# Flat connection with non-trivial holonomy? I cannot get it

maybe this is a dumb question, but I cannot understand how a principal $G$-bundle can have non-trivial holonomy with a flat connection. Maybe I'm missing something, but doesn't Ambrose-Singer theorem say that the holonomy is generated by the curvature? So if it vanishes, wouldn't holonomy be trivial?

Furthermore, why a non-flat connection can have non-trivial holonomy on contractible paths? Doesn't the lifting depends only on homotopy? Can someone, please, show me a trivial example explaining these two facts?

• Where did you come across these facts? – Jesse Madnick Jul 26 '14 at 4:43
• @JesseMadnick Well, I have never studied $G$-bundles properly, so I have "learned" just by hearing seminars and talking to people, but my main concern now is local systems, since I want to understand the geometric Langlands duality and its relations to S-duality. – user40276 Jul 26 '14 at 4:49

Look at a principal $$G$$-bundle with discrete or even finite group $$G$$. In this context, every connection is automatically flat. Those are (special) covering spaces, and holonomy will correspond to the action of the fundamental group of the base on the total space via deck transformations. The simplest example of the double cover of the circle $$S^1\to S^1,z\mapsto z^2$$ has non trivial holonomy.
EDIT 1 Concerning the Ambrose-Singer theorem you quote, looking at their original paper (thm 2, page 12), they say that the Lie subalgebra $$\mathfrak o$$ of $$\mathfrak g$$ generated by the $$\Omega(X,Y)$$, for $$X,Y$$ tangent vectors at the point $$b$$, coincides with the Lie subalgebra $$\mathfrak h$$ of the holonomy group of the fiber over $$b$$. For a flat connection $$\mathfrak o$$ is by definition $$0$$, so $$\mathfrak h=0$$ and the connected component $$\mathrm{Hol}(b)_0$$ of the holonomy group over the point $$b$$ is trivial, so that the full holonomy group $$\mathrm{Hol}(b)$$ is discrete (rather than trivial).