Product manifold: If $M_{1}=\{\left(U_{i},\varphi_{i}\right)\}$ and $M_{2}=\left\{(U_{j},\varphi_{j})\right\}$, $M_{1}\times M_{2}=\left\{(U_{i}\times U_{j}),(\varphi_{i},\psi_{j})\right\}$, and has coordinate function $(p\in M,q\in N)\to(\varphi_{i}(p),\psi_{j}(q))\in \mathbb{R}^{m+n}$

5.2.2 Vectors

The cotangent space $T_{p}^{*}M=\{\omega \mid \omega: T_{p}M\to \mathbb{R}\}$. The simplest case is $df$ for $f\in \mathcal{F}(M)$, with $df(V)=V(f)=V^{\mu} \frac{ \partial f }{ \partial x^{\mu} }\in \mathbb{R}$. Tensors $T=\otimes ^{q}T_{p}^{*}M \otimes ^{r} T_{p}M$.

5.2.6 Induced maps

5.3 Flows and Lie derivatives

The integral curve $\sigma(t,x_{0})$ of a vector $X$ which passes a point $x_{0}$ at $t=0$ is a flow generated by $X$, if $\frac{d\sigma ^{\mu}(t,x_{0})}{dt}= X^{\mu}(\sigma(t,x_{0}))$ with initial condition $\sigma ^{\mu}(0,x_{0})=x_{0}^{\mu}$. The flow $\sigma(t,x)$ is a diffeomorphism from $M$ to $M$ for fixed $t\in \mathbb{R}$, and is a commutative group: $\sigma_{t}\circ \sigma_{s}(x)=\sigma_{t+s}(x)$, $\sigma_{0}=I$, $\sigma_{-t}=\sigma_{t}^{-1}$, so for infinitesimal $\epsilon$, $\sigma_{\epsilon}^{\mu}(x)=\sigma ^{\mu}(\epsilon,x)=x^{\mu}+\epsilon X^{\mu}(x)$

Note that $\sigma_{-\epsilon}$ generates a map $(\sigma_{-\epsilon})_{*}:T_{\sigma_{\epsilon}(x)}M\to T_{x}M$, so we can move $Y|_{\sigma_{\epsilon}(x)}$ to $Y|_{x}$ and differentiate them: the Lie derivative is defined as $\mathcal{L}_{X}Y= \lim_{ \epsilon \to 0 } \frac{1}{\epsilon}[(\sigma_{-\epsilon})_{*}Y|_{\sigma_{\epsilon}(x)}-Y|_{x}]$, where $\sigma$ is the flow of $X$.

$$\begin{equation} \begin{aligned} Y|_{\sigma_{\epsilon}(x)} & = Y^{\mu}(x^{\nu}+\epsilon X^{\nu}(x)) e_{\mu} | _{x+\epsilon X} \\ & \simeq [Y^{\mu}(x)+\epsilon X^{\nu}(x) \partial_{\nu} Y^{\mu}(x)] e_{\mu}|_{x+\epsilon X} \\ \end{aligned} \end{equation}$$

where $e_{\mu}= \frac{ \partial }{ \partial x^{\mu} }$. If we map this vector defined at $\sigma_{\epsilon}(x)$ to $x$ by $(\sigma_{-\epsilon})_{*}$, we obtain

$$\begin{align} & [Y^{\mu}(x) + \epsilon X^{\lambda }(x) \partial_{\lambda}Y^{\mu}(x)] \partial_{\mu}[x^{\nu}-\epsilon X^{\nu}(x)]e_{\nu}|_{x} \\ & = [Y^{\mu}(x) + \epsilon X^{\lambda}(x)\partial_{\lambda}Y^{\mu}(x)] [ \delta ^{\nu}_{\mu} - \epsilon \partial_{\mu}X^{\nu}(x)]e_{\nu}|_{x} \\ & =Y ^{\mu}(x) e_{\mu}|_{x} + \epsilon [X^{\mu}(x)\partial_{\mu}Y^{\nu}(x)-Y^{\mu}(x)\partial_{\mu}X^{\nu}(x)]e_{\nu}|_{x} + O(\epsilon ^{2}) \end{align}$$

So $\mathcal{L}_{X}{Y}=(X^{\mu}\partial_{\mu}Y^{\nu}-Y^{\mu}\partial_{\mu}X^{\nu}) e_{\nu}$. define $[X,Y]$ by $[X,Y]f=X[Y[f]]-Y[X[f]]$, so $\mathcal{L}_{X}Y=[X,Y]$

Geometrically, the Lie bracket shows the non-commutativity of two flows. Consider two flow $\sigma(s,x),\tau(t,x)$ generated by vector fields $X,Y$, if we first move $\epsilon$ along flow $\sigma$ then move $\delta$ along $\tau$, we get the coordinate

$$\begin{aligned} \tau^\mu(\delta, \sigma(\varepsilon, x)) & \simeq \tau^\mu\left(\delta, x^\nu+\varepsilon X^\nu(x)\right) \\ & \simeq x^\mu+\varepsilon X^\mu(x)+\delta Y^\mu\left(x^\nu+\varepsilon X^\nu(x)\right) \\ & \simeq x^\mu+\varepsilon X^\mu(x)+\delta Y^\mu(x)+\varepsilon \delta X^\nu(x) \partial_\nu Y^\nu(x) . \end{aligned}$$

If we do this reversed,

$$\begin{aligned} \sigma^\mu(\varepsilon, \tau(\delta, x)) & \simeq \sigma^\mu\left(\varepsilon, x^\nu+\delta Y^\nu(x)\right) \\ & \simeq x^\mu+\delta Y^\mu(x)+\varepsilon X^\mu\left(x^\nu+\delta Y^\nu(x)\right) \\ & \simeq x^\mu+\delta Y^\mu(x)+\varepsilon X^\mu(x)+\varepsilon \delta Y^\nu(x) \partial_\nu X^\mu(x) . \end{aligned}$$

The difference is

$$\tau ^{\mu}(\delta,\sigma(\epsilon,x))-\sigma ^{\mu}(\epsilon,\tau(\delta,x)) = \epsilon\delta[X,Y]^{\mu}$$

We may define Lie derivative of a one-form $\omega\in\Omega ^{1}(M)$ along $X\in \mathcal{X}(M)$ by

$$\begin{equation} \mathcal{L}_{X}\omega \equiv \lim_{ \epsilon \to 0 } \frac{1}{\epsilon}[(\sigma_{\epsilon})^{*}\omega|_{\sigma_{\epsilon}(x)}-\omega|_{x}] \end{equation}$$

where $\omega|_{x}\in T_{x}^{*}M$ is $\omega$ at $x$. Put $\omega=\omega_{\mu}dx^{\mu}$, Repeating a similar analysis as before,

$$\begin{equation} (\sigma_{\epsilon})^{*}\omega|_{\sigma_{\epsilon}(x)} = \omega_{\mu}(x)dx^{\mu} + \epsilon[X^{\nu}(x)\partial_{\nu }\omega_{\mu}(x)+\partial_{\mu}X^{\nu}(x)\omega_{\nu}(x)]dx^{\mu} \end{equation}$$

which leads to

$$\begin{equation} \mathcal{L}_{X}\omega = (X^{\nu}\partial_{\nu}\omega_{\mu}+\partial_{\mu}X^{\nu}\omega_{\nu})dx^{\mu} \end{equation}$$

Clearly this belongs to $T_{x}^{*}(M)$. The Lie derivative of $f\in \mathcal{F}(M)$ along flow $\sigma_{s}$ generated by vector field $X$ is

$$\begin{equation} \mathcal{L}_{X}f \equiv \lim_{ \epsilon \to 0 } \frac{1}{\epsilon} [f(\sigma_{\epsilon}(x))-f(x)] = X^{\mu}(x) \frac{ \partial f }{ \partial x^{\mu} } =X[f] \end{equation}$$

5.4 Differential forms

5.4.3 Interior product, Lie derivative of forms

For $X=X^{\mu}\frac{ \partial }{ \partial x^{\mu} }$ and $\omega= (\frac{1}{r!}) \omega_{\mu_{1}\dots \mu_{r}}dx^{\mu_{1}}\wedge\dots\wedge dx^{\mu_{r}}$ we have

$$\begin{equation} i_{X}\omega = \frac{1}{(r-1)!} X^{\nu}\omega_{\nu \mu_{2}\dots \mu_{r}} dx^{\mu_{2}}\wedge\dots \wedge dx^{\mu_{r}} \end{equation}$$

For example, let $(x,y,z)$ be coordinates of $\mathbb{R}^{3}$. Then

$$\begin{equation} i_{e_{x}}(dx\wedge dy) = dy,i_{e_{x}}(dy\wedge dz)=0, i_{e_{x}}(dz\wedge dx)=-dz \end{equation}$$

The Lie derivative of a form is most neatly written with interior product: Let $\omega=\omega_{\mu}dx^{\mu}$ be a one-form,

$$\begin{equation} \begin{aligned} (di_{X}+i_{X}d)\omega & = d(X^{\mu}\omega_{\mu}) + i_{X}\left[ \frac{1}{2}(\partial_{\mu}\omega_{\nu}-\partial_{\nu}\omega_{\mu})dx^{\mu}\wedge dx^{\nu} \right] \\ & = (\omega_{\mu}\partial_{\nu}X^{\mu}+X^{\mu}\partial_{\nu}\omega_{\mu})dx^{\nu } + X^{\mu}(\partial_{\mu}\omega_{\nu}-\partial_{\nu}\omega_{\mu})dx^{\nu} \\ & = (\omega_{\mu}\partial_{\nu}X^{\mu}+X^{\mu}\partial_{\mu}\omega_{\nu})dx^{\nu} \end{aligned} \end{equation}$$

We find that $\mathcal{L}_{X}\omega=(di_{X}+i_{X}d)\omega$. This is also true for a general $r$-form.

5.5. Integration of differential forms

For an orientable manifold, take a volume element $\omega$. Function $f:M\to \mathbb{R}$ has integration over neiborhood $U_{i}$ defined by an $m$-form $f\omega$:

$$\begin{equation} \int _{U_{i}}f\omega = \int _{\varphi(U_{i})} f(\varphi_{i}^{-1}(x))h(\varphi_{i}^{-1}(x)) \, dx^{1}\dots dx^{m} \end{equation}$$

Once the integral of $f$ over $U_{i}$ is defined, the integral over whole $M$ is given with the help of the partition of unity defined now:

It follows that $f(p)=\sum_{i}f(p)\epsilon_{i}(p)=\sum_{i}f_{i}(p)$, then we maydefine the integral by

$$\begin{equation} \int _{M}f \, \omega = \sum_{i}\int _{U_{i}} f_{i}\omega \end{equation}$$

e.g. atlas of $S^{1}$ is $U_{1}=S^{1}-\{(1,0)\}$ and $U_{2}=S^{1}-\{\left(-1,0\right)\}$. $\epsilon_{1}(\theta)=\sin ^{2} \frac{\theta}{2}$ and $\epsilon_{2}(\theta) = \cos ^{2} \frac{\theta}{2}$ is a partition of unity subordinate to $\{U_{i}\}$. Let us integrate a function $f=\cos ^{2}\theta$.

$$\begin{equation} \int _{S^{1}} \, d\theta \cos ^{2}\theta = \int _{0}^{2\pi} \, d\theta \sin ^{2} \frac{\theta}{2} \cos ^{2}\theta + \int _{-\pi}^{\pi} \, d\theta \cos ^{2} \frac{\theta}{2} \cos ^{2}\theta = \pi \end{equation}$$

We have left $h$ arbitrary provided it is strictly positive so far. $h$ is multiplied by Jacobian under change of coordinates, there is no canonical way to single out the component $h$; $h=1$ in one coordinate might not imply $h=1$ in other. The situation changes if the manifold is endowed with a metric.

5.6. Lie groups and Lie algebra

5.6.2 Lie algebras

The left-invariant vector field satisfies

$$\begin{equation} L_{a^{*}}X|_{g} =\left. X^{\mu}(g) \frac{ \partial x^{\nu}(ag) }{ \partial x^{\mu}(g) } \frac{ \partial }{ \partial x^{\nu} }\right |_{ag} = \left. X^{\nu}(ag) \frac{ \partial }{ \partial x^{\nu} } \right|_{ag} \end{equation}$$

A vector $V\in T_{e}G$ defines a unique left-invariant vector field $X_{V}$ throughout $G$ by

$$\begin{equation} X_{V}|_{g} =L_{g^{*}}V,g\in G \end{equation}$$

because $X_{V}|_{ag}=L_{ag^{*}}V=(L_{a}L_{g})_{*}V=L_{a *}L_{g *}V=L_{a *}X_{V}|_{g}$. Conversely a left-invariant vector field $X$ defines a unique vector $V=X|_{e}\in T_{e}G$. Denote the set of left-invariant vector fields $\mathfrak{g}$.

We show that $\mathfrak{g}$ is closed under Lie bracket: Take two points $g,ag=L_{a}g$, apply $L_{a *}$ to Lie bracket $[X,Y]$ of $X,Y\in \mathfrak{g}$, we have

$$\begin{equation} L_{a *} [X,Y]|_{g} = [L_{a *}X|_{g},L_{a *}Y|_{g}] = [X,Y]|_{ag} \end{equation}$$

where left invariances of $X,Y$ have been used. thus $[X,Y]\in \mathfrak{g}$.

5.6.3 The one-parameter subgroup

5.6.4 Frames and structure equation

Let the set of $n$ vectors $\{V_{1},V_{2},\dots,V_{n}\}$ be a basis of $T_{e}G$ where $n=\text{dim}G$. The basis defines the set of $n$ linearly independent left-invariant vector fields $\{X_{1},\dots,X_{n}\}$ at each point $G$ defined by $X_{\mu}|_{g}=L_{g *}V_{\mu}$. Since $[X_{\mu},X_{\nu}]|_{g}$ is again an element of $g$, it can be expanded in terms of $\{X_{\mu}\}$ as

$$\begin{equation} [X_{\mu},X_{\nu}]=c_{\mu \nu}{}^{\lambda}X_{\lambda} \end{equation}$$

where $c_{\mu \nu}{}^{\lambda}$ are structure constants of Lie group $G$. If $G$ is matrix group, the LHS at $g=e$ is precisely the commutator of matrices $V_{\mu}$ and $V_{\nu}$; We may show that structure constants are indeed constants independent of $g$. Acting $L_{g *}$ to Lie bracket,

$$\begin{equation} [X_{\mu},X_{\nu}]|_{g} = c_{\mu \nu}{}^{\lambda} (e) X_{\lambda}|_{g} \end{equation}$$

which shows the $g$-independence. The structure constants determine a Lie group completely in a sense.

Introduce dual basis $\{\theta ^{\mu}\}$: $\langle\theta ^{\mu},X_{\nu}\rangle=\delta ^{\mu}_{\nu}$. We show that the dual basis satisfies Maurer-Cartan's structure equation,

$$\begin{equation} d\theta ^{\mu}= -\frac{1}{2}c_{\nu\lambda}^{\mu}\theta ^{\nu}\wedge\theta ^{\lambda} \end{equation}$$

This could be seen by:

$$\begin{equation} \begin{aligned} d\theta ^{\mu}(X_{\nu},X_{\lambda}) & =X_{\nu}[\theta ^{\mu}(X_{\lambda})]-X_{\lambda}[\theta ^{\mu}(X_{\nu})]-\theta ^{\mu}([X_{\nu},X_{\lambda}]) \\ & = X_{\nu}[\delta ^{\mu}_{\lambda}] - X_{\lambda}[\delta ^{\mu}_{\nu}] - \theta ^{\mu}(c_{\nu\lambda}^{\kappa}X_{\kappa}) \\ & =-c_{\nu\lambda}^{\mu} \end{aligned} \end{equation}$$

Define the canonical one-form or Maurer-Cartan form on $G$ $\theta:T_{g}G\to T_{e}G$ by

$$\begin{equation} \theta:X\mapsto (L_{g^{-1} *})X = (L_{g })^{-1}_{*} X,\qquad X\in T_{g}G \end{equation}$$

5.7 The action of Lie groups on manifolds

Lie groups often appears as transformations acting on a manifold, e.g. $SO(3)$ is the rotations in $\mathbb{R}^{3}$, Poincare group is the set of transformations acting on Minkowski spacetime.

examples: $SL(2,\mathbb{C})$ can be mapped to $O(1,3)$: $A\in SL(2,\mathbb{C}),x\in M_{4},Ax:=AX(x)A^{\dagger}=X(Ox),O\in O(1,3)$, where $x^{\mu}=\frac{1}{2}\mathrm{Tr}(\sigma_{\mu}X)$, $\sigma_{\mu}=(I,X,Y,Z)$. Note $A$ and $-A$ gives the same element of $O(1,3)$.

$$\begin{equation} A= \exp\left[-i \frac{\theta}{2}(n\cdot\sigma)\right]= \cos \frac{\theta}{2} I_{2} -i (n\cdot\sigma)\sin \frac{\theta}{2} \end{equation}$$

is mapped to rotation about $n$ by angle $\theta$, i.e. $O(3)$ subgroup of $O(1,3)$. $A(\theta+2\pi)=-A(\theta)$ in $SL(2,\mathbb{C})$ but they map to same element in $O(3)$. The boost along $\hat{n}$ with $v=\tanh \alpha$ is

$$\begin{equation} A=\exp\left[\frac{\alpha}{2}(n\cdot\sigma)\right] \end{equation}$$

In fact, this map is an one-to-one correspondence from $SL(2,\mathbb{C})$ to $O^{\uparrow}_{+}(1,3)=\{\Lambda\in O(1,3)|\det\Lambda=1,\Lambda_{00}>0\}$

5.7.2 Orbits, isotropy groups

The orbit of $p$ under action $\sigma$ is defined by

$$\begin{equation} Gp= \left\{\sigma(g,p)|g\in G\right\} \end{equation}$$

If the action is transitive the orbit of any $p\in M$ is $M$.

Let $G$ be a Lie group acting on $M$ transitively and $H(p)$ be an isotropy group of $p\in M$. $H(p)$ is a Lie subgroup and the coset space $G / H(p)$ is a homogeneous space. If they satisfy certain requirements, $G/H(p)$ is homeomorphic to $M$.