ defined, $\forall \, v,w \in T_xM$, $\forall \, x\in M$.

Thus, torsion is a \textbf{tensor}.

\exercisehead{10.1.7 Conlon (2008)\cite{Conl2008} }.

If $T(X,Y)=0$,
\[
T(e_i,e_j) = \Gamma^k_{ji} e_k - \Gamma^k_{ij} e_k - 0 = 0 \Longrightarrow \Gamma_{ji}^k = \Gamma^k_{ij}
\]

If $\Gamma^k_{ij} = \Gamma^k_{ji}$, $T(e_i,e_j)=0$.

\exercisehead{10.1.8, Conlon (2008)\cite{Conl2008}}

If $M\subset \mathbb{R}^m$ smoothly embedded submanifold,
$\forall \, \frac{ \partial }{ \partial x^j}, \frac{ \partial }{ \partial x^i} \in T_xM$, spanning $T_xM$, consider $\frac{ \partial }{ \partial x^j} = X^k_j \frac{ \partial }{ \partial \widetilde{x}^k}$, $\frac{ \partial }{ \partial x^i} = X_i^k(\widetilde{x}) \frac{ \partial }{ \partial \widetilde{x}^k}$

\[
\begin{gathered}
	\nabla_{\frac{\partial }{ \partial x^j} } \frac{ \partial }{ \partial x^i} = p D_{ X^k_j \frac{ \partial }{ \partial \widetilde{x}^k } } X^l_i \frac{ \partial }{ \partial \widetilde{x}^l } = p \left( X^k_j \frac{ \partial X^l_i}{  \partial \widetilde{x}^k } \frac{ \partial }{ \partial \widetilde{x}^l } \right) =   X^k_j  p \left( \frac{ \partial X^l_i}{  \partial \widetilde{x}^k } \frac{ \partial }{ \partial \widetilde{x}^l } \right)  \\
	\nabla_{\frac{\partial }{ \partial x^i} } \frac{ \partial }{ \partial x^j} =  X^k_i p \left(  \frac{ \partial X^l_j }{ \partial \widetilde{x}^k }   \frac{ \partial }{ \partial \widetilde{x}^l  } \right)
\end{gathered}
\]

If $X\in \mathfrak{X}(M)$, smooth $s:[a,b]\to M$,

then $\forall \, s(t)$,
\[
X'_{s(t)} = \nabla_{\dot{s}(t)} X \in T_{s(t)}M
\]
In fact, it's often natural to consider fields $X_{s(t)}$ along $s$, parametrized by parameter $t$, allowing
\[
X_{s(t_1)} \neq X_{s(t_2)}
\]
each of $s(t_1)=s(t_2)$.

\begin{definition}[10.1.9] Let smooth $s:[a,b] \to M$.

Vector field along $s$ is smooth $v:[a,b]\to TM$ s.t.
\[
\begin{tikzpicture}
  \matrix (m) [matrix of math nodes, row sep=7.8em, column sep=12.8em, minimum width=5.2em]
  {
   &  TM  \\
	\left[ a,b \right] & M \\
};
  \path[|->]
  (m-2-1) edge node [above] {$ v $} (m-1-2)
edge node [above] {$s$} (m-2-2)
  (m-1-2) edge node [left] {$ \pi $} (m-2-2)
  ;
\end{tikzpicture}
\]
commutes.

Note that $v\in \mathfrak{X}(s) \subset \mathfrak{X}(M)$
\end{definition}

e.g. $(Y|s)(t) = Y_{s(t)}$, restriction of $Y\in \mathfrak{X}(M) $ to $s$.

e.g. $\dot{s}(t) \in \mathfrak{X}(M)$.

$\forall \, v,w \in \mathfrak{X}(s)$, $v+w \in \mathfrak{X}(s)$,
\[
\begin{gathered}
	(fv+gv)(t) := (f(s(t)) + g(s(t)) )v(t) = f(s(t)) v(t) + g(s(t)) v(t) = (f+g)v(t)
\end{gathered}
\]
Likewise,
\[
f(v+w) = fv+fw
\]
$\mathfrak{X}(s)$ is a real vector space and $C^{\infty}[a,b]$-module.

\begin{definition}[10.1.10]
Let conection $\nabla$ on $M$.

\textbf{Associated covariant derivative} is operator
\[
\frac{\nabla}{dt} \mathfrak{X}(s) \to \mathfrak{X}(s)
\]
$\forall \, $ smooth $s$ on $M$, s.t.
\begin{enumerate}
\item $\frac{\nabla}{dt}$ $\mathbb{R}$-linear
\item $\left( \frac{\nabla}{dt} \right)(fv) = \frac{df}{dt} v+ f\frac{\nabla}{dt} v$, $\forall \, f \in C^{\infty}[a,b]$, $\forall \, v\in \mathfrak{X}(s)$
\item If $Y\in \mathfrak{X}(M)$, then
\[
\frac{\nabla}{dt} (Y|s)(t) = \nabla_{ \dot{s}(t)}Y \in T_{s(t)} M, \quad \, a\leq t \leq b
\]
\end{enumerate}
\end{definition}

\begin{theorem}[Conlon Thm. 10.1.11\cite{Conl2008} ]
	$\forall \, $ connection $\nabla$ on $M$, $\exists \, !$ \, associated covariant derivative $\frac{ \nabla }{dt}$
\end{theorem}

\begin{proof}
	Consider arbitrary coordinate chart $(U,x^1 \dots x^n)$.

	Consider smooth curve $s:[a,b]\to U$.

	Let $v\in \mathfrak{X}(s)$, $v(t) = v^i(t) \frac{ \partial }{ \partial x^i}$; $\dot{s}(t) = s^j \frac{ \partial }{ \partial x^j}$.
	\[
	\begin{gathered}
	\frac{ \nabla v}{ dt} = \frac{dv^i(t) }{dt}\frac{ \partial }{ \partial x^i} + v^i(t) \frac{ \nabla}{dt} \frac{ \partial }{ \partial x^i} = \frac{ d v^i }{dt}	\frac{ \partial }{ \partial x^i} + v^i \nabla_{ \dot{s}(t) } \frac{ \partial }{ \partial x^i } = \dot{v}^i \frac{ \partial }{ \partial x^i } + v^i\dot{s}^j \Gamma^k_{ij} \frac{ \partial }{ \partial x^k} = \left( \dot{v}^k + v^i \dot{s}^j \Gamma^k_{ij} \right)\frac{ \partial }{ \partial x^k} \end{gathered}
	\]
	This is an explicit, local formula in terms of connection, proving uniqueness.

	Existence: $\forall \, $ coordinate chart $(U,x^1\dots x^n)$, $\left( \dot{v}^k + v^i \dot{s}^j \Gamma_{ij}^k \right) \frac{ \partial }{ \partial x^k } =: \frac{ \nabla v}{dt}$.

	\[
	\begin{gathered}
	\frac{\nabla}{dt}(fv) = \dot{f}v^k + f\dot{v}^k + fv^i \dot{s}^j = \dot{f}v + f\frac{ \nabla v}{dt}
	\end{gathered}
	\]
If $f$ constant, then $\frac{\nabla}{dt}$ is $\mathbb{R}$-linear.

\end{proof}

\begin{definition}[10.1.12 Conlon (2008)\cite{Conl2008}]
	Let $(M,\nabla)$.  Let $v\in\mathfrak{X}(s)$ for smooth $s:[a,b] \to M$.

If $\frac{\nabla v}{dt} \equiv 0$ on $s$, then $v$ is \textbf{parallel } along $s$.
\end{definition}

\begin{theorem}[10.1.13]
Let $(M,\nabla)$, smooth $s:[a,b]\to M$, $c\in [a,b]$, $v_0 \in T_{s(c)}M$.

Then $\exists \, !$ parallel field $v\in \mathfrak{X}(s)$ s.t. $v(c) = v_0$.

$v$ parallel transport along $s$.
\end{theorem}
\begin{proof}
	\[
\begin{aligned}
	& \dot{s}(t) = \dot{s}^j(t) e_j \\
	&  v(t) = v^i(t) e_i  \\
& v_0 = a^i e_i
\end{aligned}
\]
\[
0 = \left( \frac{dv^k}{dt}(t) + v^i(t) \dot{s}^j(t) \Gamma^k_{ij}(s(t)) \right) e_k
\]
or equivalently
\begin{equation}
	\frac{dv^k}{dt} = - v^i \dot{s}^j \Gamma^k_{ij} , \qquad \, 1\leq k \leq n \qquad \, (10.1)
\end{equation}
with initial conditions $v^k(c) = a^k$, $1\leq k \leq n$.

By existence and uniqueness of solutions of O.D.E.

$\exists \, \epsilon > 0$ s.t. $\exists \, !$ solutions $v^k(t)$.  For $c-\epsilon < t < c+\epsilon$.

In fact, these ODEs being linear in $v^k$,  by ODE theory (Appendix C, Thm. C.4.1).

$\nexists \, $ restriction on $\epsilon$, so $\exists \, ! \, v^k(t)$ \, $\forall \, t \in [a,b]$, $1\leq k \leq n$

\end{proof}

\subsubsection{Principal bundle, vector bundle case for parallel transport}

Recall the 2 different forms or viewpoints for Lie-algebra valued 1-forms, or vector-valued 1-forms, or sections of 1-form-valued endomorphisms:
\[
\omega^k_{ \,\, i\mu} dx^{\mu} \equiv \omega^k_{\,\,i} \in \Omega^1(M,\mathfrak{gl}(n,\mathbb{F})) = \Gamma(\mathfrak{gl}(n,\mathbb{R} \otimes \left. T^* M \right|_U ) )
\]
for $\begin{aligned} & \quad \\
	& i,k = 1\dots n =\text{dim}E \\
	& \mu = 1\dots d =\text{dim}E \end{aligned}$.

Now
\[
D_X\mu = X^{\mu} D_{\frac{\partial }{ \partial x^{\mu}}}\mu = X^{\mu} \left[ \left( \frac{ \partial }{ \partial x^{\mu}} \mu^k \right) e_k + \mu^i \omega^k_{ \,\, i\mu} e_k \right] = \left( X(\mu^k) + \mu^i \omega^k_{ \,\, i}(X) \right) e_k = \left( d\mu^k(X) + \mu^i \omega^k_{ \,\, i}(X) \right) e_k
\]
So then define
\begin{equation}
	\begin{aligned}
		& D: \Gamma(E) \to \Gamma(E) \otimes \Gamma(T^*M) \\
	& D\mu  = D(\mu^i e_i) = e_k ( d\mu^k +\mu^i \omega^k_{ \,\, i}) \equiv (d+A)\mu
	\end{aligned}
\end{equation}

Also, $D$ can be defined for this case:
\[
D: \Gamma (\text{End}(E)) \to \Gamma(\text{End}E) \otimes \Gamma(T^*M)
\]
Let $\sigma = \sigma^i_{ \,\, j} e_i \otimes e^j \in \Gamma(\text{End}(E))$

\begin{equation}
\begin{gathered}
	D\sigma = D(\sigma^i_{ \,\, j} e_i ) \otimes e^j + \sigma^i_{ \,\, j} e_i \otimes D^* e^j = \left( d\sigma^k_{ \,\, j} + \sigma^i A^k_{ \,\, i} \right) e_k \otimes e^j + \sigma^i_{ \,\, j} e_i \otimes (A^*)_k^{ \,\, j} e^k = \\
	= (d\sigma^k_{ \,\, j} + \sigma^i_{ \,\, j} A^k_{ \,\, i} ) e_k \otimes e^j + \sigma^k_{ \,\, i} e_j \otimes (- A^i_{ \, \, j}) e^j = (d\sigma^k_{ \,\, j} + [A,\sigma]^k_{ \,\, j}) e_k\otimes e^j
\end{gathered}
\end{equation}
cf. Def. 4.1.4 of Jost (2011), pp. 138.

For $\mu \in \Gamma(E)$, smooth $s:[a,b] \to M$, $X(t) = \dot{s}(t)$,
\begin{equation}
D_{\dot{s}(t)}\mu = \dot{s}^{\mu}D_{\frac{ \partial }{ \partial x^{\mu}} } \mu = \dot{s}^{\mu} \left[ \frac{ \partial \mu^k }{ \partial x^{\mu}} e_k + \mu^i \omega^k_{ \, \, i\mu} e_k \right] = \left[ \dot{s}^{\mu} \frac{ \partial \mu^k}{ \partial x^{\mu} } + \dot{s}^{\mu} \mu^i \omega^k_{ \,\, i \mu} \right] e_k =\frac{d}{dt} \mu(s(t)) + \mu^i \dot{s}^{\mu} \omega^k_{ \,\, i \mu} e_k
\end{equation}

Let $D_{\dot{s}(t)}\mu=0$.  Then,
\begin{equation}
\frac{d}{dt} \mu(s(t)) = -\mu^i \dot{s}^{\mu}\omega^k_{ \,\, i\mu} e_k
\end{equation}

Recall, given vector bundle $E\xrightarrow{ \pi} N$, given $\varphi : M\to N$, then pullback
\begin{equation}
	\varphi^* E \to M
\end{equation}
i.e.

\[
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	\varphi^* E &  E  \\
	M  &  N  \\
};
\path[->]
(m-1-2) edge node [auto] {$ \varphi^*  $} (m-1-1)
edge node [auto] {$ \pi $} (m-2-2)
(m-2-1) edge node [auto]  {$  \varphi $} (m-2-2)
(m-1-1) edge node [auto] {$\psi$} (m-2-1)
;
\end{tikzpicture}
\qquad \,
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	(\varphi^* E)_x = E_{\varphi(x)}  \\
	 x\in M \\
};
 \path[|->]
	(m-2-1) edge node [auto] {$$} (m-1-1)
;
\end{tikzpicture}
\]
i.e. if $s\in \Gamma(E)$,
\[
\varphi^* s = s \circ \varphi \in \Gamma( \varphi^* E)
\]
Thus,
\[
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	\gamma^* E &  E  \\
	\left[a,b\right]  &  M  \\
};
\path[->]
(m-1-2) edge node [auto] {$ \gamma^*  $} (m-1-1)
edge node [auto] {$ \pi $} (m-2-2)
(m-2-1) edge node [auto]  {$  \gamma $} (m-2-2)
(m-1-1) edge node [auto] {$ $} (m-2-1)
;
\end{tikzpicture}
\qquad \,
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	(\varphi^* E)_c = E_{\gamma(c)}  \\
	 c\in  [a,b]\\
};
 \path[|->]
	(m-2-1) edge node [auto] {$$} (m-1-1)
;
\end{tikzpicture}
\]

For
\[
\begin{gathered}
\dot{v}^k = -v^i \dot{s}^j \Gamma^k_{ \, \, ij}   \\
v^k(c) = v_0^k \qquad \, 1 \leq k \leq m
\end{gathered}
\]
\[
\dot{v} = -v^i \dot{s}^j \Gamma_{ij}
\]
\[
\begin{gathered}
\dot{ (v+w) } = -(v^i + w^i) \dot{s}^j \Gamma_{ij}
(v+w)(c) = v(c) + w(c) = v_0 + w_0
\end{gathered}
\]
so $v+w\in \mathfrak{X}(s)$ is parallel transport of $v_0 + w_0$.

Likewise, $\forall \, a \in \mathbb{F}$, $av \in \mathfrak{X}(s)$ is the parallel transport of $av_0$.

\[
\dot{\mu}^k = - \mu^i \dot{s}^{\mu} \omega^k_{ \, \, i \mu} = -\mu^i \omega^k_{ \,\, i}(\dot{s}^{\mu})
\]

Suppose $\gamma^*E$ trivialized over $[a,b]$.

Closed interval is contractible, so this is always possible.

For chart $(U,\varphi)$,
\[
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	\gamma^* E   &  E    \\
	\left[a,b\right]  &  M  \\
};
\path[->]
(m-1-2) edge node [auto] {$ \gamma^*  $} (m-1-1)
edge node [auto] {$ \pi $} (m-2-2)
(m-2-1) edge node [auto]  {$  \gamma $} (m-2-2)
(m-1-1) edge node [auto] {$ $} (m-2-1)
;
\end{tikzpicture} \qquad \,
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
{
	\left. E \right|_U   &   U\times V  \\
	U \subset M    &    \\
};
\path[->]
(m-2-1) edge node [auto] {$ \pi^{-1}  $} (m-1-1)
edge node [auto] {$  $} (m-1-2)
(m-1-1) edge node [auto]  {$  \psi $} (m-1-2)
%(m-1-1) edge node [auto] {$ $} (m-2-1)
;
\end{tikzpicture}
\]
Consider
\[
\begin{aligned}
& \varphi:[a,b]  \times V \to \gamma^* E \\
& \varphi(t,\cdot ) = \gamma^* \circ \psi^{-1}(\gamma(t), \cdot )
\end{aligned}
\]
$\forall \, \mu \in \Gamma( \left. E\right|_{x\in M} )$, \\
$\mu = \mu^i e_i$.

$\varphi(t,e_i) = \epsilon_i$ is a basis for $\gamma^* E$.

$\forall \, \sigma \in \Gamma(\gamma^*E)$,
\[
\sigma  = \sigma^i \epsilon_i, \quad \, \sigma^i: [a,b] \to \mathbb{F}
\]
\[
\begin{gathered}
	\nabla_{ \frac{ \partial }{ \partial x^{\mu} } } \sigma = \frac{ \partial \sigma^k}{ \partial x^{\mu}  } \epsilon_k + \omega^k_{ \, \, j \mu } \sigma^j \epsilon_k = \left( \frac{ \partial \sigma ^k }{ \partial x^{\mu } } + \omega^k_{ \ , \, j \mu} \sigma^j  \right)  \epsilon_k \\
 \nabla \sigma = \epsilon_k \otimes ( d\sigma^k + \omega^k_{ \, \, j \mu} dx^{\mu} \sigma^j ) = \epsilon_k \otimes (d\sigma^k + \omega^k_{ \, \, j} \sigma^j  )  \\
 \nabla_{ \frac{d}{dt} } \sigma = \epsilon_k \otimes \left( \frac{d\sigma^k }{ dt } + \omega^k_{ \, \, j \mu } \dot{x}^{\mu} \sigma^j  \right)
\end{gathered}
\]
Now
\[
\frac{d}{dt} = \dot{x}^{\nu} \frac{ \partial }{ \partial x^{\nu } }
\]
Then $\sigma$ parallel along $\gamma$ if
\[
\frac{ d\sigma^k}{ dt} + \omega^k_{ \, \, j\mu} \dot{x}^{\mu} \sigma^j = 0
\]
\begin{definition}[3.1.4 \cite{ClSa2012}]  Parallel transport along $\gamma$ is
\begin{equation}
\begin{aligned}
	& P_{\gamma} : E_{\gamma(a)} \to E_{\gamma(b)} \\
	& P_{\gamma}(v) \mapsto \sigma(b)
\end{aligned}
\end{equation}
where $\sigma \in \Gamma(\gamma^*E)$, $\sigma$ unique and s.t. $\sigma(a)=v$.
\end{definition}

\begin{lemma}[10.1.16\cite{Conl2008}]
	holonomy
	\[
	h_s:T_xM \to T_{x_0}M
	\]
	if $\nabla$ around piecewise smooth loop $s$ is a linear transformation.
\end{lemma}

\begin{lemma}[10.1.18 Conlon (2008)\cite{Conl2008}]
	Let piecewise smooth loop $s:[a,b] \to M$ at $x_0$.

Let weak reparametrization $\widetilde{s} = s\circ r: [c,d] \to M$.

If reparametrization is orientation-preserving, then $h_{\widetilde{s}} = h_s$, \\
If reparametrization is orientation-reversing, then $h_{\widetilde{s}} = h^{-1}_s$,

\end{lemma}
\begin{proof}
	Without loss of generality,  assume smooth $s,r$
\[
\begin{aligned}
	& \widetilde{s}(\tau) = s(r(\tau)) \\
	& \widetilde{v}(\tau) = v(r(\tau))
\end{aligned}
\]
\[
\begin{aligned}
	& \widetilde{u}^j(\tau) = \frac{dt}{d\tau}(\tau) u^j(r(\tau)) \\
	& \frac{d\widetilde{v}^k}{d\tau} (\tau) = \frac{dr}{d\tau}(\tau) \frac{dv^k}{dt}(r(\tau)) \\
	& \frac{d\widetilde{v}^k }{ d\tau} = -\widetilde{v}^i \widetilde{u}^j \Gamma^k_{ij}
\end{aligned}
\]
since \[
\begin{gathered}
	\frac{dv^k}{dt} = -v^i u^j \Gamma^k_{ij} ; \qquad \, 1\leq k \leq n \\
v^k(c) = a^k; \qquad \, 1\leq k \leq a
\end{gathered}
\]
\[
\frac{dr}{d\tau} \frac{dv^k}{ dt} = -v^i \frac{dr}{d\tau} u^j \Gamma^k_{ \, \, ij} = \frac{d\widetilde{v}^k}{d\tau} = -\widetilde{v}^i \widetilde{u}^j \Gamma^k_{ \, \, ij}
\]
Thus, if $r(c) = a$, $r(d)=b$
\[
h_{\widetilde{s}}(v_0) = \widetilde{v}(d) = v(b) = h_s(v_0)
\]
If $r(c)=a$, $r(d)=b$, then
\[
\widetilde{v}(c) = v(b) = h_s(v_0)
\]
and
\[
h_{\widetilde{s}}(h_s(v_0)) = h_{\widetilde{s}}(v(b)) = \widetilde{v}(d) = v(a) = v_0
\]
At this point, I will switch to my notation because it clarified to me, at least, what was going on, in that a holonomy $h_s$ is \emph{invariant} under orientation-preserving reparametrization, and its inverse is well-defined.

For $\widetilde{s} = s\circ t:[c,d] \to M$, \\
piecewise smooth $t$ is reparametrized, i.e.
\begin{equation}
	t:[c,d] \to [a,b]
\end{equation}
Now,
\[
\begin{gathered}
	\frac{d}{d\tau}\widetilde{s}(\tau) = \frac{d}{d\tau} \widetilde{s}(t(\tau)) = \dot{s}(t) \frac{dt}{d\tau}(\tau) \equiv \dot{s} \frac{dt}{d\tau} \\
v^k(t) = v^k(t(\tau)) = v^k(\tau) \\
 \frac{dv^k}{d\tau}(t(\tau) ) = \frac{dv^k}{dt} \frac{dt}{ d\tau } = \frac{dt}{d\tau}(-v^i( \tau) \dot{s}^j(t) \Gamma^k_{ \,\, ij} ) = -v^i(\tau) \frac{d\widetilde{s}^j }{ d\tau} \Gamma^k_{ \, \, ij}
\end{gathered}
\]
Consider
\[
h_s(v_0) = v(b)
\]
If $\begin{aligned} & \quad \\
	& t(c) = a \\
	& t(d)=b \end{aligned}$,
\[
h_{\widetilde{s}}(v_0) = \widetilde{v}(d) = v(t(d)) = v(b) = h_s(v_0)
\]
If $\begin{aligned} & \quad \\
	& t(c) = b \\
	& t(d)=a \end{aligned}$,
\[
\begin{gathered}
h_{\widetilde{s}}(h_s(v_0) )  = h_{\widetilde{s}}( v(b)) = h_{\widetilde{s}}(v(t(c))) = h_{\widetilde{s}}(\widetilde{v}(c)) = \\
	= \widetilde{v}(d) = v(t(d)) = v(a) = v_0
\end{gathered}
\]
Thus,
\[
\boxed{ h_{\widetilde{s}}= h_s^{-1} }
\]

\end{proof}

I am working through Conlon (2008) \cite{Conl2008} , Clarke and Santoro (2012) \cite{ClSa2012}, and Schreiber and Waldorf (2007) , concurrently, for holonomy.

\part{Complex Manifolds}

EY : 20170123 I don't see many good books on Complex Manifolds for physicists other than Nakahara's.  I will supplement this section on Complex Manifolds with external links to the notes of other courses that I found useful to myself.

\href{http://www.caramdir.at/uploads/math/piii-cm/complex-manifolds.pdf}{Complex Manifolds - Lecture Notes}
Koppensteiner (2010) \cite{Kopp2010}

\href{http://www.staff.science.uu.nl/~vando101/MRIlectures.pdf}{Lectures on Riemannian Geometry, Part II: Complex Manifolds by Stefan Vandoren}

Vandoren (2008) \cite{Vand2008}

\part{Jets, Jet bundles, $h$-principle, $h$-Prinzipien}

cf. Eliashberg and Misahchev (2002) \cite{ElMi2002}

cf. Ch. 1 Jets and Holonomy, Sec. 1.1 Maps and sections of Eliashberg and Misahchev (2002) \cite{ElMi2002}.

Visualize $f:\mathbb{R}^n \to \mathbb{R}^q$ as graph $\Gamma_f \subset \mathbb{R}^n \times \mathbb{R}^q$.

Consider this graph as image of $\begin{aligned} & \quad \\
	& \mathbb{R}^n \to \mathbb{R}^n \times \mathbb{R}^q \\
	& x \mapsto (x,f(x)) \end{aligned}$, i.e.

$\begin{aligned} & \quad \\
	& \mathbb{R}^n \to \mathbb{R}^n \times \mathbb{R}^q \\
	& x \mapsto (x,f(x)) \end{aligned}$ is called section (by mathematicians), \\
is called \emph{field} or $\mathbb{R}^q$-valued field (by physicists).

cf. Ch. 1 Jets and Holonomy, Sec. 1.2 Coordinate definition of jets of Eliashberg and Misahchev (2002) \cite{ElMi2002}.

\begin{definition}[$r$-jet]
Given (smooth) $f: \mathbb{R}^n \to \mathbb{R}^q$, given $x\in \mathbb{R}^n$.

$r$-jet of $f$ at $x$ - sequence of derivatives of $f$, up to order $r$, $\equiv$
\begin{equation}
J^r_f(x) = (f(x),f'(x) \dots f^{(r)}(x) )
\end{equation}
\end{definition}

$f^{(q)}$ consists of all partial derivatives $D^{\alpha}f$, $\alpha= (\alpha_1\dots \alpha_n)$, $|\alpha| = \alpha_1 + \dots + \alpha_n =s$, ordered lexicographically.

e.g. $q=1$, \\
$f: \mathbb{R}^n \to \mathbb{R}$.   \\
1-jet of $f$ at $x = J_f^1(x) = (f(x), f^{(1)}(x))$.
\[
f^{(1)}(x) = \lbrace D^{\alpha}f| \alpha = (\alpha_1 \dots \alpha_n), |\alpha| = \alpha_1  + \dots + \alpha_n  =1 \rbrace = \left( \frac{ \partial f}{ \partial x^1 }, \frac{ \partial f}{ \partial x^2 } , \dots \frac{ \partial f}{ \partial x^n } \right)
\]
Let $d_r = d(n,r) = $ number of all partial derivatives $D^{\alpha}$ of order $r$ of function $\mathbb{R}^n \to \mathbb{R}$.

Consider $r$-jet $J^r_f(x)$ of map $f:\mathbb{R}^n\to \mathbb{R}^q$ as pt. of space $\mathbb{R}^q \times \mathbb{R}^{qd_1} \times \mathbb{R}^{qd_2} \times \dots \times \mathbb{R}^{qd_r} = \mathbb{R}^{q N_r}$, where $N_r = N(n,r) = 1 + d_1 + d_2 + \dots + d_r$, i.e.
\[
J_f^r(x) = (f(x),f^{(1)}(x) , \dots f^{(r)}(x) ) \in \mathbb{R}^q \times \mathbb{R}^{qd_1} \times  \dots \times \mathbb{R}^{qd_r} = \mathbb{R}^{q N_r}
\]

\exercisehead{1}

Given order $r$, consider $n$-tuple of (positive) integers $(r_1,r_2\dots r_n)$ s.t. $r_1 + r_2 + \dots + r_n = r$, and $r_k\geq 0$. \\
 Imagine $r_k = $ occupancy number, num ber of balls in $k$th cell.  $(r_1\dots r_n)$ describes a positive ocnfiguration of occupancy numbers, with indistinguishable balls; 2 distributions are distinguishable only if corresponding $n$-tuples $(r_1 \dots r_n)$ not identical.

Represent balls by stars, and indicate $n$ cells by $n$ spaces between $n+1$ bars.

With $n+1$ bars, $r$ stars, 2 bars are fixed. $n-1$ bars and $r$ stars to arrange linearly, so a total of $n-1+r$ objects to arrange.  $r$ stars indistinguishable amongst themselves, so choose $r$ out of $n-1+r$ to be stars.
\begin{equation}
\Longrightarrow d_r = d(n,r)=\binom{n-1+r}{r}
\end{equation}

Use \emph{induction} (cf. \href{http://www.cs.columbia.edu/~cs4205/files/CM4.pdf}{Ch. 4 Binomial Coefficients}).
\[
\begin{aligned}
	& N_0 = N(n,0) = \binom{n-1+0}{0} = 1 \\
	& N_1 = N(n,1) = 1+ \binom{n-1+1}{1} = 1 + n = \frac{ (n+1)!}{n! 1!}
\end{aligned}
\]
Induction step:
\[
N_{r-1} = N(n,r-1) = \sum_{k=1}^{r-1} d_k + 1 = \binom{ n+r-1}{r-1}
\]
and so
\[
\begin{gathered}
N_r = N(n,r) = \sum_{k=1}^r d_k + 1 = \sum_{k=1}^r \binom{n-1+k}{k} + 1 = \sum_{k=1}^{r-1} \binom{ n-1 +k}{k} + \binom{n-1+r}{r} + 1 = \\
	 =  \binom{n+r-1}{r-1} + \binom{n-1+r}{r} = \frac{ (n+r-1)! }{ (r-1)! n!} + \frac{ (n-1+r)! }{ r! (n-1)! } = \frac{ (n+r)!}{n!r!} = \binom{n+r}{r}
\end{gathered}
\]
\[
\begin{tikzpicture}
  \matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
  {
    \mathbb{R}^{qN_r} &  \\
    \mathbb{R}^n & \mathbb{R}^q \\
};
  \path[->]
  (m-2-1) edge node [auto] {$ J_f^r $} (m-1-1)
          edge node [auto] {$f $} (m-2-2)
  ;
\end{tikzpicture}   \quad \quad \, \begin{tikzpicture}
  \matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
  {
   J_f^r(x)  &  \\
    x & f(x) \\
};
  \path[|->]
  (m-2-1) edge node [auto] {$ J_f^r $} (m-1-1)
          edge node [auto] {$f $} (m-2-2)
  ;
\end{tikzpicture}
\]

\begin{definition}[space of $r$-jets]
space of $r$-jects of maps $\mathbb{R}^n \to \mathbb{R}^q$ or space of $r$-jets of sections $\mathbb{R}^n \to \mathbb{R}^n \times \mathbb{R}^q \equiv $
\begin{equation}
J^r(\mathbb{R}^n, \mathbb{R}^q) = \mathbb{R}^n \times \mathbb{R}^{qN_r} \equiv \mathbb{R}^n \times \mathbb{R}^q \times \mathbb{R}^{qd_1} \times \mathbb{R}^{qd_2} \times \dots \times \mathbb{R}^{qd_r}
\end{equation}
e.g. $J^1(\mathbb{R}^n, \mathbb{R}^q) = \mathbb{R}^n \times \mathbb{R}^q \times M_{q\times n}$, where $M_{q\times n} = \mathbb{R}^{qn}$ is the space of $(q\times n)$-matrices.
\end{definition}

\part{Morse Theory}

\section{Morse Theory introduction from a physicist}

I needed some physical motivation to understand Morse theory, and so I looked at Hori, et. al. \cite{Hori2003}.

cf. pp. 43, Sec. 3.4 Morse Theory, from Ch. 3. Differential and Algebraic Topology of Hori, et. al. \cite{Hori2003}.

Consider smooth $f:M \to \mathbb{R}$, with non-degenerate critical points.

If no critical values of $f$ between $a$ and $b$ ($a<b$), then subspace on which $f$ takes values less than $a$ is deformation retract of subspace where $f$ less than $b$, i.e.
\[
\lbrace x \in M | f(x) < b\rbrace \times [0,1] \xrightarrow{ F } \lbrace x \in M | f(x) < b\rbrace
\]
$\forall \, x \in M$ s.t. $f(x) < b$,
\[
\begin{aligned}
  & F(x,0)  = x \\
  & F(x,1) \in \lbrace x \in M | f(x) < a \rbrace
  \end{aligned} \qquad \qquad \, \text{ and } F(a',1) = a' \qquad \, \forall \, a' \in M \text{ s.t. } f(a') < a
\]

To show this, consider $-\nabla f/|\nabla f|^2$

Morse lemma: $\forall \, $ critical pt. $p$ s.t. $\exists \, $ choice of coordinates s.t.
\begin{equation}
  f  = - (x_1^2 + x_2^2 + \dots + x_{\mu}^2) + x_{\mu + 1}^2 + \dots + x_n^2
\end{equation}
where $f(p)=0$ and $p$ is at origin of these coordinates.

\begin{itemize}
\item difference between
  \[
f^{-1}(\lbrace x \leq -\epsilon \rbrace) , \, f^{-1}(\lbrace x \leq + \epsilon \rbrace)
\]
can be determined by local analysis and only depends on $\mu$, $\mu \equiv $ ``Morse index'' $=$ number of negative eigenvalues of Hessian of $f$ at critical pt.

Answer: \\

\[
f^{-1}(\lbrace x \leq + \epsilon \rbrace) \text{ can be obtained from } f^{-1}(\lbrace x \leq -\epsilon \rbrace) \text{ by ``attaching $\mu$-cell'' along boundary $f^{-1}(0)$ }
\]

\item ``attaching $\mu$-cell to $X$ mean, take
  $\mu$-ball $B_{\mu} = \lbrace |x| \leq 1 \rbrace$ in $\mu$-dim. space, \\
  identity pts. on boundary $S^{\mu-1}$ with pts. in the space $X$, through \\
  cont. $f : S^{\mu-1} \to X$, i.e. take
  \[
X \coprod B_{\mu}
\]
with $x\sim f(x)$ \, $\forall \, x \in \partial B_{\mu} = S^{\mu -1}$.

\item find homology of $M$,

  $f$ defines chain complex $C_f^*$, $k$th graded piece $C^{\alpha_k}$, $\alpha_k$ is number of critical pts. with index $k$.
  \begin{equation}
\begin{aligned}
  & \partial : C_p^k \to C^{k-1}_p \\
  & \partial x_a = \sum_b \Delta_{a,b} x_b
  \end{aligned}
    \end{equation}
  where $\Delta_{a,b} :=$ signed number of lines of gradient flow from $x_a$ to $x_b$, $b$ labels pts. of index $k-1$.

  Gradient flow line is path $x(t)$ s.t. $\dot{x} = \nabla (f)$, with $\begin{aligned} & \quad  \\
    & x(-\infty) = x_a \\
    & x(+\infty) = x_b \end{aligned}$

\item  To define this number ($\Delta_{a,b} $?), construct moduli space of such lines of flow (???) \\
  by intersecting outward and inward flowing path spaces from each critical point, and then show this moduli space is oriented, 0-dim. manifold (pts. with signs)

\item $\partial^2=0$ proof

  $\partial$, boundary of space of paths connecting critical points, whose index differs by $2 =$ union over compositions of paths between critical pts. whose index differs by $1$.

  $\Longrightarrow$ coefficients of $\partial^2$ are sums of signs of pts. in $0$-dim. space, which is boundary of $1$-dim. space.

  These signs must therefore add to $0$, so $\partial^2=0$.

  \end{itemize}

Hori, et. al. \cite{Hori2003} is good for physics, but there isn't much thorough, step-by-step explanations of the math.  I will look at Hirsch (1997) \cite{MHirsch1997} and Shastri (2011) \cite{AShastri2011} at the same time.

\subsection{Introduction, definitions of Morse Functions, for Morse Theory}

cf. Ch. 6, Morse Theory of Hirsch (1997) \cite{MHirsch1997}, Section 1. Morse Functions, pp. 143-

Recall for $TM$, $T_xM \xrightarrow{\varphi}\mathbb{R}^n$.  \\
Cotangent bundle $T^*M$ defined likewise:
\[
T^*_xM \xrightarrow{ \varphi } \text{ dual vector space } (\mathbb{R}^n)^* = L(\mathbb{R}^n,\mathbb{R})
\]
i.e.
\[
T^*M = \bigcup_{x\in M} (M_x^*) \qquad \qquad \, M_x^* = L(M_x,\mathbb{R})
\]
If chart $(\varphi, U)$ on $M$, natural chart on $T^*M$ is
\[
\begin{aligned}
  & T^*U \to \varphi(U) \times (\mathbb{R}^n)^* \\
  & \lambda \in M_x^* \mapsto (\varphi(x), \lambda \varphi_x^{-1} )
  \end{aligned}
\]

Projection map
\[
\begin{aligned}
  & p : T^* \to M \\
  & M_x^* \mapsto x
  \end{aligned}
\]
Let $C^{r+1}$ map, $1\leq r \leq \omega$, $f:M \to \mathbb{R}$, $\forall \, x \in M$, linear map $T_x f :M_x \to \mathbb{R}$ belongs to $M_x^*$
\[
T_xf = Df_x \in M_x^*
\]
Then
\[
\begin{aligned}
  & Df:M \to T^*M \\
  &  x\mapsto Df_x = Df(x)
  \end{aligned}
\]
is $C^r$ section of $T^*M$.

\begin{definition}
  \textbf{critical point} $x$ of $f$ is zero of $Df$, i.e. \begin{equation}
Df(x) = 0
  \end{equation} of vector space $M_x^*$.
  \end{definition}

Thus, set of critical pts. of $f$ is counter-image of submanifold $Z^* \subset T^*M$ of zeros.  \\
Note $Z^*\approx M$, codim. of $Z^*$ is $n=\text{dim}M$.

\begin{definition}
  \textbf{Morse function} $f$ if $\forall \, $ critical pts. of $f$ are nondegenerate.
  \end{definition}

Note set of critical pts. closed discrete subset of $M$.  \\
Let open $U \subset \mathbb{R}^n$, let $C^2$ map $g:U\to \mathbb{R}$, \\
critical pt. $p\in U$ nondegenerate iff
\begin{itemize}
\item linear $D(Dg)(p):\mathbb{R}^n \to (\mathbb{R}^n)^*$ bijective
\item identify $L(\mathbb{R}^n, (\mathbb{R}^n)^*)$ with space of bilinear maps $\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$, $\Longrightarrow$ equivalent to condition that symmetric bilinear $D^2g(p) : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ non-degenerate
\item $n\times n$ \emph{Hessian matrix}
  \[
\left[ \frac{ \partial^2 g}{ \partial x^i  \partial x^j }(p) \right]
\]
has rank $n$
\end{itemize}

Hessian of $g$ at critical pt. $p$ is quadratic form $H_pf$ associated to bilinear form $D^2g(p)$
\[
\Longrightarrow H_pf(y) =D^2g(p)(y,y) = \sum_{i,j} \frac{ \partial^2g}{ \partial x^i \partial x^j}(p)y^i y^j
\]

Let open $V \subset \mathbb{R}^n$, suppose $C^2$ diffeomorphism $h: V\to U$.

Let $q=h^{-1}(p)$, so $q$ is critical pt. of $gh:V\to \mathbb{R}$.

\begin{tikzpicture}
  \matrix (m) [matrix of math nodes, row sep=3.8em, column sep=4.8em, minimum width=2.2em]
  {
    \mathbb{R}^n & \mathbb{R} \\
     \mathbb{R}^n & \\
};
  \path[->]
  (m-1-1) edge node [above] {$H_q(gh) $} (m-1-2)
          edge node [auto] {$Dh(q) $} (m-2-1)
  (m-2-1) edge node [below] {$H_pg $} (m-1-2)
  ;
\end{tikzpicture}

(quadratic) form $(H_pf)$ invariant under diffeomorphisms.

Let $C^2$ $f:M\to \mathbb{R}$.  \\
$\forall \,$ critical pt. $x$ of $f$, define \\
Hessian quadratic form
\[
\begin{aligned}
  & H_xf : M_x \to \mathbb{R} \\
  & H_xf : M_x \xrightarrow{ D\varphi_x } \mathbb{R}^n \xrightarrow{ H_{\varphi(x)}(f\varphi^{-1} ) } \mathbb{R}
\end{aligned}
\]
where $\varphi$ is any chart at $x$.

Thus, critical pt. of a $C^2$ real-valued function nondegenerate iff associated Hessian quadratic form is nondegenerate.

Let $Q$ nondegenerate quadratic form on vector space $E$.

$Q$ negative definite on subspace $F \subset E$ if $Q(x)<0$ whenever $x\in F$ nonzero.

Index of $Q \equiv \text{Ind}Q$, is largest possible dim. of subspace on which $Q$ is negative definite.

cf. 1.1. Morse's Lemma of  Ch. 6, pp. 145, Morse Theory of Hirsch (1997) \cite{MHirsch1997}

\begin{lemma}[Morse's Lemma]
  Let $p\in M$ be nondegenerate critical pt. of index $k$ of $C^{r+2}$ map $f:M\to \mathbb{R}$, $1\leq r \leq \omega$.

  Then $\exists \, C^r$ chart $(\varphi,U)$ at $p$ s.t.
  \begin{equation}
    f\varphi^{-1}(u_1 \dots u_n) = f(p) - \sum_{i=1}^k u_i^2 + \sum_{i = k+1}^n u_i^2
  \end{equation}
  \end{lemma}

Let ${\,}^TQ \equiv Q^T$ denote tranpose of matrix $Q$.

\begin{lemma}
  Let $A = \text{diag}\lbrace a_1 , \dots , a_n \rbrace$ diagonal $n\times n$ matrix, with diagonal entries $\pm 1$.

  Then $\exists \, $ neighborhood $N$ of $A$ in vector space of symmetric $n\times n$ matrices, $C^{\infty}$ map
  \begin{equation}
P:N \to GL(n,\mathbb{R})
  \end{equation}
  s.t. $P(A)=I$, and if $P(B) = Q$, then $Q^T BQ = A$
\end{lemma}

\begin{proof}
  Let $B = [b_{ij}]$ be symmetri matrix near $A$ s.t. $b{11} \neq 0$ and $b_{11}$ has same sign as $a_1$.

    Consider $x=Ty$ where
    \[
\begin{aligned}
&  x_1 = \left[ y_1 - \frac{b_{12}}{b_{11}} y_2 - \dots - \frac{b_{1n}}{b_{11}} y_n \right] / \sqrt{ |b_n | } \\
& x_k = y_k \text{ for } k = 2, \dots n
  \end{aligned}
    \]
  \end{proof}

\section{Lagrange multipliers}

From \emph{wikipedia:Lagrange multiplier}, \url{https://en.wikipedia.org/wiki/Lagrange_multiplier},
find local minima (maxima), pt. $a\in N$, s.t. $\exists \, $ neighborhood $U$ s.t. $f(x) \geq f(a)$ ($f(x) \leq f(a)$) \, $\forall \, x\in U$.

For $f:U\to \mathbb{R}$, open $U\subset \mathbb{R}^n$, find $x\in U$ s.t. $D_xf \equiv Df(x) =0$, check if Hessian $H_x f<0$.

Maxima may not exit since $U$ open.

References:

\href{http://www.math.uni.wroc.pl/~karch/analiza_nieliniowa/18_mnozniki-lagrangea.pdf}{Relative Extrema and Lagrange Multipliers}

Other interesting links:

\href{http://oai.cwi.nl/oai/asset/2552/2552A.pdf}{The Lagrange Multiplier Rule on Manifolds and Optimal Control of nonlinear systems}

\part{Classical Mechanics applications}

cf. Arnold, Kozlov, Neishtadt (2006) \cite{AKN2006}.

If known forces $\mathbf{F}_1 \dots \mathbf{F}_n$ acts on points, then
\begin{equation}
\sum_{i=1}^n \langle m_i \ddot{ \mathbf{r}}_i - \mathbf{F}_i , \mathbf{\xi}_i \rangle = 0
\end{equation}
cf. Eq. (1.26) of Arnold, Kozlov, Neishtadt (2006) \cite{AKN2006}, where $\xi_1 , \dots \xi_n$ are arbitrary tangent vectors to $M$, $\mathbf{\xi}_i, \dots \mathbf{\xi}_n \in TM$.

$\sum_{i=1}^n \langle m_i \ddot{ \mathbf{r}}_i - \mathbf{F}_i , \mathbf{\xi}_i \rangle$ called "general equation of dynamics" or d'Alembert-Lagrange principle.

\end{multicols*}

\begin{thebibliography}{9}

\bibitem{Fell1968}
William Feller.  \textbf{An Introduction to Probability Theory and Its Applications}, Vol. 1, 3rd Edition.  1968.  ISBN-13: 978-0471257080

\bibitem{JLee2009}
Jeffrey M. Lee. \textbf{Manifolds and Differential Geometry}, \emph{Graduate Studies in Mathematics} Volume: 107, American Mathematical Society, 2009. ISBN-13: 978-0-8218-4815-9

\bibitem{JLee2012}
John Lee, \textbf{Introduction to Smooth Manifolds} (Graduate Texts in Mathematics, Vol. 218), 2nd edition, Springer,  2012, ISBN-13: 978-1441999818

\bibitem{VGuilleminAPollack2010}
Victor Guillemin, Alan Pollack. \textbf{Differential Topology}. American Mathematical Society. 2010. ISBN-13: 978-0821851937
\url{https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwjG96q9z63JAhWMLYgKHempDoMQFggmMAE&url=http\%3A\%2F\%2Fwww.mat.unimi.it\%2Fusers\%2Fdedo\%2Ftop\%2520diff\%2FGuillemin-Pollack_Differential\%2520topology.pdf&usg=AFQjCNF5imOH5xeXRSK60qzM7zT97sdIsw}

\bibitem{AShastri2011}
Anant R. Shastri. \textbf{Elements of Differential Topology}. CRC Press. 2011. ISBN-13: 978-0415339209

\bibitem{MHirsch1997}
Morris W. Hirsch, \textbf{Differential Topology} (Graduate Texts in Mathematics), Graduate Texts in Mathematics (Book 33), Springer (September 16, 1997). ISBN-13: 978-0387901480

\bibitem{AMS2008}
P.-A. Absil, R. Mahony, and R. Sepulchre.  \textbf{Optimization Algorithms on Matrix Manifolds}.  Princeton University Press.  2008.  ISBN 978-0-691-13298-3
\url{https://press.princeton.edu/titles/8586.html}

\bibitem{Pras1996}
Agostino Pr\'{a}staro.  \textbf{Geometry of PDEs and Mechanics}.  World Scientific Publishing Co.  1996.  QC125.2.P73 1996  530.1'55353--dc20.

Agostino Pr\'{a}staro.  \emph{Dipartimento di Metodi e Modelli Matematici per le Scienze Applicate, Universit\`{a} degli Studi di Roma ``La Sapienza''}.

\bibitem{Kopp2010}
  Clemens Koppensteiner.  \emph{Complex Manifolds: Lecture Notes}.  \url{http://www.caramdir.at/uploads/math/piii-cm/complex-manifolds.pdf}

\bibitem{Vand2008}
Stefan Vandoren. \textbf{Lectures on Riemannian Geometry, Part II: Complex Manifolds}.  \url{http://www.staff.science.uu.nl/~vando101/MRIlectures.pdf}

\bibitem{Hori2003}
Kentaro Hori (Author, Editor), Sheldon Katz (Editor), Albrecht Klemm (Editor), Rahul Pandharipande (Editor), Richard Thomas (Editor), Cumrun Vafa (Editor), Ravi Vakil (Editor), Eric Zaslow (Editor).  \textbf{Mirror Symmetry} (Clay Mathematics Monographs, V. 1).  Clay Mathematics Monographs (Book 1).  American Mathematical Society (August 19, 2003).  ISBN-10: 0821829556  ISBN-13: 978-0821829554  \url{https://web.archive.org/web/20060919020706/http://math.stanford.edu/~vakil/files/mirrorfinal.pdf}

\bibitem{Conl2008}
Lawrence Conlon.  \textbf{Differentiable Manifolds} (Modern Birkhäuser Classics).  2nd Edition.  Birkhäuser; 2nd edition (October 10, 2008).  ISBN-13: 978-0817647667

\bibitem{ClSa2012}
Andrew Clarke and Bianca Santoro.  \emph{Holonomy Groups in Riemannian Geometry}.  \href{https://arxiv.org/pdf/1206.3170.pdf}{arXiv:1206.3170 [math.DG]}

\bibitem{ScWa2007}
Urs Schreiber and Konrad Waldorf.  \emph{Parallel Transport and Functors}.  \href{https://arxiv.org/pdf/0705.0452.pdf}{arXiv:0705.0452 [math.DG]}

\bibitem{ElMi2002}
Y. Eliashberg, N. Mishachev, \emph{Introduction to the $h$-principle}, Grad. Studies in Math. Vol 48 (AMS 2002)

\bibitem{AKN2006}
Vladimir I. Arnold, Valery V. Kozlov, Anatoly I. Neishtadt.  \textbf{Mathematical Aspects of Classical and Celestial Mechanics}.  Third Edition.  Springer.

\end{thebibliography}

\end{document}


\documentclass[12pt]{article}
\usepackage{fullpage}
\usepackage{amsthm}
\usepackage{amsfonts,amsmath, amssymb,latexsym,mathrsfs}
\usepackage[margin=1.15in]{geometry}
\usepackage{enumitem}
\setlength{\parindent}{0pt}
\usepackage{tikz-cd}
\usepackage{fancyhdr}

\theoremstyle{definition}
\newtheorem{thm}{Theorem}[section]

\newtheorem{prop}{Proposition}[section]

\theoremstyle{definition}
\newtheorem{definition}{Definition}[section]

\theoremstyle{remark}
\newtheorem*{remark}{Remark}

\theoremstyle{definition}
\newtheorem{example}{Example}[section]

\theoremstyle{definition}
\newtheorem{lem}{Lemma}[section]

\theoremstyle{definition}
\newtheorem{cor}{Corollary}[section]

\date{}
\title{Chapter 7: Integration}

\begin{document}
\maketitle

\section{Integration Techniques}

In this section, we will focus more on how to find the family of antiderivatives that is determined by the function we provided, since if we know such family, then given an initial value of $F(x)$, we can recover the specific antiderivative that we need.

\subsection{Guess and Check method}

Frankly speaking, this is gambling based on experience.\\
Although it is one of the fundamental method we use to solve problems, for example, it appears in also cryptography with the name Bruce force attack, and it also appears in computer science with the name Brute force search. However, as you probably know, this is usually not the most systematic or the most efficient way to the solution, and it usually requires either experience on the work or machine that do not rest.

\subsection{Integration using Substitution}
Exercise as motivation: \textsection 6.4 35,37.

Similarly, the chain rule of derivative is also extremely useful in finding integral.

Theoretically, integration using substitution is using the philosophy that given an integration as $\int f(g(x)) g'(x) dx$, we will have $F(g(x))+C$ as its derivative, which is due to $\frac{d}{dx}(F(g(x))) = f(g(x))g'(x)$, and doing integration on both side will give us the ''substitution rule'' we want.

Technically, you want to recognize the ''$g(x)$'' part in your function, then substitude $g(x)$ as $w$, with the rule $dw=w'(x)dx =\frac{dw}{dx} dx$ (naively, we do have $\frac{dw}{dx} dx =dw$! However, this does not make much sense here which if you are particular interesting, you can ask me in person.)

Now let us consider an example, to see how is it actually applies.
Personally, my favorite almost trivial example is \[\int^{\infty}_0 e^{-x}dx=-\int_{0}^{-\infty}e^w -dw=\int^{0}_{-\infty}e^w dw=1.\text{ (Here I let }w=-x)\]

I will explain why is this one interesting in the next subsection, but here there is a warning you may see, \textbf{Check the up and low bound for your integration!} Let us see another example to see this clearer.

\[\int_{0}^{\pi/4}\frac{\tan^3 \theta}{\cos^2 \theta}d \theta = \int_{0}^{1}w^3dw=\frac{1}{4},\text{ ( Here I take } w = \cos \theta.)\]

Fast question: Can I do the following integrals? Why?$$(1) \int_{0}^{\pi/2}\frac{\tan^3 \theta}{\cos^2 \theta}d \theta  \text{ or } (2)\int_{-\pi/4}^{\pi/4}\frac{\tan^3 \theta}{\cos^2 \theta}d \theta$$

Suggested Problems: $\mathsection$7.1 some ex, 81,89,90-96,97,99,109,111,114,125,141,145,147

\subsection{Integration using partial fractions}

Small break after the introduction to substitution, we have this partial fractions.

Using the method of the substitution, we can already compute many weird integration like $\int \frac{2}{x-10} dx$.

Now, how about $\int \frac{1}{x^2-1} dx$? We are out of knowledge about how to do it right now, but there is an important method we ususally can apply, which is the partial fractions.

The idea here is to break $\frac{1}{x^2-1}$ as a sum: $$\frac{1}{x^2-1}=\frac{1}{2(x-1) }+(-\frac{1}{2(x+1)})$$

But then now we can compute the integration in the question, since we can definitely compute $\int \frac{1}{2(x-1) } dx$ and $\int \frac{1}{2(x+1) }dx$.

If you run into repeated factors? The most general case (example): $$\frac{p(x)}{(x+c_1)^2(x+c_2)(x^2+c_3)}=\frac{A}{x+c_1}+\frac{B}{(x+c_1)^2}+\frac{C}{x+c_2}+ \frac{Dx+E}{x^2+c_3}$$

\subsection{Integration by parts}
If we call the integration using substitution the masterpiece established on the Chain Rule, then I have to say integration by parts is the same thing established on product rule.

Similar to the substitution, we reverse our thoughts here.

What is exactly happen in the integration by parts is that we are just undoing the product rules. Therefore, the key part is recognize your product rule carefully,

Consider the product rule:
\[(uv)'=u'v+uv'=u'v+uv' \]
By some twisted, we will have the following formula for $uv'$ which we have not seen the advantages yet.
\[uv'=(uv)'−u'v\]
But now, integrate both sides will give us:
\[\int udv=\int(uv)′−\int vdu= uv - \int v du\], now if you remember what we talked about in the substitution chapter, you will immediately recognize that this is what we want.

Traditional Example:\[\int \ln(x) dx\] \[\int \cos^2 x\]

Suggested Problems: $\mathsection$ 7.2 49,53,55,65,69,73

\subsection{Trigonometry Substitution}

If the above method does not work and you have terms of $A\pm Bx^{2}$, then we will do trig substitution.

If you have terms of $A-Bx^{2}$, you should try substitute $x=\sqrt{\frac{A}{B}}\sin \theta$.\\
If you have terms of $A+Bx^{2}$, you should try substitute $x=\sqrt{\frac{A}{B}}\tan \theta$.

Since the relation, $\sin^{2}\theta+\cos^{2}\theta=1$ and $\tan^{2}\theta+1=\sec^{2}\theta$ would help you simplify the computation and thus will be helpful.

\section{Integration Approximation}

Estimate the integral is based on the fact that after taking limits of Left/Right/Mid/Trapezoidal Riemann Sum, they are gonna equals and will give the actual integration value, thus if we just compute the Riemann sums in finite many block, we will have a value that is pretty much close to the actual integration, and we can refine our calculation by taking more blocks.

Reminder: we have then four ways of estimating an integral using a Riemann Sum:
\begin{enumerate}
\item LEFT(n)
\item RIGHT(n)
\item MID(n)
\item TRAP(n)
\end{enumerate}
Think about what they are?
\textbf{Think about Pictures!}

Is your estimation over/under?\\

\textbf{Again, Pictures!}(It probably will never make sense without picture.)
\begin{enumerate}
\item If the graph of $f$ is increasing on $[a,b]$, then $$LEFT(n)\leq \int^b_a f(x) dx \leq RIGHT(n)$$
\item If the graph of $f$ is decreasing on $[a,b]$, then $$RIGHT(n)\leq \int^b_a f(x) dx \leq LEFT(n)$$
\item If the graph of $f$ is concave up on $[a,b]$, then $$MID(n)\leq \int^b_a f(x) dx \leq TRAP(n)$$
\item If the graph of $f$ is concave down on $[a,b]$, then $$TRAP(n)\leq \int^b_a f(x) dx \leq MID(n)$$

\end{enumerate}

Suggested Problems: $\mathsection$7.5	1,3,5,16,17,25

\end{document}