Limit Points, Boundary Points, and Sequential Limits

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In this lecture, we study the topological nature of some familiar notions from analysis such as limit points and limits of sequences.

Throughout this lecture, we assume that the nonempty set $S$ is a topological space.

Definition. A point $x\in S$ is called a limit point of $A\subset S$ if for any open set $U$ containing $x$, $(U-\{x\})\cap A\ne\emptyset$. The set of limiting points of $A$ is called the derived set of $A$ and is denoted by $A’$.

Exercise. If $A\subset B\subset S$, then show that $A’\subset B’$.

Definition. Let $S$ be a space. $x\in S$ is called a boundary point of $A\subset S$ if for any open set $U$ containing $x$, $U\cap A\ne\emptyset$ and $U\cap(S\setminus A)\ne\emptyset$. The set of boundary points of $A$ is called the boundary of $A$ and is denoted by $B(A)$.

In Lecture 1, we studied the notion of the closure of a set. The following theorem relates the closure, limit points and boundary points of a set.

Theorem. Let $A\subset S$. Then
$$\bar A=A\cup A’=A\cup B(A).$$

Proof. First we show that $\bar A=A\cup B(A)$. Clearly, $A\cup B(A)\subset\bar A$. If $x\not\in\bar A$. Then there exists an open set $U$ containing $x$ such that $U\cap A=\emptyset$. This implies that $x\not\in A$ and $x\not\in B(A)$.

Now we show that $\bar A=A\cup A’$. Clearly, $A\cup A’\subset\bar A$. If $x\not\in A\cup A’$, then there exists an open set $U$ containing $x$, $(U-\{x\})\cap A=\emptyset$. Since $x\not\in A$, $U\cap A=\emptyset$. Thus, $x\not\in\bar A$.

Note that $A\cup A’=A\cup B(A)$ does not necessarily mean that $A’\subset B(A)$ or $B(A)\subset A’$ as shown in the following example.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$ and let $A=(0,1)\cup\{2\}$. Then $\bar A=[0,1]\cup\{2\}$, $A’=[0,1]$, and $B(A)=\{0,1,2\}$.

Definition. Let $A,B\subset S$. $A$ is dense in $B$ if $B\subset\bar A$. We say $A$ is dense in $S$ if $\bar A=S$.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$. Let $A=(a,b)$ and $B=[a,b]$. Then $A$ is dense in $B$.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$. Since $\bar{\mathbb{Q}}=\mathbb{R}$, $\mathbb{Q}$ is dense in $\mathbb{R}$.

Definition. $A\subset S$ is said to be nowhere dense in $S$ if $\bar A$ contains no member of $\tau\setminus\{\emptyset\}$.

Example. Consider the Euclidean space $(\mathbb{R},\xi)$. Since $\bar{\mathbb{Z}} =\mathbb{Z}$ contains no open interval, $\mathbb{Z}$ is nowhere dense in $\mathbb{R}$.

Exercise. If $p\geq 2$ is an integer, a $p$-adic rational is a real number $r=\frac{k}{p^n}$ for some nonnegative integer $k$ and positive integer $n$. Show that the set of $p$-adic rationals in $I=[0,1]$ is dense in $I$.

Definition. $A\subset S$ is said to be perfect if $A$ is closed and $A\subset A’$.

Exercise. The Cantor ternary set $K$ is the set of all $x\in [0,1]$ having a ternary expansion $x=\frac{t_1}{3}+\frac{t_2}{3^2}+\cdots+\frac{t_n}{3^n}+\cdots$ with $t_n\ne 1$, for all $n\in\mathbb{N}$. Intuitively, $K$ can be thought of as the set obtained from $[0,1]$ following the successive removal of all open middle thirds. Show that $K$ is uncountable, perfect, and nowhere dense in $[0,1]$.

Definition. Let $\{x_n\}_{n\in\mathbb{N}}$ be a sequence in $S$ and $x\in S$. We say that $\{x_n\}_{n\in\mathbb{N}}$ converges to $x$ and write $x_n\rightarrow x$ if for any open set $U$ containing $x$, there exists a natural number $N$ such that $x_n\in U$ for all $n\geq N$. As a sequential limit, $x$ is also denoted by $\displaystyle\lim_{n\to\infty}x_n$.

Note that a sequence may have no limits, a unique limit, or several limits, depending upon the topology on $S$.

Example. Let $\tau$ be the cofinite topology on $\mathbb{R}$ i.e.
$$\tau=\{\emptyset\}\cup\{U\subset \mathbb{R}: \mathbb{R}\setminus U\ \mbox{is finite}\}.$$
Let $x_n=n$, $n\in\mathbb{N}$. Let $x\in\mathbb{R}$ and $x\in U\in\tau$. Then since $\mathbb{R}\setminus U$ is finite, there exists a natural number $N$ such that $x_n\in U$ for all $n\geq N$. Hence, $x_n\rightarrow x$ for all $x\in\mathbb{R}$.

Theorem. Let $A\subset S$ and $x\in S$.

  1. If $\{x_n\}_{n\in\mathbb{N}}$ is a sequence in $A$ such that $x_n\rightarrow x$, then $x\in\bar A$.
  2. If $\{x_n\}_{n\in\mathbb{N}}$ is a sequence of distinct points in $A$ such that $x_n\rightarrow x$, then $x\in A’$.

Proof.

  1. Assume the hypothesis. Let $U$ be an open set containing $x$. Then there exists a nutural number $N$ such that $x_n\in U$ for all $n\geq N$. Since $\{x_n\}_{n\in\mathbb{N}}\subset A$, $G\cap A\ne\emptyset$. Hence, $x\in\bar A$.
  2. Left as an exercise.

A limit point is not necessarily a sequential limit, and a sequential limit is not necessarily a limit point as seen in the following example.

Example. Let $S=\{a,b,c\}$ and $\tau=\{\emptyset,\{a,b\},\{c\},S\}$. Let $x_1=a$, $x_2=b$, and $x_n=c$ for all $n\geq 3$. Clearly, $x_n\rightarrow c$. However, $c$ cannot be a limit point since $c\in\{c\}\in\tau$ and $(\{c\}-\{c\})\cap\{a,b,c\}=\emptyset$. $a$ and $b$ are limit points but they cannot be sequential limits since $a,b\in\{a,b\}\in\tau$ and $x_n\notin\{a,b\}$ for all $n\geq 3$.

Exercise. Let $\tau=\{\emptyset\}\cup\{(a,\infty):a\in\mathbb{R}\}\cup\{\mathbb{R}\}$. Verify that $\tau$ is a topology on $\mathbb{R}$ and establish in $(\mathbb{R},\tau)$ the following sequential convergence and divergence:

  1. If $x_n=n$, for each $n\in\mathbb{N}$, then $x_n\rightarrow x$, for all $x\in\mathbb{R}$.
  2. If $x_n=-n$, for each $n\in\mathbb{N}$, then $\{x_n\}_{n\in\mathbb{N}}$ does not converge in $\mathbb{R}$.
  3. If $x_n=(-1)^n$, for each $n\in\mathbb{N}$, then $x_n\rightarrow x$, for all $x\leq -1$.

Bases and Subbases

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In this lecture, we study on how to generate a topology on a set from a family of subsets of the set. The idea is pretty much similar to basis of a vector space in linear algebra.

Definition. Let $(S,\tau)$ be a space and $\mathcal{B}\subset 2^S$. $\mathcal{B}$ is called a base for $\tau$ if
$$\tau=\{\emptyset\}\cup\left\{U\subset S:U=\bigcup_{\alpha\in\Lambda}B_\alpha,B_\alpha\in\mathcal{B},\forall\alpha\in\Lambda\right\}.$$
Clearly $\tau$ is base for itself.

Theorem 1. Let $S\ne \emptyset$ and $\mathcal{B}\subset 2^S$. $\mathcal{B}$ is a base for a topology on $S$ if and only if

1. $S=\bigcup\{B:B\in\mathcal{B}\}$.

2. For any $B_1,B_2\in\mathcal{B}$ with $x\in B_1\cap B_2$, there exists $B_3\in\mathcal{B}$ such that $x\in B_3\subset B_1\cap B_2$.

Proof. Suppose that $\mathcal{B}$ is a base for a topology $\tau$ on $S$. Since $S\in\tau$, it should be a union of members of $\mathcal{B}$, so $S\subset\bigcup\{B:B\in\mathcal{B}\}$ and hence $S=\bigcup\{B:B\in\mathcal{B}\}$. Let $B_1,B_2\in\mathcal{B}$ with $x\in B_1\cap B_2$. Since $B_1,B_2\in\tau$, $B_1\cap B_2\in\tau$. So $B_1\cap B_2=\bigcup_{\alpha\in\Lambda}B_\alpha$ where $B_\alpha\in\mathcal{B}$ for $\alpha\in\Lambda$. Since $x\in B_1\cap B_2$, $x\in B_\alpha$ for some $\alpha\in\Lambda$. Set $B_3=B_\alpha$. Then we are done.

Suppose that 1 and 2 hold. Let
$$\tau=\{\emptyset\}\cup\left\{U\subset S:U=\bigcup_{\alpha\in\Lambda}B_\alpha,B_\alpha\in\mathcal{B},\forall\alpha\in\Lambda\right\}.$$ Then clearly $\emptyset,S\in\tau$. So O1 is satisified. Let $U_\alpha\in\tau$, $\alpha\in\Lambda$. For each $x\in\bigcup_{\alpha\in\Lambda}U_\alpha$, there exist $B_x\in\mathcal{B}$ such that $x\in B_x\subset\bigcup_{\alpha\in\Lambda}U_\alpha$, so $\bigcup_{\alpha\in\Lambda}U_\alpha=\bigcup_{x\in \bigcup_{\alpha\in\Lambda}U_\alpha}B_x\in\tau$. Hence O2 is satisfied. Let $U_1,U_2\in\tau$ and $x\in U_1\cap U_2$. Then there exist $B_1,B_2\in\mathcal{B}$ such that $x\in B_1\subset U_1$ and $x\in B_2\subset U_2$. Since $x\in B_1\cap B_2$, by property 2 there exists $B_3\in\mathcal{B}$ such that $x\in B_3\subset B_1\cap B_2\subset U_1\cap U_2$. Thus $G_1\cap G_2$ can be expressed as a union of members of $\mathcal{B}$. Hence $U_1\cap U_2\in\tau$ and O3 is satisfied. Therefore $\mathcal{B}$ is a base for a topology $\tau$ on $S$.

Example. Let $\mathbb{R}$ be the set of real numbers and $\mathcal{B}_1=\{(a,b): a,b\in\mathbb{R}, a<b\}$. Then $\mathcal{B}_1$ is a base for the Euclidean topology (also called the interval topology) $\xi$ on $\mathbb{R}$.

Exercise: Prove that $\mathcal{B}_1$ is a base for a topology.

Example. Let $\mathcal{B}_2=\{[a,b): a,b\in\mathbb{R}, a<b\}$. Then $\mathcal{B}_2$ is a base for a topology $\mathcal{L}$ on $\mathbb{R}$. $\mathcal{L}$ is called the lower limit topology on $\mathbb{R}$. Similarly $\mathcal{B}_3=\{(a,b]:a,b\in\mathbb{R}, a<b\}$ is a base for a topology $\mathcal{U}$ on $\mathbb{R}$ called the upper limit topology. $(\mathbb{R},\mathcal{L})$ and $(\mathbb{R},\mathcal{U})$ have the same topological properties. However, $(\mathbb{R},\xi)$ and $(\mathbb{R},\mathcal{L})$ have different topological properties.

Exercise: Prove that $\mathcal{B}_2$ is a base for a topology.

Proposition. Let $(S,\tau)$ be a space and $\mathcal{B}$ a base for $\tau$. Let $U\subset S$. Then $U\in\tau$ if and only if for any $x\in U$, there exists $B\in\mathcal{B}$ such that $x\in B\subset U$.

Proof. Left as an exercise for the readers.

Definition. Let $\mathcal{B}_1,\mathcal{B}_2\subset 2^S$. Then $\mathcal{B}_1$ and $\mathcal{B}_2$ are said to be equivalent bases if they are bases for the same topology on $S$. If $\mathcal{B}_1$ and $\mathcal{B}_2$ are equivalent bases, we write $\mathcal{B}_1\sim\mathcal{B}_2$.

Theorem. Let $(S,\tau)$ be a space. Let $\mathcal{B}_1$ be a base for $\tau$ and $\mathcal{B}_2\subset 2^S$. Suppose that the following conditions 1 and 2 are satisfied:

1. If $x\in B_1\in\mathcal{B}_1$, then there exists $B_2\in\mathcal{B}_2$ such that $x\in B_2\subset B_1$.

2. If $x\in B_2\in\mathcal{B}_2$, then there exists $B_1\in\mathcal{B}_1$ such that $x\in B_1\subset B_2$.

Then $\mathcal{B}_2$ is also a base for $\tau$ and hence $\mathcal{B}_1\sim\mathcal{B}_2$.

Proof. Let $U$ be a nonempty open set. Then $U=\bigcup_{\alpha\in\Lambda}B_\alpha$, $B_\alpha\in\mathcal{B}_1$, $\alpha\in\Lambda$. If $x\in U$ then $x\in B_\alpha$ for some $\alpha\in\Lambda$. By condition 1, there exists $B\in\mathcal{B}_2$ such that $x\in B\subset B_\alpha\subset U$. So for any $x\in U$ there exists $B_x\in\mathcal{B}_2$ such that $x\in B_x\subset U$, and hence $U=\bigcup_{x\in U}B_x$. Conversely, if $U$ is a union of members of $\mathcal{B}_2$, then by condition 2 $U$ is expressed as a union of members of $\mathcal{B}_1$, and so $U\in\tau$. Therefore, $\mathcal{B}_2$ is also a base for the topology $\tau$.

Example. Consider $\mathbb{R}^2$. The (Euclidean) distance between $(x_1,y_1)$ and $(x_2,y_2)$ in $\mathbb{R}^2$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. For any $(x_0,y_0)\in\mathbb{R}^2$ and $\epsilon>0$, let
$$B((x_0,y_0);\epsilon):=\{(x,y)\in\mathbb{R}^2: \sqrt{(x-x_0)^2+(y-y_0)^2}<\epsilon\}.$$ The collection
$$\mathcal{B}_1=\{B((x_0,y_0);\epsilon): (x_0,y_0)\in\mathbb{R}^2, \epsilon>0\}$$
is a base for a topology $\xi^2$ on $\mathbb{R}^2$, called the Euclidean topology on $\mathbb{R}^2$. An equivalent base is
$$\mathcal{B}_2=\{\{(x,y)\in\mathbb{R}^2: a<x<b, c<y<d\}:a,b,c,d\in\mathbb{R}\},$$
the collection of all open rectangular regions in the plane $\mathbb{R}^2$.

Exercise: Prove that $\mathcal{B}_1$ and $\mathcal{B}_2$ are equivalent bases for a topology on $\mathbb{R}^2$.

Definition. Let $(S,\tau)$ be a space and $\mathcal{S}\subset 2^S$. Let
$$\mathcal{B}=\{F_1\cap\cdots\cap F_n: F_1,\cdots,F_n\in\mathcal{S}\}$$
i.e. the collection of finite intersections of members of $\mathcal{S}$. $\mathcal{S}$ is called a subbase for $\tau$ if $\mathcal{B}$ is a base for $\tau$.

Theorem. If $\mathcal{S}\subset 2^S$ and $\bigcup\{U:U\in\mathcal{S}\}=S$, then $\mathcal{S}$ is a subbase for a unique topology $\tau$ on $S$.

Proof. We prove this theorem using Theorem 1. Let $\mathcal{B}=\{F_1\cap\cdots\cap F_n: F_1,\cdots,F_n\in\mathcal{S}\}$. Clearly $\mathcal{S}\subset\mathcal{B}$, so $\mathcal{B}\ne\emptyset$. Let $x\in S$. Since $\bigcup\{U:U\in\mathcal{S}\}=S$, there exists $U\in\mathcal{S}$ such that $x\in U$. Since $U\in\mathcal{B}$, $x\in\bigcup\{B:B\in\mathcal{B}\}$, so $S\subset \bigcup\{B:B\in\mathcal{B}\}$ i.e. the condition 1 in Theorem 1 is satisfied. Let $B_1,B_1\in\mathcal{B}$ with $x\in B_1\cap B_2$. Then $B_1=F_1\cap\cdots\cap F_m$ and $B_2=G_1\cap\cdots\cap G_n$ for some $F_1,\cdots,F_m,G_1,\cdots,G_n\in\mathcal{S}$. Put $B_3=B_1\cap B_2$. Then $B_3\in\mathcal{B}$ and $x\in B_3$. Hence the condition 2 in Theorem 1 is also satisfied. This completes the proof.

Example. Let $\mathcal{S}=\{(-\infty,b): b\in\mathbb{R}\}\cup\{(a,\infty):a\in\mathbb{R}\}$. Then $\mathcal{S}$ is a subbase for the Euclidean topology $\xi$ on $\mathbb{R}$.

Exercise. Prove that $\mathcal{S}$ in the above example is a subbase for the Euclidean topology $\xi$ on $\mathbb{R}$.

Open and Closed Sets

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In the first lecture, we study open sets and closed sets which are the building blocks of topology. Let us begin with the definition of open sets and topology.

Definition. Let $S$ be a nonempty set and $\tau\subset 2^S$ such that

O1. $\emptyset, S\in\tau$.

O2. If $U_\alpha\in\tau$ for each $\alpha\in\Lambda$, then $\bigcup_{\alpha\in\Lambda}
U_\alpha\in\tau$.

O3. If $U_i\in\tau$ for each $i=1,\cdots,n$, then $\bigcap_{i=1}^nU_i\in\tau$.

Then $\tau$ is called a topology on $S$, the elements of $\tau$ are called open sets, and the ordered pair $(S,\tau)$ is called a topological space or simply a space.

Example. Let $S=\{a,b\}$. Then there are four possible topologies on $S$. They are
\begin{align*}
\tau_1&=\{\emptyset,S\},\\
\tau_2&=\{\emptyset,\{a\},S\},\\
\tau_3&=\{\emptyset,\{b\},S\},\\
\tau_4&=\{\emptyset,\{a\},\{b\},S\}.
\end{align*}

Exercise. Let $S=\{a,b,c\}$. Find all possible topologies on $S$. There are exactly 29 of them.

Definition. Let $S\ne\emptyset$. The smallest topology on $S$ is $\{\emptyset,S\}$ and is called the indiscrite topology. A space with the indiscrete topology is called an indiscrete space. The largest topology on $S$ is the power set $2^S$ and is called the discrete topology. A space with the discrete topology is called a discrete space.

Definition. Let $(S,\tau)$ be a space and $A\subset S$. $A$ is said to be closed if its compliment is open i.e. $A^c=S\setminus A\in\tau$.

Theorem. Let $(S,\tau)$ be a space. Then

C1. $\emptyset,S$ are closed.

C2. If $A_\alpha\subset S$ is closed for each $\alpha\in\Lambda$, then $\bigcap_{\alpha\in\Lambda}A_\alpha$ is closed.

C3. If $A_i\subset S$ is closed for each $i=1,\cdots,n$, then $\bigcup_{i=1}^nA_i$ is closed.

Proof. C1 is trivial. C2 and C3 can be easily shown by using De Morgan’s laws.

Remark. One may also define a topology using closed sets instead of open sets. Let $S\ne\emptyset$. Let $\mathcal{F}\subset 2^S$ satisfying C1, C2, C3:

C1. $\emptyset,S\in\mathcal{F}$.

C2. If $A_\alpha\in\mathcal{F}$ for each $\alpha\in\Lambda$, then $\bigcap_{\alpha\in\Lambda}A_\alpha\in\mathcal{F}$.

C3. If $A_i\subset S\in\mathcal{F}$ for each $i=1,\cdots,n$, then $\bigcup_{i=1}^nA_i\in\mathcal{F}$.

Let $\tau=\{U\subset S: S\setminus U\in\mathcal{F}\}$. Then $\tau$ is a topology on $S$.

Definition. Let $(S,\tau)$ be a space and $A\subset S$. The closure of $A$ is the smallest closed set containing $A$, that is
$$\bar A=\{F:A\subset F\ \mbox{and}\ F\ \mbox{is closed}\}.$$
Clearly $A\subset S$ is closed in $S$ if and only if $A=\bar A$.

Theorem. Let $(S,\tau)$ be a space and $A\subset S$. Then $x\in\bar A$ if and only if for any open set $U$ containing $x$, $U\cap A\ne\emptyset$.

Proof. Let $x\in\bar A$ and $U$ be an open set containing $x$. Suppose that $U\cap A=\emptyset$. Then $A\subset S\setminus U$ and $S\setminus U$ is closed. Since $\bar A$ is the smallest closed set containing $A$, $x\in\bar A\subset S\setminus U$. This is a contradiction. Hence $U\cap A\ne \emptyset$.

Assume that for any open set $U$ containing $x$, $U\cap A\ne \emptyset$. If $x\not\in\bar A$, then $x\in S\setminus\bar A=\cup\{S\setminus F: A\subset F\ \mbox{and}\ F\ \mbox{is closed}\}$. So there is a closed set $F$ containing $A$ such that $x\in S\setminus F$. Since $S\setminus F$ is open, this is a contradiction. Hence, $x\in\bar A$.

Exercise. Let $A,B$ be subsets of a space. Show that if $A\subset B$ then $\bar A\subset\bar B$.

Exercise. Let $A,B$ be subsets of a space. Prove or disprove:

  1.  $\overline{A\cup B}=\bar A\cup\bar B$.
  2. $\overline{A\cap B}=\bar A\cap\bar B$.

Exercise. Let $(S,\tau)$ be a space and $A_\alpha\in 2^S$ for each $\alpha\in\Lambda$. Show that

  1. $\overline{\bigcap_{\alpha\in\Lambda}A_\alpha}\subset\bigcap_{\alpha\in\Lambda}\bar A_\alpha$.
  2. $\overline{\bigcup_{\alpha\in\Lambda}A_\alpha}\supset\bigcup_{\alpha\in\Lambda}\bar A_\alpha$.

Give examples to show that the inclusions in 1 and 2 cannot be replaced by equality.


Parallel Transport, Holonomy, and Curvature

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Let $\gamma: [0,1]\longrightarrow M$ be a path. Using connection $\nabla$, one can consider the notion of moving a vector in $L_{\gamma(0)}$ to $L_{\gamma(1)}$ without changing it. This is parallel transporting a vector from $L_{\gamma(0)}$ to $L_{\gamma(1)}$. The change is measured relative to $\nabla$, so if $\xi(t)\in L_{\gamma(t)}$ is moving without changing, it must satisfy the differential equation
$$\nabla_{\dot{\gamma}(t)}\xi=0,$$
where $\dot{\gamma}(t)$ is the tangent vector field to the curve $\gamma(t)$. The image of $\gamma(t)$ is covered by the $U_\alpha$’s on which $L$ has nowhere vanishing sections $s_\alpha$’s. Since $\gamma([0,1])$ is compact, the image of $\gamma$ is covered by only finitely many of such open sets. Let $U_\alpha$ be one of such open sets and assume that it contains $\gamma(0)$. Then $\xi|_{U_\alpha}(t)=\xi_\alpha(\gamma(t))s_\alpha(\gamma(t))$ where $\xi_\alpha: U_\alpha\longrightarrow\mathbb{C}$.
\begin{align*}
\nabla_{\dot{\gamma}(t)}\xi&=d\xi_\alpha(\dot{\gamma}(t))s_\alpha(\gamma(t))+A_\alpha(\dot{\gamma}(t))\xi_\alpha(\dot{\gamma}(t))s_\alpha(\gamma(t))\\
&=\left(\frac{d\xi_\alpha}{dt}(\gamma(t))+A_\alpha(\dot{\gamma}(t))\xi_\alpha(\dot{\gamma}(t))\right)s_\alpha(\gamma(t)).
\end{align*}
$\nabla_{\dot{\gamma}(t)}\xi=0$ implies that
$$\frac{d\xi_\alpha}{dt}=-A_\alpha(\dot{\gamma}(t))\xi_\alpha.$$
The solution of this equation is given by
$$\xi_\alpha(t)=\xi_\alpha(\gamma(0))\exp\left(-\int_0^tA_\alpha(\dot{\gamma}(u))du\right).$$
The standard existence and uniqueness theorems (Frobenius’ theorem) tell that parallel transport defines an isomorphism $L_{\gamma(0)}\cong L_{\gamma(t)}$ for any $\gamma(t)\in U_\alpha$. Suppose that the path $\gamma$ is covered by finitely many open sets $U_\alpha$, $U_{\alpha_1}$, $U_{\alpha_2}$, $\cdots$, $U_{\alpha_n}$ as shown in the following figure.

As discussed, we know that $L_{\gamma(0)}\cong L_{\gamma(t_1)}$. Using $\xi_{\alpha_1}(\gamma(t_1))=\xi_\alpha(\gamma(t_1))$ as the initial condition, we also find $\xi|_{U_{\alpha_1}}(t)=\xi_{\alpha_1}(\gamma(t))s_{\alpha_1}(\gamma(t))$ where
$$\xi_{\alpha_1}(t)=\xi_{\alpha_1}(\gamma(t_1))\exp\left(-\int_{t_1}^tA_{\alpha_1}(\dot{\gamma}(u))du\right).$$
This implies that  $L_{\gamma(t_1)}\cong L_{\gamma(t_2)}$. Continuing this process, we obtain $L_{\gamma(t_2)}\cong L_{\gamma(t_3)}$, $\cdots$, $L_{\gamma(t_n)}\cong L_{\gamma(1)}$. Since the relation $\cong$ is transitive, we have
$$L_{\gamma(0)}\stackrel{P_\gamma}{\cong}L_{\gamma(1)}.$$
In general, $P_\gamma$ depends on $\gamma$ and $\nabla$. Now we are particularly interested in the case when $\gamma:[0,1]\longrightarrow M$ is a loop i.e. $\gamma(0)=\gamma(1)$. Then we can define the holonomy $\mathrm{hol}(\nabla,\gamma)$ of the connection $\nabla$ along the loop $\gamma$ by
$$P_\gamma(s)=\mathrm{hol}(\nabla,\gamma)s$$
for any nowhere vanishing section $s\in L_{\gamma(0)}$. So, what is really the meaning of the holonomy? In Euclidean space (the world we are familiar with), we can move a  vector without changing its direction and magnitude by parallel translation. That is, in Euclidean space parallel translation is parallel transport. So, we do not distinguish vectors that have the same direction and magnitude in Euclidean space. In a curved manifold, there is no such parallel translation and parallel transport is considered relative to the connection $\nabla$ as we discussed above. For those who live in a manifold with connection $\nabla$, they will not know the difference when a vector is parallel transported relative to $\nabla$ along a loop. The initial vector and the one that comes back to the initial point after parallel transport must coincide. However, in our perspective (for those who live in Euclidean space) we notice a difference between them. The holonomy measures such a difference.

Since $\gamma$ is a loop, both $\gamma(0)$ and $\gamma(1)$ belong to the same open set, say $U_\alpha$.
\begin{align*}
P_\gamma(\xi(0))&=\xi(1)\\
&=\xi_\alpha({\gamma(1)})\exp\left(-\oint_{\gamma}A_\alpha(\dot{\gamma}(u))du\right)s(\gamma(1))\\
&=\xi_\alpha({\gamma(0)})\exp\left(-\oint_{\gamma}A_\alpha(\dot{\gamma}(u))du\right)s(\gamma(0)).
\end{align*}
On the other hand, $\xi(0)=\xi_\alpha(\gamma(0))s(\gamma(0))$. So
\begin{align*}
P_\gamma(\xi(0))&=P_\gamma(\xi_\alpha(\gamma(0))s(\gamma(0)))\\
&=\xi_\alpha(\gamma(0))P_\gamma(s(\gamma(0))).
\end{align*}
Hence, we see that
$$P_\gamma(s(\gamma(0)))=\exp\left(-\oint_{\gamma}A_\alpha(\dot{\gamma}(u))du\right)s(\gamma(0))$$
and that the holonomy is given by
$$\mathrm{hol}(\nabla,\gamma)=\exp\left(-\oint_\gamma A_\alpha\right).$$
If $\gamma$ is the boundary of a disk, then by Stokes’ theorem we have
\begin{align*}
\mathrm{hol}(\nabla,\gamma)&=\exp\left(-\int_DdA_\alpha\right)\\
&=\exp\left(-\int_D F\right)\ \ \ \ \ \ \ (1)
\end{align*}
where $D$ is the interior of the disk.

Proposition. If $L\stackrel{\pi}{\longrightarrow}M$ is a line bundle with connection $\nabla$ and $\Sigma$ is a compact submanifold of $M$ with boundary loop $\gamma=\partial M$, then
$$\mathrm{hol}(\nabla,\gamma)=\exp\left(-\int_\Sigma F\right).\ \ \ \ \ \ \ (2)$$

Proof. By compactness, we can triangulate $\Sigma$ so that each of the triangles is in some $U_\alpha$. Then we apply (1) to each triangle and the holonomy up and down the interior edges cancels to give the required result.

Remark. Clearly holonomy is a gauge invariant quantity. In gauge theory, (2) is called a Wilson line or a Wilson loop. It is important to note that the gauge connection may be constructed from the collection of Wilson loops up to gauge transformation.

Example. [Parallel Transport on the 2-Sphere] In this example, we calculate the holonomy of the standard connection on $TS^2$. Before we proceed, let us take look at the figure below.

It clearly shows that the holonomy is $e^{i\theta}$ since the discrepancy between the initial vector and the parallel transported vector along the loop is given by a rotation by angle $\theta$. Recall that
\begin{align*}
\frac{\partial}{\partial\theta}&=(-\sin\theta\sin\phi,\cos\theta\cos\phi,0),\\
\frac{\partial}{\partial\phi}&=(\cos\theta\cos\phi,\sin\theta\cos\phi,-\sin\phi).
\end{align*}
The unit normal vector field $\hat n$ is computed to be
\begin{align*}
\hat n&=(\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi)\\
&=\sin\phi\frac{\partial}{\partial\phi}\times\frac{\partial}{\partial\theta}.
\end{align*}
Consider a nowhere vanishing section
$$s=(-\sin\theta,\cos\theta,0).$$
Then
$$ds=(-\cos\theta,-\sin\theta,0)d\theta.$$
\begin{align*}
\nabla s&=\pi(dx)\\
&=ds-\langle ds,\hat n\rangle\hat n\\
&=(-\cos\theta\cos^2\phi,-\sin\theta\cos^2\phi,\sin\theta\cos\phi)d\theta\\
&=\cos\phi\hat n\times s d\theta\\
&=i\cos\phi sd\theta.
\end{align*}
The last expression is obtained by the definition of scalar multiplication
$$(\alpha+i\beta)v=\alpha v+\beta\cdot u\times v$$
for $\alpha,\beta\in\mathbb{C}$ and $u,v\in T_\ast S^2$, as seen here. So the connection 1-form is
$$A=i\cos\phi d\theta$$
and the curvature is
\begin{align*}
F&=dA\\
&=-i\sin\phi d\phi\wedge d\theta.
\end{align*}
Note that $\sin\phi d\phi\wedge d\theta$ is the area form on $S^2$, so
$$F=-i\mathrm{area}.$$
The area of the region bounded by the loop is
$$\int_0^\theta\int_0^{\frac{\pi}{2}}\sin\phi d\phi d\theta=\theta.$$
Therefore, the holomony is
$$\mathrm{hol}(\nabla,\gamma)=e^{i\theta}$$
as we already know.

 

References:

[1] M. Murray, Notes on Line Bundles

Sections of a Line Bundle II: Gauge Potential, Gauge Transformation, and Field Strength

1 Reply

A connection on a line bundle can be defined in a pretty much similar fashion to a connection on a manifold that is discussed here since sections are like vector fields. Let $L\longrightarrow M$ be a line bundle. A connection $\nabla$ is a bilinear map which maps a pair $(X,s)$ of a tangent vector field $X$ on $M$ and a section $s: M\longrightarrow L$ to a section $\nabla_Xs$ such that
\begin{align*}
\nabla_{fX+gY}s&=f\nabla_Xs+g\nabla_Ys\ (\mbox{linearity})\\
\nabla_Xfs&=df(X)s+f\nabla_Xs\ (\mbox{Leibniz rule})
\end{align*}
where $X,Y\in\mathfrak{X}(M)$, $f,g\in C^\infty (M)$ and $s:M\longrightarrow L$ is a section. Denote by $\Gamma(M,L)$ the set of sections $M\longrightarrow L$. If we omit specifying tangent vector field on which $\nabla$ acts, Liebniz rule can be written as
$$\nabla fs=df\otimes s+f\nabla s$$
where the tensor product $\otimes$ is evaluated as
$$df\otimes s(X(m),m)=df(X(m))s(m)$$
for $m\in M$.

Example. Trivial bundle $L=M\times\mathbb{C}$.

Let $\nabla$ be a general connection. Let $s$ be a nowhere vanishing section. Define a 1-form $A’$ on $M$ by $\nabla s=A’\times s$. If $\xi\in\Gamma(M,L)$ then $\xi=fs$ for some $f:M\longrightarrow\mathbb{C}$. By Leibniz,
\begin{align*}
\nabla\xi&=df\otimes s+f\nabla s\\
&=(df+fA’)s.
\end{align*}
Recall that every section of a trivial bundle looks like $s(x)=(x,g(x))$ for some function $g: M\longrightarrow\mathbb{C}$. By identifying sections with functions, the ordinary differentiation $d$ of functions defines a connection. More specifically,
$$ds:=dg\otimes s.$$
Now,
\begin{align*}
\nabla s-ds&=A’\otimes s-dg\otimes s\\
&=(A’-dg)\otimes s.
\end{align*}
Let $A:=A’-dg$. Then $A$ is a 1-form on $M$ and
$$\nabla s=ds+A\otimes s.$$
Hence, all connections on $L$ are of the form
$$\nabla=d+A$$
where $A$ is a 1-form on $M$.

Let $L\stackrel{\pi}{\longrightarrow}M$ be a line bundle and $s_\alpha: U_\alpha\longrightarrow L$ local nowhere vanishing sections, Define a one-form $A_\alpha$ on $U_\alpha$ by $\nabla s_\alpha=A_\alpha\otimes s_\alpha$. $A_\alpha$ is called a connection one-form (in differential geometry) or a gauge potential (in physics) on $U_\alpha$. If $\xi\in\Gamma(M,L)$ then $\xi|_{U_\alpha}=\xi_\alpha s_\alpha$ where $\xi_\alpha : U_\alpha\longrightarrow\mathbb{C}$. By Leibniz rule,
\begin{align*}
\nabla\xi|_{U_\alpha}&=d\xi_\alpha\otimes s_\alpha+\xi_\alpha\nabla s_\alpha\\
&=( d\xi_\alpha+A_\alpha\xi_\alpha)\otimes s_\alpha.
\end{align*}
Since each fibre $L_m$ is a one-dimensional complex vector space, the transition map would be $g_{\alpha\beta}: U_\alpha\cap U_\beta\longrightarrow\mathrm{GL}(1,\mathbb{C})\cong\mathbb{C}^\times$, where $\mathbb{C}^\times$ is the multiplicative group of non-zero complex numbers. The transition maps satisfy
\begin{equation}
\label{eq:transition}
s_\alpha=g_{\alpha\beta}s_\beta. \ \ \ \ \ (1)
\end{equation}
The collection of functions $\xi_\alpha$ defines a section $\xi$ if on any intersection $U_\alpha\cap U_\beta\ne\emptyset$, $\xi_\alpha=g_{\alpha\beta}\xi_\beta$. The transition map $g_{\alpha\beta}$ gives rise to the change of coordinates. Since $s_\alpha$ and $s_\beta$ are related by (1) on $U\alpha\cap U_\beta\ne\emptyset$,
$$\nabla s_\alpha=(dg_{\alpha\beta})\otimes s_\beta+g_{\alpha\beta}\nabla s_\beta.$$
Since $\nabla s_\alpha=A_\alpha\otimes s_\alpha$,
\begin{align*}
A_\alpha\otimes s_\alpha&=(dg_{\alpha\beta})\otimes s_\beta+g_{\alpha\beta}\nabla s_\beta\\
&=(dg_{\alpha\beta})\otimes s_\beta+g_{\alpha\beta}A_\beta\otimes s_\beta\\
&=(dg_{\alpha\beta}+g_{\alpha\beta}A_\beta)\otimes s_\beta.
\end{align*}
So, we obtain
\begin{equation}
\label{eq:gauge}
A_\alpha=g^{-1}_{\alpha\beta}dg_{\alpha\beta}+A_\beta.\ \ \ \ \ \ (2)
\end{equation}
In physics, this is the gauge transformation for electromagnetism. The converse is also true, namely if $\{A_\alpha\}$ is a collection of 1-forms satisfying (2) on $U_\alpha\cap U_\beta$, then there exists a connection $\nabla$ such that $\nabla s_\alpha=A_\alpha\otimes s_\alpha$. First define $\nabla s_\alpha=A_\alpha\otimes s_\alpha$ for each nowhere vanishing  section $s_\alpha: U_\alpha\longrightarrow L$. On $U_\alpha\cap U_\beta\ne\emptyset$, by (1)
\begin{align*}
\nabla s_\alpha&=\nabla(g_{\alpha\beta}s_\beta)\\
&=dg_{\alpha\beta}\otimes s_\beta+g_{\alpha\beta}\nabla s_\beta.
\end{align*}
This must coincide with $A_\alpha\otimes s_\alpha$. By the gauge transformation (2)
\begin{align*}
A_\alpha\otimes s_\alpha&=g^{-1}_{\alpha\beta}dg_{\alpha\beta}\otimes s_\alpha+A_\beta\otimes s_\alpha\\
&=dg_{\alpha\beta}\otimes(g^{-1}_{\alpha\beta}s_\alpha)+A_\beta\otimes(g_{\alpha\beta}s_\beta)\\
&=dg_{\alpha\beta}\otimes s_\beta+g_{\alpha\beta}A_\beta\otimes s_\beta\\
&=\nabla s_\alpha.
\end{align*}
For $\xi\in\Gamma(M,L)$, $\nabla s_\alpha$ is linearly extended to $\nabla\xi$.

Next discussion requires some knowledge of differential forms, wedge product and exterior derivative. If you are not so familiar with these, please study them before you continue. One good source is Barrett O’Neil’s Elementary Differential Geometry [2].

Let $F_\alpha$ be the two-form
$$F_\alpha=dA_\alpha.$$
Physically $F_\alpha$ is the field strength relative to the section (field) $s_\alpha: U_\alpha\longrightarrow L$. Recall that on $U_\alpha\cap U_\beta\ne\emptyset$ the gauge potentials $A_\alpha$ and $A_\beta$ are related by the gauge transformation (2). If $F_\alpha$ and $F_\beta$ do not agree on $U_\alpha\cap U_\beta$, it would be a physically awkward situation. The following proposition tells us that it will not happen.

Proposition. If $s_\beta: U_\beta\longrightarrow L$ is another local section where $U_\alpha\cap U_\beta\ne\emptyset$, then $F_\alpha=F_\beta$.

Proof. \begin{align*}
F_\alpha&=dA_\alpha\\
&=d(g^{-1}_{\alpha\beta}dg_{\alpha\beta}+A_\beta)\\
&=dg^{-1}_{\alpha\beta}\wedge dg_{\alpha\beta}+g^{-1}_{\alpha\beta}d(dg_{\alpha\beta})+dA_\beta\\
&=-g^{-1}_{\alpha\beta}(dg_{\alpha\beta})g^{-1}_{\alpha\beta}\wedge dg_{\alpha\beta}+dA_\beta\\
&=dA_\beta=F_\beta.
\end{align*}
From second line to third line, $d(dg_{\alpha\beta})=d^2g_{\alpha\beta}=0$ and $dg^{-1}_{\alpha\beta}=-g^{-1}_{\alpha\beta}(dg_{\alpha\beta})g^{-1}_{\alpha\beta}$ (which is obtained from  $g^{-1}_{\alpha\beta}g_{\alpha\beta}=I$) have been used.

Physically speaking the proposition says that the field strength is invariant under gauge transformation. The two-forms agree on the intersection of two open sets in the cover and hence define a global two-form. It is denoted by $F$ and is also called the curvature of the connection $\nabla$ in differential geometry.

Remark. In a principal G-bundle with a Lie group $G$, the transition map is given by $g_{\alpha\beta}:U_\alpha\cap U_\beta\longrightarrow G$and the connection 1-forms (gauge potentials) $A_\alpha$ take values in $\mathfrak{g}$, the Lie algebra of $G$. The gauge transformation is given by
$$A_\alpha=g^{-1}dg_{\alpha\beta}+g^{-1}_{\alpha\beta}A_\alpha g_{\alpha\beta}.$$
The curvature (field strength) $F$ is invariant under the gauge transformation and is given by
$$F=dA_\alpha+[A_\alpha,A_\alpha].$$
Note that for each pair of tangent vector fields $(X,Y)$, $F$ is evaluated as
$$F(X,Y)=dA_\alpha(X,Y)+[A_\alpha(X),A_\alpha(Y)].$$

References:

[1] M. Murray, Notes on Line Bundles

[2] B. O’Neill, Elementary Differential Geometry, Academic Press, 1966