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Reading an article I faced with the following theorem, please give me a reference to a proof of the fact which is stated without any reference in the article. Is it a well-known fact?

Theorem. Let $E \overset{\pi}{\to} M$ be a complex line bundle over a Riemann surface $M$ with boundary $\partial M$. Let $\nabla^1$ and $\nabla^2$ be connections on $E$ with connection 1-forms $A_1$ and $A_2$. Then the following statements are equal:

  1. There exists a unitary bundle isomorphism $F \colon E \to E$ such that $F = Id$ on $E|_{\partial M}$ and $\nabla^1 = F^* \nabla^2 F$.

  2. Let $\gamma_1$, $\ldots$, $\gamma_M$ be non-homotopically equivalent loops in $M$, non-homotopically equivalent to any boundary component and let $\omega_1$, $\ldots$, $\omega_M$ be a dual basis of $H^1(M,\partial M)$, i.e. $\int_{\gamma_i} \omega_j = \delta_{ij}$. Then $$ A_1 - A_2 = 2\pi \sum\limits_{k=1}^M n_k \omega_k + df, \quad f|_{\partial M} = 0, \; n_k \in \mathbb Z. $$

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  • $\begingroup$ I think the first part may be in Wells - Differential analysis on complex manifolds. $\endgroup$ Commented Nov 25, 2013 at 18:39

2 Answers 2

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From (1) to (2):

Your bundle isomorphism is a (unitary) gauge transformation. Thus the difference in connections (= difference in connection 1-forms) is "pure gauge," i.e.,

$A_1 - A_2 = g^{-1} dg = \omega.$

Since this new connection is pure gauge, it will have no holonomy, since it can be "gauged away."

But holonomy around the loop $\gamma$ is just parallel transport:

$hol^\omega(\gamma) = \exp{\int_\gamma -\omega} = 1.$

We know the integral is 1 from triviality of the pure gauge connection. Now it is clear that $\int_\gamma \omega \in 2\pi\mathbb{Z}$. Use the fact that the loops $\gamma_i$ form a (relative) homology basis. Thus $\omega = \sum 2\pi n_i \omega_i$ upto an exact form, say $df$. We know we may take $f|_{\partial M} = 0$, since since the gauge transformation $F$ is the identity there.

Going the other direction, you may construct the unitary gauge transformation from the information given in (2).

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I think it's an exercise proved as follows.

A unitary bundle isomorphism is given by $g:M\rightarrow S^1$ with $g|_{\partial M} =Id$. Thus $A_1 -A_2 =g^{-1} dg =2\pi\sqrt{-1}f$, where $g=e^{2\pi\sqrt{-1}f}$, and $f$ is a multi-valued function, which represents a class in $H^1(M,\partial M)$.
Integrate over $\gamma_1,\ldots,\gamma_M$ to determine this class.

It basically follows from $H^1(M,\partial M)\cong[(M,\partial M), (S^1,1)]$

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