The Structure Group of a Bundle

For line bundles, the transition function $c_ {UV}$ is realized by general linear groups $GL(n)$. In a general bundle, it may be possible to choose the transition function $c_ {VU}$ such that they all lie in a specific Lie group $G$,

\[c_ {UV}: U\cap V \to G\]

we then say that $G$ is the structure group of the bundle.

Let $M$ be an oriented Riemannian surface ($\text{dim}_ {\mathbb{R}}=2$ manifold), $U,V$ be different patches which support orthonormal, oriented frame $e_ {U},e_ {V}$ of tangent vectors. Then the transition functions reduce to $SO(2)$ matrices. We say that the orthonormal frames have allowed us to reduce the structure group from $GL(N)$ to $SO(N)$.

Complex Line Bundles

We have seen that the matrix $J$ resembles the imaginary unit $i$ in many ways, such as $J^{2}=-1$. Given a two dimension real vector space, $J$ actually allows us to turn it into a one-dimensional complex vector space. We will try to apply this idea in the context of vector bundles.

Given an oriented surface $M$, and let $M_ {p}$ be the tangent plane at $p$ of the surface, and $e_ {1,2}$ be the frames of $M_ {p}$. Define the action of $J$ on the plane as rotating it by a right angle in the positive sense, namely

\[J: M_ {p} \to M_ {p}, \quad \; e_ {1}\to e_ {2},\; e_ {2}\to-e_ {1}\]

then naturally $J^{2}=-1$. $J$ is globally defined, because it makes sense in an overlapping region $U \cap V$, since if the frames in $U$-coordinates and $V$-coordinates are connected by some transition function $c_ {UV}$, then rotating all the coordinates by the right angle preserves the same transition function. Thus, $J$ allows us to consider each fiber in $TM^{2}$ as a complex line, the first component of the frame,

\[e_ {1} \in M_ {p} \approx \mathbb{R}^{2}\]

can be considered as a complex basis vector

\[e:= e_ {1} \in M_ {p}^{2}\approx \mathbb{C}^{1}\]

and the action of pure imaginary number $i$ is given by

\[i e := J e_ {1} = e_ {2}.\]

In term of the complex bases, the complex coordinates can be chosen as

\[e^ {U} = e^{U}_ {1}, \quad e^{V} = e^{V}_ {1}, \quad etc.\]

The transition function at point $p$ is now given by

\[c_ {UV}(p) = e^{i\alpha(p)}.\]

The tangent bundle to an oriented Riemannian surface can be considered as a complex line bundle! The structure group of this bundle is now $U(1)$, the unitary group in one variable!

The next natural question to ask is, what is the connection of the complex line bundle. It should be uniquely given by the connection of the two dimensional real vector space. In terms of the orthonormal frames $e^{U},e^{V},$ etc., we have

\[\nabla e_ {i} =e_ {j} \otimes \omega^{j}_ {\;\;i} = e_ {j} \otimes \omega_ {ji},\quad \text{Euclidean metric}.\]

Recall that $\omega_ {ij}$ is anti-symmetric. We have

\[\nabla e_ {1}=e_ {2}\omega_ {21}.\]

The connection for a complex line bundle should be a complex-number-valued 1-form, we will denote it by $\omega$ too. Let $e$ (without the index) be the complex line bundle connection, we have

\[\nabla e = e \omega, \quad \omega \in \mathbb{C}^{1}.\]

Since $e_ {2}=ie_ {1}$, we have

\[\nabla e = \nabla e_ {1} = e_ {2}\otimes \omega_ {21}=ie_ {1}\omega_ {21}=e_ {1} i\omega_ {21}= e \otimes (i\omega_ {21}).\]

Thus we have $\omega = i\omega_ {21}$.

One last thing. For $\omega=i\omega_ {21}$ be truly a connection of the complex line bundle (defined by the 2D real vector field), $\nabla$ defined by $\omega$ should commute with any constant number, namely

\[\nabla c \psi = c \nabla \psi, \quad c = \text{const}.\]

This is of course true for any real constant, but we have a special constant, namely $i$, which is given by $J$ in the 2D real vector version. To check it, we just need to make sure this property holds for the basis,

\[\nabla i e = i \nabla e.\]

This is much easier to check, we have

\[\nabla ie = \nabla i e_ {1} = \nabla e_ {2} = e_ {1} \omega_ {12} = i e \otimes i\omega_ {21} = i \nabla e.\]

Next natural question to ask is, what is the curvature of the complex line bundle? We have

\[\theta = d\omega+\omega \wedge \omega = d\omega = d(-i\omega_ {12}) = -id\omega_ {12} = -i\theta_ {12}.\]

To be specific,

\[\theta = -i\theta_ {12} = -i K \sigma^{1}\wedge \sigma^{2}\]

where $K$ is the Gauss-Riemann curvature $R^{12}_ {\;\;\;12}$ of $M^{2}$.