Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
To me (maybe this is not the full picture), flatness of the connection means that holonomy only depends on the homotopy class of the path, so you get a homomorphism $\pi_1(M) \to GL(L_x)$. I'll think about it some more. –  Dylan Moreland Jul 21 '11 at 17:11
    
@Dylan: OK thanks that then actually answers it I think. I should have provided more context! $L$ is associated to the principal $G$ bundle $\tilde M \to M$ where $\tilde M$ is the universal cover of $M$ so $G = \pi_1(M)$. So $\alpha$ then must be this map $\pi_1(M) \to GL(L_x)$. You should post this as an answer so I can accept it. –  Eric O. Korman Jul 21 '11 at 23:59
    
That makes more sense. I didn't think it was comprehensive enough to be an answer, but will post it as such. I also have some notes of Kevin Costello's on this theorem that I can upload. –  Dylan Moreland Jul 22 '11 at 1:35
    
Yea I would appreciate those notes. Thanks! –  Eric O. Korman Jul 22 '11 at 2:03
    
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, as everything seems to be. –  Dylan Moreland Jul 26 '11 at 15:27
add comment

2 Answers

up vote 1 down vote accepted

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 $\pi_1(M, x) \to GL(L_x)$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.