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% file: DGDT_dump.tex
% Differential Geometry, Differential Topology, in unconventional ``grande'' format; fitting a widescreen format
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\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}
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\begin{multicols*}{2}
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\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{Cohomology; Stoke's Theorem}
\section{Stoke's Theorem}
\begin{theorem}[Stoke's Theorem]
Let $M$ be oriented, smooth $n$-manifold with boundary, \\
let $\omega$ be a compactly supported smooth $(n-1)$-form on $M$, or if $\omega \in A_c^{n-1}(M)$, \\
Then
\begin{equation}
\int_M d\omega = \int_{\partial M} \omega
\end{equation}
If $\partial M = \emptyset$, then $\int_{\partial M} \omega = 0$
$\int_{\partial M} \omega$ interpreted as $\int_{\partial M} i^*_{\partial M} \omega = \int_{\partial M} i^*\omega$ so
\begin{equation}
\int_M d\omega = \int_{\partial M} i^*(\omega)
\end{equation}
where inclusion $i: \partial M \hookrightarrow M$
\end{theorem}
\begin{proof}
Begin with very special case: \\
Suppose $M = \mathbb{H}^n$ (upper half space), $\partial M = \mathbb{R}^{n-1}$ \\
$\omega$ has compact support, so $\exists \, R >0$ s.t. $\text{supp}\omega \subseteq $ rectangle $A=[-R,R] \times \dots \times [-R, R] \times [0,R]$.
$\forall \, \omega \in A_c^{n-1}(\mathbb{H}^n)$
\begin{equation}
\omega = \sum_{j=1}^n (-1)^{j-1} f_j dx^1 \wedge \dots \wedge \widehat{dx}^j \wedge \dots \wedge dx^n \equiv \sum_{i=1}^n \omega_i dx^1 \wedge \dots \wedge \widehat{dx}^i \wedge \dots \wedge dx^n
\end{equation}
with Conlon (2008) \cite{Conl2008} and John Lee (2012) \cite{JLee2012}'s notation, respectively, and where $f_j$ has compact support.
\[
i^*\omega = (f_1 \circ i) dx^2 \wedge \dots \wedge dx^n \in A_c^{n-1}(\partial \mathbb{H}^n)
\]
\[
\begin{aligned}
d\omega & = \sum_{i=1}^n d\omega_i \wedge dx^1 \wedge \dots \wedge \widehat{dx}^i \wedge \dots \wedge dx^n = \sum_{i,j=1}^n \frac{\partial \omega_i}{ \partial x^j} dx^j \wedge dx^1 \wedge \dots \wedge \widehat{dx}^i \wedge \dots \wedge dx^n = \\
& = \sum_{i=1}^n (-1)^{i-1} \frac{ \partial \omega_i}{ \partial x^i } dx^1 \wedge \dots \wedge dx^n
\end{aligned}
\]
i.e. (for another notation)
\[
d\omega = \left( \sum_{j=1}^n \frac{\partial f_j}{ \partial x^j} \right) dx^1 \wedge \dots \wedge dx^n \in A_c^n(\mathbb{H}^n)
\]
\[
d\omega = \left( \sum_{j=1}^n \frac{\partial f_j}{ \partial x^j} \right) dx^1 \wedge \dots \wedge dx^n \in A_c^n(\mathbb{H}^n)
\]
\[
\int_{\mathbb{H}^n} d\omega = \sum_{i=1}^n (-1)^{i-1} \int_A \frac{ \partial \omega_i}{ \partial x^i } dx^1 \wedge \dots \wedge dx^n = \sum_{i=1}^n (-1)^{i-1} \int_0^R \int_{-R}^R \dots \int_{-R}^R dx^1 \dots dx^n \frac{ \partial \omega_i}{\partial x^i}(x)
\]
We can change order of integration in each term so to do $x^i$ integration first.
By fundamental thm. of calculus, terms for which $i\neq n$ reduce to
\[
\begin{gathered}
\sum_{i=1}^{n-1} (-1)^{i-1} \int_0^R \int_{-R}^R \dots \int_{-R}^R \frac{\partial \omega_i}{\partial x^i}(x) dx^1 \dots dx^n = \sum_{i=1}^{n-1} (-1)^{i-1} \int_0^R \int_{-R}^R \dots \int_{-R}^R \frac{\partial \omega_i}{\parital x^i}(x) dx^i dx^1 \dots \widehat{dx}^i \dots dx^n = \\
= \sum_{i=1}^{n-1} (-1)^{i-1} \int_0^R \int_{-R}^R \dots \int_{-R}^R [\omega_i(x)]_{x^i = -R}^{x^i = R} dx^1 \dots \widehat{dx}^i \dots dx^n = 0
\end{gathered}
\]
because we've chosen $R$ large enough that $\omega =0 $ when $x^i = \pm R$.
\end{proof}
\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}
\begin{definition}[metric]
Let smooth manifold $M$, $\text{dim}M = n$, $\forall \, p \in M$, $\exists \, $ vector space $T_pM$, and so for
\begin{equation}
\begin{aligned}
& g_p (T_pM)^2 \to \mathbb{R} \\
& g_p : (X_p,Y_p) \mapsto g_p(X_p,Y_p) \in \mathbb{R}
\end{aligned}
\end{equation}
with $g_p$ being bilinear, symmetric (in $X_p,Y_p$), nondegenerate (i.e. if $g_p(X_p,Y_p)=0$, then $X_p$ or $Y_p=0$)
Note that
\[
g\in \Gamma((TM \otimes TM)^*)
\]
and that for $\begin{aligned} & \quad \\
& X = X^i \frac{ \partial }{ \partial x^i } \\
& Y = Y^i \frac{ \partial }{ \partial x^i } \end{aligned}$
so
\[
g(X,Y) = g_{ij} X^i Y^j
\]
\end{definition}
Now for
\[
\begin{gathered}
\begin{aligned}
& F: M \to N \\
& F: x \mapsto y = y(x)
\end{aligned} \qquad \, \begin{aligned}
& DF \equiv F_* : T_p M \to T_{F(p)}N \\
& DF : X_p \mapsto (DF) (X^j \frac{ \partial }{ \partial x^j }) = X^j \frac{\partial y^i}{\partial x^j} \frac{ \partial }{ \partial y^i}
\end{aligned}
\end{gathered}
\]
\[
\begin{gathered}
(F^* g')(X,Y) = (F^*g')(X^i \frac{ \partial }{ \partial x^i } , Y^j \frac{ \partial }{ \partial x^j} ) = (F^*g')_{ij} X^i Y^j = g'(F_*X, F_*Y) = \\
=
\end{gathered}
\]
\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)\cite{ScWa2007}, concurrently, for holonomy.
\section{Path Groupoid of a smooth manifold; generalization of paths}
cf. Schreiber and Waldorf (2007)\cite{ScWa2007}.
\begin{definition}[path]
\textbf{path} is a smooth map $\gamma:[0,1] \to M$, between 2 pts. $x,y \in M$, \\
which has a sitting instant; i.e. number $0 < \epsilon < \frac{1}{2}$ s.t.
\begin{equation}
\gamma(t) = \begin{cases} x & \text{ for } 0 \leq t < \epsilon \\
y & \text{ for } 1 - \epsilon < t \leq 1 \end{cases}
\end{equation}
Denote the set of such paths by $PM$,
\begin{equation}
PM \equiv \lbrace \gamma \in \Gamma(M) | \text{ smooth } \gamma : [0,1] \to M \text{ s.t. } \exists \, 0 < \epsilon < \frac{1}{2} \text{ s.t. } \begin{cases} x & \text{ for } 0 \leq t < \epsilon \\
y & \text{ for } 1 - \epsilon < t \leq 1 \end{cases} \rbrace
\end{equation}
\end{definition}
cf. Def. 2.1. of Schreiber and Waldorf (2007)\cite{ScWa2007}
Define \emph{composition}: \\
Given paths $\gamma_1, \gamma_2$; $\begin{aligned} & \qquad \\
& \gamma_1(0) = x \\
& \gamma_1(1) = y \end{aligned}$, \, $\begin{aligned} & \qquad \\
& \gamma_2(0) = y \\
& \gamma_2(1) = z \end{aligned}$, \\
define composition to be path
\begin{equation}
\begin{gathered}
\gamma_2 \circ \gamma_1 \\
(\gamma_2 \circ \gamma_1)(t) := \begin{cases} \gamma_1(2t) & \text{ for } 0 \leq t \leq \frac{1}{2} \\
\gamma_2(2t-1) & \text{ for } \frac{1}{2} \leq t \leq 1 \end{cases}
\end{gathered}
\end{equation}
$\gamma_2 \circ \gamma_1$ smooth since $\gamma_1, \gamma_2$ both constant near gluing pt., due to sitting instants $\epsilon_1, \epsilon_2$, respectively.
Define \emph{inverse}:
\begin{equation}
\begin{gathered}
\gamma^{-1} : [0,1] \to M \\
\gamma^{-1}(t) := \gamma(1-t)
\end{gathered}
\end{equation}
(so that $\gamma^{1}(t) = \begin{cases} y & \text{ for } 1-\epsilon < 1-t \leq 1 \text{ or } 0 \leq t < \epsilon \\
x & \text{ for } 0 \leq 1 - t < \epsilon \text{ or } 1 - \epsilon < t \leq 1 \end{cases}$)
\begin{definition}[thin homotopy equivalent]
2 paths $\gamma_1$, $\gamma_2$ s.t. $\begin{aligned} & \quad \\ & \gamma_1(0) = \gamma_2(0) = x \\
& \gamma_1(1) = \gamma_2(1) = y \end{aligned}$, $\gamma_1, \gamma_2$ are thin homotopy equivalent, \\
if $\exists \, $ smooth $h: [0,1] \times [0,1] \to M$ s.t.
\begin{enumerate}
\item $\exists \, 0 < \epsilon < \frac{1}{2} $ with
\begin{enumerate}
\item $\begin{aligned} & \qquad \\
& h(s,t) = x \text{ for } 0 \leq t < \epsilon \\
& h(s,t) = y \text{ for } 1-\epsilon < t \leq 1 \end{aligned}$
\item \item $\begin{aligned} & \qquad \\
& h(s,t) = \gamma_1(t) \text{ for } 0 \leq s < \epsilon \\
& h(s,t) = \gamma_2(t) \text{ for } 1-\epsilon < s \leq 1 \end{aligned}$
\end{enumerate}
\item differential of $h$ has at most rank 1 everywhere, i.e.
\begin{equation}
\text{rank}(\left. dh\right|_{(s,t)}) \leq 1 \qquad \, \forall \, (s,t) \in [0,1]\times [0,1]
\end{equation}
\end{enumerate}
\end{definition}
cf. Def. 2.2. of Schreiber and Waldorf (2007)\cite{ScWa2007}
$\begin{aligned} & \qquad \\
& h(s,t) = \gamma_1(t) \text{ for } 0 \leq s < \epsilon \\
& h(s,t) = \gamma_2(t) \text{ for } 1-\epsilon < s \leq 1 \end{aligned}$ is the homotopy from $\gamma_1$ to $\gamma_2$, i.e. $\begin{aligned} & \quad \\
& h(0,t) = \gamma_1(t) \\
& h(1,t) = \gamma_2(t) \end{aligned}$ \\
and define an equivalence relation on $PM$.
Note that for $h:[0,1] \times [0,1] \to M$,
\[
\left. (Dh) \right|_{(s,t)} = \left[ \frac{ \partial h^i }{ \partial s} , \frac{\partial h^i }{ \partial t} \right]
\]
\begin{equation}
\begin{gathered}
P^1 M \equiv \text{ set of thin homotopy classes of paths, i.e. } \\
P^1 M = \lbrace [ \gamma] | \gamma_1 \in PM, \text{ if } \exists \, \text{ smooth } h:[0,1]\times [0,1] \to M \text{ s.t. } h \text{ thin homotopy of } \gamma_1 \text{ and } \gamma_2, \gamma_1 \sim \gamma_2 \rbrace
\end{gathered}
\end{equation}
$\text{pr}:PM \to P^1M$ is projection to classes.
Denote thin homotopy class of path $\gamma$, $\begin{aligned} & \qquad \\
& \gamma(0) = x \\
& \gamma(1) = y \end{aligned}$, by $\overline{\gamma}$, or $[\gamma]$.
\subsection{Reparametrization of thin homotopies}
Let $\beta: [0,1] \to [0,1]$, $\begin{aligned} & \quad \\
& \beta(0) = 0 \\
& \beta(1) = 1 \end{aligned}$.
Then $\forall \, $ path $\gamma$, $\begin{aligned} & \quad \\
& \gamma(0) = x \\
& \gamma(1) = y \end{aligned}$, $\gamma \circ \beta$ is also a path $\begin{aligned} & \quad \\
& \gamma\circ \beta(0) = x \\
& \gamma\circ \beta(1) = y \end{aligned}$ and
\begin{equation}
h(s,t) := \gamma(t\beta(1-s) + \beta(t) \beta(s)
\end{equation}
defines a homotopy from $\gamma$ to $\gamma \circ \beta$.
\[
\gamma_1 \circ \gamma_2 \in PM \xmapsto{\text{pr}} [\gamma_1 \circ \gamma_2] = [\gamma_1][ \gamma_2] \in P^1M
\]
Composition of thin homotopy classes of paths obeys following rules:
\begin{lemma}\label{Eq:CompositionOfThinHomotopyClassesOfPaths}
$\forall$ path $\gamma$, $\begin{aligned} & \quad \\
& \gamma(0) = x \\
& \gamma(1) = y \end{aligned}$
\begin{enumerate}
\item $\overline{\gamma} \circ \overline{\text{id}_x} = \overline{\gamma} = \overline{\text{id}_y} \circ \overline{\gamma} \equiv [\gamma]1_x = [\gamma] = 1_y [\gamma] $
\item for paths $\gamma'$; $\begin{aligned} & \quad \\
& \gamma'(0) = y \\
& \gamma'(1) = z \end{aligned}$, \quad $\begin{aligned} & \quad \\
& \gamma''(0) = z \\
& \gamma''(1) = w \end{aligned}$
\begin{equation}
(\overline{\gamma}'' \circ \overline{\gamma}') \circ \overline{\gamma} = \overline{\gamma}'' \circ (\overline{\gamma}' \circ \overline{\gamma}) \equiv ( [\gamma''][\gamma'])[\gamma] = [\gamma'']([\gamma'][\gamma])
\end{equation}
\item $\overline{\gamma} \circ \overline{\gamma}^{-1} = \overline{\text{id}_y}$ and $\overline{\gamma^{-1}} \circ \overline{\gamma} = \overline{\text{id}_x}$ $\equiv [\gamma] [\gamma^{-1}] = 1_y \text{ and } [\gamma^{-1}][\gamma] = 1_x$
\end{enumerate}
\end{lemma}
cf. Lemma 2.3. of Schreiber and Waldorf (2007)\cite{ScWa2007}
\begin{definition}[path groupoid]
$\forall \, $ smooth manifold $M$, \\
consider category whose set of objects is $M$, \\
whose set of morphisms is $P^1M$, where class $[\gamma]$, $\begin{aligned} & \qquad \\
& [\gamma](0) = x \\
& [\gamma](1) = y \end{aligned}$ is a morphism from $x$ to $y$ and \\
composition $[\gamma_1 ][\gamma_2] = [\gamma_1 \circ \gamma_2] \in P^1M$
Lemma \ref{Eq:CompositionOfThinHomotopyClassesOfPaths} are axioms of a category, 3rd. property says $\forall \,$ morphism is invertible.
Hence, we've defined a groupoid, called \textbf{path groupoid} of $M$, $\mathcal{P}_1(M)$.
\end{definition}
So
\[
\begin{gathered}
\text{Obj}(\mathcal{P}_1(M)) = M \\
\text{Mor}(\mathcal{P}_1(M)) = P^1M
\end{gathered}
\]
$\forall \, $ smooth $f:M \to N$, denote functor $f_*$
\begin{equation}
f_* : \mathcal{P}_1(M) \to \mathcal{P}_1(N)
\end{equation}
with
\[
\begin{gathered}
f_*(x) = f(x) \\
(f_*)([\gamma]) := [f\circ \gamma]
\end{gathered}
\]
If $\gamma \sim \gamma'$, for $f\circ \gamma$, $f\circ \gamma'$, \\
$f\circ h(s,t) $ with $\begin{aligned} & \quad \\
& f\circ h(0,t) = f\circ \gamma(t) \\
& f\circ h(1,t) = f\circ \gamma'(t) \end{aligned}$, \\
so $f\circ h$ is a thin homotopy between $f\circ \gamma$, $f\circ \gamma'$ and so $[f\circ \gamma]$ \emph{well-defined}.
\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 (