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%Nakahara_GTP-solutions.tex
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\title{
Solutions to \emph{Geometry,Topology,Physics} by Mikio Nakahara, 2003.
}
\author{
Ernest Yeung
}
\date{zima 2012}
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}
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\maketitle
\href{https://github.com/ernestyalumni/mathphysics/blob/master/LaTeX_and_pdfs/Nakahara_GTP-solutions.tex}{Permanent home} of this \verb|Nakahara GTP-solutions.tex| and \verb|.pdf| files on \verb|github|, \url{https://github.com/ernestyalumni/mathphysics/blob/master/LaTeX_and_pdfs/Nakahara_GTP-solutions.tex}. \textbf{Go here} for the latest version; if you came here from anywhere else (such as Google Drive), then the version of solutions most likely will be outdated.
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In the beginning of 2017, I received a very generous donation from a reader from Norway who found these notes useful, through \emph{PayPal}. If you find these notes useful, feel free to donate directly and easily through \href{https://www.paypal.com/cgi-bin/webscr?cmd=_donations&business=ernestsaveschristmas%2bpaypal%40gmail%2ecom&lc=US&item_name=ernestyalumni&currency_code=USD&bn=PP%2dDonationsBF%3abtn_donateCC_LG%2egif%3aNonHosted}{PayPal}, which won't go through a 3rd. party such as indiegogo, kickstarter, patreon. Otherwise, under the \emph{open-source MIT license}, feel free to copy, edit, paste, make your own versions, share, use as you wish.
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%-----------------------------------%-----------------------------------%-----------------------------------%-----------------------------------%-----------------------------------
Solutions for \emph{Geometry, Topology, and Physics}. Mikio Nakahara. Institute of Physics Publishing. 2003. ISBN 0 7503 0606 8
%-----------------------------------%-----------------------------------%-----------------------------------%-----------------------------------%-----------------------------------
\section{Quantum Physics}
\subsection{Analytical mechanics}
\subsubsection{Newtonian mechanics}
\subsubsection{Lagrangian formalism}
$\mathcal{L}$ independent of coordinate $q_k$; $q_k$ cyclic. \\
$q_k(t) \to q_k(t) + \delta q_k(t)$
\begin{equation}
S[q(t), \dot{q}(t) ] = \int_{t_i}^{t_f} L(q,\dot{q}) dt \quad \quad \quad \, (1.3)
\end{equation}
\[
\delta S = \int_{t_i}^{t_f} \sum_k \delta q_k \left( \frac{ \partial L}{ \partial q_k} - \frac{d}{dt} \frac{ \partial L}{ \partial \dot{q}_k } \right) + \sum_k \left[ \delta_k \frac{ \partial L}{ \partial \dot{q}_k} \right]_{t_i}^{t_f} = 0
\]
Note that
\[
- \int \delta q_k \frac{d}{dt} \frac{ \partial L}{ \partial \dot{q}_k } + \delta q_k \left. \frac{ \partial L}{ \partial \dot{q}_k} \right|_{t_i}^{t_f} = \int \delta \dot{q}_k \frac{ \partial L}{ \partial \dot{q}_k }
\]
with $p_k = \frac{ \partial L}{ \partial \dot{q}_k}$.
\begin{equation}
\Longrightarrow \delta q_k(t_i) p^k(t_i) = \delta q_k(t_f) p^k(t_f)
\end{equation}
since $t_i,t_f$ arbitrary, $\delta q_k(t) p^k(t)$ independent of $t$ and hence conserved.
\subsubsection{Hamiltonian formalism}
\exercisehead{1.1} $A = A(q,p), B(q,p)$ defined on phase space of $H = H(q,p)$
\[
\begin{aligned}
[A, c_1 B_1 + c_2 B_2 ] & = \partial_{q_k} A \partial_{p_k} (c_1 B_1 + c_2 B_2) - \partial_{p_k} A \partial_{q_k} ( c_1 B_1 + c_2 B_2) = \\
& = c_1 \partial_{q_k} A \partial_{p_k} B_1 - c_1 \partial_{p_k} A \partial_{q_k} B_1 + c_2 \partial_{q_k} A \partial_{p_k} B_2 - c_2 \partial_{p_k} A \partial_{q_k} B_2 = \\
& = c_1 [ A,B_1] + c_2 [A,B_2]
\end{aligned}
\]
\[
[A,B] = \partial_{q_k} A \partial_{p_k} B - \partial_{p_k} A \partial_{q_k} B = - (\partial_{q_k} B \partial_{p_k} A - \partial_{p_k} B \partial_{q_k} A ) = - [B,A]
\]
\[
\begin{aligned}
[[A,B],C] & = \partial_{q_k} ( \partial_{q_l} A \partial_{p_l} B - \partial_{p_l} A \partial_{q_l} B ) \partial_{p_k} C - \partial_{p_k} ( \partial_{q_l} A \partial_{p_l} B - \partial_{p_l} A \partial_{q_l} B ) \partial_{q_k} C = \\
& = \partial^2_{q_k q_l} A \partial_{p_l} B \partial_{p_k} C - \partial^2_{q_k q_l} B \partial_{p_k} C \partial_{p_l} A + \\
& \phantom{ = } + \partial^2_{q_k p_l} B \partial_{q_l} A \partial_{p_k} C - \partial^2_{q_k p_l} A \partial_{q_l} B \partial_{p_k} C + \\
& \phantom{ = } + \partial^2_{p_k p_l} A \partial_{q_k} C \partial_{q_l} B - \partial^2_{p_k p_l } B \partial_{q_k} C \partial_{q_l} A + \\
& \phantom{ = } + \partial^2_{q_l p_k} B \partial_{q_k} C \partial_{p_l}A - \partial^2_{q_l p_k} A \partial_{q_k} C \partial_{p_l} B
\end{aligned}
\]
\[
\Longrightarrow [[A,B], C ] + [[C,A], B] + [[B,C], A] = 0
\]
\begin{equation}
\frac{dA}{dt} = \sum_k \left( \frac{ dA}{ dq_k } \frac{ dq}{ dt} + \frac{dA}{ dp_k } \frac{dp_k}{dt} \right) = \sum_k \left( \frac{dA}{ dq_k} \frac{ \partial H}{ \partial p_k} - \frac{ dA}{ d p_k} \frac{ \partial H}{ \partial q_k} \right) = [A,H] \quad \quad \quad \, (1.23)
\end{equation}
\subsection{ Canonical quantization }
\subsubsection{Hilbert space, bras, and kets}
\subsubsection{Axioms of canonical quantization}
A3. Poisson bracket in classical mechanics is replaced by commutator.
\begin{equation}
[ \widehat{A}, \widehat{B} ] \equiv \widehat{A} \widehat{B} - \widehat{B} \widehat{A} \quad \quad \quad \, (1.31)
\end{equation}
\subsection*{Problems}
\problemhead{1.1} $H = \int d^n x \left[ \frac{1}{2} \left( \frac{ \partial \phi }{ \partial t} \right)^2 + \frac{1}{2} ( \nabla \phi )^2 + V(\phi) \right]$ \\
If $\phi$ time independent, $H[\phi] = H_1[\phi] + H_2[\phi]$
\[
\begin{aligned}
& H_1[\phi] \equiv \frac{1}{2} \int d^n x ( \nabla x )^2 \\
& H_2[\phi] \equiv \int d^n x V(\phi)
\end{aligned}
\]
\begin{enumerate}
\item[(1)] $\phi(x) \to \phi(\lambda x)$
\[
\begin{gathered}
(\nabla \phi)^2 = \partial_j \phi \partial^j \phi = \lambda^{-2} \partial_i \phi \partial^i \phi = \lambda^{-2} (\nabla \phi)^2 \\
\frac{ \partial }{ \partial x^i} = \frac{ \partial y^i}{ \partial x^i } \frac{ \partial}{ \partial y^i } = \lambda \frac{ \partial }{ \partial y^i } \\
d^n y = \lambda^n d^n x
\end{gathered}
\]
\item[(2)] $
\begin{aligned}
& H_1[\phi] \to \lambda^{n-2} H_1[\phi] \\
& H_2[\phi] \to \lambda^n H_2[\phi]
\end{aligned}$
\[
\Longrightarrow \partial_{\lambda} H = (n-2) H_1 + n H_2 = 0 \text{ or } (2-n) H_1 - nH_2 = 0
\]
\item[(3)]
\end{enumerate}
\subsubsection{ Heisenberg equation, Heisenberg picture and Schr\"{o}dinger picture}
\subsubsection{Wavefunction}
\subsubsection{Harmonic oscillator}
\subsection{ Path integral quantization of a Bose particle}
\subsubsection{Path integral quantization}
\subsubsection{Imaginary time and partition function}
\subsubsection{Time-ordered product and generating functional }
\subsection{ Harmonic oscillator}
\subsubsection{Transition amplitude}
\subsubsection{Partition function}
\subsection{ Path integral quantization of a Fermi particle }
\subsection{ Quantization of a scalar field}
\subsection{ Quantization of a Dirac field}
\subsection{ Gauge theories }
\subsubsection{ Abelian gauge theories}
\begin{align}
& \nabla \cdot B = 0 & (1.241a) \\
& \frac{ \partial B}{ \partial t } + \nabla \times E = 0 & (1.241b) \\
& \nabla \cdot E = \rho & (1.241c) \\
& \frac{ \partial E}{ \partial t} - \nabla \times B = -j & (1.241d)
\end{align}
\[
A_{\mu} = ( - \phi, A )
\]
\[
\begin{aligned}
& B = \nabla \times A \\
& E = \frac{ \partial A}{ \partial t} - \nabla \phi \quad \quad \quad \, (1.242)
\end{aligned}
\]
\exercisehead{1.11}
\[
\begin{gathered}
\nabla \cdot B = 0 = \partial_i B_i = \partial_i (\nabla \times A)_i = \partial_i \epsilon_{ijk} \partial_j A_k = \epsilon_{ijk} \partial_i \partial_j A_k = \\
= \partial_1 \partial_2 A_3 - \partial_1 \partial_3 A_2 + \partial_2 \partial_3 A_1 - \partial_2 \partial_1 A_3 + \partial_3 \partial_1 A_2 - \partial_3 \partial_2 A_1 = \partial_1 F_{23} + \partial_2 F_{31} + \partial_3 F_{12} = 0
\end{gathered}
\]
Watch out for convention.
\[
\partial_{\mu} = \left( \frac{ \partial }{ \partial t }, \nabla \right)
\]
\[
\begin{gathered}
\frac{ \partial B}{ \partial t} + \nabla \times E = \partial_0 \epsilon_{ijk} \partial_j A_k + \epsilon_{ijk} \partial_j E_k = \epsilon_{ijk} \partial_0 \partial_j A_k + \epsilon_{ijk} \partial_j \left( \partial_0 A_k - \partial_k \phi \right) = \epsilon_{ikj} \partial_0 \partial_k A_j + \epsilon_{ijk} \partial_j \partial_0 A_k + \epsilon_{ijk} \partial_j \partial_k A_0 = \\
= \epsilon_{ikj} \partial_0 \partial_k A_j + \epsilon_{ijk} \partial_j \partial_0 A_k + \epsilon_{ikj} \partial_j \partial_k A_0 =
\end{gathered}
\]
Fix $i$.
\[
\begin{gathered}
\Longrightarrow = \partial_0 \partial_k A_j - \partial_0 \partial_j A_k + \partial_j \partial_0 A_k - \partial_k \partial_0 A_j + \partial_j \partial_k A_0 - \partial_k \partial_j A_0
\end{gathered}
\]
\exercisehead{1.12}
\[
\begin{gathered}
\mathcal{L} = (D^{\mu} \phi)^{\dag} ( D_{\mu} \phi ) + m^2 \phi^{\dag} \phi \\
\begin{aligned}
& \phi' = e^{- i e \alpha(x)} \phi \\
& (\phi')^{\dag} = \phi^{\dag} e^{ i e \alpha(x) }
\end{aligned} \\
A_{\mu}' = A_{\mu} - \partial_{\mu} \alpha(x) \quad \quad \, (1.257)
\end{gathered}
\]
Now easily
\[
\phi^{\dag} e^{i e \alpha(x) } e^{-i e \alpha(x) } \phi = \phi^{\dag} \phi
\]
\[
\begin{gathered}
\begin{aligned}
& D_{\mu} = \partial_{\mu} - ie A_{\mu} \\
& D_{\mu} \phi = \partial_{\mu} \phi - ie A_{\mu} \phi
\end{aligned} \\
\partial_{\mu} ( e^{-i e \alpha} \phi ) - i e( A_{\mu} - \partial_{\mu} \alpha ) e^{-i e \alpha } \phi = - e \partial_{\mu} \alpha \phi e^{-i e \alpha } + e^{-i e \alpha} \partial_{\mu} \phi - ie A_{\mu} e^{-i e \alpha} \phi + i e \partial_{\mu} \alpha e^{-ie \alpha } \phi = e^{-i e \alpha (x) } ( D_{\mu} \phi )
\end{gathered}
\]
\[
\begin{gathered}
\begin{aligned}
& D^{\mu} \phi = \partial^{\mu} \phi - i e A^{\mu} \phi \\
& (D^{\mu} \phi)^{\dag} = \partial^{\mu} \phi^{\dag} + i e A^{\mu} \phi^{\dag}
\end{aligned} \\
(D^{\mu} \phi)^{\dag} \to \partial^{\mu} ( \phi^{\dag} e^{ie \alpha} ) + ie(A^{\mu} - \partial^{\mu} \alpha )\phi^{\dag} e^{ie \alpha} = \partial^{\mu} \phi^{\dag} e^{ie \alpha } + \phi^{\dag} ( ie \partial^{\mu} \alpha ) e^{ie \alpha} + ie A^{\mu} \phi^{\dag} e^{ie \alpha} - ie \partial^{\mu} \alpha \phi^{\dag} e^{ie \alpha} = (D^{\mu} \phi)^{\dag} e^{ie \alpha }
\end{gathered}
\]
Clearly $\mathcal{L}$ is invariant.
\subsubsection{ Non-Abelian gauge theories }
\subsubsection{ Higgs fields }
\subsection{ Magnetic monopoles}
\subsection{ Instantons }
\subsubsection{ Introduction}
\subsubsection{ The (anti-)self-dual solution }
\section{Mathematical Preliminaries}
\subsection{Maps}
\subsubsection{Definitions}
\exercisehead{2.1}
$D = [-\pi/2, \pi/2]$, \, $R = [-1.1]$ \, $f(x) = \sin{(x)}$ is bijective on $D$ to $R$
\exercisehead{2.2}
$f:x \to x^2$, $g: \to \exp{x}$
\[
f \circ g(x) = exp{(2x)} \, \quad \, g \circ f(x) = \exp{x^2}
\]
\exercisehead{2.3} Consider $f(x) = f(y)$
\[
gf(x) = x = gf(y) = y \Longrightarrow x =y
\]
$f$ injective.
\[
\forall \, x \in X, \, x = g\circ f(x) = g(f(x)) = g(y) \text{ since } f:X\to Y. \text{ so } \, \exists \, y \in Y, \, \text{ s.t. } g(y) = x
\]
$g$ surjective.
\exercisehead{2.4}
\[
\begin{aligned}
& a^{-1}: E^n \to E^n \\
& a^{-1}(x) = x - a \\
& a^{-1}a(x) = (x+a) - a = x \\
& aa^{-1}(x) = (x-a) + a = x
\end{aligned} \quad \quad \quad \, \begin{aligned}
R^{-1}(x) = R^T x
& R^{-1}R(x) = R^T R x = 1 x = x \\
& RR^{-1}(x) = RR^Tx = 1x = x \\
& (R^T R)_{ij} = R^T_{ik} R_{kj} = R_{ki} R_{kj} = \delta_{ij} \\
& (RR^T)_{ij} = R_{ik} R^T_{kj} = R_{ik} R_{jk} = R^T_{ki} R^T_{kj} = \delta_{ij}
\end{aligned}
\]
20120306 check above.
\[
\begin{aligned}
& (R,a)(x) = Rx + a \\
& (R,a)^{-1}(x) = R^T(x-a) \\
& (Ra)^{-1} Ra(x) = R^T(Rx + a- a) = x \\
& (Ra)(Ra)^{-1}(x) = R(R^T(x-a)) + a = x
\end{aligned}
\]
\subsubsection{Equivalence relation and equivalence class}
\exercisehead{2.5} $m \sim m$ since $\frac{m}{2} = \frac{m}{2}$ \\
$m \sim n$. Then $n\sim m$ since remainder of $n$ by 2 is the same as the remainder of $m$ divided by 2. \\
$m \sim n$, $n\sim p$. $m\sim p$ since remainder of $m$ by 2 is the same as the remainder of $m$ by 2 which is the same as the remainder of $p$ by 2.
\exercisehead{2.6} $H = \lbrace \tau \in \mathcal{C} | Im{\tau} \geq 0 \rbrace$ \\
$SL(2,\mathbb{Z}) \equiv \lbrace \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) | a,b, c, d \in \mathbb{Z}, \, ad - bc = 1 \rbrace$ \\
$\tau \sim \tau'$ if $\exists \, A = \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \in SL(2,\mathbb{Z})$ s.t. $\tau' = \frac{ a\tau + b}{ c\tau + d }$ \\
Note that
\[
\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} \tau \\ 1 \end{matrix} \right) = \left( \begin{matrix} a\tau + b \\ c \tau + d \end{matrix} \right)
\]
\[
\tau \sim \tau \text{ since } \left( \begin{matrix} 1 & \\ & 1 \end{matrix} \right) \left( \begin{matrix} \tau \\ 1 \end{matrix} \right) = \left( \begin{matrix} \tau \\ 1 \end{matrix} \right)
\]
If $\tau \sim \tau'$ s.t. $\tau' = \frac{ a\tau + b}{ c\tau +d }$. Consider $\left( \begin{matrix} d & - b \\ -c & a \end{matrix} \right)$ \\
$\tau' \sim \tau$ since
\[
\left( \begin{matrix} d & -b \\ -c & a \end{matrix} \right) \left( \begin{matrix} \tau' \\ 1 \end{matrix} \right) = \left( \begin{matrix} d\tau' - b \\ -c\tau' + a \end{matrix} \right) \quad \quad \quad \, \frac{ d \tau' - b }{ -c\tau' + a } = \frac{ d \left( \frac{ a \tau + b }{ c\tau + d } \right) - b }{ -c \left( \frac{ a\tau + b }{ c\tau + d } \right) + a } = \frac{ ad \tau - bc \tau }{ ad-bc } = \tau
\]
Given $\tau \sim \tau'$, $\tau' \sim \tau''$
\[
\begin{aligned}
& \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} \tau \\ 1 \end{matrix} \right) = \left( \begin{matrix} \tau' \\ 1 \end{matrix} \right) \\
& \left( \begin{matrix} e & f \\ g & h \end{matrix} \right) \left( \begin{matrix} \tau' \\ 1 \end{matrix} \right) = \left( \begin{matrix} \tau'' \\ 1 \end{matrix} \right)
\end{aligned} \quad \quad \Longrightarrow \left( \begin{matrix} e & f \\ g & h \end{matrix} \right) \left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} \tau \\ 1 \end{matrix} \right) = \left( \begin{matrix} \tau'' \\ 1 \end{matrix} \right) = \left( \begin{matrix} a e + cf & be + df \\ ag + ch & bg + d h \end{matrix} \right) \left( \begin{matrix} \tau \\ 1 \end{matrix} \right)
\]
\[
(ae + cf) (bg - dh) - (ag+ch) (be +df) = abeg + adeh + bcfg + cdfh - abeg - adfg - bceh - cdfh = 1
\]
Example 2.6. \\
$g \sim g'$ if $\exists \, h \in H$ s.t. $g' = gh$ \\
$[g] = \lbrace gh | h \in H \rbrace \equiv gH$ (left) coset. \\
quotient space $\equiv G/H$ \\
if $H$ normal subgroup of $G$, $ghg^{-1} \in H$, \, $\forall \, g \in G$, $h \in H$, then $G/H$ quotient group. \\
Since
\[
(g')^{-1} h g' = h'' \in H \quad \, \Longrightarrow hg' = g'h''
\]
\[
\Longrightarrow ghg'h' = gg' h'' h' = gg'h'''
\]
$[g][g'] = [gg']$ well-defined. \\
Note $[e]$, $[g]^{-1} = [g^{-1}]$
\exercisehead{2.7} $a,b \in G$ conjugate to each other, $a\simeq b$, if $\exists \, g \in G$ s.t. $b = gag^{-1}$
$a\simeq a$ since $a = eae^{-1} = a$ \\
If $a\simeq b$, $g^{-1}bg = a$ and $g^{-1} \in G$, so $b\simeq a$. \\
If $a \simeq b$ and $b\simeq c$, then $c = hbh^{-1}$, so $c = hg a g^{-1} h^{-1}$ and $(hg)^{-1} = g^{-1}h^{-1}$.
\subsection{Vector spaces}
\subsubsection{Vectors and vector spaces}
\subsubsection{Linear maps, images and kernels}
\begin{theorem}[2.1] If linear $f: V \to W$,
\[
\text{dim}{V} = \text{dim}{ (\text{ker}{f}) } + \text{dim}{ (\text{im}{f} ) }
\]
\end{theorem}
\begin{proof}
$\text{ker}{f}$, $\text{im}{f}$ vector spaces. \\
Let basis of $\text{ker}{f}$ be $\lbrace g_1 \dots g_r \rbrace$ \\
\phantom{Let } basis of $\text{im}{f}$ be $\lbrace h_1' \dots h_s' \rbrace$ \\
$\forall \, i \, (1 \leq i \leq s)$, $h_i \in V$ s.t. $f(h_i) = h_i'$ \\
Consider $\lbrace g_1 \dots g_r, h_1 \dots h_s \rbrace$ \\
Let arbitrary $v\in V$ \\
$f(v) \in \text{im}{f}$, so $f(v) = c^i h_i' = c^i f(h_i)$ \\
by linearity $f(v- c^i h_i) = 0$, \\
so $v - c^i h_i \in \text{ker}{f}$ \\
so $\forall \, $ arbitrary $v\in V$, $v$ linear combination of $\lbrace g_1 \dots g_r, h_1 \dots h_s \rbrace$
\end{proof}
\exercisehead{2.8}
\begin{enumerate}
\item[(1)] If $x_1, x_2 \in ker{f}$, then for $x_1 + x_2$
\[
\text{ by linearity of $f$, } f(x_1 + x_2) = f(x_1) + f(x_2) = 0 + 0 = 0 \quad \, \Longrightarrow x_1 + x_2 \in ker{f}
\]
\[
f(cx_1) = cf(x_1) = 0 \quad \, \Longrightarrow cx_1 \in ker{f}
\]
Under closure of addition and multiplication, $ker{f}$ is a vector space. \\
If $w_1, \, w_2 \in im{f}$,
\[
\begin{aligned} w_1 & = f(x_1) \\ w_2 & = f(x_2) \end{aligned} \text{ for some } x_1, x_2 \in V
\]
\[
w_1 + w_2 = f(x_1) + f(x_2) = f(x_1 + x_2) \in im{f} \text{ since $f$ linear and since } x_1 + x_2 \in V
\]
\[
cw_1 = cf(x_1) = f(cx_1) \in im{f} \text{ since } cx_1 \in V
\]
\item[(2)] If $f:V \to V$ isomorphism, $f$ bijective. $f(0)=0$ always for a linear map. Consider $x \in ker{f}$. So $f(x) =0$. $f(x) = f(0)$, so $x=0$. \\
Since $f$ linear, if $f(x) = f(y)$, $f(x) - f(y) = f(x-y) =0$. Since $ker{f} =0$, $x-y =0$ so $x=y$. $f$ bijective so $f$ an isomorphism.
\end{enumerate}
\subsubsection{Dual vector space}
\[
f(\mathbf{v}) = f_i \alpha^i(v^j \mathbf{e}_j ) = f_i v^j \alpha^i(\mathbf{e}_j) = f_i v^i \quad \quad \, (2.12)
\]
Use notation $\langle, \rangle : V^* \times V \to K$ \\
$f: V \to W$ \\
$g: W \to K$ \\
$g\in W^*$ \\
$g\circ f : V \to K$ (Note $g\circ f$ on $V$, key observation) \\
$g\circ f \equiv h \in V^*$
\[
h(\mathbf{v}) \equiv g(f(\mathbf{v})) \quad \, \mathbf{v} \in V \quad \, (2.13)
\]
Given $g \in W^*$, $f:V \to W$ induces map $h\in V^*$ \\
$f^*: W^* \to V^*$ \\
$f^*:g \mapsto h = f^*(g) = g\circ f$ \quad \, $h$ is the pullback of $g$ by $f^*$
\exercisehead{2.9}
Given $f_i = A_i^{\, k} e_k$,
\[
\alpha^j f_i = A_i^{\, k} \alpha^j e_k = A_i^{\, j}
\]
\[
\beta^j A_j^{ \, i} = \beta^j \alpha^i f_j = \alpha^i \Longrightarrow \alpha^i = \beta^j A_j^{\, i }
\]
\subsubsection{Inner product and adjoint}
isomorphism $g:V \to V^*$, $g\in GL(m,K)$
\[
g:v^j \to g_{ij} v^j \quad \quad \, (2.14)
\]
$g(v_1, v_2) \equiv \langle gv_1, v_2 \rangle$
\[
g(v_1, v_2) = v_1^{\, i} g_{ji} v_2^{\, j} \quad \quad \, (2.16)
\]
$W = W(n,\mathbb{R})$, $\lbrace f_{\alpha} \rbrace$ basis $G:W \to W^*$ \\
Given $f:V \to W$ \\
adjoint of $f$, $\widetilde{f}$
\[
G(w,fv) = g(v,\widetilde{f}w) \quad \quad \, (2.17)
\]
where $v\in V$, $w\in W$
\[
w^{\alpha} G_{\alpha \beta} f^{\beta}_{\, i} v^i = v^i g_{ij} \widetilde{f}^j_{\, \alpha} w^{\alpha} \quad \quad \, (2.18)
\]
$G_{\alpha \beta} f^{\beta}_{\, i} = g_{ij} \widetilde{f}^j_{\, \alpha} $
\[
\widetilde{f} = g^{-1} f^t G^t \quad \quad \, (2.19)
\]
\exercisehead{2.10} (cf. wikipedia, ``Rank'') Consider a $m\times n$ matrix $A$ with column rank $A$ (maximum number of linearly independent column vectors of $A$). \\
$dim$. of column space of $A = r$. Then let $c_1 \dots c_r$ basis. \\
Place $c_i$'s as column vectors to form $m \times r$ matrix $C = [ c_1 \dots c_r ]$ \\
$\exists \, r \times m$ matrix $R$ s.t. $A = CR$. \quad \, $r_{ij}$ \, $\begin{aligned} & \quad \\ i & = 1 \dots r \\ j & = 1 \dots m \end{aligned}$ \\
$A=CR$, so $\forall \, $ row vector of $A$ is a linear combination of row vectors of $R$, so row space of $A$ contained in row space of $R$. \\
row rank $A \leq $ row rank $R$. \\
$R$ has $r$ rows, $a_{ij} = c_{ik} r_{kj}$ \, $j= 1 \dots n$. row rank $R \leq r = $ column rank $A$ \\
row rank $A \leq $ column rank $A$ \\
row rank $A^T \leq $ column rank $A^T$
$\Longrightarrow $ row rank $A = $ column rank $A$ or $\operatorname{rank}{(A)} = \operatorname{rank}{(A^T)}$ \\
Likewise for $NfM$ by following the same arguments.
\exercisehead{2.11} \begin{enumerate}
\item[(a)] $g(v_1,v_2) = \overline{v}_1^{ \, i } g_{ij} v_2^{\, j }$ \\
\[
\overline{g(v_2,v_1)} = \overline{ \overline{v}_2^{ \, i} g_{ij} v_1^{ \, j} } = v_2^{ \, i } \overline{g}_{ij} \overline{v}_1^{\, j} = \overline{v}_1^{\, j} g_{ji} v_2^{\, i} = g(v_1,v_2)
\]
\[
g(v,\widetilde{f}w) = \overline{v}^i g_{ij} \widetilde{f}_{jk} w^k = \overline{G(w,fv)} = \overline{ \overline{w}^{\alpha} G_{\alpha \beta} f_{\beta \gamma} v^{\gamma} } = w^{\alpha} \overline{G}_{\alpha \beta} \overline{f}_{\beta \gamma} \overline{v}^{\gamma} = \overline{G}_{k\beta} \overline{f}_{\beta i} \overline{v}^i w^k
\]
\[
\Longrightarrow g_{ij} \widetilde{f}_{jk} = \overline{G}_{k \beta} \overline{f}_{\beta i } = f^{\dag}_{i \beta} G_{\beta k}^{\dag}
\]
\[
\Longrightarrow \widetilde{f} = g^{-1} f^{\dag} G^{\dag}
\]
\item[(b)]
\end{enumerate}
\subsubsection{Tensors}
tensor $T$ of type $(p,q)$ maps $p$ dual vectors and $q$ vectors to $\mathbb{R}$
\begin{equation}
T: \overset{p}{\bigotimes} V^* \overset{q}{\bigotimes} V \to \mathbb{R}
\end{equation}
\exercisehead{2.12} $\begin{aligned} & \quad \\ & f:V \to W \\ & f(v) = w \end{aligned}$, so $f(v) \in W$. \\
Then tensor identified with dual vector of $W^*$. Since $f:V$, Then $(1,1)$.
\subsection{Topological spaces}
\subsubsection{Definitions}
\exercisehead{2.13}
$\tau_{\mathbb{R}} = \lbrace (a,b) | a,b \in \lbrace \mathbb{R}, \pm \infty \rbrace \rbrace$ \\
$\bigcap_{n=1}^{\infty} (a,b + \frac{1}{n} ) = (a,b]$. Then $\lbrace b \rbrace \in \tau_{\mathbb{R}}$, $\forall \, b \in \mathbb{R}$. \\
So then $\forall \, $ subset $Y \subset X$ is open. Discrete topology.
\subsubsection{Continuous maps}
\exercisehead{2.14} If $\begin{aligned} & \quad \\
& f: \mathbb{R} \to \mathbb{R} \\
& f(x) = x^2 \end{aligned}$ then $(-\epsilon, + \epsilon) \mapsto [0, \epsilon^2]$
\subsubsection{Neighborhoods and Hausdorff spaces}
\exercisehead{2.15} Let $X = \lbrace \text{John}, \text{Paul}, \text{Ringo}, \text{George}\rbrace$. \\
\[
\begin{aligned} & \quad \\
& U_0 = \emptyset \\
& U_1 = \lbrace \text{John} \rbrace \\
& U_2 = \lbrace \text{John}, \text{Paul} \rbrace \\
& U_3 = X \end{aligned} \quad \quad \quad \, \begin{aligned} & U_1 \bigcup U_2 = U_2 \\ & U_1 U_2 = U_1 \end{aligned}
\]
Consider John and Ringo. Only neighborhood with open set is $X$ for Ringo. Then $XU_{\text{John}} \neq \emptyset$
\exercisehead{2.16} $\forall \, a, b$, consider $\left( \frac{3a - b}{2}, \frac{a+b}{2} \right)$, $\left( \frac{b+a}{2}, \frac{3b - a}{2} \right)$
\subsubsection{Closed set}
\subsubsection{Compactness}
\subsubsection{Connectedness}
\subsection{Homeomorphisms and topological invariants}
\subsubsection{Homeomorphisms}
\subsubsection{Topological invariants}
\exercisehead{2.18} $f:S^1 \to E$ \\
$\begin{aligned} & \quad \\
& f(x,y) = (ax,by) \\
& f^{-1}(x,y) = \left( \frac{x}{a}, \frac{y}{b} \right) \end{aligned}$ \quad \, $ff^{-1} = f^{-1}f = 1$ bijective and cont.
\subsubsection{Homotopy type}
\subsubsection{Euler characteristic: an example}
\section{Homology Groups}
\subsection{Abelian groups}
\subsubsection{Elementary group theory}
e.g. $f: \mathbb{Z} \to \mathbb{Z}_2 = \lbrace 0 , 1 \rbrace$ \\
$\begin{aligned}
& f(2n) = 0 \\
& f(2n+1) = 1 \end{aligned}$ \\
homomorphism $f(2m+1 +2n) = f(2(m+n) + 1) = 1 = 1 + 0 = f(2m+1) + f(2n)$ \\
$k\mathbb{Z} \equiv \lbrace kn | n \in \mathbb{Z} \rbrace$, $k \in \mathbb{N}$ subgroup of $\mathbb{Z}$, $\mathbb{Z}_2 = \lbrace 0 ,1 \rbrace$ not a subgroup. \\
Let $H$ subgroup of $G$, $\forall \, x, y \in G$, $x\sim y$ if $x- y \in H$ \\
group operation in $G/H$ naturally induced: $[x] + [y] = [x+y]$ \\
$G/H$ group since $H$ always a normal subgroup of $G$. \\
\quad $aH = Ha \, \forall \, a \in G$ \\
\quad \, if $aH = Ha$ \, $\forall \, a \in G$, normal indeed. \\
\quad $aH = a + x - y = x - y + a = Ha$ (since $G$ abelian) \\
If $H = G$, $0-x \in G$, \, $\forall \, x \in G$, $G/G = \lbrace 0 \rbrace$ \\
If $H = \lbrace 0 \rbrace$, $G/H = G$ since $x-y = 0$ iff $x=y$ \\
Ex. 3.1. Let us work out the quotient group $\mathbb{Z}/ k\mathbb{Z}$. \\
$km - kn = k (m-n) \in k \mathbb{Z}$ \, $[km ] = [kn]$ \\
$\forall \, j$, \, $1 \leq j \leq k-1$, $(km + j) - (kn +j) = k(m-n) \in k \mathbb{Z}$. $[km + j ] = [kn +j ]$ \\
$\forall \, j, l$, $0\leq j , l \leq k -1$, $j\neq l$, $(km+j) - (kn +l) = k(m-n) + (j-l ) \notin k\mathbb{Z}$. Never belong to the same equivalence class. \\
$\Longrightarrow \mathbb{Z}. k \mathbb{Z} = \lbrace [0], \dots, [k-1] \rbrace$. \\
Define isomorphism $\begin{aligned} & \quad \\
& \varphi : \mathbb{Z}/ k\mathbb{Z} \to \mathbb{Z}_k \\
& \varphi([j]) = j \end{aligned}$, then $\mathbb{Z}/ k\mathbb{Z} \simeq \mathbb{Z}_k$
\begin{lemma}[3.1] Let $f:G_1 \to G_2$ homomorphism. Then
\begin{enumerate}
\item[(a)] $\ker{f} = \lbrace x | x \in G_1, f(x) = \rbrace$ subgroup of $G$. Note: $\ker{f}$ \textbf{normal subgroup} of $G_1$, \\
$\begin{gathered}
f(gxg^{-1}) = f(g)f(x) f(g^{-1}) = 1 \, \forall \, g \in G, \, x\in \ker{f} \, \Longrightarrow \ker{f} \text{ normal subgroup }
\end{gathered}$
\item[(b)] $\text{im}{f} = \lbrace x | x \in f(G_1) \subset G_2 \rbrace$ subgroup of $G_2$
\end{enumerate}
\end{lemma}
\begin{proof}
\begin{enumerate}
\item[(a)] Let $x,y \in \ker{f}$ \\
$xy \in \ker{f}$ since $f(xy) = f(x) f(y) = 1\cdot 1 = 1$ \\
Note $1 \in \ker{f}$ since $f(1) = f(1\cdot 1 ) = f(1) f(1) \Longrightarrow f(1) = 1$ \\
$x^{-1} \in \ker{f}$ since $f(x^{-1}\cdot x) = f(x^{-1})f(x) = f(x^{-1}) 1 = f(1) = 1$ \, $f(x^{-1})= 1$
\item[(b)] Let $\begin{aligned} & \quad \\
& y_1 = f(x_1) \\
& y_2 = f(x_2) \end{aligned}$ \quad \, $\begin{aligned} & \quad \\
& y_1,y_2 \in \text{im}(f) \\
& x_1, x_2 \in G_1 \end{aligned}$ \quad \, \\
$\begin{gathered}
y_1 y_2 = f(x_1) f(x_2) = f(x_1x_2) \quad x_1 x_2 \in G_1 \quad y_1 y_2 \in \text{im}{f} \\
1 \in \text{im}{f} \text{ since } f(1) = 1 \, (\text{or } f(x_1) = f(x_1 \cdot 1) = f(x_1) f(1); \, \text{ so } f(1) = 1 \text{ and } 1 \in \text{im}{f}) \\
1 = f(xx^{-1}) = f(x) f(x^{-1}) = yf(x^{-1}) \Longrightarrow f(x^{-1}) = y^{-1} \text{ and } y^{-1} \in \text{im}{f} \text{ since } x^{-1} \in G
\end{gathered}$
\end{enumerate}
\end{proof}
\begin{theorem}[3.1] (Fundamental Thm. of homomorphism)
\[
G_1/ \ker{f} \simeq \text{im}{f}
\]
\end{theorem}
\begin{proof} By Lemma 3.1, both sides are groups. \end{proof}
Define. $\begin{aligned} & \quad \\
& \varphi: G_1/ \ker{f} \to \text{im}{f} \\
& \varphi([x]) = f(x) \end{aligned}$. \quad \, $\varphi$ well-defined since $\forall \, x' \in [x]$, $\exists \, h \in \ker{f}$ s.t. $ \begin{gathered} x' = x + h \\
f(x') = f(x) f(h) = f(x) \end{gathered}$ \\
\[
\varphi([x]+ [u]) = \varphi([x+y]) = f(x+y) = f(x) + f(y) = \varphi([x]) + \varphi([y])
\]
$\varphi$ 1-to-1: if $\varphi([x]) = \varphi([y])$, then $f(x) = f(y)$ or $f(x) - f(y) = f(x-y) = 0$. $x-y \in \ker{f}$ so $[x] = [y]$ \\
$\varphi$ onto: if $y \in \text{im}{f}$, $\exists \, x \in G_1$ s.t. $f(x) = y = \varphi([x])$
\subsubsection{Finitely generated Abelian groups and free Abelian groups}
\begin{lemma}[3.2] Let $G$ be free Abelian group of rank $r$,
\[
G = \lbrace n_1 x_1 + \dots + n_r x_r | n_i \in \mathbb{Z}, \, 1 \leq i \leq r, \, n_1 x_1 + \dots + n_r x_r = 0 \text{ only if } n_1 = \dots = n_r = 0 \rbrace \equiv \text{ free Abelian group of rank $r$ }
\]
Let subgroup $H$. Choose $p$ generators $x_1 \dots x_p$ of $r$ generators of $G$ so generate $H$.
\[
H \simeq k_1 \mathbb{Z} \oplus \dots \oplus k_p \mathbb{Z} \text{ and } H \text{ rank } p
\]
\end{lemma}
\begin{proof}
\[
\begin{gathered}
f: \underbrace{\mathbb{Z} \oplus \dots \oplus \mathbb{Z} }_{m} \to G \\
f(n_1 \dots n_m) = n_1 x_1 + \dots + n_m x_m \quad \, (\text{surjective homomorphism})
\end{gathered}
\]
From Thm. 3.1. $ \underbrace{\mathbb{Z} \oplus \dots \oplus \mathbb{Z}}_{ m } / \ker{f} \simeq G$ \\
$\ker{f}$ subgroup of $\underbrace{ \mathbb{Z} \oplus \dots \oplus \mathbb{Z} }_{m}$, so Lemma 3.2, $\ker{f} \simeq k_1 \mathbb{Z} \mathbb{Z} \oplus \dots \oplus k_p \mathbb{Z}$
\[
G \simeq \underbrace{ \mathbb{Z} \oplus \dots \oplus \mathbb{Z} }_{m} / \ker{f} \simeq \underbrace{ \mathbb{Z} \oplus \dots \oplus \mathbb{Z} }_{m} /(k_1 \mathbb{Z} \oplus \dots \oplus k_p \mathbb{Z} ) \simeq \underbrace{ \mathbb{Z} \oplus \dots \oplus \mathbb{Z} }_{m-p} \oplus Z_{k_1} \oplus \dots \oplus \mathbb{Z}_{k_p}
\]
\end{proof}
\subsection{Simplexes and simplicial complexes}
\subsubsection{Simplexes}
number of $q$-faces in $r$-simplex is $\binom{r+1}{q+1}$ \\
$r+1$ \, $p_0 \dots p_r$ pts., choose $p_{i_0} \dots p_{i_q}$ pts.
\subsubsection{Simplicial complexes and polyhedra}
Example 3.5. Fig. 3.5(b) not a triangulation of a cylinder. \\
$\begin{aligned} & \quad \\
& \sigma_2 = \langle p_0 p_1 p_2 \rangle \\
& \sigma_2' = \langle p_2 p_3 p_0 \rangle \end{aligned}$ \\
$\sigma_2 \sigma_2' \neq \emptyset$\\
$\sigma_2 \sigma_2' = \langle p_0 \rangle \bigcup \langle p_2 \rangle$ \quad \, $\sigma_2 \sigma_2'$ is not an actual face.
\subsection{Homology groups of simplicial complexes}
\subsubsection{Oriented simplexes}
\subsubsection{Chain group, Cycle group and boundary group}
\begin{definition}[3.2] $r$-chain group $C_r(K)$ of simplicial complex $K$ is free Abelian group generated by oriented $r$-simplexes of $K$ element of $C_r(K)$ is $r$-chain.\end{definition}
Let $I_r$ $r$- simplexes in $K$, $\sigma_{r,i}$ ($1 \leq i \leq I_r$)
\begin{equation}
c = \sum_{i=1}^{I_r} c_i \sigma_{r,i} \quad \, c_i \in \mathbb{Z}, \text{ coefficients of $c$ } \quad \quad \quad \, (3.15)
\end{equation}
addition of 2 $r$-chains $\begin{aligned} & \quad \\
& c = \sum_i c_i \sigma_{r,i} \\
& c' = \sum_i c_i' \sigma_{r,i} \end{aligned}$ \quad \quad \, $c + c' = \sum_i (c_i + c'_i) \sigma_{r,i} \quad \quad \, (3.16)$ \\
inverse $-c =\sum_i ( - c_i) \sigma_{r,i}$
\begin{equation}
C_r(K) \simeq \underbrace{\mathbb{Z} \oplus \dots \oplus \mathbb{Z}}_{ I_r} \quad \quad \, (3.17) \text{ free Abelian group of rank $I_r$}
\end{equation}
$0$-simplex has no boundary: $\partial_0 p_0 = 0$ \quad \, (3.18) \\
Fig. 3.7(a) oriented 1 simplex.
\[
\partial_1 (p_0 p_1) + \partial_1 (p_1 p_2) = p_1 - p_0 + p_2 - p_1 = p_2 - p_0 = \partial_1 (p_2 p_0)
\]
Fig. 3.7(b) triangle.
\[
\partial_1 (p_0 p_1) + \partial_1 (p_1 p_2) + \partial_1 (p_2 p_0) = p_1 - p_0 + p_2 - p_1 + p_0 - p_2 = 0
\]
Let $\sigma_r(p_0 \dots p_r)$ oriented $r$-simplex.
\begin{equation}
\partial_r \sigma_r \equiv \sum_{i=0}^r (-1)^i (p_0 p_1 \dots \widehat{p}_i \dots p_r ) \quad \quad \quad (3.20)
\end{equation}
$K \equiv n$-dim. simplicial complex. \\
chain complex $C(K)$. \\
$i$ inclusion map $i: 0 \hookrightarrow C_n(K)$
\begin{equation}
0 \xrightarrow{i} C_n(K) \xrightarrow{\partial_n} C_{n-1}(K) \xrightarrow{\partial_{n-1}} \dots \xrightarrow{\partial_2} C_1(K) \xrightarrow{\partial_1} C_0(K) \xrightarrow{\partial_0} 0 \quad \quad \, (3.23)
\end{equation}
\begin{definition}[3.3] If $c \in C_r(K)$, $\partial_r c = 0$, $c$ \, $r$-cycle.
\[
\boxed{ Z_r(K) = \lbrace c | \partial_r c =0 \rbrace = \ker{\partial_r} \equiv r-\text{cycle group} }
\]
if $r =0$, $\partial_0 c =0$, $Z_0(K) = C_0(K)$
\end{definition}
\begin{definition}[3.4] If $\exists \, d \in C_{r+1}(K)$, $c = \partial_{r+1} d \quad \quad \quad \, (3.25)$, $c$ \, $r$-boundary
\[
\boxed{ B_r(K) = \text{im}{\partial_{r+1}} \equiv r-\text{boundary group} \quad \quad \quad \, B_n(K) =0 }
\]
\end{definition}
\begin{theorem}[3.3]
\begin{equation}
B_r(K) \subset Z_r(K) \quad \, (\subset C_r(K)) \quad \quad \quad \, (3.27)
\end{equation}
\end{theorem}
\begin{proof}
\[
\forall \, c \in B_r(K), \, \exists \, d \in c_{r+1}(K) \, \text{ s.t. } c = \partial_{r+1}d \quad \quad \, \partial_r c = \partial_r \partial_{r+1} d = 0 \, (\text{Lemma } 3.3, \, \partial^2 d =0) \quad \, c \in Z_r(K)
\]
\end{proof}
\subsubsection{Homology groups}
\exercisehead{3.1}
$K = \lbrace p_0, p_1 \rbrace$. $I_r = I_0 = 2$ \, $0-$simplexes. $c= c_1 p_0 + c_2 p_1$ \, $0-$chains. \\
\[
\begin{gathered}
c_0(K) \simeq \mathbb{Z} \oplus \mathbb{Z} \\
\partial_0 c = 0 \quad \, \forall \, c \in C_0(K) \quad \, C_0(K) = Z_r(K) \\
B_0(K) = \text{im}{\partial_1} =0 \\
\Longrightarrow H_0(K) = \mathbb{Z} \oplus \mathbb{Z}
\end{gathered}
\]
if $r\neq 0$, $Z_r(K) = 0$ since there are no other simplexes than $0-$simplexes. $H_r(K) =0$
\begin{equation}
\Longrightarrow H_r(K) = \begin{cases} \mathbb{Z} \oplus \mathbb{Z} & (r=0) \\
\lbrace 0 \rbrace & (r\neq 0) \end{cases} \quad \quad \quad \, (3.34)
\end{equation}
Ex. 3.7. $k = \lbrace p_0, p_1, (p_0 p_1) \rbrace$ \\
\[
\begin{aligned}
& C_0(K) = \lbrace i p_0 + j p_1 | i,j \in \mathbb{Z} \rbrace \\
& C_1(K) = \lbrace k(p_0 p_1) | k \in \mathbb{Z}
\end{aligned}
\]
$B_1(K) = 0$, since $(p_0 p_1)$ not a boundary of any simplex in $K$ i.e. $\exists \, d \in K$ s.t. $\partial_2 d = (p_0 p_1)$, since $\nexists \, $ 2-simplex
\[
H_1(K) = Z_1(K)/ B_1(K) = Z_1(K)
\]
If $z = m(p_0 p1_) \in \mathbb{Z}_1(K)$,
\[
\partial_1 z = m (p_1- p_0) = 0 \Longrightarrow m = 0 \quad \, Z_1(K) = 0
\]
\begin{equation}
H_1(K) = 0 \quad \quad \quad \, (3.35)
\end{equation}
$Z_0(K) = C_0(K)$ \\
Define surjective (onto) homomorphism. $\begin{aligned} & \quad \\
& f: Z_0(K) \to \mathbb{Z} \\
& f(ip_0 + jp_1) = i+ j \end{aligned}$
\[
\begin{gathered}
\ker{f} = f^{-1}(0) = B_0(K) \\
\partial_1(k(p_0 p_1)) = kp_1 - k p_0 \text{ and } f(kp_1 - k p_0) = 0
\end{gathered}
\]
Thm. 3.1. $Z_0(K)/\ker{f} \simeq \text{im}{f} = \mathbb{Z}$.
\begin{equation}
H_0(K) = Z_0(K)/ B_0(K) \simeq \mathbb{Z} \quad \quad \quad \, (3.30)
\end{equation}
Ex. 3.8. Triangulation of $S^1$ (\textbf{triangulation of circle})
\[
\begin{gathered}
K = \lbrace p_0, p_1, p_2, (p_0 p_1), (p_1 p_2), (p_2 p_0) \rbrace \\
B_1(K) = 0 \quad \, (\text{no 2-simplices in $K$}) \quad \quad \, H_1(K) = Z_1(K)/B_1(K) = Z_1(K) \\
\partial_1 z = \partial_1( i (p_0 p_1) + j (p_1 p_2) + k (p_2 p_0)) = i (p_1 - p_0 + j (p_2 - p_1) + k(p_0 - p_2) = (k-i) p_0 + (i-j) p_1 + (j-k) p_2 = 0 \\
\Longrightarrow i =j = k \\
Z_1(K) = \lbrace i ((p_0 p_1) + (p_1p_2) + (p_2 p_0) ) | i \in \mathbb{Z} \rbrace \simeq \mathbb{Z} \Longrightarrow H_1(K) = Z_1(K) \simeq \mathbb{Z} \quad \quad \, (3.37)
\end{gathered}
\]
\[
\begin{gathered}
Z_0(K) = C_0(K) \\
B_0(K) = \lbrace \partial_1 [ l (p_0 p_1) + m (p_1 p_2) + n (p_2 p_0) ] | l,m,n \in \mathbb{Z} \rbrace = \lbrace (n-l) p_0 + (l-m) p_1 + (m-n) p_2 | l, m ,n \in \mathbb{Z} \rbrace
\end{gathered}
\]
\exercisehead{3.2} Let $K = \lbrace p_0, p_1, p_2, p_3, (p_0 p_1), (p_1 p_2), (p_2 p_3), (p_3 p_0) \rbrace$ \\
$B_1(K) = 0 \quad \, (KC_2(K) = \emptyset)$ \\
$H_1(K) = Z_1(K)$
\[
\partial_1 ( i (p_0 p_1 ) + j (p_1 p_2) + k (p_2 p_0) + l (p_3 p_0) ) = (l- i) p_0 + (i-j) p_1 + (j-k) p_2 + (k-l) p_3
\]
$\partial_1 c = 0$ if $i = l =j =k$, $Z_1(K) \simeq \mathbb{Z}$. $H_1(K) \simeq \mathbb{Z}$ \\
$Z_0(K) = C_0(K)$ \quad $(\partial_0 p_i = 0)$ \\
Define surjective homomorphism $\begin{aligned} & \quad \\
& f: C_0(K) \mapsto \mathbb{Z} \\
& f(ip_0 + j p_1 + k p_2 + l p_3) = i + j + k + l \end{aligned}$
\[
B_0(K) = \lbrace ap_0 + bp_1 + cp_2 + dp_3 | d = -(a+b+c) \rbrace \quad \quad \quad \, \begin{aligned}
& a = l - i \\
& b = i - j \\
& c = j-k \\
& d = k- l = - (a+b+c) \end{aligned}
\]
\[
\begin{gathered}
\Longrightarrow B_0(K) = \ker{f} \\
H_0(K) = Z_0(K)/B_0(K) = C_0(K)/\ker{f} \simeq \text{im}{f} = \mathbb{Z}
\end{gathered}
\]
\exercisehead{3.3} $B_2(K) =0$ \quad $KC_2(K) = \emptyset$
\[
\begin{gathered}
\partial_2 c_2 = \partial_2 ( i (p_0 p_1 p_2) + j (p_0 p_1 p_3) + k (p_0 p_2 p_3) + l (p_1 p_2 p_3) ) = \\
= ( i + l ) (p_1 p_2) + ( -i + k ) (p_0 p_2) + ( i + j ) (p_0 p_1) + ( -j + -k ) (p_0 p_3) + ( j -l ) (p_1 p_3) + ( k + l ) (p_2 p_3) = 0 \\
\Longrightarrow i =k = -l = -j \Longrightarrow Z_2(K) \simeq \mathbb{Z} \text{ so } \boxed{ H_2(K) \simeq \mathbb{Z} }
\end{gathered}
\]
\[
\begin{gathered}
\partial_1 c_1 = \partial_1 (a(p_0p_1) + b(p_0p_2) + c(p_0p_3) + d(p_1p_2) + e(p_1p_3) + f(p_2p_3)) = \\
= (-a-b-c) p_0 + (a-d-e) p_1 + (b+d-f) p_2 + (c+e+f) p_3 = 0
\end{gathered}
\]
By linear algebra,
\[
\left[ \begin{matrix} -1 & - 1 & -1 & & & \\
1 & & & -1 & -1 & \\
& 1 & & 1 & & -1 \\
& & 1 & & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} a \\ b \\ c \\ d \\ e \\ f \end{matrix} \right] = 0 \Longrightarrow \left[ \begin{matrix} 1 & & & -1 & -1 & \\
& 1 & 1 & 1 & 1 & \\
& 1 & & 1 & & -1 \\
& & 1 & & 1 & 1 \end{matrix} \right] \Longrightarrow \begin{aligned} & a = d + e \\
& b = -d + f \\
& c = -e -f \end{aligned}
\]
If
\[
\begin{aligned}
& a = i + j \\
& b = -i + k \\
& c = -j -k \\
& d = i + l \\
& e = j - l \\
& f = k + l
\end{aligned}
\]
Then $B_1(K) = Z_1(K)$. Then $Z_1(K)/B_1(K) = 0$ by algebra. $\boxed{ H_1(K) = 0 }$ \\
Let $\begin{aligned} & \quad \\
& f: Z_0(K) \to \mathbb{Z} \\
& f(ip_0 + jp_1 + k p_2 + l p_3) = i + j +k + l \end{aligned}$ \\
$\ker{f} = B_0(K)$ since if
\[
\begin{aligned}
& i = - ( a + b+ c) \\
& j = a- d -e \\
& k = b + d - f \\
& l = c + e+ f
\end{aligned}
\]
then $i+j + k + l =0$
\[
Z_0(K)/ \ker{f} \simeq \text{im}{f} = \mathbb{Z} \Longrightarrow \boxed{ H_0 \simeq \mathbb{Z}}
\]
\subsubsection{Computation of $H_0(K)$}
\begin{theorem}[3.5] Let $K$ be \emph{connected} simplicial complex. Then
\begin{equation}
H_0(K) \simeq \mathbb{Z} \quad \quad \quad \, (3.43)
\end{equation}
\end{theorem}
\subsubsection{More homology computations}
Example 3.10. Fig. 3.8. triangulation of M\"{o}bius strip. \\
$B_2(K) =0$ \, $\partial_2 z =0$ each $(p_0 p_2), (p_1 p_4), (p_2 p_3), (p_4p_5), (p_3 p_1), (p_5p_0)$ appear only once. $Z_2(K)=0$
\begin{equation}
H_2(K)= 0 \quad \quad \quad \quad \, (3.44)
\end{equation}
$H_1(K) = Z_1(K)/B_1(K)$. $\ker{\partial_1}$ consists of closed loops. (closes on itself so $\partial_1 c_1 =0$) \\
Note $z \sim z'$ if $z - z' = \partial_2 c_2 \in B_1(K)$ \\
easily verify, $H_1(K)$ generated by just $[z]$. $H_1(K) = \lbrace i [z] | i \in \mathbb{Z} \rbrace \simeq \mathbb{Z} \quad \quad \, (3.45)$ \\
Example 3.11. projective plane $\mathbb{R}P^2$ \\
Example $H_2(K)$ from a slightly different view pt. Add all 2-simplexes in $K$ with same coefficient
\[
z \equiv \sum_{i=1}^{10} m \sigma_{2,i}, \quad \quad \, m \in \mathbb{Z}
\]
Observe that each 1-simplex of $K$ is a common face of exactly 2 2-simplices.
\begin{equation}
\partial_2 z = 2m (p_3 p_5) + 2m (p_5 p_4) + 2m (p_4 p_3) \quad \quad \quad \, (3.47)
\end{equation}
$Z_2(K) =0$ since $\partial_2 z = 0$ if $m=0$ \\
Note, any 1-cycle homologous to a multiple of $z = (p_3 p_5) + (p_5 p_4) + (p_4 p_3)$ \\
even multiples of $z$ is a boundary of 2-chain by Eq. 3.4.7.
\begin{equation}
H_1(K) = \lbrace [z] | [z] + [z] \sim [0] \rbrace \simeq \mathbb{Z}_2 \quad \quad \quad \, (3.48)
\end{equation}
This example shows that a homology group is not necessarily free Abelian. \\
Example 3.12. surface of the torus has no boundary, but it's not a boundary of some 3-chain. \\
Thus, $H_2(T^2)$ freely generated by 1 generator, surface itself, $H_2(T^2) \simeq \mathbb{Z}$ \\
closed loops $a,a'$ homologous since $a'-a = \partial d$ bounds shaded area Fig. 3.10 (could think of $a$ running backwards) \\
See Figure 3.12, triangulation of the Klein bottle (clear from there). \\
inner 1 simplices cancel out to leave only outer 1-simplexes (1 side of the ``square'')
\[
\begin{gathered}
\begin{aligned}
z & = \sum m \sigma_{2,i} \\
\partial_2 z & = -2ma \end{aligned} \\
a = (p_0 p_1 ) + (p_1 p_2 ) + ( p_2 p_0)
\end{gathered}
\]
$\partial_2z =0$ if $m=0$
\begin{equation}
H_2(K) = Z_2(K) \simeq 0 \quad \quad \quad \, (3.50)
\end{equation}
\[
b = (p_0 p_3) + (p_3 p_4) + (p_4p_0)
\]
1-cycle \, $\begin{aligned} & \quad \\
c_1 & = ia + ib \\
\partial_1 c_1 & = 0 \text{ closed } \end{aligned}$ \\
Now $\partial_2 z = + 2 m a$ so
\[
2ma \sim 0
\]
Thus, $H_1(K)$ generated by 2 cycles $a,b$ s.t. $a+a =0$ (remember $a$ is ``glued backwards'')
\begin{equation}
H_1(K) = \lbrace i [a] + j[b] | i,j \in \mathbb{Z} \rbrace \simeq \mathbb{Z}_2 \oplus \mathbb{Z} \quad \quad \, (3.51)
\end{equation}
\subsection{General properties of homology groups}
\subsubsection{Connectedness and homology groups}
\begin{theorem}[3.6] Let $K = \bigcup_{k=1}^N K_i$, $K_i K_j \neq \emptyset$. $K_i$ connected components. \\
Then
\[
H_r(K) = \bigoplus_{i=1}^N H_r(K_i)
\]
\end{theorem}
\begin{proof}
$C_r(K) = \bigoplus_{i=1}^N C_r(K_i)$ (rearrange cycles) \\
since $Z_r(K_i) \supset B_r(K_i)$, $H_r(K_i)$ well-defined.
\end{proof}
\subsubsection{Structure of homology groups}
\begin{definition}[3.6] Let $K$ simplicial complex.
\begin{equation}
b_r(K) \equiv \dim{H_r(K;\mathbb{R})} \quad (\text{$i$th Betti number}) \quad \quad \quad \, (3.56)
\end{equation}
\end{definition}
\begin{theorem}[3.7] (The Euler-Poincar\`{e} theorem) Let $K$ $n-$dim. simplicial complex, $I_r$ number of $r$-simplexes in $K$.
\begin{equation}
\chi(K) \equiv \sum_{i=0}^n (-1)^r I_r = \sum_{r=0}^n (-1)^r b_r(K) \quad \quad \quad \, (3.57)
\end{equation}
\end{theorem}
\problemhead{3.1}
Let $S^2$ with $h$ handles and $q$ holes $=X$
\[
C_3(X) \bigcap X= \emptyset \text{ so } B_2(X) = 0 \quad \, Z_2(X) \simeq \mathbb{Z} \, (1 \text{ surface }) \, \boxed{ H_2(X) \simeq \mathbb{Z} \quad \, b_2(X) = 1 }
\]
$h$ handles. Think of $T^2$. \\
$q$ holes. For orientable case, the first hole does not change the homology, i.e.
\[
\partial c_1 =0 \text{ but for } d = \sum_{\sigma_2 \in C_2(X)} m\sigma_2, \, \partial d = c_1 \quad \, c_1 = 0
\]
\[
\boxed{ H_1(X) = \mathbb{Z}^{2h + q -1} }
\]
The sphere is connected (and note compact). $\boxed{ H_0(X) = \mathbb{Z} }$
\problemhead{3.2}
Each cross cap hole contributes to $H_1(X)$ a 1 cycle $z_q$ s.t. $2z_q \sim 0$. Thus,
\[
\boxed{ H_1(X) = \mathbb{Z}_2^q }
\]
Otherwise,
\[
\boxed{ \begin{aligned}
& H_2(X) = \mathbb{Z} \\
& H_0(X) = \mathbb{Z} \end{aligned} }
\]
\section{Homotopy Groups}
\section{Manifolds}
\subsection{Manifolds}
\subsubsection{Heuristic introduction}
\subsubsection{Definitions}
\subsubsection{Examples}
\exercisehead{5.1}
\[
S^n = \lbrace ( x_1 \dots x_{n+1} \rbrace \in \mathbb{R}^{n+1} | \sum_{i=1}^{n+1} x_i^2 = 1 \rbrace \subset \mathbb{R}^{n+1}
\]
Let $\begin{aligned} & \quad \\
& N = (0 \dots 0, 1) \\
& S = (0 \dots 0, -1) \end{aligned}$ \quad \, $x\in S^n$. Consider $t(x-N) + N = tx + (1-t)N$ when $x_{n+1} =0$
\[
\begin{gathered}
tx_{n+1} + (1-t) = 0 \text{ or } tx_{n+1} + -1 + t = 0 \\
\Longrightarrow \frac{ 1}{ 1- x_{n+1} } = t \quad \left( \text{ or } \frac{1}{ 1 + x_{n+1} } \right)
\end{gathered}
\]
\[
\begin{aligned}
& \pi_1: S^n - N \to \mathbb{R}^n \\
& \pi_1(x_1 \dots x_{n+1}) = \left( \frac{x_1}{ 1 - x_{n+1} } \dots \frac{x_n}{ 1 - x_{n+1} } , 0 \right) \\
& \pi_2 : S^n -S \to \mathbb{R}^n \\
& \pi_2(x_1 \dots x_{n+1}) = \left( \frac{x_1}{ 1 + x_{n+1}} \dots \frac{x_n}{ 1 + x_{n+1} }, 0 \right)
\end{aligned}
\]
Note, for $y_i = \frac{x_i}{ 1 - x_{n+1}}$
\[
\begin{gathered}
y_1^2 + \dots +y_n^2 = |y|^2 = \frac{1- x_{n+1}^2}{ (1-x_{n+1})^2 } = \frac{1+ x_{n+1}}{ 1- x_{n+1}} \text{ or } x_{n+1} = \frac{ |y|^2 - 1 }{ |y|^2 + 1 } \\
x_i = y _i (1- x_{n+1}) = \frac{2y_i}{ 1 + |y|^2 }
\end{gathered}
\]
\[
\begin{aligned}
& \pi_1^{-1}: \mathbb{R}^n \to S^n - N \\
& \pi_1^{-1}(y_1 \dots y_n) = \left( \frac{2y_1}{ 1 + |y|^2 } \dots \frac{2y_n}{1+ |y|^2 } , \frac{ |y|^2 - 1 }{ |y|^2 + 1 } \right) \\
& \pi_2^{-1}(y_1 \dots y_n) = \left( \frac{2y_1}{ 1 + |y|^2 } \dots \frac{2y_n}{1+ |y|^2 } , \frac{ 1 - |y|^2 }{ |y|^2 + 1 } \right)
\end{aligned}
\]
$\pi_1,\pi_2$ diff. injective, and $(S^n - N ) \bigcup (S^n-S) = S^n$ \\
Consider $(S^n - N)(S^n-S) = S^n - N \bigcup S$
\[
\begin{aligned}
& \pi_1 \pi_2^{-1}(y_1 \dots y_n) = \left( \frac{y_1 }{ |y|^2} \dots \frac{y_n}{|y|^2} , 0 \right) \\
& \pi_2 \pi_1^{-1}(y_1 \dots y_n) = \left( \frac{y_1 }{ |y|^2} \dots \frac{y_n}{|y|^2} , 0 \right)
\end{aligned}
\]
since, for example,
\[
\begin{gathered}
\frac{ \frac{2y_i }{ 1 + |y|^2} }{ 1 - \frac{ 1 - |y|^2}{ 1 + |y|^2 }} = \frac{y_i }{ |y|^2 }
\end{gathered}
\]
$\pi_1 \pi_2^{-1}$ bijective and $C^{\infty}$, $\pi_1 \pi_2^{-1}$ diffeomorphism. \\
$\lbrace (S^n-N, \pi_1), (S^n - S, \pi_2) \rbrace$ \, $C^{\infty}$ atlas or differentiable structure.
\subsection{The calculus on manifolds}
\subsubsection{Differentiable maps}
\exercisehead{5.2} If $f:M\to N$ smooth, \\
Consider $\begin{aligned} & \quad \\
& (U'_a, \varphi'_a) \in \mathcal{A} \\
& (V'_b, \psi'_b)\in\mathcal{B}\end{aligned}$ \, s.t. $f(U_a') \subseteq V_b'$ \\
Consider
\[
\psi'_b f(\varphi_a')^{-1} = \psi_b'(\psi_{\beta}^{-1} \psi_{\beta}) f (\varphi_{\alpha}^{-1} \varphi_{\alpha}) (\varphi'_a)^{-1} = (\psi_b'\psi_{\beta}^{-1})(\psi_{\beta} f \varphi_{\alpha}^{-1})(\varphi_{\alpha} \varphi'_a)^{-1}
\]
$\begin{aligned} & \quad \\
& U_{\alpha} U_a' \neq \emptyset \\
& V_{\beta}V'_b \neq \emptyset \end{aligned}$ since $\begin{gathered} \quad \\
x \in U_{\alpha}, U_a' \\
f(x) \in V_{\beta} V'_{\beta} \end{gathered}$ \\
Then $(\psi_b' \psi_{\beta}^{-1}), (\varphi_{\alpha} \varphi'_a)^{-1}$ , \, $C^{\infty}$ (def. of atlas) and $(\psi_{\beta} f \varphi_{\alpha}^{-1})$ \, $C^{\infty}$ (given). \\
So $\psi_b' f(\varphi_a')^{-1}$ \, $C^{\infty}$
\subsubsection{Vectors}
curve $c:I \to M$ \\
\phantom{curve } $f:M \to \mathbb{R}$
\[
\frac{d}{dt} \left. f(c(t)) \right|_{t=0} \quad \quad \quad \, (5.18)
\]
for $p = c(0) \in (U, \varphi) = (U, x^1 \dots x^n)$, \, $fc = f\varphi^{-1}\varphi c$
\[
\frac{d}{dt} \left. f(c(t)) \right|_{t=0} = \frac{ \partial (f\varphi^{-1}) }{ \partial x^{\mu} }(x) \left. \frac{ d (x^{\mu}(c)) }{ dt}(t) \right|_{t=0} = \frac{ \partial f}{ \partial x^{\mu}} \left. \frac{dx^{\mu}(c(t)) }{dt} \right|_{t=0} = \left. \frac{ \partial f}{ \partial x^{\mu} } \frac{d c^{\mu}}{dt} \right|_{t=0} \quad \quad \, (\text{abuse of notation})
\]
Note that abuse of notation: $ \left. \frac{ dx^{\mu} }{dt}(c(t)) \right|_{t=0 } \equiv \left. \frac{d x^{\mu} ( \varphi{ (c(t)) }) }{dt} \right|_{t=0}$
$ \left. \frac{df(c(t))}{dt} \right|_{t=0}$ obtained by differential operator $X$ to $f$
\[
X = X^{\mu}\left( \frac{ \partial }{ \partial x^{\mu} } \right) \quad \quad \, \left( X^{\mu} = \left. \frac{dx^{\mu}(c(t))}{dt} \right|_{t=0} \right) \quad \quad \quad \, (5.20)
\]
\[
\left. \frac{df(c(t))}{ dt} \right|_{t=0} = X^{\mu} \left( \frac{ \partial f}{ \partial x^{\mu} } \right) = X[f] \quad \quad \quad \, (5.21)
\]
Introduce equivalence class of curves in $M$. \\
curves $c_1(t) \sim c_2(t)$ if
\begin{enumerate}
\item[(1)] $c_1(0) = c_2(0) = p$
\item[(2)] $
\left. \frac{dx^{\mu}{ (c_1(t)) } }{ dt} \right|_{t=0} = \left. \frac{dx^{\mu}(c_2(t)) }{ dt} \right|_{t=0}
$
\end{enumerate}
Identify tangent vector $X$ with $[c(t)]$.
$T_p M$, tangent space of $M$ at $p$, all $[c]$ at $p \in M$
Use Sec. 2.2's theory of vector spaces to analyze $T_pM$
\hrulefill
Recall curve $c(t) : \mathbb{R} \to M$, \\
\phantom{Recall curve } $f: M \to \mathbb{R}$
Now tangent vector at $c(0)$ was, recall
\[
\left. \frac{df(c(t))}{ dt} \right|_{t=0} = \frac{ \partial f}{ \partial x^{\mu}} \left. \frac{ dx^{\mu}(c(t))}{ dt} \right|_{t=0} = \left. \frac{dx^{\mu}(c(t))}{ dt} \right|_{t=0} \frac{ \partial f}{ \partial x^{\mu} }
\]
Let $X^{\mu} = \left. \frac{dx^{\mu}(c(t))}{ dt} \right|_{t=0}$
\[
\begin{gathered}
X[f] \equiv \left. \frac{df(c(t))}{ dt} \right|_{t=0} \\
\Longrightarrow X = x^{\mu} \frac{ \partial}{ \partial x^{\mu}} \end{gathered}
\]
So tangent vector $X$ is spanned by $\frac{ \partial }{ \partial x^{\mu} }$
Suppose $a^{\mu} \frac{ \partial }{ \partial x^{\mu} } =0$
\[
a^{\mu} \frac{ \partial }{ \partial x^{\mu}} x^{\nu} = a^{\mu} \delta_{\mu}^{ \, \, \nu } = a^{\nu} = 0
\]
So $\lbrace \frac{ \partial }{ \partial x^{\mu}} \rbrace$ linearly independent.
$\lbrace \frac{ \partial }{ \partial x^{m}} \rbrace$ a basis.
\hrulefill
(cf. wikipedia) Consider short $\varphi : U \to \mathbb{R}^n$, $p\in U$, define map $ \begin{aligned} & \quad \\
& ( d\varphi)_p : T_x M \to \mathbb{R}^n \\
& (d\varphi)_p( [c(0 ) ]) = \frac{d}{dt} ( \varphi ( c(0 )) ) \end{aligned}$
Consider $c(0) = p = c(t)$ \, $\forall \, t $ s.t. $c(t) \in U$. \\
Surely $\varphi(c(t)) = $ const. (for $\varphi$ is a well-defined function) \\
$\frac{d}{dt} \varphi(c(t)) = 0$. For $[c(0)] = [p]$, $ (d\varphi)_p ([p ]) = 0$ \, $\forall \, $ chart $\varphi$ \, $(d\varphi)_p$ injective. \\
Consider $a^i e_i \in \mathbb{R}^n$
\[
a^i = \frac{d}{dt} ( \varphi^i(c(0)) ) \equiv \left. \frac{d}{dt} x^i(c(t)) \right|_{t=0}
\]
$f: M \to \mathbb{R} \in C^{\infty}(M)$ if $f\varphi^{-1} \in C^{\infty}$. $\forall \, $ chart $\varphi : U \to \mathbb{R}^n$
derivation at $p$ is linear $D : C^{\infty}(M) \to \mathbb{R}$ \, ( or $\begin{aligned} & \quad \\
& X: C^{\infty}(M) \to \mathbb{R} \\
& X[f] =\left. \frac{df(c(t)) }{ dt} \right|_{t=0} \end{aligned}$ ). These derivations form a vector space, tangent space $T_pM$.
\[
\left. \frac{df(c(t))}{dt} \right|_{t=0} = \frac{ \partial g}{ \partial x^{\mu} }(x^{\nu}) \frac{dx^{\mu}(c(t)) }{ dt} = \frac{dx^{\mu}(c(t)) }{dt} \frac{ \partial}{ \partial x^{\mu} } f
\]
$g(x^{\nu}) = (f\cdot \varphi^{-1})(x^{\nu})$
Consider $p\in U_i U_j$, \, $\begin{aligned} & \quad \\
& x = \varphi_i(p) \\
& y = \varphi_j(p) \end{aligned}$
\[
\begin{gathered}
X = X^{\mu} \frac{ \partial }{ \partial x^{\mu} } = Y^{\nu} \frac{ \partial }{ \partial y^{\nu}} \\
Y^{\nu} = X^{\mu} \frac{ \partial y^{\nu} }{ \partial x^{\mu} } \quad \quad \quad \, (5.23)
\end{gathered}
\]
\subsubsection{One-forms}
$\begin{aligned}
& \omega \in T_p^*M \\
& \omega : T_p M \to \mathbb{R} \end{aligned}$ \quad \textbf{cotangent vector or 1-form}. \\
\textbf{ differential } $df \in T_p^*M$ on $V \in T_pM$
\[
\langle df , V \rangle = V[f] = V^{\mu} \frac{ \partial f}{ \partial x^{\mu} } \in \mathbb{R} \quad \quad \quad \, (5.24)
\]
arbitrary 1-form $\omega$
\[
\omega = \omega_{\mu} dx^{\mu} \quad \quad \quad \, (5.26)
\]
inner product defined
\[
\begin{aligned}
& \langle \, , \, \rangle : T_p^*M \times T_pM \to \mathbb{R} \\
& \langle \omega, V \rangle = \omega_{\mu} V^{\mu} \langle dx^{\mu}, \frac{ \partial}{ \partial x^{\nu} } \rangle = \omega_{\mu} V^{\mu} \quad \quad \, (5.27)
\end{aligned}
\]
For $p \in U_iU_j$, $\begin{aligned} & \quad \\
& x = \varphi_i(p) \\
& y = \varphi_j(p) \end{aligned}$ \\
$\begin{aligned}
& \quad \\
& dy^{\nu} \in T_p^*M \text{ so } \\
& dy^{\nu} = \omega_{\mu} dx^{\mu} \end{aligned}$ so
\[
\begin{gathered}
\langle dy^{\nu} , \frac{ \partial }{ \partial x^{\gamma} } \rangle = \delta^{\mu \gamma} \frac{ \partial y^{\nu} }{ \partial x^{\mu} } = \frac{ \partial y^{\nu} }{ \partial x^{\gamma} } = \omega_{\mu} \langle dx^{\mu}, \partial_{x^{\gamma}} \rangle = \omega_{\gamma} \\
\Longrightarrow dy^{\nu} = \frac{ \partial y^{\nu} }{ \partial x^{\mu} } dx^{\mu}
\end{gathered}
\]
\[
\omega = \omega_{\mu} dx^{\mu} = \psi_{\nu} dy^{\nu} = \psi_{\nu} \frac{ \partial y^{\nu}}{ \partial x^{\mu} } dx^{\mu} \text{ or } \omega_{\mu} = \psi_{\nu} \frac{ \partial y^{\nu }}{ \partial x^{\mu}}
\]
\subsubsection{Tensors}
$\tau^q_{r, p }(M)$ set of type $(q,r)$ tensors at $p\in M$.
element of $\tau^q_{r, p }(M)$
\[
T = T^{\mu_1 \dots \mu_q }_{\phantom{\mu_1 \dots \mu_q} \nu_1 \dots \nu_r } \frac{ \partial }{ \partial x^{\mu_1} } \dots \frac{ \partial}{ \partial x^{\mu_q} } dx^{\nu_1 } \dots dx^{\nu_r} \quad \quad \quad (5.29)
\]
$T: \otimes^q T_p^*M \otimes^r T_pM \to \mathbb{R}$
Let $\begin{aligned} & \quad \\
& V_i = V_i^{\mu} \frac{ \partial }{ \partial x^{\mu }} \quad \quad ( 1 \leq i \leq r) \\
& \omega_i = \omega_{i \mu} dx^{\mu} \quad \quad ( 1 \leq i \leq q ) \end{aligned}$
\[
T(\omega_1 \dots \omega_q, V_1 \dots V_r) = T^{\mu_1 \dots \mu_q }_{\phantom{\mu_1 \dots \mu_q} \nu_1 \dots \nu_r } \omega_{1 \mu_1} \dots \omega_{q \mu_q} V_1^{\nu_1} \dots V_r^{\nu_r}
\]
%\[
%T(\omega_1 \dots a_i \omega_i + b_i \psi_i \dots \omega_q; V_1 \dots c_j V_j + d_j W_j \dots V_r) = T^{\mu_1 \dots \mu_q}_{\phantom{\mu_1 \dots \mu_q} \nu_1 \dots nu_r } \omega_{1 \mu_1} \dots ( a_i \omega_{i \mu_i} + b_i \psi_{i \mu_i} ) \dots \omega_{q \mu_q} V_1^{\nu_1} \dots \omega_j V_j^{\nu_j}
%\]
\[
\begin{gathered}
T(\omega_1 \dots a_i \omega_i + b_i \psi_i \dots \omega_q, V_1 \dots V_r) = T^{\mu_1 \dots \mu_q}_{\phantom{\mu_1 \dots \mu_q} \nu_1 \dots \nu_r } \omega_{1 \mu_1} \dots (q_i \omega_{i \mu_i} + b_i \psi_{i \mu_i} ) \dots \omega_{q \mu_q} , V_1^{\nu_1} \dots V_r^{\nu_r} ) = \\
= a_i T^{\mu_1 \dots \mu_q}_{\phantom{\mu_1 \dots \mu_q} \nu_1 \dots \nu_r } \omega_{1 \mu_1} \dots \omega_{i \mu_i} \dots \omega_{q \mu_q} , V_1^{\nu_1} \dots V_r^{\nu_r} + b_i T^{\mu_1 \dots \mu_q}_{\phantom{\mu_1 \dots \mu_q} \nu_1 \dots \nu_r } \omega_{1 \mu_1} \dots \psi_{i \mu_i} \dots \omega_{q \mu_q} , V_1^{\nu_1} \dots V_r^{\nu_r} = \\
= a_i T(\omega_1 \dots \omega_i \dots \omega_q, V_1 \dots V_r) + b_i T(\omega_1 \dots \psi_i \dots \omega_q, V_1 \dots V_r)
\end{gathered}
\]
\[
\begin{gathered}
T(\omega_1 \dots \omega_q, V_1 \dots c_j V_j + d_j W_j \dots V_r) = T^{ \mu_1 \dots \mu_q}_{\phantom{\mu_1 \dots \mu_q} \nu_1 \dots \nu_r} \omega_{1\mu_1} \dots \omega_{q \mu_q} V_1^{\nu_1} \dots (c_j V_j^{\nu_j} + d_j W_j^{\nu_j} \dots V_r = \\
= c_j T^{\mu_1 \dots \mu_q}_{\phantom{\mu_1 \dots \mu_q} \nu_1 \dots \nu_r } \omega_{1 \mu_1} \dots \omega_{q \mu_q} V_1^{\nu_1} \dots V_j^{\nu_j} \dots V_r + d_j T^{\mu_1 \dots \mu_q}_{\phantom{\mu_1 \dots \mu_q} \nu_1 \dots \nu_r } \omega_{1 \mu_1} \dots \omega_{q \mu_q} V_1^{\nu_1} \dots W_j^{\nu_j} \dots V_r = \\
= c_j T(\omega_1 \dots \omega_q, V_1 \dots V_j \dots V_r) + d_j T(\omega_1 \dots \omega_q, V_1 \dots W_j \dots V_r)
\end{gathered}
\]
\subsubsection{Tensor fields}
vector field - vector assigned smooth $\forall \, p \in M$ \\
\quad i.e. $V$ vector field if $V[f] \in \mathcal{F}(M)$ \quad \, $\forall \, f \in \mathcal{F}(M)$ \\
tensor field of type $(q,r)$, \, $\tau^q_{r,p}(M)$ \, $\forall \, p \in M$ \\
$\tau_1^0(M)$ set of dual vector fields $\equiv \Omega^1(M)$ \\
$\tau_0^0(M) = \mathcal{F}(M)$
\subsubsection{ Induced maps}
smooth $f:M \to N$ \\
differential $f_*:T_pM \to T_{f(p)}N$ \\
By def. of tangent vector as directional derivative along a curve, \\
\quad if $g \in \mathcal{F}(N)$, $fg \in \mathcal{F}(M)$ \\
vector $V \in T_pM$ acts on $gf$ to give number $V[gf]$ \\
define $\begin{aligned} & \quad \\
& f_* V \in T_{f(p)}N \\
& (f_*V)[g] = V[gf] \quad \quad \, (5.31) \end{aligned}$ \\
or for $\begin{aligned} & \quad \\
& (U,\varphi) \subset M \\
& (V, \psi) \subset N \end{aligned}$
\[
(f_*V)[g\psi^{-1}(y) ] \equiv V[gf\varphi^{-1}(x) ] \quad \quad \quad \, (5.32)
\]
Let $V = V^{\mu} \frac{ \partial }{ \partial x^{\mu} }$, \, $f_* V = W^{\alpha} \frac{ \partial }{ \partial y^{\alpha} }$
\[
W^{\alpha} \frac{ \partial }{ \partial y^{\alpha} } [ g\psi^{-1}(y) ] = V^{\mu} \frac{ \partial }{ \partial x^{\mu} }[gf\varphi^{-1}(x) ]
\]
With $y = \psi(f(p))$, \\
take $g= y^{\alpha}$
\[
W^{\alpha} = V^{\mu} \frac{ \partial y^{\alpha} }{ \partial x^{\mu} } \quad \quad \quad \, (5.33)
\]
$f(p) = f\varphi^{-1}(x)$
\[
V^{\mu} \frac{ \partial }{ \partial x^{\mu} }[ y^{\alpha} f \varphi^{-1}(x) ] = V^{\mu} \frac{ \partial }{ \partial x^{\mu} }[y^{\alpha} ]
\]
\hrulefill
20121030
Consider smooth $\begin{aligned} & \quad \\
& f: M \to N \\
& f_* : T_p M \to T_{f(p)}N \end{aligned}$
\[
\begin{aligned}
& (U , \varphi) \subset M \quad \quad \, \varphi = x^{\mu} \\
& (V, \psi ) \subset N \quad \quad \, \psi = y^{\nu} \end{aligned}
\]
\[
\begin{aligned}
& \psi f \varphi^{-1} : \varphi(U) \subseteq \mathbb{R}^m \to \mathbb{R}^n \\
& \psi f\varphi^{-1} \equiv f^{\nu}(x^{\mu })
\end{aligned} \quad \quad \, \begin{aligned}
& X \in T_p M \\
& Y \in T_{f(p)} N \end{aligned} \quad \quad \, \begin{aligned}
X & = X^{\mu} \frac{ \partial }{ \partial x^{\mu } } \\
Y & = Y^{\nu} \frac{ \partial }{ \partial y^{\nu } } \end{aligned}
\]
\[
\begin{aligned}
& f_* X \in T_{f(p)} N \\
& f_* X = Y^{\nu} \frac{ \partial }{ \partial y^{\nu }}
\end{aligned}
\]
\[
\begin{aligned}
& g \in \mathcal{F}(N) \\
& g: N \to \mathbb{R}
\end{aligned} \quad \quad \,
\begin{aligned}
& g\psi^{-1}: \psi(N) \to \mathbb{R} \\
& g\psi^{-1} \equiv g(y^{\nu} )
\end{aligned}
\]
\[
\begin{aligned}
& gf : M \to \mathbb{R} \\
& gf \in \mathcal{F}(M)
\end{aligned}
\]
\[
\begin{aligned}
& (f_* X)[g] \equiv X[gf] \\
& (f_* X)[g] = Y^{\nu} \frac{ \partial g}{ \partial y^{\nu }}(y) = X^{\mu} \frac{ \partial }{ \partial x^{\mu }}[gf] = X^{\mu} \frac{ \partial }{ \partial x^{\mu } } g( f^{\nu}(x) ) = X^{\mu} \frac{ \partial g}{ \partial y^{\nu} }(y) \frac{ \partial f^{\nu} }{ \partial x^{\mu} }(x)
\end{aligned}
\]
\[
\Longrightarrow \boxed{ Y^{\nu} = X^{\mu } \frac{ \partial f^{\nu }}{ \partial x^{\mu }} = X^{\mu} \frac{ \partial y^{\nu} }{ \partial x^{\mu }} }
\]
where
\[
\begin{aligned}
& gf = g \psi^{-1} \psi f \varphi^{-1} \\
& g\psi^{-1} \psi f \varphi^{-1} : \varphi(U) \to \mathbb{R} \\
& g\psi^{-1} \psi f \varphi^{-1} = g(f^{\nu}(x) )
\end{aligned}
\]
\exercisehead{5.3}
$\begin{aligned} & \quad \\
f : & M \to N \\
g : & N \to P \\
gf : & M \to P \end{aligned}$ \quad \, Consider $\begin{aligned} & \quad \\
p \in M , & (U, \varphi) \subset M \\
q = f(p) \in N , & (V, \psi) \subset N \\
r = g(q) \in P , & (W, \chi) \subset P \end{aligned}$ \quad \, $\begin{aligned} & \quad \\
& V \in T_pM \\
& W \in T_qN \\
& X \in T_{g(q)}P \end{aligned}$ \quad \, $\begin{aligned} & \quad \\
& V = V^{\alpha} \frac{ \partial }{ \partial x^{\alpha} } \\
& W = W^{\beta} \frac{ \partial }{ \partial y^{\beta} } \\
& X = X^{\gamma} \frac{ \partial }{ \partial z^{\gamma} } \end{aligned}$
\[
\begin{gathered}
\begin{aligned}
& f_*V \in T_{f(p)}N \\
& f_*V[h] = V[hf] \\
& f_*V[h\psi^{-1}(y)] = V[hf\varphi^{-1}(x)]
\end{aligned} \quad \quad \, \begin{aligned}
& g_*W \in T_{g(q)}P \\
& g_*W[k] = W[kg] \\
& g_*W[k\chi^{-1}(z)] = W[kg\psi^{-1}(y)]
\end{aligned} \quad \quad \, \begin{aligned}
& gf_*V \in T_{gf(p)}P \\
& gf_*V[l] = V[lgf] \\
& gf_*V[l\chi^{-1}(z)] = V[lgf\varphi^{-1}(x)]
\end{aligned} \\
\end{gathered}
\]
\[
\Longrightarrow g_*(f_*V)[l] = f_* V[lg] = V[lgf] = gf_*V[l]
\]
In coordinates,
\[
\begin{gathered}
g_*(f_*V)[l\chi^{-1}(z) ] = f_*V[lg\psi^{-1}(y)] = W^{\alpha} \frac{ \partial }{ \partial y^{\alpha}}[ lg\psi^{-1}(y)] = V^{\mu} \frac{ \partial y^{\alpha} }{ \partial x^{\mu} } \frac{ \partial }{ \partial y^{\alpha} } [lg\psi^{-1}(y)] = V^{\mu} \frac{ \partial (y^{\alpha}f\varphi^{-1}(x)) }{ \partial x^{\mu} } \frac{ \partial }{ \partial y^{\alpha} } [lg\psi^{-1}(y) ] \\
gf_*V[l \chi^{-1}(z) ] = V[lgf\varphi^{-1}(x) ] = V^{\alpha} \frac{ \partial }{ \partial x^{\alpha} }( lgf\varphi^{-1}(x)) \\
\Longrightarrow \frac{ \partial }{ \partial x^{\mu} }( lgf\varphi^{-1}(x)) = \frac{ \partial (y^{\alpha}f\varphi^{-1}(x)) }{ \partial x^{\mu} } \frac{ \partial }{ \partial y^{\alpha} } [lg\psi^{-1}(y) ]
\end{gathered}
\]
Chain rule is reobtained. \\
\[
\begin{aligned}
& f: M \to N \\
& f^* : T^*_{f(p)}N \to T^*_p M \quad \quad \, \text{ \textbf{ pull back }}
& \langle f^* \omega, V \rangle = \langle \omega, f_* V \rangle \\
& \begin{aligned}
& \omega \in T^*_{f(p)}N \\
& V \in T_p M \end{aligned}
\end{aligned}
\]
tensor of type $(0,r)$ $f^*: \tau^0_{r, f(p)}(N) \to \tau^0_{r,p}(M)$
\hrulefill
20121030
pullback
\[
\begin{gathered}
\begin{aligned}
& f: M \to N \\
& f^* : T^*_{f(p)} N \to T^*_p M \end{aligned} \quad \quad \, \begin{aligned} & X \in T_pM \\
& X = X^{\mu} \frac{ \partial }{ \partial x^{\mu } } \end{aligned} \\
\begin{aligned}
& (U , \varphi ) \subset M \quad \, \varphi = x^{\mu} \\
& ( V, \psi ) \subset N \quad \quad \, \psi = y^{\nu } \end{aligned} \quad \quad \,
\begin{aligned}
& \omega \in T^*_{f(p)} N \\
& \omega = \omega_{\alpha}dy^{\alpha}
\end{aligned}
\end{gathered}
\]
\[
\begin{aligned}
& f^* \omega \in T^*_pM \text{ so } \\
& f^* \omega = \psi_{\beta} dx^{\beta}
\end{aligned}
\]
\[
\begin{gathered}
\langle f^* \omega, X \rangle = \langle \psi_{\beta} dx^{\beta}, X^{\mu} \frac{ \partial }{ \partial x^{\mu }} \rangle = \psi_{\mu} X^{\mu} = \langle \omega, f_* X \rangle = \langle \omega_{\alpha} dy^{\alpha}, Y^{\nu} \frac{ \partial }{ \partial y^{\nu }} \rangle = \\
= \langle \omega_{\alpha} dy^{\alpha}, X^{\mu} \frac{ \partial y^{\nu }}{ \partial x^{\mu} } \frac{ \partial }{ \partial y^{\nu} } \rangle = \omega_{\nu} X^{\mu} \frac{ \partial y^{\nu }}{ \partial x^{\mu }}
\end{gathered}
\]
\[
\boxed{ \psi_{\mu} = \omega_{\nu} \frac{ \partial y^{\nu }}{ \partial x^{\mu }} = \frac{ \partial y^{\nu }}{ \partial x^{\mu }} \omega_{\nu} }
\]
\exercisehead{5.4}
\[
\begin{aligned}
& \omega = \omega_{\alpha} dy^{\alpha} \in T^*_{f(p)}N \\
& f^*\omega = \xi_{\mu} dx^{\mu} \in T_p^* M
\end{aligned}
\quad \quad \, \begin{aligned}
& V \in T_p M \\
& V = V^{\mu} \frac{ \partial }{ \partial x^{\mu}}
\end{aligned}
\]
\[
\begin{gathered}
\langle f^* \omega , V \rangle = \langle \xi_{\mu} dx^{\mu}, V^{\nu} \frac{ \partial }{ \partial x^{\nu}} \rangle = \xi_{\mu} V^{\mu} = \\
= \langle \omega, f_* V \rangle = \langle \omega_{\alpha} dy^{\alpha} , W^{\beta} \frac{ \partial }{ \partial y^{\beta}} \rangle = \langle \omega_{\alpha} dy^{\alpha} , V^{\mu} \frac{ \partial y^{\beta}}{ dx^{\mu} } \frac{ \partial }{ \partial y^{\beta}} \rangle = \omega_{\alpha} V^{\mu} \frac{ \partial y^{\beta}}{ \partial x^{\mu} } \delta^{\alpha}_{ \, \beta} = \omega_{\alpha} V^{\mu} \frac{ \partial y^{ \alpha} }{ \partial x^{\mu} } = \\
\Longrightarrow \boxed{ \xi_{\mu} = \omega_{\alpha} \frac{ \partial y^{\alpha} }{ \partial x^{\mu} } }
\end{gathered}
\]
\exercisehead{5.5} \[
(gf)^*: T^*_{ gf(p)}P \to T^*_p M
\]
\[
\begin{gathered}
\langle (gf)^* \chi, V \rangle = \langle \chi, (gf)_* V \rangle = \langle \chi, g_* f_*V \rangle = \langle g^* \chi, f_* V \rangle = \langle f^* g^* \chi, V \rangle \\
\Longrightarrow (gf)^* = f^* g^*
\end{gathered}
\]
with
\[
\begin{aligned}
& \begin{aligned}
& g^* : T^*_{g(q)}P \to T^*_q N \\
& \langle g^* \omega , W \rangle = \langle \omega, g_* W \rangle
\end{aligned} \quad \,
& \begin{aligned}
& f^* : T^*_{f(p)}N \to T^*_p M \\
& \langle f^* \nu , V \rangle = \langle \nu, f_* V \rangle
\end{aligned}
\end{aligned}
\]
\subsubsection{ Submanifolds}
\begin{definition}[5.3] (Immersion, submanifold, embedding) \\
Let smooth $f: M \to N $ \\
$\text{dim}{M} \leq \text{dim}{N}$
\begin{enumerate}
\item[(a)] immersion $f$ if $f_* : T_pM \to T_{f(p)} N$ injection (1-to-1) i.e.
\[
\text{rank}{f_*} = \text{dim}{M}
\]
\item[(b)] embedding $f$ if $f$ injection and immersion \\
$f(M)$ submanifold of $N$.
\end{enumerate}
\end{definition}
\subsection{ Flow and Lie derivatives }
Let vector field $X$ in $M$. \\
integral curve $x(t)$ of $X$ is a curve in $M$, tangent vector at $x(t)$ is $\left. X \right|_x$
\[
\frac{dx^{\mu}}{ dt} = X^{\mu}(x(t))
\]
where $x^{\mu}$ is $\mu$th component of $\varphi(x(t))$, $X = X^{\mu} \frac{ \partial }{ \partial x^{\mu }}$
Note the abus of notation: $x$ used to denote a pt. in $M$ as well as its coordinates.
Let $\sigma(t,x_0)$ integral curve of $X$ which passes a pt. $x_0$ at $t=0$.
\begin{gather}
\frac{d}{dt} \sigma^{\mu}(t,x_0) = X^{\mu}(\sigma(t,x_0)) \quad \quad \, (5.40a) \\
\sigma^{\mu}(0, x_0) = x_0^{\mu} \quad \quad \, (5.40b) \text{ initial condition }
\end{gather}
$\sigma : \mathbb{R} \times M \to M$. Flow generated by $X \in \mathcal{X}(M)$ s.t.
\[
\sigma(t, \sigma^{\mu}(s,x_0) ) = \sigma( t+ s ,x_0)
\]
The previous was true because of the following. From uniqueness of ODEs.
\[
\begin{gathered}
\begin{gathered}
\frac{d}{dt} \sigma^{\mu}(t, \sigma^{\mu}(s,x_0) ) = X^{\mu}( \sigma(t, \sigma^{\mu}(s,x_0)) ) \\
\sigma^{\mu}(0 , \sigma(s,x_0) ) = \sigma(s,x_0) \end{gathered} \\
\begin{gathered}
\frac{d}{dt} \sigma^{\mu}( t+ s ,x_0) = \frac{d}{d (t+s) } \sigma^{\mu}(t+s , x_0 ) = X^{\mu}(\sigma(t+s, x_0)) \\
\sigma( 0 + s, x_0) = \sigma(s,x_0)
\end{gathered}
\end{gathered}
\]
\begin{theorem}[5.1] $\forall \, x \in M$, $\exists \, $ differentiable $\sigma: \mathbb{R} \times M \to M$ s.t.
\begin{enumerate}
\item[(i)] $\sigma(0,x) = x$
\item[(ii)] $t \mapsto \sigma(t,x)$ solution of (5.40a), (5.40b)
\item[(iii)] $\sigma(t, \sigma^{\mu}(s,x)) = \sigma(t+s,x)$
\end{enumerate}
\end{theorem}
Example 5.9. Let $M = \mathbb{R}^2$, $X((x,y)) = -y \frac{ \partial}{\partial x} + x \frac{ \partial }{ \partial y }$
\[
X((x,y))[f] = -y \frac{ \partial f}{ \partial x} + x \frac{ \partial f}{ \partial y } = \frac{df(\sigma(t)) }{ dt} = \frac{ dx^{\mu}(c(t)) }{ dt} \frac{ \partial f}{ \partial x^{\mu} }
\]
\[
\begin{gathered}
\begin{aligned}
\frac{dx}{dt} & = -y \\
\frac{dy}{dt} & = x
\end{aligned} \quad \quad \, \frac{ d^2y }{dt^2} = -y \Longrightarrow \begin{aligned} y & = Ac(t) = Bs(t) \\ x & = - As(t) + Bc(t) \end{aligned} \quad \quad \, \Longrightarrow \begin{aligned} A & = y \\ B & = x \end{aligned} \\
\begin{aligned}
x & = x \cos{(t) } -y \sin{(t) } \\
y & = x \sin{(t) } + y \cos{(t)}
\end{aligned}
\end{gathered}
\]
\exercisehead{5.7}
\[
\begin{gathered}
\begin{gathered}
X = y \frac{ \partial }{ \partial x } + x \frac{ \partial }{ \partial y} \\
X[f] = y \frac{ \partial f}{ \partial x} + x \frac{ \partial f}{ \partial y} = \frac{ dx^{\mu} }{dt} \frac{ \partial f}{ \partial x^{\mu} }
\end{gathered} \\
\begin{aligned}
\frac{dx}{dt} & = y \\
\frac{dy}{dt} & = x
\end{aligned} \quad \quad \, \begin{aligned}
\ddot{x} & = x \\
\ddot{y} & = y \end{aligned} \quad \quad \, \begin{aligned}
x & = Ae^t + Be^{-t} \\
y & = Ae^t - Be^{-t} \end{aligned}
\end{gathered}
\]
\subsubsection{ One-parameter group of transformations}
For fixed $t\in \mathbb{R}$, flow $\sigma(t,x)$ is a diffeomorphism from $M$ to $M$, $\sigma_t : M \to M$ \\
$\sigma_t$ made into a commutative group by
\begin{enumerate}
\item[(i)] $\sigma_t(\sigma_s(X)) = \sigma_{t+s}(X)$ i.e. $\sigma_t \circ \sigma_s = \sigma_{t+s}$
\item[(ii)] $\sigma_0 = 1$
\item[(iii)] $\sigma_{-t} = (\sigma_t)^{-1}$
\end{enumerate}
Under action $\sigma_{\epsilon}$, infinitesimal $\epsilon$, from (5.40a), (5.40b)
\[
\sigma_{\epsilon}^{\mu}(x) = \sigma^{\mu}(\epsilon, x) = x^{\mu} + \epsilon X^{\mu}(x)
\]
So vector field $X$ is the infinitesimal generator of transformation $\sigma_{\epsilon}$
Given a vector field $X$, corresponding flow $\sigma$ of referred to as exponentiation of $X$
\[
\sigma^{\mu}(t,x) = \exp{ (tX) } x^{\mu} \quad \quad \quad \, (5.43)
\]
Since
\[
\begin{gathered}
\sigma^{\mu}(t,x) = x^{\mu} + t \left. \frac{d}{ds} \sigma^{\mu}(s,x) \right|_{s=0} + \frac{t^2}{2!} \left. \left( \frac{d}{ds} \right)^2 \sigma^{\mu}(s,x) \right|_{s=0} + \dots = \left[ 1 + t \frac{d}{ds} + \frac{t^2}{2!} \left( \frac{d}{ds} \right)^2 + \dots \right] \left. \sigma^{\mu}(s,x) \right|_{s=0} = \\
= \exp{ \left( t \frac{d}{ds} \right) } \left. \sigma^{\mu}(s,x) \right|_{s =0 } \quad \quad \, (5.44)
\end{gathered}
\]
Properties
\begin{enumerate}
\item[(i)] $\sigma(0,x) = x = \exp{ ( 0X) } x $ \quad \quad \quad \, (5.45a)
\item[(ii)] $\frac{ d\sigma (t,x) }{ dt} = X\exp{(tX)}x = \frac{d}{dt} [ \exp{ (tX) } x ]$ \quad \quad \quad \, (5.45b)
\item[(iii)] $\sigma(t,\sigma(s,x) ) = \sigma(t, \exp{(sX)} x) = \exp{ (tX)} \exp{(sX) }x = \exp{ [ (t+s) X ] }x = \sigma(t+s,x)$
\end{enumerate}
\subsubsection{ Lie derivatives }
Let $\sigma(t,x)$, $\tau(t,x)$ be 2 flows generated by vector fields $X,Y$
\[
\begin{aligned}
& \frac{ d\sigma^{\mu}(s,x) }{ ds} = X^{\mu}(\sigma(s,x)) \quad \quad \quad \, (5.46a) \\
& \frac{d\tau^{\mu}(t,x) }{ dt} = Y^{\mu}(\tau(t,x) ) \quad \quad \quad \, (5.46b)
\end{aligned}
\]
evaluate change of $Y$ along $\sigma(s,x)$ \\
Compare $Y$ at $x$ and at $x' = \sigma_{\epsilon}(x)$ nearby \\
But components of $Y$ at 2 pts. belong to different tangent spaces $T_pM, T_{\sigma_{\epsilon}(x) } M$ \\
map $\left. Y \right|_{\sigma_{\epsilon}(x) }$ to $T_xM$ by
\[
( \sigma_{-\epsilon })_* : T_{\sigma_{ \epsilon(x)} } M \to T_x M
\]
Lie derivative of vector field $Y$ along flow $\sigma$ of $X$
\begin{equation}
\mathcal{L}_x Y = \lim_{\epsilon \to 0 } \frac{1}{ \epsilon } \left[ ( \sigma_{-\epsilon })_* \left. Y \right|_{\sigma_{\epsilon(x)} } - \left. Y \right|_x \right] \quad \quad \quad \, (5.47)
\end{equation}
\[
\begin{gathered}
(\sigma_{-\epsilon})_* : T_{\sigma_{ \epsilon }(x) } M \to T_x M \\
((\sigma_{-\epsilon})_* Y ) [g] = Y [ g(\sigma_{-\epsilon} ) ]
\end{gathered}
\]
\exercisehead{5.8 }
\[
\begin{gathered}
\begin{gathered}
(\sigma_{\epsilon})_* : T_{ \sigma_{-\epsilon }(x) } M \to T_x M \\
\mathcal{L}_X Y = \lim_{ \epsilon \to 0 } \frac{1}{ \epsilon } \left[ \left. Y \right|_x - \left. (\sigma_{\epsilon })_* Y \right|_{ \sigma_{ - \epsilon}(x) } \right]
\end{gathered} \\
\begin{gathered}
(\sigma_{\epsilon})_* : T_{ x } M \to T_{\sigma_{\epsilon}(x)} M \\
\mathcal{L}_X Y = \lim_{ \epsilon \to 0 } \frac{1}{ \epsilon } \left[ \left. Y \right|_{\sigma_{\epsilon}(x)} - \left. (\sigma_{\epsilon })_* Y \right|_{ x } \right]
\end{gathered}
\end{gathered}
\]
Let $(U, \varphi)$ be a chart with coordinates $x$ \\
Let $ \begin{aligned} & \quad \\
& X = X^{\mu} \frac{ \partial }{ \partial x^{\mu } } \\
& Y = Y^{\mu} \frac{ \partial }{ \partial x^{\mu} } \end{aligned}$
\[
\begin{gathered}
\sigma_{ \epsilon}(x) = x^{\mu} + \epsilon X^{\mu}(x) \\
\left. Y \right|_{ \sigma_{\epsilon }(x) } = \left. Y^{\mu}(x^{\nu} + \epsilon X^{\nu}(x)) \right|_{x + \epsilon X} \simeq \left. \left[ Y^{\mu}(x) + \epsilon X^{\nu}(x) \partial_{\nu} Y^{\mu }(x) \right] e_{\mu} \right|_{x + \epsilon X}
\end{gathered}
\]
$\lbrace e_{\mu} = \frac{ \partial }{ \partial x^{\mu} } \rbrace$ is the coordinate basis.
Consider Fig. 5.12. Note the need to ``pullback'' vector $Y$ to $x$.
$(\sigma_{-\epsilon})_*$ at $\sigma_{\epsilon}(x)$ to $x$.
Recall that
\[
\begin{aligned}
& \sigma_{-\epsilon} : M \to M \\
& (\sigma_{- \epsilon})_* : T_{ \sigma_{\epsilon}(x) } M \to T_x M \\
& ((\sigma_{-\epsilon})_* Y)[g] = Y[g(\sigma_{-\epsilon }) ]
\end{aligned}
\]
\[
\begin{gathered}
\left. Y \right|_{ \sigma_{\epsilon}(x) } = \left. Y^{\mu} (x^{\nu} + \epsilon X^{\nu}(x) ) \frac{ \partial }{ \partial x^{\mu} } \right|_{x + \epsilon X} \simeq \left. \left[ Y^{\mu}(x) + \epsilon X^{\nu} \partial_{\nu} Y^{\mu}(x) \right] \frac{ \partial }{ \partial x^{\mu} } \right|_{x + \epsilon X } \\
\text{ Now } \left. (\sigma_{-\epsilon})_* Y \right|_{ \sigma_{\epsilon}(x) } = W^{\alpha} \frac{ \partial }{ \partial x^{\alpha }} = Y^{\mu} \frac{ \partial }{ \partial x^{\mu}} \sigma_{-\epsilon}^{\alpha} \frac{ \partial }{ \partial x^{\alpha }} \\
\Longrightarrow W^{\alpha} =\left. Y^{\mu} \right|_{ \sigma_{\epsilon}(x) } \frac{ \partial }{ \partial x^{\mu }} \sigma^{\alpha}_{-\epsilon } = \\
= (Y^{\mu}(x) + \epsilon X^{\nu} \partial_{\nu} Y^{\mu}(x)) \frac{ \partial }{ \partial x^{\mu} } (x^{\alpha} - \epsilon X^{\alpha} ) = [ Y^{\mu}(x) + \epsilon X^{\nu} \partial_{\nu} Y^{\mu}(x)] ( \delta_{\mu}^{ \, \, \alpha} - \epsilon \partial_{\mu} X^{\alpha} ) = \\
= Y^{\alpha}(x) + \epsilon X^{\nu} \partial_{\nu} Y^{\alpha}(x) - \epsilon Y^{\mu} \partial_{\mu} X^{\alpha} + \mathcal{O}(\epsilon^2 )
\end{gathered}
\]
So the first order term is
\[
\epsilon (X^{\nu} \partial_{\nu} Y^{ \alpha}(x) - Y^{\nu} \partial_{\nu} X^{\alpha} )
\]
\exercisehead{5.9}
Given $\begin{aligned}
& \quad \\
& X = X^{\mu} \frac{ \partial }{ \partial x^{\mu }} \\
& Y = Y^{\mu} \frac{ \partial }{ \partial x^{\mu }}\end{aligned}$ \\
Lie bracket $[X,Y]$. $[X,Y]f = X[Y[f]] - Y[X[f]]$.
\[
\begin{gathered}
[X,Y] f = \left[ X^{\nu} \frac{ \partial Y^{\mu}}{ \partial x^{\nu }} - Y^{\nu} \frac{ \partial X^{\mu} }{ \partial x^{\nu }} \right] \frac{ \partial f }{ \partial x^{\mu }} \\
\Longrightarrow \boxed{ [X,Y] = \left[ X^{\mu} \frac{ \partial Y^{\nu }}{ \partial x^{\mu} } - Y^{\mu} \frac{ \partial X^{\nu }}{ \partial x^{\mu} } \right] \frac{ \partial }{ \partial x^{\nu }} }
\end{gathered}
\]
This is the \textbf{local form of the Lie bracket}
$\Longrightarrow \mathcal{L}_XY = [X,Y]$ (5.49b). $[X,Y]$ indeed 1st.-order derivative and indeed a vector field.
\exercisehead{5.10}
\begin{enumerate}
\item[(a)] bilinearity
Want: $\begin{aligned}
& [X, c_1 Y_1 + c_2 Y_2 ] = c_1[X,Y_1] + c_2 [ X,Y_2 ] \\
& [c_1 X_1 + c_2 X_2, Y ] = c_1 [X_1, Y] + c_2 [ X_2, Y ]
\end{aligned}$
\[
\begin{aligned}
[X, c_1 Y_1 + c_2 Y_2 ] & = X^{\mu} \frac{ \partial}{ \partial x^{\mu} } ( c_1 Y_1 + c_2 Y_2)^{\nu} - ( c_1 Y_1 + c_2 Y_2)^{\mu} \frac{ \partial X^{\nu} }{ \partial x^{\mu} } = \\
& = c_1 \left( X^{\mu} \frac{ \partial Y_1^{\nu} }{ \partial x^{\mu} } - Y_1^{\mu} \frac{ \partial X^{\nu }}{ \partial x^{\mu }} \right) + c_2 \left( X^{\mu} \frac{ \partial Y_2^{\nu} }{ \partial x^{\mu} } - Y_2^{\mu} \frac{ \partial X^{\nu }}{ \partial x^{\mu} } \right) = c_1 [ X,Y_1] + c_2[ X,Y_2] \\
[c_1 X_1 + c_2 X_2, Y ] & = (c_1 X_1 + c_2 X_2)^{\mu} \frac{ \partial Y^{\nu}}{ \partial x^{\mu} } - Y^{\mu} \frac{ \partial }{ \partial x^{\mu} }( c_1 X_1 + c_2 X_2)^{\nu} = \\
& = c_1 \left( X_1^{\mu} \frac{ \partial Y^{\nu}}{ \partial x^{\mu} } - Y^{\mu} \frac{ \partial X_1^{\nu}}{ \partial x^{\mu }} \right) + c_2 \left( X_2^{\mu} \frac{ \partial Y^{\nu }}{ \partial x^{\mu }} - Y^{\mu} \frac{ \partial X_2^{\nu }}{ \partial x^{\mu}} \right) = c_1 [X_1, Y] + c_2 [ X_2, Y ]
\end{aligned}
\]
\item[(b)] \[
[Y,X] = Y^{\mu} \frac{ \partial X^{\nu}}{ \partial x^{\mu} } - X^{\mu} \frac{ \partial Y^{\nu }}{ \partial x^{\mu }} = - \left( X^{\mu} \frac{ \partial Y^{\nu }}{ \partial x^{\mu} } - Y^{\mu} \frac{ \partial X^{\nu}}{ \partial x^{\mu }} \right) = - [ X,Y]
\]
\item[(c)] Want:
\[
[[X,Y],Z] + [[Z,X],Y] + [[Y,Z], X] = 0
\]
Now
\[
\begin{gathered}
[[X,Y],Z] \\
[X,Y]^{\mu} \frac{ \partial Z^{\nu}}{ \partial x^{\mu }} - Z^{\mu} \frac{ \partial}{ \partial x^{\mu}} [X,Y]^{\nu} = (X^a \partial_a Y^{\mu} - Y^a \partial_a X^{\mu} ) \partial_{\mu}Z^{\nu} - Z^{\mu} ( \partial_{\mu}X^a \partial_a Y^{\nu} + X^a \partial^2_{\mu a} Y^{\nu} - \partial_{\mu} Y^a\partial_a X^{\nu} - Y^a \partial^2_{\mu a} X^{\nu} ) = \\
= X^a \partial_a Y^{\mu} \partial_{\mu} Z^{\nu} - Y^a \partial_a X^{\mu} \partial_{\mu} Z^{\nu} - Z^{\mu} \partial_{\mu} X^a \partial_a Y^{\nu} -Z^{\mu}X^a \partial^2_{\mu a} Y^{\nu} + Z^{\mu} \partial_{\mu} Y^a \partial_a X^{\nu} + Z^{\mu} Y^a \partial^2_{\mu a} X^{\nu}
\end{gathered}
\]
Likewise,
\[
\begin{gathered}
% \begin{gathered}
[[Z,X],Y]^{\nu} = Z^a \partial_a X^{\mu} \partial_{\mu} Y^{\nu} - X^a \partial_a Z^{\mu} \partial_{\mu} Y^{\nu} - Y^{\mu} \partial_{\mu} Z^a \partial_a X^{\nu} - Y^{\mu} Z^a \partial^2_{\mu a} X^{\nu} + Y^{\mu} \partial_{\mu} X^a \partial_a Z^{\nu} + Y^{\mu} X^a \partial^2_{\mu a} Z^{\nu} \\
%\end{gathered}
[[Y,Z],X]^{\nu} = Y^a \partial_a Z^{\mu} \partial_{\mu} X^{\nu} - Z^a \partial_a Y^{\mu} \partial_{\mu} X^{\nu} - X^{\mu} \partial_{\mu} Y^a \partial_a Z^{\nu} - X^{\mu} Y^a \partial^2_{\mu a} Z^{\nu} + X^{\mu} \partial_{\mu} Z^a \partial_a Y^{\nu} + X^{\mu} Z^a \partial^2_{\mu a} Y^{\nu}
\end{gathered}
\]
All the 18 terms cancel.
\end{enumerate}
\subsection{ Differential forms }
symmetry operation on tensor \quad
$\omega \in \tau^0_{r,p}(M)$
\begin{equation}
P \omega(v_1 \dots v_r) \equiv \omega(v_{P(1)} \dots v_{P(r)} ) \quad \quad \quad (5.59)
\end{equation}
$v_i \in T_p M$, $P \in S_r$, symmetry group of order $r$
\[
\begin{gathered}
\omega(e_{\mu_1} \dots e_{\mu_r} ) = \omega_{\mu_1 \dots \mu_r} \\
P\omega(e_{\mu_1} \dots e_{\mu_r} ) = \omega_{ \mu_{P(1)} \dots \mu_{ P(r)} }
\end{gathered}
\]
symmetrizer $\mathcal{S}$, $\omega \in \tau_{r,p}^0(M)$
\[
S_{\omega} = \frac{1}{r!} \sum_{P \in S_r} P\omega \quad \quad \quad (5.60)
\]
anti-symmetrizer $\mathcal{A}$
\[
\mathcal{A} \omega = \frac{1}{r!} \sum_{ P \in S_r} \text{sgn}{(P)} P\omega
\]
\subsubsection{ Definitions}
\begin{definition}[5.4] diff. form or $r$-form, totally antisymmetric tensor of type $(0,r)$ \\
define wedge product $\wedge$ of $r$ \, 1-forms by the totally antisymm. tensor product
\begin{equation}
dx^{\mu_1} \wedge dx^{\mu_2} \wedge \dots \wedge dx^{\mu_r} = \sum_{P\in S_r} \text{sgn}{(P)} dx^{\mu} \quad \quad \quad (5.62)
\end{equation}
\end{definition}
% \subsubsection{ Definitions }
%\begin{definition}[5.4] differential form of order $r$ or $r$-form is a totally antisymmetric tensor of type $(0,r)$
%\end{definition}
%define
%\begin{equation}
% dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r} = \sum_{ p \in S_r} \text{sgn}{(p)} dx^{\mu_{p(1)}} \wedge \dots \wedge dx^{\mu_{p(r)}} \quad \quad \quad (5.62)
%\end{equation}
e.g.
\[
\begin{gathered}
dx^{\mu} \wedge dx^{\nu} = dx^{\mu} \otimes dx^{\nu} - dx^{\nu} \otimes dx^{\mu} \\
dx^{\lambda} \wedge dx^{\mu} \wedge dx^{\nu} = dx^{\lambda} dx^{\mu} dx^{\nu} + dx^{\nu} dx^{\lambda} dx^{\mu} + dx^{\mu} dx^{\nu} dx^{\lambda} - dx^{\lambda} dx^{\nu} dx^{\mu} - dx^{\nu} dx^{\mu} dx^{\lambda} - dx^{\mu} dx^{\lambda} dx^{\nu}
\end{gathered}
\]
\begin{enumerate}
\item[(i)]
\item[(ii)] \[
\begin{gathered}
dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r} = \text{sgn}{ (P)} dx^{\mu_{P(1)} } \wedge \dots \wedge dx^{\mu_{ P(r)} } \\
\sum_{ Q \in S_r} \text{sgn}{ (Q)} dx^{\mu_{Q(1) } } dx^{\mu_{Q(2) } } \ldots dx^{\mu_{Q(r)}} = \sum_{Q \in S_r} \text{sgn}{ (Q) } ( \text{sgn}(P))^2 dx^{\mu_{Q(P(1)) }} \dots dx^{\mu_{Q(P(r)) } } = \\
= ( \text{sgn}{ P } ) \sum_{Q \in S_r } \text{sgn}{(Q)} \text{sgn}{ P } dx^{\mu_{Q(P(1)) }} \ldots dx^{ \mu_{Q(P(r)) } }
\end{gathered}
\]
\end{enumerate}
vector space of $r$-forms at $p\in M$ by $\Omega_p^r(M)$ \\
set of $r$-forms (5.62) forms basis of $\Omega_p^r(M)$
\[
\omega \in \Omega_p^r(M)
\]
\begin{equation}
\omega = \frac{1}{r!} \omega_{\mu_1 \dots \mu_r} dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r} \quad \quad \quad (5.63)
\end{equation}
$\omega_{\mu_1 \dots \mu_r}$ totally antisymmetric, reflecting antisymmetry of basis
$\binom{m}{r}$ choices of $(\mu_1 \dots \mu_r)$ out of $(1 \dots n)$ in (5.62)
\[
\text{dim}{ \Omega_p^r(M) } = \binom{m}{r}
\]
since $\binom{m}{r} = \binom{m}{m-r}$, $\Omega_p^r(M) \simeq \Omega_p^{m-r}(M)$
Let $\begin{aligned} & \quad \\ & \omega \in \Omega_p^q(M) \\ & \xi \in \Omega_p^r(M) \end{aligned}$
action of $(q+r)$-form $\omega \wedge \xi $ on $q+r$ vectors
\begin{equation}
(\omega \wedge \xi )(v_1 \dots v_{q+r} ) = \frac{1}{q! r!} \sum_{ p \in S_{q+r}} \text{sgn}{(P)} \omega( v_{p(1)} \dots v_{p(q)} ) \xi(v_{p(q+1)} \dots v_{ p(q+r) } ) \quad \quad \quad (5.65)
\end{equation}
with this product, define
\begin{equation}
\Omega_p^*(M) \equiv \bigoplus_{k=0}^m \Omega_p^k(M) \quad \quad \quad (5.66)
\end{equation}
\exercisehead{5.13} \[
\begin{aligned}
& x = r\cos{\theta} \\
& y = r\sin{\theta}
\end{aligned} \quad \quad \quad \begin{aligned} & r = \sqrt{ x^2 + y^2 } \\ & \theta = \arctan{ \left( \frac{y}{x} \right) } \end{aligned}
\]
\[
\begin{gathered}
\begin{aligned}
& \partial_r x = c_{\theta} \\
& \partial_r y = s_{\theta}
\end{aligned} \quad \quad \begin{aligned}
& \partial_{\theta} x = -rs_{\theta} \\
& \partial_{\theta} y = rc_{\theta}
\end{aligned} \quad \quad \begin{aligned}
\partial_x r = \frac{x}{r} \quad \quad \partial_y r = \frac{y}{r} \\
\begin{aligned}
& \partial_x \theta = \frac{ \frac{-y}{x^2} }{ 1 + \frac{y^2}{x^2} } = \frac{-y}{ x^2 + y^2 } \\
& \partial_y \theta = \frac{x}{ x^2 + y^2 }
\end{aligned}
\end{aligned} \\
\begin{aligned}
dx \wedge dy = (c_{\theta} dr + - rs_{\theta} d\theta ) \wedge (s_{\theta}dr + rc_{\theta} d\theta ) = rc_{\theta}^2 dr \wedge d\theta + rs_{\theta}^2 dr \wedge d\theta = rdr \wedge d\theta
\end{aligned}
\end{gathered}
\]
\exercisehead{5.14}
\[
\xi \wedge \xi = (v_1 \wedge \dots \wedge v_q) \wedge (v_1 \wedge \dots \wedge v_q) = (-1)^q v_1 \wedge ( v_1 \wedge \dots \wedge v_q) \wedge ( v_2 \wedge \dots \wedge v_q) = (-1)^{q^2} \xi \wedge \xi = -\xi \wedge \xi \text{ if $q$ odd }
\]
\subsubsection{ Exterior derivatives }
\begin{definition}[5.5] exterior derivatives
\[
\begin{aligned}
& d_r : \Omega^r(M) \to \Omega^{r+1}(M) \\
& \omega = \frac{1}{r!} \omega_{ \mu_1 \dots \mu_r} dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r} \\
& d_r \omega = \frac{1}{r!} \left( \frac{ \partial }{ \partial x^{\nu }} \omega_{\mu_1 \dots \mu_r} \right) dx^{\nu} \wedge dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r} \quad \quad \quad (5.68)
\end{aligned}
\]
\end{definition}
Example 5.10. in 3-dim.
\[
\begin{aligned}
& \omega_0 = f(x,y,z) \\
& \omega_1 = \omega_x(x,y,z) dx + \omega_y(x,y,z)dy + \omega_z(x,y,z) dz \\
& \omega_2 = \omega_{xy}(x,y,z) dx \wedge dy + \omega_{yz}(x,y,z) dy \wedge dz + \omega_{zx}(x,y,z) dz \wedge dx \\
& \omega_3 = \omega_{xyz}(x,y,z) dx \wedge dy \wedge dz
\end{aligned}
\]
Recall
\[
\omega = \frac{1}{r!} \omega_{ \mu_1 \dots \mu_r} dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}
\]
e.g.
\[
\omega_{12} dx^1 \wedge dx^2 + \omega_{13} dx^1 \wedge dx^3 + \omega_{21} dx^2 \wedge dx^1 + \dots = (\omega_{12} - \omega_{21} ) dx^1 \wedge dx^2 + \dots
\]
$\omega_{\mu_1 \dots \mu_r}$ itself must be antisymmetrized.
\[
d\omega = \frac{1}{r!} \left( \frac{ \partial \omega_{ \mu_1 \dots \mu_r} }{ \partial x^{\nu} } \right) dx^{\nu} \wedge dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}
\]
\begin{enumerate}
\item[(i)]
\[
d\omega_0 = \frac{ \partial f}{ \partial x} dx + \frac{ \partial f}{ \partial y} dy + \frac{ \partial f}{ \partial z} dz
\]
\item[(ii)]
\[
\begin{aligned}
d\omega_1 & = \frac{1}{1!} \left( \frac{ \partial \omega_x}{ \partial y} dy \wedge dx + \frac{ \partial \omega_x}{ \partial z} dz \wedge dx + \frac{ \partial \omega_y}{ \partial x} dx \wedge dy + \frac{ \partial \omega_y}{ \partial z} dz \wedge dy + \frac{ \partial \omega_z}{ \partial x} dx \wedge dz + \frac{ \partial \omega_z}{ \partial y} dy \wedge dz \right) = \\
& = \left( \frac{ \partial \omega_y }{ \partial x} - \frac{ \partial \omega_x}{ \partial y} \right) dx \wedge dy + \left( \frac{ \partial \omega_z}{ \partial y} - \frac{ \partial \omega_y}{ \partial z} \right) dy \wedge dz + \left( \frac{ \partial \omega_x}{ \partial z} - \frac{ \partial \omega_z}{ \partial x} \right) dz \wedge dx
\end{aligned}
\]
\item[(iii)]
\[
\begin{aligned}
d\omega_2 & = \frac{1}{2!} \left( \frac{ \partial \omega_{xy} }{ \partial z} dz \wedge dx \wedge dy + \frac{ \partial \omega_{yx} }{ \partial z} dz \wedge dy \wedge dx + \frac{ \partial \omega_{yz} }{ \partial x} dx \wedge dy \wedge dz + \frac{ \partial \omega_{zy} }{ \partial x} dx \wedge dz \wedge dy + \dots \right) = \\
& = \left( \frac{ \partial \omega_{yz} }{ \partial x} + \frac{ \partial \omega_{zx} }{ \partial y} + \frac{ \partial \omega_{xy} }{ \partial z} \right) dx \wedge dy \wedge dz
\end{aligned}
\]
\end{enumerate}
\exercisehead{5.15} 20130929
$\begin{aligned} & \quad \\
& \xi \in \Omega^q(M) \\
& \omega \in \Omega^r(M) \end{aligned}$ \quad \quad $\xi \wedge \omega = \xi_{i_1 \dots i_q} \omega_{j_1 \dots j_r} dx^{i_1} \wedge \dots \wedge dx^{i_q} \wedge dx^{j_1} \wedge \dots \wedge dx^{j_r}$
\[
\begin{gathered}
d(\xi \wedge \omega) = \frac{1}{ (q+r)!} \frac{ \partial ( \xi_{i_1 \dots i_q} \omega_{j_1 \dots j_r} )}{ \partial x^{\nu} }dx^{\nu} \wedge dx^{i_1} \wedge \dots \wedge dx^{i_q} \wedge dx^{j_1} \wedge \dots \wedge dx^{j_r} = \\
= \frac{1}{(q+r)!} ( \partial_{\nu} \xi_{i_1 \dots i_q} \omega_{j_1 \dots j_r} + \xi_{i_1 \dots i_q} \partial_{\nu} \omega_{j_1 \dots j_r} ) dx^{\nu} \wedge dx^{i_1} \wedge \dots = \\
= \frac{1}{ (q+r)!} \lbrace \partial_{\nu} \xi_{i_1 \dots i_q} \omega_{j_1 \dots j_r} dx^{\nu} \wedge dx^{i_1} \wedge \dots \wedge dx^{i_q} \wedge dx^{j_1} \wedge \dots \wedge dx^{j_r} + \\
+ \xi_{i_1 \dots i_q} \partial_{\nu} \omega_{j_1 \dots j_r} dx^{i_1} \wedge \dots \wedge dx^{i_q} \wedge dx^{\nu} \wedge dx^{j_1} \wedge \dots \wedge dx^{j_r} (-1)^q \rbrace
\end{gathered}
\]
\[
\begin{gathered}
d\xi \wedge \omega + (-1)^q \xi \wedge d\omega = \\
= \frac{1}{q!} \frac{ \partial \xi_{i_1 \dots i_q} }{ \partial x^{\nu} } dx^{\nu} \wedge dx^{i_1} \wedge \dots \wedge dx^{i_q} \wedge \omega_{j_1 \dots j_r} dx^{j_1} \wedge \dots \wedge dx^{j_r}
\end{gathered}
\]
decomposable forms
\[
\begin{aligned}
& \xi = f dx_{i_1} \wedge \dots \wedge dx_{i_q} = fdx_I \\
& \omega = gdx_{j_1} \wedge \dots \wedge dx_{j_q} = gdx_I
\end{aligned}
\]
\[
\begin{gathered}
d(\xi \wedge \omega) = d(fdx_I \wedge g dx_J) = d(fg) \wedge dx_2 \wedge dx_J = (fdg = gdf ) \wedge dx_2 \wedge dx_J = \\
= df \wedge dx_I \wedge g dx_J + (-1)^q fdx_I \wedge dg \wedge dx_J = d\xi \wedge \omega + \xi \wedge d\omega
\end{gathered}
\]
$\begin{aligned}
& X = X^{\mu} \partial_{x^{\mu }} \in \mathcal{X}{(M)} \\
& Y = Y^{\nu} \partial_{x^{\nu }}\end{aligned}$ \quad \quad \quad $\omega = \omega_{\mu} dx^{\mu} \in \Omega^1(M)$
\[
\begin{gathered}
X[\omega(Y)] - Y[\omega(X)] - \omega([X,Y]) = \frac{ \partial \omega_{\mu} }{ \partial x^{\nu }} ( X^{\nu} Y^{\mu} - X^{\mu} Y^{\nu} ) \\
[X,Y] = [X^{\mu} \partial_{x^{\mu}} Y^{\nu} - Y^{\mu} \partial_{x^{\mu}} X^{\nu} ] \partial_{x^{\nu} } \\
\omega([X,Y]) = \omega_{\nu} ( X^{\mu} \partial_{\mu} Y^{\nu} - Y^{\mu} \partial_{\mu} X^{\nu} ) \\
\begin{aligned}
& X[\omega(Y)] = X^{\mu} \partial_{\mu} ( \omega_{\nu} Y^{\nu} ) = X^{\mu} \partial_{\mu} \omega_{\nu} Y^{\nu} + X^{\mu} \omega_{\nu} \partial_{\mu} Y^{\nu} \\
& Y[\omega(X)] = Y^{\mu} \partial_{\mu} (\omega_{\nu} X^{\nu} ) = Y^{\mu} \partial_{\mu} \omega_{\nu} X^{\nu} + Y^{\mu} \omega_{\nu} \partial_{\mu} X^{\nu}
\end{aligned} \\
\Longrightarrow X[\omega(Y)] - Y[\omega(X) ] - \omega([X,Y]) = X^{\mu} \partial_{\mu} \omega_{\nu} Y^{\nu} - Y^{\mu} \partial_{\mu} \omega_{\nu} X^{\nu} = \partial_{\mu} \omega_{\nu} (X^{\mu} Y^{\nu} - Y^{\mu} X^{\nu })
\end{gathered}
\]
Suppose
\[
d\omega(X_1 \dots X_{p+1}) = \sum_{i=1}^r (-1)^{i+1} X_i \omega(X_1 \dots \widehat{X}_i \dots X_{i+1}) + \sum_{ i < j } (-1)^{i+1} \omega([X_i, X_j], X_1 \dots \widehat{X}_i \dots \widehat{X}_j \dots X_{i+1}) \quad \quad \quad (5.71)
\]
for $r$-form $\omega \in \Omega^r(M)$
\[
\begin{gathered}
\omega = \frac{1}{r!} \omega_{\mu_1 \dots \mu_r} dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r} \\
d\omega = \frac{1}{r!} \left( \frac{ \partial}{ \partial x^{\nu} }\omega_{\mu_1 \dots \mu_r} \right) dx^{\nu} \wedge dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}
\end{gathered}
\]
\subsubsection{ Interior product and Lie derivative of forms }
interior product $i_X : \Omega^r(M) \to \Omega^{r-1}(M)$ where $X \in \chi(M)$ \\
Let $\omega \in \Omega^r(M)$, define
\begin{equation}
i_X \omega(X_1 \dots X_{r-1}) \equiv \omega(X,X_1 \dots X_{r-1} ) \quad \quad \quad \, (5.78)
\end{equation}
with
\[
\begin{aligned}
& X = X^{\mu} \frac{ \partial}{ \partial x^{ \mu} } \\
& \omega = \frac{1}{r!} \omega_{\mu_1 \dots \mu_r} dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}
\end{aligned}
\]
\begin{equation}
i_X \omega = \frac{1}{ (r-1)!} X^{\nu} \omega_{\nu \mu_2 \dots \mu_r} dx^{\mu_2} \wedge \dots \wedge dx^{\mu_r} = \frac{1}{r!} \sum_{s=1}^r X^{\mu_s} \omega_{ \mu_1 \dots \mu_s \dots \mu_r }(-1)^{s-1} dx^{\mu_1} \wedge \dots \wedge \widehat{ dx^{\mu_s} } \wedge \dots \wedge dx^{\mu_r} \quad \quad \quad \, (5.79)
\end{equation}
cf. wikipedia
$i_X: \Omega^p(M) \to \Omega^{p-1}(M)$ \quad \quad \, $i_X$ a map that sends a $p$-form $\omega$ to the $(p-1)$ form $i_X \omega$
\[
(i_X \omega)(X_1 \dots X_{p-1}) = \omega(X,X_1 \dots X_{p-1} )
\]
$\alpha$ 1-form $i_X\alpha = \alpha(X)$
for $\begin{aligned} & \quad \\
& \beta \in \Omega^p(M) \\
& \gamma \in \Omega^q(M) \end{aligned}$
\[
i_X(\beta \wedge \gamma) = (i_X \beta) \wedge \gamma + (-1)^p \beta \wedge (i_X \gamma )
\]
\[
i_X \omega = \frac{1}{ (p-1)!} X^i \omega_{ii_2 \dots i_p} dx^{i_2} \wedge \dots \wedge dx^{i_p} = \frac{1}{ p!} \sum_{s=1}^p X^{i_s} \omega_{ i_1 \dots i_s \dots i_p }(-1)^{s-1} dx^{i_1} \wedge \dots \wedge \widehat{dx^{i_s}} \wedge \dots \wedge dx^{i_r}
\]
\[
\begin{aligned}
& i_{\partial_x}(dx \wedge dy) = \frac{1}{ (2-1)!} dy = dy \\
& i_{\partial_x}(dy \wedge dz) = 0 \\
& i_{\partial_x}( dz \wedge dx) = \delta^i_{ \, \, 1 } \epsilon_{31} dx^3 = -dz
\end{aligned}
\]
%%%%%%%% interior product on 1-form
Let $\omega$ 1-form. $i_X \omega = \omega(X)$
\[
\begin{gathered}
(di_X + i_X d)\omega = d(X^{\mu} \omega_{\mu} ) + i_X [ \frac{1}{2} (\partial_{\mu} \omega_{\nu} - \partial_{\nu} \omega_{\mu} ) dx^{\mu} \wedge dx^{\nu} ] = \\
= (\omega_{\mu} \partial_{\nu} X^{\mu} + X^{\mu} \partial_{\nu} \omega_{\mu} ) dx^{\nu} + \frac{1}{2} ( X^{\mu} \partial_{\mu} \omega_{\nu} dx^{\nu} - X^{\nu} \partial_{\nu} \omega_{\mu} dx^{\mu} )
\end{gathered}
\]
using, recall,
\[
i_X \omega = \frac{1}{ (p-1)!} X^i \omega_{i i_2 \dots i_p } dx^{i_2} \wedge \dots \wedge dx^{i_p}
\]
Recall (5.55): $\mathcal{L}_X\omega = (X^{\nu} \partial_{\nu} \omega_{\mu} + \partial_{\mu} X^{\nu} \omega_{\nu} ) dx^{\mu}$
\begin{equation}
\mathcal{L}_X \omega = (di_X + i_X d) \omega \quad \quad \quad \, (5.80)
\end{equation}
for $r$-form, $\omega = \frac{1}{r!} \omega_{\mu_1 \dots \mu_r} dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r}$
\begin{equation}
\mathcal{L}_X\omega = \lim_{ \epsilon \to 0} \frac{1}{\epsilon} ( (\sigma_{\epsilon})^* \left. \omega \right|_{ \sigma_{\epsilon}(X) } - \left. \omega \right|_X ) = X^{\nu} \frac{1}{r!} \partial_{\nu} \omega_{ \mu_1 \dots \mu_r} dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r} + \sum_{s=1}^r \partial_{\mu_s} X^{\nu} \frac{1}{r!} \omega_{ \mu_1 \dots (s \to \nu ) \dots \mu_r} dx^{\mu_1} \wedge \dots \wedge dx^{\mu_r} \quad \quad \, (5.81)
\end{equation}
%%%%%%% Example 5.12 % Hamiltonian phase space
Ex.5.12. $(q^{\mu}, p_{\mu})$ tangent bundle!!! \\
symplectic 2 form
\begin{equation}
\omega = dp_{\mu} \wedge dq^{\mu} \quad \quad \quad \, (5.88)
\end{equation}
1-form $\theta = q^{\mu} dp_{\mu}$ \quad \quad \, or (???) $\theta = p_{\mu} dq^{\mu}$
\[
\omega = d\theta
\]
Given $f(q,p)$ in phase space
define Hamiltonian vector field
\[
X_f = \frac{ \partial f}{ \partial p_{\mu}} \frac{ \partial }{ \partial q^{\mu} } - \frac{ \partial f}{ \partial q^{\mu} } \frac{ \partial }{ \partial p_{\mu} } \quad \quad \, (5.91)
\]
\[
i_{X_f} \omega = \frac{ - \partial f}{ \partial p_{\mu}} dp_{\mu} - \frac{ \partial f}{ \partial q^{\mu} } dq^{\mu} = -df
\]
\[
i_X \omega = \frac{1}{ (p-1)!} X^i \omega_{ ii_2 \dots i_p} dx^{i_2} \wedge \dots \wedge dx^{i_p}
\]
Consider vector field generated by Hamiltonian
\begin{equation}
X_H = \frac{ \partial H }{ \partial p_{\mu}} \frac{ \partial }{ \partial q^{\mu} } - \frac{ \partial H}{ \partial q^{\mu} } \frac{ \partial }{ \partial p_{\mu} } \quad \quad \, (5.92)
\end{equation}
Hamilton's eqns. of motion.
\[
\begin{aligned}
& \dot{q}^{\mu} = \frac{ \partial H}{ \partial p_{\mu} } \\
& \dot{p}_{\mu} = \frac{ - \partial H}{ \partial q^{\mu }}
\end{aligned} \quad \quad \, (5.93) \quad \quad \text{(Hamilton's eqn. of motion)}
\]
\begin{equation}
X_H = \dot{q}^{\mu} \frac{ \partial }{ \partial q^{\mu} } + \dot{p}_{\mu} \frac{ \partial }{ \partial p_{\mu} } = \frac{d}{dt} \quad \quad \quad (5.94)
\end{equation}
symplectic 2-form $\omega$ left-invariant along flow generated by $X_H$
\[
\mathcal{L}_{X_H}\omega = d(i_{X_H} \omega ) + i_{X_H}(d\omega) = d(i_{X_H} \omega ) = - d^2 H = 0
\]
used $(di_X + i_X d) \omega = \mathcal{L}_X \omega \quad \quad (5.82)$
Conversely, if $X$ satisfies $\mathcal{L}_X \omega=0$, $\exists \, $ Hamiltonian $H$ s.t. Hamilton's eqn. of motion is satisfied along the flow generated by $X$
from $\mathcal{L}_X \omega = d(i_X \omega ) = 0$ and hence by Poincar\'{e}'s lemma, $\exists \, H(q,p)$ s.t.
\[
i_X \omega = -dH
\]
\subsection{ Integration of differential forms }
\subsubsection{ Orientation }
integration of differential form over manifold $M$ defined only when $M$ is ``orientable''
Let $M$ connected $m$-dim. differential manifold \\
$\forall \, p \in M$, $T_pM$ spanned by basis $\lbrace \frac{ \partial }{ \partial x^{\mu} } \rbrace$, $x^{\mu}$ local coordinate on chart $U_i \ni p$ \\
Let $U_j$ another chart s.t. $U_i \cap U_j \neq \emptyset$ with local cordinates $y^{\alpha}$. \\
If $p \in U_i \cap U_j$, $T_pM$ spanned by either $\lbrace e_{\mu} \rbrace = \lbrace \frac{ \partial }{ \partial x^{\mu} }$, or $\lbrace \widetilde{e}_{\alpha} \rbrace = \lbrace \frac{ \partial }{ \partial y^{\alpha} } \rbrace$
\begin{equation}
\frac{ \partial }{ \partial y^{\alpha} } = \frac{ \partial x^{\mu} }{ \partial y^{\alpha}} \frac{ \partial }{ \partial x^{\mu} } \quad \quad \, (5.97)
\end{equation}
If $J = \text{det}{ \left( \frac{ \partial x^{\mu} }{ \partial y^{\alpha}} \right)} >0$ on $U_i \cap U_j$, $\lbrace e_{\mu} \rbrace$, $\lbrace \widetilde{e}_{\alpha} \rbrace$ define same orientation on $U_i \cap U_j$
If $J <0$, opposite orientation.
\begin{definition}[5.6]
$M$ connected manifold covered by $\lbrace U_i \rbrace$ \\
manifold $M$ orientable if $\forall \, $ overlapping charts $U_i, \, U_j$, $\exists \, $ local coordinates $\begin{aligned} & \quad \\
& \lbrace x^{\mu} \rbrace \text{ for } U_i \\
& \lbrace y^{\alpha} \rbrace \text{ for } U_j \end{aligned}$ s.t. $J = \text{det}{ \left( \frac{ \partial x^{\mu} }{ \partial y^{\alpha}} \right) } > 0$
\end{definition}
If $M$ nonorientable, $J$ can't be positive in all interactions of charts.
If $m$-dim. $M$ orientable, $\exists \, m$-form $\omega$ s.t. $\omega\neq 0$ \\
This $m$-form $\omega$ is volume element \\
2 vol. elements $\omega, \omega'$ equivalent if $\exists \, $ strictly positive $h \in \mathcal{F}(M)$ s.t. $\omega = h\omega'$ \\
take $m$-form
\begin{equation}
\omega = h(p) dx^1 \wedge \dots \wedge dx^m \quad \quad \quad \, (5.98)
\end{equation}
with positive-definite $h(p)$ on chart $(U,\varphi)$, $x = \varphi(p)$
If $M$ orientable, extend $\omega$ throughout $M$ s.t. component $h$ positive definite on chart $U_i$ \\
If $M$ orientable, $\omega$ vol. element.
Let $p\in U_i \cap U_j \neq \emptyset$, $\begin{aligned} & \quad \\
& x^{\mu} \text{ coordinates of $U_i$ } \\
& y^{\alpha} \text{ coordinates of $U_j$ } \end{aligned}$
\[
\begin{gathered}
\omega = h(p) \frac{ \partial x^1}{ \partial y^{\mu_1} } dy^{\mu_1} \wedge \dots \wedge \frac{ \partial x^m}{ \partial y^{\mu_m} }dy^{\mu_m} = h(p) \text{det}{ \left( \frac{ \partial x^{\mu} }{ \partial y^{\nu} } \right) } dy^1 \wedge \dots \wedge dy^m \\
\text{det}{ \left( \frac{ \partial x^i }{ \partial y^j } \right) } = \sum_{ \sigma \in S_n} \text{sgn}{ (\sigma)} \frac{ \partial x^1}{ \partial y^{\sigma_1} } \dots \frac{ \partial x^m}{ \partial y^{\sigma_m}} = \sum^m_{ i_1\dots i_m = 1} \epsilon_{i_1 \dots i_m} \frac{ \partial x^1}{ \partial y^{i_1} } \dots \frac{ \partial x^m}{ \partial y^{i_m} } = \epsilon_{i_1 \dots i_m } \frac{ \partial x^1}{ \partial y^{i_1} } \dots \frac{ \partial x^m}{ \partial y^{i_m}} \\
\frac{ \partial x^1}{ \partial y^{i_1}} \dots \frac{ \partial x^m}{ \partial y^{i_m} } dy^{i_1} \wedge \dots \wedge dy^{i_m} = \frac{ \partial x^1}{ \partial y^{i_1}} \dots \frac{ \partial x^m}{ \partial y^{i_m}} \epsilon^{i_1 \dots i_m} dy^1 \wedge \dots \wedge dy^{i_m} = \text{det}{ \left( \frac{ \partial x^i}{ \partial y^j} \right) } dy^1 \wedge \dots \wedge dy^{i_m}
\end{gathered}
\]
\subsubsection{ Integration of forms }
integration of function $f:M \to \mathbb{R}$ over oriented $M$ \\
take vol. element $\omega$ \\
in coordinate neighborhood $U_i$, \, $x=\varphi(p)$, $p\in U_i$
\begin{equation}
\int_{U_i} f\omega \equiv \int_{\varphi{ (U_i)} } f(\varphi_i^{-1}(x)) h(\varphi_i^{-1}(x)) dx^1 \dots dx^m \quad \quad \, (5.100)
\end{equation}
\begin{definition}[5.7]
open covering $\lbrace U_i \rbrace$ of $M$ s.t. $\forall \, p \in M$, $p$ covered by a finite number of $U_i$. $M$ paracompact.
\end{definition}
If diff. $\epsilon_i(p)$ s.t. \begin{enumerate}
\item[(i)] $0\leq \epsilon_i(p) \leq 1$
\item[(ii)] $\epsilon_i(p) =0 $ if $p \notin U_i$
\item[(iii)] $\epsilon_1(p) + \epsilon_2(p) + \dots = 1$ \quad \, $\forall p \in M$
\end{enumerate}
$\lbrace \epsilon(p) \rbrace$ partition of unity subordinate to covering $\lbrace U_i \rbrace$
from (iii),
\begin{equation}
f(p) = \sum_i f(p) \epsilon_i(p) = \sum_i f_i(p) \quad \quad \quad \, (5.101)
\end{equation}
$f_i(p) \equiv f(p) \epsilon_i(p)$, \, $f_i(p) =0$ \, $\forall \, p \notin U_i$
Hence, $\forall \, p \in M$, paracompactness ensures in (5.101), $f_i(p) < \infty $, finite, in sum $\sum_i$
define
\begin{equation}
\int_M f \omega \equiv \sum_i \int_{U_i} f_i \omega \quad \quad \quad \, (5.102)
\end{equation}
Although a different atlas $\lbrace (V_i, \psi_i)\rbrace$ gives different coordinates and different partition of unity, integral defined by (5.102) same.
Ex. 5.13. $S^1$. $\begin{aligned} & \quad \\ & U_1 = S^1 - \lbrace (1,0) \rbrace \\ & U_2 = S^1 - \lbrace (-1,0) \rbrace \end{aligned}$ \quad \quad \, $\begin{aligned} & \quad \\ & \epsilon_1(\theta) = \sin^2{ \left( \frac{ \theta}{2} \right) } \\ \epsilon_2(\theta) = \cos^2{ \left( \frac{\theta}{2} \right) } \end{aligned}$ \quad \quad \, $\epsilon_1 + \epsilon_2 = 1 $ on $S^1$
$f= \cos^2{\theta}$
\[
\begin{aligned}
& \int_0^{2\pi} d\theta \cos^2{\theta} = \pi \\
& \int_S^1 d\theta \cos^2{\theta} = \int_0^{2\pi } d\theta \sin^2{\frac{ \theta}{2} } \cos^2{\theta} + \int_{-\pi}^{\pi} d\theta \cos^2{ \frac{ \theta}{2} } \cos^2{\theta} = \frac{ \pi}{2} + \frac{\pi}{2} = \pi
\end{aligned}
\]
\subsection{ Lie groups and Lie algebras }
\subsubsection{ Lie groups }
Take $x,y,z \in \mathbb{R} -0$ s.t. $xy=z$ \quad \quad $xy=z$ \quad \, $\frac{ \partial z}{ \partial x } = y \neq 0$
\exercisehead{5.19}\begin{enumerate}
\item[(a)] $\mathbb{R}^+ = \lbrace x \in \mathbb{R} | x > 0 \rbrace$ \quad \quad \, $\frac{ \partial }{ \partial x} x^{-1} = -x^{-2} \neq 0 $ \, diff.
\item[(b)]
\[
\begin{gathered}
\partial_x z = \partial_x ( x+y) = 1 \\
\partial_x( x^{-1} ) = \partial_x (-x) = -1
\end{gathered}
\] \quad diff.
\item[(c)] \[
\begin{gathered}
(a,b) + (x,y) = (a+x, b+y) \quad \quad \, Dg =Dg(x) = \left[ \begin{matrix} a & \\ & b \end{matrix} \right] \\
(x,y)^{-1} = (-x, -y) \quad \quad \, D(x,y)^{-1} = \left[ \begin{matrix} -1 & \\ & -1 \end{matrix} \right]
\end{gathered}
\]
\end{enumerate}
Lorentz group
\[
O(1,3) = \lbrace M \in GL(4,\mathbb{R}) | M \eta M^T = \eta \rbrace \quad \quad \eta = \text{diag}{ ( -1, 1, 1, 1 ) }
\]
\exercisehead{5.20} 20130801
\[
\text{det}{ M \eta M^T } = (\text{det}{M})^2 \text{det}{\eta} = \text{det}{ \eta} \quad \quad \, (\text{det}{M} )^2 = 1 \quad \quad \, \text{det}{M} = \pm 1
\]
if $UMU^T = \text{diag}{ ( \lambda_1, \lambda_2, \lambda_3, \lambda_5) } = \lambda_1 \lambda_2 \lambda_3 \lambda_4 = (\text{det}{U})^2 \text{det}{M} = (\pm 1) (\text{det}{U} )^2$
$(0,0)$ entry of $M\eta M^T$
\[
-m_0^2 + m_1^2 + m_2^2 + m_3^2 = 1
\]$M$ unbounded so $O(1,3)$ noncompact
\begin{theorem}{5.2} Every closed subgroup $M$ of a Lie group $G$ is a Lie subgroup
\end{theorem}
e.g. $O(n)$, $SL(n,\mathbb{R})$, $SO(n)$ Lie subgroups of $GL(n,\mathbb{R})$
$SL(n,\mathbb{R})$ closed subgroup, consider $ \begin{aligned} & f: GL(n, \mathbb{R}) \to \mathbb{R} \\
& A \mapsto \text{det}{A} \end{aligned}$
$f$ cont. $\lbrace 1 \rbrace$ closed $f^{-1}(1) = SL(n,\mathbb{R})$ so $SL(n,\mathbb{R})$ closed. By Thm. 5.2., $SL(n, \mathbb{R}) $ Lie subgroup.
Let $G$ Lie group. \\
\phantom{Let } $H$ Lie subgroup.
Define $g\sim g'$ if $\exists \, h \in H$ s.t. $g' = gh$
\[
[g] = \lbrace gh | h \in G \rbrace
\]
coset space $G /H$ is a manifold (not necessarily a Lie group)
if $H$ normal subgroup of $G$, i.e. $ghg^{-1} \in H$, \, $\forall \, \begin{aligned} & \\
& g \in G \\
& h \in H \end{aligned}$, then $G/H$ Lie groupp \\
take $[g], [g'] \in G/H$ \\
Let $gh, g'h'$ representative of $[g], [g']$, resp.
\[
\begin{gathered}
[g][g'] = gh g'h' = gg' h'' h' \in [gg'] \\
g^{-1} h \\
[g][g^{-1} ] = gh g^{-1} h' = gg^{-1} h'' h' = e h'' h' = h'' h' \in [e]
\end{gathered}
\]
\subsubsection{ Lie algebras }
left-translation \\
$L_g:G \to G$ \\
$L_gh = gh = x^{ik}{(g)} x^{kj}{(h)} = x^{ij}{(gh)}$ \\
$L_eh = eh = x^{ik}{ (e)} x^{kj}{ (h)} = \delta^{ik} x^{kj}{(h)} = x^{ij}{(h)} = 1h = h$ \\
$L_ge = ge = x^{ik}{(g)}x^{kj}{(e)} = x^{ik}{(g)} \delta^{kj} = x^{ik}{(g)} = g1 = g $ \\
$L_{g*} : T_hG \to T_{gh}G$
Pushforward? \\
Recall local coordinate form of pushforward
\[
\begin{aligned}
& X \equiv X_h \in T_h G \\
& X_h = X_h^{ij} \left. \frac{ \partial }{ \partial x^{ij} } \right|_h
\end{aligned}
\]
\[
L_{g*}X_h \equiv L_{g*} X = X_h^{kl} \frac{ \partial x^{ij}{ (gh) } }{ \partial x^{kl}{(h)} } \left. \frac{ \partial }{ \partial x^{ij} } \right|_{gh} = X^{ij}_{gh} \left. \frac{ \partial }{ \partial x^{ij} } \right|_{gh}
\]
Note that $x^{ik}{ (g)} x^{kj}{ (h)} = x^{ij}{ (gh)}$
\[
\Longrightarrow \frac{ \partial x^{ij}{(gh)} }{ \partial x^{kl}{ (h)} } = x^{im}{(g)} \delta^{km} \delta^{jl} = x^{ik}{ (g)} \delta^{jl} = \begin{cases} 0 & \text{ if } j\neq l \\ x^{ik} & \text{ if } j =l \end{cases}
\]
so
\[
\begin{gathered}
L_{g*} X_h = X_h^{kl} \frac{ \partial x^{ij}{(gh)} }{ \partial x^{kl}{(h)} } \left. \frac{ \partial }{ \partial x^{ij} } \right|_{gh} = X^{kl}_h x^{ik}{(g)} \delta^{jl} \left. \frac{ \partial }{ \partial x^{ij} } \right|_{gh} = X^{kj}_h x^{ik}{(g)} \left. \frac{ \partial }{ \partial x^{ij}} \right|_{gh} = x^{ik}{(g)} X^{kj}_h \left. \frac{ \partial }{ \partial x^{ij}} \right|_{gh}
\end{gathered}
\]
santiy check: $L_{e*}X_h = eX_h = X_h$
Consider left-invariant vector fields \\
\quad $L_{g*} X = X$ \, $\forall \, g$ (cf. wikipedia) \\
\quad $L_{g*} \left. X \right|_h = X_{gh}$ (cf. Nakahara) \\
\exercisehead{5.21}
\begin{equation}
L_{a^*} \left. X \right|_g = X^{\mu}(g) \frac{ \partial x^{\nu}(ag) }{ \partial x^{\mu}(g) } \left. \frac{ \partial }{ \partial x^{\nu} } \right|_{ag} = x^{\nu}(ag) \left. \frac{ \partial }{ \partial x^{\nu }} \right|_{ag} \quad \quad \, (5.110)
\end{equation}
\[
\begin{gathered}
L_{a*} \left. X \right|_g = \left. X \right|_{ag}
\end{gathered} \quad \quad \quad \begin{gathered}
L_a g = ag \\
\begin{aligned}
y^i & = y^i(x^j ) = a_{ij} x^j \\
\partial_j y^i & = a_{ij}
\end{aligned}
\end{gathered}
\]
\[
d(y(x)) X^{\mu}(g) \left. \frac{ \partial }{ \partial x^{\mu} } \right|_g = X^{\mu}(g) \left. \frac{ \partial }{ \partial x^{\mu} } \right|_g y^{\nu}
\]
Recall that
\[
\begin{aligned}
& V = V^{\mu} \frac{ \partial }{ \partial x^{\mu}} \\
& f_* V = W^{\alpha} \frac{ \partial }{ \partial y^{\alpha }}
\end{aligned} \quad \quad \, W^{\alpha} = V^{\mu} \frac{ \partial y^{\alpha }}{ \partial x^{\mu}}
\]
\[
L_{a*} \left. X \right|_g = X^{\mu}(g) \frac{ \partial y^{\nu}(ag) }{ \partial x^{\mu} } \left. \frac{ \partial }{ \partial y^{\nu} } \right|_{ag} = \left. X \right|_{ag} = Y^{\nu} \left. \frac{ \partial }{ \partial y^{\nu}} \right|_{ag}
\]
$V \in T_e G$ defines unique left invariant vector field $X_V$
\begin{equation}
\left. X_V \right|_g = L_{g*} V \quad \quad g \in G \quad \quad \quad (5.111)
\end{equation}
$\mathbb{g} \equiv $ set of left invariant vector fields on $G$ \\
$T_eG \to \mathbb{g}$ is an isomorphism \\
$V \mapsto X_V$ \\
$\mathbb{g} \subset \chi(G)$ \\
Lie Bracket (Sec. 5.3) also defined on $\mathbb{g}$
\[
\begin{gathered}
g, ag = L_ag \in G \\
X,Y \in \mathbb{g}
\end{gathered}
\]
\begin{equation}
\left. L_{a*}[ X,Y] \right|_g = [ L_{a*} \left. X \right|_g, L_{a*} \left. Y \right|_g ] = \left. [X,Y] \right|_{ag} \quad \quad \quad (5.112)
\end{equation}
so $[X,Y] \in \mathbb{g}$ \\
e.g. $GL(n,\mathbb{R})$ coordinates given by $n^2$ entries $x^{ij}$ of the matrix.
\[
\begin{aligned}
& g = \lbrace x^{ij}(g) \rbrace
& a = \lbrace x^{ij}(a) \rbrace
\end{aligned} \, \in GL(n, \mathbb{R}) \quad \quad \, L_a g = ag = x^{ik}(a) x^{kj}(g)
\]
take $V = V^{ij} \left. \frac{ \partial }{ \partial x^{ij} } \right|_e \in T_eG$
\begin{equation}
\begin{aligned}
\left. X_V \right|_g & = L_{g^*} V = V^{ij} \left. \frac{ \partial }{ \partial x^{ij} } \right|_e x^{kl}(g) x^{lm}(e) \left. \frac{ \partial }{ \partial x^{km}} \right|_g = v^{ij} x^{kl}(g) \delta_i^l \delta^m_j \left. \frac{ \partial }{ \partial x^{km}} \right|_g = V^{ij} x^{ki}(g) \left. \frac{ \partial }{ \partial x^{kj} } \right|_g = \\
& = x^{ki}(g) V^{ij} \left. \frac{ \partial }{ \partial x^{kj} } \right|_g = (gV)^{kj} \left. \frac{ \partial }{ \partial x^{kj} } \right|_g \quad \quad \quad (5.113)
\end{aligned}
\end{equation}
\[
\begin{aligned}
& V = V^{ij} \left. \frac{ \partial }{ \partial x^{ij} } \right|_e \\
& W = W^{ij} \left. \frac{ \partial }{ \partial x^{ij} } \right|_e
\end{aligned} \quad \quad \quad
\]
\[
\left. [X_V, X_W ] \right|_g = x^{ki}(g) V^{ij} \left. \frac{ \partial }{ \partial x^{kj} } \right|_g x^{ca}(g) W^{ab} \left. \frac{ \partial }{ \partial x^{cb} } \right|_g - (V \leftrightarrow W ) =
\]
\begin{equation}
= x^{ij}(g) [ V^{jk} W^{kl} - W^{jk} V^{kl} ] \left. \frac{ \partial }{ \partial x^{il} } \right|_g = ( g[V,W[)^{ij} \left. \frac{ \partial }{ \partial x^{ij} } \right|_g \quad \quad \quad (5.114)
\end{equation}
\begin{equation}
\Longrightarrow L_{g*}V = gV \quad \quad \quad (5.115)
\end{equation}
\begin{equation}
\left. [ X_V, X_W ] \right|_g = L_{g*}[V,W] = g[V,W] \quad \quad \quad (5.116)
\end{equation}
\begin{definition}[5.11]
$\mathbb{g} \equiv $ set of left-invariant vector fields with Lie bracket $[ \, , \, ] : \mathbb{g} \times \mathbb{g} \to \mathbb{g}$ \quad \, Lie algebra of Lie group $G$
\end{definition}
\hrulefill
20141117 EY recap: Recapping,
\[
\begin{gathered}
\begin{aligned}
& L_a : G \to G \\
& L_a g = ag \end{aligned} \quad \quad \, \begin{aligned}
& R_a : G \to G \\
& R_a g = ga \end{aligned}
\end{gathered}
\]
$L_a$, $R_a$ are diffeomorphisms; indeed $L_{a^{-1}} = (L_a)^{-1}$, $R_{a^{-1}} = (R_a)^{-1}$ \quad \, $\forall \, a \in G$
$\forall \, X_g = T_gG$, \\
locally $X_g = X_g^i \frac{ \partial }{ \partial g^i }$ \\
$L_{a*} : T_gG \to T_{ag}G$ \\
locally, $L_{a*} X_g = X^i_g \frac{ \partial (ag)^j}{ \partial g^i} \frac{ \partial}{ \partial (ag)^j}$ (because that's what pushforwards do)
Note the abuse of notation above.
\textbf{if} $X_g$ is \textbf{left-invariant},
\[
L_{a*} X_g = X_{ag}
\]
this implies
\[
X^i_g \frac{ \partial (ag)^j}{ \partial g^i} = X^j_{ag}
\]
Now isomorphisms can be shown so that
\[
T_1G = \mathfrak{g} = \lbrace X | X \in TG, \, \forall \, a,g \in G, L_{a*} X_g = X_{ag} \rbrace \text{ i.e. set of left-invariant vector fields}
\]
indeed, for instance, $\forall \, V \in T_1G$, locally $V = V^i \left. \frac{ \partial }{ \partial x^i} \right|_1$,
\[
\begin{gathered}
L_{g*} V = X \\
L_{a*}X = L_{a*} L_{g*} V = L_{ag*} V = X_{ag}
\end{gathered}
\]
uniqueness can be shown in either cases, note.
\emph{Specialize} to the case of $G = GL(n)$
\[
\begin{aligned}
& (L_ag)^{ij} = a^{ik} g^{kj} \\
& X_g = X^{ij}_g \frac{ \partial }{ \partial g^{ij} } \\
& L_{a*} X_g = X^{ij}_g \frac{ \partial (ag)^{kl} }{ \partial g^{ij}} \frac{ \partial }{ \partial (ag)^{kl} } = X^{ij}_g a^{ki} \frac{ \partial }{ \partial (ag)^{kj} } = (aX_g)^{kj} \frac{ \partial }{ \partial (ag)^{kj}}
\end{aligned}
\]
where I used the following calculation:
\[
(ag)^{kl} = a^{km} g^{ml} \Longrightarrow \frac{ \partial (ag)^{kl}}{ \partial g^{ij}} = a^{ki} \delta^{jl}
\]
If $X_g$ left invariant,
\[
aX_g=X_{ag}
\]
\hrulefill
e.g. $\mathbf{so}(n)$ Lie algebra of $SO(n)$
e.g. 5.15
\begin{enumerate}
\item[(a)] $G = \mathbb{R}$ \quad \quad \, define $La : x \mapsto x + a$, \quad \, left invariance field $X = \frac{ \partial }{ \partial x }$
\[
L_{a*} \left. X \right|_x = \frac{ \partial (a+x) }{ \partial x} \frac{ \partial }{ \partial (a+ x) } = \frac{ \partial }{ \partial (x+a) } = \left. X \right|_{x+a}
\]
$X = \frac{ \partial }{ \partial \theta}$ \quad unique left vector field on $G = SO(2) = \lbrace e^{i \theta} | 0 \leq \theta \leq 2 \pi \rbrace$
\[
\frac{ \partial ( \phi + \theta ) }{ \partial \theta } \frac{ \partial }{ \partial (\phi + \theta ) } = \frac{ \partial }{ \partial ( \phi + \theta ) }
\]
\item[(b)] Let $\mathbf{gl}{ (n, \mathbb{R})}$ \quad curves $\begin{aligned} & \quad \\
& c : (-\epsilon , \epsilon ) \\
& c(0 ) = 1 \end{aligned}$ $ \, \to GL(n, \mathbb{R})$ \quad \quad $c(s) = 1 + sA + \mathcal{O}(s^2)$ \quad near $s=0$, $A$ \, $n\times n$ matrix of real entries
\end{enumerate}
\subsubsection{The one-parameter subgroup }
\subsubsection{Frames and structure equation }
$\lbrace V_1 \dots V_n \rbrace$ basis of $T_1G = \mathfrak{g}$
\begin{equation}
[X_a, X_b] = c_{ab}^c X_c \quad \quad \quad \, (5.133)
\end{equation}
From local form of the Lie bracket, cf. Exercise 5.9
\[
[X,Y] = \left( X^{\mu} \frac{ \partial Y^{\nu}}{ \partial x^{\mu }} - Y^{\mu} \frac{ \partial X^{\nu}}{ \partial x^{\mu} } \right) \frac{ \partial }{ \partial x^{\nu}}
\]
\[
[V_a,V_b] = \left( V_a^i \frac{ \partial V_b^j}{ \partial x^i} - V_b^i \frac{ \partial V_a^j}{ \partial x^i} \right) \frac{ \partial }{ \partial x^j} = c^c_{ab}V_c = c^c_{ab} X_c^j \frac{ \partial }{ \partial x^j}
\]
dual basis to $\lbrace X_a \rbrace$, $\lbrace \theta^a \rbrace$ s.t. $\langle \theta^a, X_b \rangle = \delta^a_{ \, \, b}$ \\
dual basis satisfies \textbf{Maurer-Cartan's structure equation}
\[
d\theta^a(X_b,X_c) = X_b \delta^a_{ \, \, c} - X_c \delta^a_{ \, \, b} - \theta^a([X_b,X_c]) = X_b\delta^a_{ \, \, c} - X_c \delta^a_{ \, \, b} - \theta^a(c^d_{bc} X_d ) = X_b \delta^a_{ \, \, c} - X_c \delta^a_{ \, \, b} - c^a_{bc} = -c^a_{bc}
\]
where I used $d\omega(X,Y) = X\omega(Y) - Y\omega(X) - \omega([X,Y])$, cf. wikipedia ``exterior derivative'', etc.
\begin{equation}
d\theta^a = \frac{-1}{2}c^a_{bc} \theta^b \wedge \theta^c \quad \quad \quad \, (5.136)
\end{equation}
%canonical 1-form, \textbf{Maurer-Cartan form on $G$ }
%\begin{equation}
%\begin{aligned}
% & \theta : T_gG \to T_e G \\
% & \theta : X \mapsto (L_{g^{-1}})_* X = (L_g)^{-1}_* X, \quad \, X \in \, T_gG
%\end{aligned} \quad \quad \quad \, (5.137)
%\end{equation}
define Lie algebra valued 1-form $\theta: T_gG \to T_1G$, \textbf{canonical 1-form} or \textbf{Maurer-Cartan form} on $G$ \\
\[
\theta \in \Omega^1(G;\mathfrak{g})
\]
\begin{equation}
\theta : X \mapsto (L_{g^{-1}})_* X = (L_g)^{-1}_* X \text{ where } X \in T_gG \quad \quad \quad \, (5.137)
\end{equation}
\begin{theorem}[5.3]
\begin{enumerate}
\item[(a)] canonical 1-form is $\theta = V_a \otimes \theta^a$, where $\lbrace V_a \rbrace$ basis of $T_1G = \mathfrak{g}$, $\lbrace \theta^a \rbrace$ dual basis of $T^*_gG$
EY : 20141117 here's where Nakahara has a mistake, $\theta$ isn't at $e$ but at $g$
\item[(b)] where \\
$d\theta = V_a\otimes d\theta^a$ and
\begin{equation}
[ \theta \wedge \theta ] \equiv [ V_{a} , V_{b} ] \otimes \theta^{a} \wedge \theta^{b} \quad \quad \quad \, (5.140)
\end{equation}
\end{enumerate}
\end{theorem}
\begin{proof}
\begin{enumerate}
\item[(a)] $\forall \, Y = Y^a X_a \in T_gG$
\[
\theta(Y) = (L_{g^{-1}})_* Y = (L_g)^{-1}_* Y = Y^a(L_{g^{-1}})_* X_a = Y^a (L_{g^{-1}})_* (L_g)_* V_a = Y^a V_a
\]
On the other hand
\[
(V_a \otimes \theta^a)(Y) = V_a \otimes \theta^a(Y^b X_b) = V_a (Y^b(\theta^a(X_b))) = Y^a V_a
\]
EY : 20141117 note that Nakahara, I believe, made a mistake with thinking $\theta^a$ is a dual basis at $e$, not $g$
Thus
\[
\boxed{ \theta = V_a \otimes \theta^a }
\]
\item[(b)] Now $[V_a,V_b] = c^c_{ab}V_c$
\[
\begin{aligned}
& d\theta + \frac{1}{2} [ \theta \wedge \theta ] = 0 \\
& \frac{1}{2} [ \theta \wedge \theta ] = \frac{1}{2} [V_a,V_b] \otimes \theta^a \wedge \theta^b = \frac{1}{2} V_c \otimes c^c_{ab} \theta^a \wedge \theta^b \\
& d\theta = V_c \otimes d\theta^c = V_c \otimes \frac{-1}{2} c^c_{ab} \theta^a \wedge \theta^b
\end{aligned}
\]
\[
\Longrightarrow \boxed{ d\theta + \frac{1}{2} [ \theta \wedge \theta ] = 0 }
\]
\end{enumerate}
\end{proof}
\hrulefill
20141117 EY's recap:
$c^{c}_{ab}$ independent of $g \in G$
$\forall \, g\in G$, $\forall \, \theta^a$ in dual basis for $T_g^*G$ (also $\mathfrak{g}^*$), $\theta^a \in T_g^*G$, \\
$\forall \, d\theta^a \in \Omega^2_g(G)$, then
\[
d\theta^a = \frac{-1}{2} c^a_{bc} \theta^b \wedge \theta^c
\]
is satisfied, \textbf{Maurer-Cartan's structure equation}.
\hrulefill
\subsection{ The action of Lie groups on manifolds }
\subsubsection{ Definitions}
\subsubsection{ Orbits and isotropy groups }
\subsubsection{ Induced vector fields }
\subsubsection{ The adjoint representation }
\section{de Rham Cohomology Groups}
$r$-form $\omega$ in $\mathbb{R}^r$
\[
\omega = a(x) dx^1 \wedge dx^2 \wedge \dots \wedge dx^r
\]
define integration of $\omega$ over $\overline{\sigma}_r$
\begin{equation}
\int_{ \overline{\sigma}_r } \omega \equiv \int_{ \overline{\sigma}_r } a(x) dx^1 dx^2 \dots dx^r \quad \quad \quad \, (6.2)
\end{equation}
\[
\int_{ \overline{\sigma}_2} \omega = \int_{ \overline{\sigma}_2 } dx dy = \int_0^1 dx \int_0^{1-x} dy = \frac{1}{2}
\]
\[
\int_0^1 dx \int_0^{1-x}dy \int_0^{1-y-x} dz = \int_0^1 dx \int_0^{1-x} dy (1-y-x) = \frac{1}{6}
\]
Let smooth $f:\sigma_r \to M$
$s_r = f(\sigma_r) \subset M$ (singular) $r$-simplex in $M$
define integration of $r$-form $\omega$ over $r$-chain in $M$
\begin{equation}
\int_{s_r} \omega = \int_{ \overline{\sigma}_r } f^* \omega \quad \quad \quad \, (6.6)
\end{equation}
general $r$-chain $c = \sum_i a_i s_{r,i} \in C_r(M)$
\begin{equation}
\int_c \omega = \sum_i a_i \int_{s_{r,i} } \omega \quad \quad \quad \, (6.7)
\end{equation}
\subsection{ Stokes' theorem }
\begin{theorem}[Stokes' thm.]
$\begin{aligned}
& \omega \in \Omega^{r-1}(M) \\
& c\in C_r(M)
\end{aligned}$
then
\begin{equation}
\int_c d\omega = \int_{\partial c} \omega \quad \quad \quad \, (6.8)
\end{equation}
\end{theorem}
\begin{proof}
$c$ linear combination of $r$-simplexes \\
suffices to prove (6.8) for $r$-simplex $s_r$ in $M$
Let $f:\overline{\sigma}_r \to M$ s.t. $f(\overline{\sigma}_r) = s_r$
\[
\int_{s_r} d\omega = \int_{ \overline{\sigma}_r} f^*(d\omega) = \int_{\overline{\sigma}_r} d(f^* \omega)
\]
using (5.75)
Also we have
\[
\int_{\partial s_r} \omega = \int_{ \partial \overline{\sigma}_r } f^* \omega
\]
\end{proof}
\subsubsection{ Preliminary consideration }
\subsubsection{ Stokes' theorem }
\subsection{ de Rham cohomology groups }
\section{Riemannian Geometry}
\subsection{ Riemannian manifolds and pseudo-Riemannian manifolds }
\subsubsection{ Metric tensor }
\begin{definition}[7.1]
\begin{enumerate}
\item[(i)] $ g_p(U,V) = g_p(V,U) $
\end{enumerate}
\end{definition}
Since $g \in \tau_2^0(M)$ (2 covariant indices, type (0,2) tensor)
Recall from Ch. 5, 5.2.3, 1-forms,
$df \in T_p^* M$ on $V \in T_p M$ defined.
\[
\langle df, V \rangle = V[f] = V^{\mu} \frac{ \partial f}{ \partial x^{\mu} } \in \mathbb{R}
\]
If $\exists \, $ metric $g$
\[
g_p : T_p M \otimes T_p M \to \