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

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  1. Pingback: Sections of a Line Bundle I | MathPhys Archive

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