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I am reading a paper that says $L$ is a flat complex $G$-line bundle over $M$ with holonomy $\alpha$. Here $G$ is an abelian Lie group and $\alpha$ is a character of $G$. I have two questions:

  1. If the bundle is flat then isn't its holonomy trivial?
  2. I'm a little familiar with holonomy being a group so whats it mean for the holonomy to be a character? The bundle is constructed as the associated line bundle to a principal $G$ bundle using the representation $\alpha$. So is saying that the holonomy is $\alpha$ just repeating this fact? If so, how does it connect to the other definition of holonomy?

Any references where this stuff is talked about in some detail would be highly appreciated.

Thanks!

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    Sure! Will do it the next time I get near a scanner. I offered because I've never seen this mentioned in a book, although perhaps it is in [Kobayashi-Nomizu](http://books.google.com/books/about/Foundations_of_Differential_Geometry.html?id=jhWkPwAACAAJ), as everything seems to be.2011-07-26

2 Answers 2

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Flatness of the connection implies that parallel transport along each path $\gamma$ depends only upon the path homotopy class of $\gamma$. In particular, you get a representation $\pi_1(M, x) \to GL(L_x)$.

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As others pointed out integration the two form $F_{A}=dA+A\wedge A$give you a map $\pi_{1}(M)\rightarrow F$where $F$ is the fibre lying over the base point. Here we are working with the associated vector bundle of $E$, so locally it has the form $(g,m,v)\rightarrow (1,m,\alpha(g)v)$and the fibre is one dimensional. But here $L$ is no longer a $G$ bundle anymore, all we can say is a loop in $M$'s image must land in $L$, and a shift by $\alpha(h)$ will act by conjugation on $v$ as $\alpha(h)\alpha(g)v\alpha(h)^{-1}$. However we know $G$ is abelian, and $L$ is one dimensional. So regardless of initial point in $L_{x}$ we have a unique map $\pi_{1}(M)\rightarrow L_{x}$ determined by $\alpha$. I do not think $\pi_{1}(M)\rightarrow GL(L_{x})$ holds unless one proceed from the associated line bundle to the associated $G$ bundle again. Indeed to say $\alpha$ is the character of $G$ means differently from $\alpha$ is an homomorphism from $\pi_{1}(M)$ to $GL(L_{x})$, since $\pi_{1}(M)\not =G$ is totally possible.