How do I prove that gauge-equivalence classes of $U(1)$ connections on a line bundle $L\to M$ are determined uniquely by pairs $(\alpha,F)$, where $$\alpha\in\text{Hom}(\pi_1(M),U(1)),~~~~F\in \Omega^2(M)?$$
$\begingroup$
$\endgroup$
4
-
$\begingroup$ you can find it in the books of geometric quantization $\endgroup$– user21574Commented Mar 4, 2016 at 5:04
-
$\begingroup$ what are some books you recommend? $\endgroup$– David RobertsCommented Mar 4, 2016 at 5:23
-
$\begingroup$ @jol $\alpha$ is not well-defined if the connection is not flat, see my comment below. $\endgroup$– Sebastian GoetteCommented Mar 4, 2016 at 10:59
-
$\begingroup$ Holonomy around cycles is enough to determine a unique bundle with connection (possibly non-flat). $\endgroup$– user40276Commented Mar 4, 2016 at 20:29
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
The 2-form $F$ has to be closed. Then you choose an open covering $U_i$ on which $F$ has a primitive $\theta_i$, i.e. $d \theta_i = F$. Now you try to glue together the trivial bundles $U_i \times U(1)$ with connection $\theta_i$. In order to do this, the curvature form has to be integral.
Finally, equivalence classes of flat connections are parametrized by $Hom(\pi_1(M), U(1)$. Every such flat bundle arises from a homomorphism $\lambda: \pi_1(M) \to U(1)$ as the associated bundle $\tilde{M} \times_\lambda U(1)$.
A classical reference for this material is Woddhouse 1997 "Geometric Quantization", Proposition 8.3.1
-
3$\begingroup$ The procedure sketched in the first paragraph gives a solution, but only up to a possible twist by a flat line bundle. To detect this, instead of $\alpha$ (which is well-defined only if the connection is flat), you need to specify the holonomy along a fixed set of loops generating $H_1(M)$. Remember on one hand that holonomy is gauge invariant, hence well-defined in our context. On the other hand, if the curvature is nonzero, you can always choose a representative of each class in $\pi_1(M)$ such that you obtain any prescribed value for the holonomy. $\endgroup$ Commented Mar 4, 2016 at 10:56
-
$\begingroup$ @Sebastian Goette, thank you , good comment $\endgroup$– user21574Commented Mar 4, 2016 at 13:30