Monthly Archives: November 2012

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

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
\nabla_{fX+gY}s&=f\nabla_Xs+g\nabla_Ys\ (\mbox{linearity})\\
\nabla_Xfs&=df(X)s+f\nabla_Xs\ (\mbox{Leibniz rule})
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,
\nabla\xi&=df\otimes s+f\nabla s\\
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,
$$ds:=dg\otimes s.$$
\nabla s-ds&=A’\otimes s-dg\otimes s\\
&=(A’-dg)\otimes s.
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
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,
\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.
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
s_\alpha=g_{\alpha\beta}s_\beta. \ \ \ \ \ (1)
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 $\nabla s_\alpha=A_\alpha\otimes s_\alpha$,
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.
So, we obtain
A_\alpha=g^{-1}_{\alpha\beta}dg_{\alpha\beta}+A_\beta.\ \ \ \ \ \ (2)
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)
\nabla s_\alpha&=\nabla(g_{\alpha\beta}s_\beta)\\
&=dg_{\alpha\beta}\otimes s_\beta+g_{\alpha\beta}\nabla s_\beta.
This must coincide with $A_\alpha\otimes s_\alpha$. By the gauge transformation (2)
A_\alpha\otimes s_\alpha&=g^{-1}_{\alpha\beta}dg_{\alpha\beta}\otimes s_\alpha+A_\beta\otimes s_\alpha\\
&=dg_{\alpha\beta}\otimes s_\beta+g_{\alpha\beta}A_\beta\otimes s_\beta\\
&=\nabla s_\alpha.
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
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*}
&=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\\
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
Note that for each pair of tangent vector fields $(X,Y)$, $F$ is evaluated as


[1] M. Murray, Notes on Line Bundles

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

Sections of a Line Bundle I

A section of a line bundle is like a vector field. It is a map $s: M\longrightarrow L$ such that $s(m)\in L_m$ or $\pi\circ s(m)=m$. Section of a line bundle is one-to-one.

Example. For the trivial bundle $L=M\times\mathbb{C}$,  every section $s$ looks like $s(x)=(x,f(x))$ for some function $f$.

Example. For a tangent bundle $TM$, sections are vector fields.
s: M&\longrightarrow TM\\
x&\longmapsto v_x\in T_xM
For the tangent bundle $TS^2$ (minitwistor space) over $S^2$, one can think of a section as a map $s: S^2\longrightarrow TS^2$ such that $\langle s(x),x\rangle=0$ for each $x\in S^2$.

Proposition. A line bundle $L$ is trivial if and only if it has a nowhere vanishing section.

Proof. Suppose that $L$ is trivial. Let $\varphi: L\longrightarrow M\times\mathbb{C}$ be the trivialization. Then $s: M\longrightarrow L$ defined by $s(m)=\varphi^{-1}(m,1)$ is a nowhere vanishing section. Conversely, if $s$ is a nowhere vanishing section, define a trivialization $M\times\mathbb{C}\longrightarrow L$ by $(m,\lambda)\longmapsto\lambda s(m)$. This is an isomorphism.

Physically sections are fields and if we cannot differentiate fields, we cannot do physics. Let $L\stackrel{\pi}{\longrightarrow}M$ be a line bundle and $s:M\longrightarrow L$ a section. Let $\gamma:(-\epsilon,\epsilon)\longrightarrow M$ be a path through $\gamma())=m$. The conventional definition of the derivative of $s$ would be
$$\lim_{t\to 0}\frac{s(\gamma(t))-s(\gamma(0))}{t}.$$
However, this definition makes no sense because $s(\gamma(t))\in L_{\gamma(t)}$ and $s(\gamma(0))\in L_{\gamma(0)}=L_m$ and we cannot perform the required subtraction
So, we need to devise a way to differentiate sections of a line bundle. To get a clue, we need to review what we already know and maybe start from there since we cannot create something from nothing. At least we learned how to differentiate vector fields in Euclidean space, say $\mathbb{R}^3$. Let $X$ be a vector field in $\mathbb{R}^3$. The covariant derivative $\nabla_vX$ of $X$ in the direction of the tangent vector $v\in T_p\mathbb{R}^3$ is
&=\lim_{t\to 0}\frac{X(p+tv)-X(p)}{t}.
At this moment, one may say “Wait a minute! We have already examined the definition and know that it does not work for sections of a line bundle.” I know and please be patient. We haven’t got a clue yet and something useful may come out of this along the way. Since we cannot create the derivative of a section out of thin air, it is still important to review what we already know. We can naturally extend the above definition to the covariant derivative $\nabla_XY$ of a vector field $Y$ with respect to a vector field $X$. The covariant derivative $\nabla$ satisfies the following properties:

  1. $\nabla_{f X+gZ}Y=f\nabla_XY+g\nabla_ZY$;
  2. $\nabla_XfY=(Xf)Y+f\nabla_XY$ where $Xf$ denotes the directional derivative $Xf=\sum_{i=1}^n\alpha^i\frac{\partial f}{\partial x^i}$.

The first property is linearity and the second property is Leibniz rule. These are the most basic rules that you would expect from differentiation. Denote by $\mathfrak{X}(\mathbb{R}^3)$ the set of all tangent vector fields on $\mathbb{R}^3$. The covariant derivative may be regarded as a bilinear map $\nabla:\mathfrak{X}(\mathbb{R}^3)\times\mathfrak{X}(\mathbb{R}^3)\longrightarrow\mathfrak{X}(\mathbb{R}^3)$ satisfying the properties 1 and 2 and we write $\nabla(X,Y)$ as $\nabla_XY$. This gives us a clue on how to define the derivative of a section. It turns out that there isn’t a unique way to differentiate a section for there can be many different maps $\nabla$ satisfying the properties 1 and 2. In fact, one needs to make a choice. Such a choice of differentiation is called a connection.

Definition. Let $M$ be a differentiable manifold of dimension $n$ and $\mathfrak{X}(M)$ the set of all tangent vectors on $M$. A connection on $M$ is a bilinear map $\nabla:\mathfrak{X}(M)\times\mathfrak{X}(M)\longrightarrow\mathfrak{X}(M)$ such that
\nabla_{f X+gZ}Y&=f\nabla_XY+g\nabla_ZY\\
where $\nabla(X,Y)$ is written as $\nabla_XY$.

The one we defined here is a way of differentiating (connection) of vector fields on a differentiable manifold, but we still have not defined a way of differentiating sections of a line bundle. But we now have a much clearer picture about it. Before we continue, let us briefly discuss differentials because they are closely related to directional derivative. For each $i=1,\cdots,n$, the differential 1-form $dx^i$ is a 1-form on $T_\ast M$ such that
$$dx^i\left(\frac{\partial}{\partial x^j}\right)=\frac{\partial x^i}{\partial x^j}=\delta_{ij}$$
where $\delta_{ij}$ is the Kronecker’s delta. That is, $dx^i$ is is the dual vector of the tangent vector $\frac{\partial}{\partial x^i}$, $i=1,\cdots,n$ and that the  $dx^i$, $i=1,\cdots,n$ form the standard basis for the cotangent space $T^\ast M$. For any tangent vector field $X=\sum_{j=1}^n\alpha^j\frac{\partial}{\partial x^j}$,
So if we define the differential of $f$ by
$$df:=\sum_{i=1}^n\frac{\partial f}{\partial x^i}dx^i,$$
$$df(X)=\sum_{i=1}^n\frac{\partial f}{\partial x^i}dx^i(X)=Xf.$$
Thus the Leibniz rule can be also written as

Let $\gamma: (-\epsilon,\epsilon)\longrightarrow M$ be a path with $\gamma(0)=p$. On a local coordinate neighborhood $(U(p),\varphi)$, $\gamma(t)$ is written as $\gamma(t)=(x^1(t),\cdots,x^n(t))$ and
$$\frac{d\gamma}{dt}(0)=\sum_{i=1}^n\frac{dx^i}{dt}(0)\left(\frac{\partial}{\partial x^i}\right)_p\in T_p(M).$$
df\left(\frac{d\gamma}{dt}(0)\right)&=\sum_{i=1}^n\left(\frac{\partial}{\partial x^i}\right)_pdx^i\left(\frac{d\gamma}{dt}(0)\right)\\
&=\sum_{i=1}^n\left(\frac{\partial}{\partial x^i}\right)_p\frac{dx^i}{dt}(0)\\
For the tangent vector field $\dot{\gamma}(t)=\frac{d\gamma}{dt}$, we have

We will discuss connection on a line bundle in the second part of this lecture. I would like to end this lecture with a physical motivation for considering bundles.  The fields in physics are usually given by map $\phi:M\longrightarrow\mathbb{C}^n$ where $M$ is the spacetime. In quantum mechanics, particles are described by so-called complex-valued wave functions (also called state function) $\phi: M\longrightarrow\mathbb{C}$. Due to the Uncertainty Principle, one cannot pinpoint the exact location of a particle. The best thing one can do is to measure a probable location of the particle. The probability of a particle in the state  $\phi$ to be discovered in the region $U\subset M$ is
To define probability, all we need to know is that $\phi(x)$ takes its value in $\mathbb{C}$ with Hermitian product, and there is no reason for this to be the same vector space for all values of $\phi(x)$. Functions like $\phi$ which are the generalization of complex-valued functions are called sections of vector bundles.


[1] M. Murray, Notes on Line Bundles

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

Line Bundles

Simply speaking, a line bundle is a complex vector bundle such that each fibre $F_x$ is a one-dimensional complex vector space i.e. one-dimensional vector space over the complex field $\mathbb{C}$. More specifically,

Definition. A complex line bundle over a manifold $M$ is a manifold $L$ and a smooth onto map $\pi: L\longrightarrow M$ such that

  1. For each $m\in M$, $\pi^{-1}(m)=L_m$ is a one-dimensional complex vector space.
  2. For each $m\in M$, there exists an open neighborhood $U(m)\subset M$ such that $\pi^{-1}(U(m))\stackrel{\varphi}{\cong}U(m)\times\mathbb{C}$ (here $\cong$ means “is homeormorphic to” as usual) and $\varphi(L_m)\subset\{m\}\times\mathbb{C}$. Moreover, $\varphi|_{L_m}:L_m\longrightarrow\{m\}\times\mathbb{C}$ is a linear isomorphism.

Example. The Trivial Bundle $M\times\mathbb{C}$.

Example. If $u\in S^2$, the tangent plane at $u$ is identified with
$$T_uS^2=\{v\in\mathbb{R}^3:\langle u,v\rangle=0\}.$$
We can make this 2-dimensional real vector space a 1-dimensional complex vector space by defining
$$(a+i\beta)v:=\alpha v+\beta\cdot u\times v.$$
So, the tangent bundle $TS^2$ is a line bundle. $TS^2$ as a complex line bundle is called the mini-twistor space and it plays an important role in the study of BPS monopoles in physics.

Example. Let $\Sigma\subset\mathbb{R}^3$ be a surface. If $x\in\Sigma$ and $\hat n_x$ is a unit normal, then $T_x\Sigma=\hat n_x^\perp$ (the orthogonal complement of $\hat n_x$). We make this a 1-dimensional complex vector space by defining
$$(\alpha+i\beta)v=\alpha v+\beta\hat n_x\times v.$$
So, the tangent bundle $T\Sigma$ is a line bundle.

Example. [Hopf Bundle] Let $\mathbb{C}P^1$ be the set of all lines through the origin in $\mathbb{C}^2$. Denote the line through the vector $z=(z^0,z^1)$ by $[z]=[z^0,z^1]$. Define two open sets $U_i$, $i=0,1$ by
$$U_i=\{[z^0,z^1]:z^i\ne 0\},\ i=0,1$$
and $\psi_i:U_i\longrightarrow\mathbb{C}$ by
$$\psi_0([z])=\frac{z^1}{z^0},\ \psi_1([z])=\frac{z^0}{z^1}.$$
Then $\mathbb{C}P^1$ is a complex manifold of dimension 1. As a manifold $\mathbb{C}P^1$ is diffeomorphic to $S^2$. An explicit diffeomorphism $S^2\longrightarrow\mathbb{C}P^1$ is given by
Define a line bundle $H\subset\mathbb{C}^2\times\mathbb{C}P^1$ over $\mathbb{C}P^1$ by
$$H=\{(\omega,[z]): \omega=\lambda z\ \mbox{for some}\ \lambda\in\mathbb{C}\setminus\{0\}\}.$$
Define a projection $\pi:H\longrightarrow\mathbb{C}P^1$ by $\pi(\omega,[z])=[z]$. The fibre $H_{[z]}=\pi^{-1}([z])$ is the set $\{(\lambda z,[z]):\lambda\in\mathbb{C}\setminus\{0\}\}$ which is identified with the line $[z]$ through the vector $z$. The fibre $H_{[z]}$ can be made to a 1-dimensional complex vector space by
\alpha(\omega,[z])+\beta(\omega’,[z])&:=(\alpha\omega+\beta\omega’,[z]),\ \alpha,\beta\in\mathbb{C}\setminus\{0\},\\


[1] M. Murray, Notes on Line Bundles

Vector Bundles

Let $M$ be a differentiable manifold of dimension $n$. Consider an atlas $\mathcal{U}=\{U_\alpha\}_{\alpha\in\mathcal{A}}$ along with coordinates $x_\alpha^1,\cdots,x_\alpha^n$ in $U_\alpha$. For $x=(x_\alpha^1(x),\cdots,x_\alpha^n(x))\in U_\alpha$, a tangent vector is given by
$$v=\sum_{j=1}^nv_\alpha^j\frac{\partial}{\partial x_\alpha^j}.$$
If $x\in U_\alpha\cap U_\beta$, then $v$ is also written as
$$v=\sum_{j=1}^nv_\beta^j\frac{\partial}{\partial x_\beta^j}.$$
Here, the change of coordinates is given by
$$v_\beta^j=vx_\beta^j=\sum_{k=1}^nv_\alpha^k\frac{\partial x_\beta^j}{\partial x_\alpha^k}.$$
For $x\in U_\alpha\cap U_\beta$ and $f=(f^1,\cdots,f^n)\in\mathbb{R}^n$, define
h_{\alpha\beta}(x)(f)&=\left(\sum_{k=1}^n\frac{\partial x_\beta^1}{\partial x_\alpha^k}f^k,\cdots,(\sum_{k=1}^n\frac{\partial x_\beta^n}{\partial x_\alpha^k}f^k\right)\\
\frac{\partial x_\beta^1}{\partial x_\alpha^1} & \cdots & \frac{\partial x_\beta^1}{\partial x_\alpha^n}\\
\vdots & \ddots & \vdots\\
\frac{\partial x_\beta^n}{\partial x_\alpha^1} & \cdots & \frac{\partial x_\beta^n}{\partial x_\alpha^n}
Hence $h_{\alpha\beta}:U_\alpha\cap U_\beta\longrightarrow\mathrm{Aut(\mathbb{R}^n)}$. The resulting bundle over $M$ with fibre $F=\mathbb{R}^n$ is called the tangent bundle of $M$ and is denoted by $TM$. Note that $TM$ is the set of all tangent vectors of $M$ i.e. $TM=\bigcup_{x\in M}T_xM$. For each $x\in U_\alpha$, the fibre $\pi^{-1}(x)$ of $x\in M$ is $T_xM\cong\{x\}\times\mathbb{R}^n$. The local trivialization map $h_\alpha:\pi^{-1}(U_\alpha)\longrightarrow U_\alpha\times\mathbb{R}^n$ is given by
$$h_\alpha(v)=(x,(v_\alpha^1,\cdots,v_\alpha^n)),\ v\in T_xU_\alpha(=T_xM),\ x\in U_\alpha.$$

A fibre bundle $(E,M,F,\pi)$ is called a vector bundle over $M$ if each fibre $F_x$ of $x\in M$ is a vector space. So, a tangent bundle is a vector bundle. The tangent bundle $TM$ is a differentiable manifold of dimension $2n$ with local coordinates in $\pi^{-1}(U_\alpha)$ being $(x_\alpha^1,\cdots,x_\alpha^n,v_\alpha^1,\cdots,v_\alpha^n)$. The Jacobian is given by
\frac{\partial x_\beta^1}{\partial x_\alpha^1} & \cdots & \frac{\partial x_\beta^1}{\partial x_\alpha^n}\\
\vdots & \ddots & \vdots\\
\frac{\partial x_\beta^n}{\partial x_\alpha^1} & \cdots & \frac{\partial x_\beta^n}{\partial x_\alpha^n}
\end{pmatrix}:U_\alpha\cap U_\beta\longrightarrow\mathrm{GL}(n,\mathbb{R}).
Let $g_{\alpha\beta}=J(x_\beta^1,\cdots,x_\beta^n;x_\alpha^1,\cdots,x_\alpha^n)$ Then $g_{\alpha\beta}$ satisfies
g_{\beta\alpha}(x)&=g_{\alpha\beta}^{-1}(x),\ x\in U_\alpha\cap U_\beta;\\
g_{\alpha\beta}(x)g_{\beta\gamma}(x)g_{\gamma\alpha}(x)&=I_n,\ x\in U_\alpha\cap U_\beta\cap U_\gamma.
For $x\in U_\alpha\cap U_\beta$ and $f\in\mathbb{R}^n$, $h_{\alpha\beta}(x)(f)=g_{\alpha\beta}\cdot f$. So, $\mathrm{GL}(n,\mathbb{R})$ acts on the fibre $\mathbb{R}^n$. The map $g_{\alpha\beta}$ itself is often called a transition map.

If the transition map $h_\alpha\beta$ is the group action of a Lie group $G$ on the fibre $F$, the fibre bundle $(E,M,F,\pi)$ is called a $G$-bundle and the Lie group $G$ is called a structure group. The tangent bundle $TM$ is also a G-bundle with structure group $\mathrm{GL}(n,\mathbb{R})$.

Differentiable Manifolds and Tangent Spaces

In $\mathbb{R}^n$, there is a globally defined orthonormal frame
$$E_{1p}=(1,0,\cdots,0)_p,\ E_{2p}=(0,1,0,\cdots,0)_p,\cdots,E_{np}=(0,\cdots,0,1)_p.$$
For any tangent vector $X_p\in T_p(\mathbb{R}^n)$, $X_p=\sum_{i=1}^n\alpha^iE_{ip}$. Note that the coefficients $\alpha^i$ are the ones that distinguish tangent vectors in $T_p(\mathbb{R}^n)$. For a differentiable function $f$, the directional derivative $X_p^\ast f$ of $f$ with respect to $X_p$ is given by
$$X_p^\ast f=\sum_{i=1}^n\alpha^i\left(\frac{\partial f}{\partial x_i}\right).$$
We identify each $X_p$ with the differential operator
$$X_p^\ast=\sum_{i=1}^n\alpha^i\frac{\partial}{\partial x_i}:C^\infty(p)\longrightarrow\mathbb{R}.$$
Then the frame fields $E_{1p},E_{2p},\cdots,E_{np}$ are identified with
$$\left(\frac{\partial}{\partial x_1}\right)_p,\left(\frac{\partial}{\partial x_2}\right)_p,\cdots,\left(\frac{\partial}{\partial x_n}\right)_p$$
respectively. Unlike $\mathbb{R}^n$, we cannot always have a globally defined frame on a differentiable manifold. So it is necessary for us to use local coordinate neighborhoods that are homeomorphic to $\mathbb{R}^n$ and the associated frames $\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2},\cdots,\frac{\partial}{\partial x_n}$.

Example. The points $(x,y,z)$ are represented in terms of the spherical coordinates $(\phi,\theta)$ as
$$x=\sin\phi\cos\theta,y=\sin\phi\sin\theta,z=\cos\phi,\ 0\leq\phi\leq\pi,\ 0\leq\theta\leq 2\pi.$$
By chain rule, one finds the standard basis $\frac{\partial}{\partial\phi},\frac{\partial}{\partial\theta}$ for $T_\ast S^2$:
\frac{\partial}{\partial\phi}&=\cos\phi\cos\theta\frac{\partial}{\partial x}+\cos\phi\sin\theta\frac{\partial}{\partial y}-\sin\phi\frac{\partial}{\partial z},\\
\frac{\partial}{\partial\theta}&=-\sin\phi\sin\theta\frac{\partial}{\partial x}+\sin\phi\cos\theta\frac{\partial}{\partial y}.
The frame field is not globally defined on $S^2$ since $\frac{\partial}{\partial\theta}$ at $\phi=0,\pi$. More generally, the following theorem holds.

Frame field on 2-sphere

Theorem. [Hairy Ball Theorem] If $n$ is even, a non-vanishing $C^\infty$ vector field on $S^n$ does not exist i.e. a $C^\infty$ vector field on $S^n$ must take zero value at some point of $S^n$.

The Hairy Ball Theorem tells us why we have ball spots on our heads. It can be also stated as “you cannot comb a hairy ball flat.” There may also be a meteorological implication of this theorem. It may implicate that there must be at least one spot on earth where there is no wind at all. No-wind spot may be the eye of a hurricane. So, as long as there is wind (and there always is) on earth, there must be a hurricane somewhere at all times.

It has been known that all odd-dimensional spheres have at least one non-vanishing $C^\infty$ vector field and that only spheres $S^1, S^3, S^7$ have a $C^\infty$ field of basis. For instance, there are three mutually perpendicular unit vector fields on $S^3\subset\mathbb{R}^4$ i.e. a frame field: Let $S^3=\{(x^1,x^2,x^3,x^4)\in\mathbb{R}^4: \sum_{i=1}^4(x^i)^2=1\}$. Then
X&=-x^2\frac{\partial}{\partial x^1}+x^2\frac{\partial}{\partial x^2}+x^4\frac{\partial}{\partial x^3}-x^3\frac{\partial}{\partial x^4},\\
Y&=-x^3\frac{\partial}{\partial x^1}-x^4\frac{\partial}{\partial x^2}+x^1\frac{\partial}{\partial x^3}+x^2\frac{\partial}{\partial x^4},\\
Z&=-x^4\frac{\partial}{\partial x^1}+x^3\frac{\partial}{\partial x^2}-x^2\frac{\partial}{\partial x^3}+x^1\frac{\partial}{\partial x^4}
form an orthonormal basis of $C^\infty$ vector fields on $S^3$.

Fibre Bundles

A fibre bundle is an object $(E,M,F,\pi)$ consisting of

  1. The total space $E$;
  2. The base space $M$ with an open covering $\mathcal{U}=\{U_\alpha\}_{\alpha\in\mathcal{A}}$;
  3. The fibre $F$ and the projection map $E\stackrel{\pi}{ \longrightarrow}M$.

The simplest case is $E=M\times F$. In this case, the bundle is called a trivial bundle. In general the total space may be too complicated for us to understand, so it would be nice if we can always find smaller parts that are simple enough for us to understand such as trivial bundles. For this reason, we want the fibre bundle to have the additional property: For each $U_\alpha\in\mathcal{U}$, there exists a homeomorphism $h_\alpha : \pi^{-1}(U_\alpha)\longrightarrow U_\alpha\times F$. Such a homeomorphism $h_\alpha$ is called a local trivialization. For each $x\in M$, $F_x:=\pi^{-1}(x)$ is homeomorphic to $\{x\}\times F$. $F_x$ is called the fibire of $x$.

Let $x\in U_\alpha\cap U_\beta$. Then $F_x^\alpha\subset\pi^{-1}(U_\alpha)$ and $F_x^\beta\subset\pi^{-1}(U_\beta)$ may not be the same. However, the two fibres are homeomorphic. For each $x\in M$, denote by $h_{\alpha\beta}(x)$ the homeomorphism from $F_x^\alpha$ to $F_x^\beta$. Then for each $x\in M$, $h_{\alpha\beta}(x)\in\mathrm{Aut}(F)$ where $\mathrm{Aut}(F)$ is the group of homeomorphisms from $F$ to itself i.e. the automorphism group of $F$. The map $h_{\alpha\beta}: U_\alpha\cap U_\beta\longrightarrow\mathrm{Aut}(F)$ is called a transition map. Note that for $U_\alpha,U_\beta\in\mathcal{U}$ with $U_\alpha\cap U_\beta\ne\emptyset$, $h_\alpha\circ h_\beta^{-1}:(U_\alpha\cap U_\beta)\times F\longrightarrow (U_\alpha\cap U_\beta)\times F$ satisfies
$$h_\alpha\circ h_\beta^{-1}(x,f)=(x,h_{\alpha\beta}(x)f)$$
for any $x\in U_\alpha\cap U_\beta$, $f\in F$

The Representation of $\mathrm{SU}(2)$

Let $\mathcal{H}_j$ be the space of polynomial functions on $\mathbb{C}^2$ that are homogemeous of degree $2j$. An element in $\mathcal{H}_j$  is a polynomial in complex variables $x$ and $y$ that is a linear combination of polynomials $x^py^q$ where $p+q=2j$. $\mathcal{H}_j$ has dimension $2j+1$ since it has a basis given by
For any $g\in\mathrm{SU}(2)$, let $U_j(g)$ be the linear transformation of $\mathcal{H}_j$ given by
for $f\in\mathcal{H}_j$ and $v\in\mathbb{C}^2$. Then $U_j$ is a representation: $U_j(I)$ is the identity. For any $g,h\in\mathrm{SU}(2)$,
for $f\in\mathcal{H}_j$ and $v\in\mathbb{C}^2$.

Physicists call $U_j$ spin-$j$ representation. Since $2j+1$ has to be a positive integer, we have spin-0 representation, spin-$\frac{1}{2}$ representation, spin-1 representation, etc. It is interesting to see the correspondence between spin-$j$ representation and particles. The only known spin-0 particle is Higgs-boson, the so-called God particle, which is responsible for giving masses to bosons. The Higgs-boson appears to have been discovered recently by the LHC (Large Hadron Collider) at CERN. Spin-$\frac{1}{2}$ particles are fermions which include all quarks and leptons. Spin-1 particles are gauge bosons (force-carrying particles) such as photons, W bosons, Z bosons, gluons. Curiously there are currently no spin-$\frac{3}{2}$ particles predicted in particle physics. The hypothetical gravitons are believed to be spin-$2$ particles.

Proposition. The spin-0 representation of $\mathrm{SU}(2)$ is equivalent to the trivial representation in which every element of the group acts on $\mathbb{C}$ as the identity.

Proposition. The spin-$\frac{1}{2}$ representation of $\mathrm{SU}(2)$ is equivalent to the fundamental representation in which every element $g\in\mathrm{SU}(2)$ acts on $\mathbb{C}^2$ by matrix multiplication.

Note that the $U_j$ are irreducible and that they are all of the irreducible representations.

$\mathbb{R}^3$ can be identified with the set of $2\times 2$ Hermitian matricies of the form
z & x-iy\\
x+iy & -z
1 & 0\\
0 & -1
0 & 1\\
1 & 0
0 & -i\\
i & 0
where $\sigma_1,\sigma_2,\sigma_3$ are called the Pauli spin matrices in physics. Define an inner product $\langle\ ,\ \rangle$ on the Hermitian matrices by
$$\langle X, Y\rangle=\frac{1}{2}\mathrm{tr}(XY).$$
In particular,
$$|X|^2=\frac{1}{2}\mathrm{tr}(X^2)=-\det X.$$
With this inner product, the identification is an isometry. $\mathrm{SU}(2)$ acts on $\mathbb{R}^3$ via the representation
defined by
for $U\in\mathrm{SU}(2)$ and $X\in\mathbb{R}^3$. It turns out that $X\longmapsto UXU^{-1}$ is an orientation preserving isometry of $\mathbb{R}^3$, so
Since both $U$ and $-U$ result the same isometry, the representation $\rho:\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$ is a $2:1$ map. Since $\mathrm{SU}(2)=S^3$ is simply-connected, $\rho$ is a universal convering map and that we have
The quotient group $\mathrm{SU}(2)/\mathbb{Z}_2$ is denoted by $\mathrm{PSU}(2)$ and called the projective special unitary group.

The double cover i.e. $2:1$ cover of $\mathrm{SO}(n)$ is called the spin group and is denoted by $\mathrm{Spin}(n)$. For $n>2$, $\mathrm{Spin}(n)$ is simply-connected so it is the universal cover of $\mathrm{SO}(n)$. Some examples of spin groups are
\mathrm{Spin}(1)&=\mathrm{O}(1)=\mathbb{Z}_2=\{\pm I\}\\
Note that $\mathrm{SO}(3)\subset\mathrm{GL}(3,\mathbb{R})\subset\mathrm{GL}(3,\mathbb{C})$, so for any $g\in\mathrm{SU}(2)$, $\rho(g):\mathbb{C}^3\longrightarrow\mathbb{C}^3$. Hence, $\rho$ is in fact equivalent to the spin-1 representation of $\mathrm{SU}(2)$.

In quantum mechanics, unitary representation is particularly important. Let $\mathcal{H}$ be the Hilbert space of states derived from a quantum mechanical system. Let $\rho$ be a representation of a Lie group $G$ on $\mathcal{H}$ i.e. $\rho:G\longrightarrow\mathrm{GL}(\mathcal{H})$. $\rho$ is called a unitary representation if
for all $g\in G$ and $\psi,\phi\in\mathcal{H}$. Intuitively each $\rho(g)$ may be understood as a rotation. For example, say $G=\mathrm{SO}(3)$. First rotating the particle by some amount $h\in\mathrm{SO}(3)$ and then rotating it by some amount $g\in\mathrm{SO}(3)$ should have the same effect as rotating it by the amount $gh\in\mathrm{SO}(3)$. That is $\rho(g)\rho(h)=\rho(gh)$. This tells why we need a representation in quantum mechanics. In quantum mechanics, the inner product $\langle\ ,\ \rangle$ measures probability. For instance if a particle is in the state $\psi$, then $\langle\psi,\phi\rangle$ is the probability of finding the particle in the state $\phi$. Rotating a particle amounts to a change of coordinates and the state of a particle should not depend on a change of coordinates. Hence, in quantum mechanics we require representation to be unitary.

Lastly I would like to mention the difference between bosons and fermions in terms of representation. Let $f\in\mathcal{H}_j$, the spin-$j$ representation space. Then
since $f$ is a homogeneous polynomial of degree $2j$. This implies that
Hence, $U_j$ maps both $I$ and $-I$ to the identity if $j$ is an integer, while it does not if $j$ is a half-integer.


[1] John Baez, Javier P. Muniain, Gauge Fields, Knots and Gravity, World Scientific 1994

[2] Brian C. Hall, Lie Groups, Lie Algebras, and Representations, An Elementary Introduction, Springer-Verlag 2003