Is the holonomy group for general (not necessarily Riemannian) compact manifolds compact?
I believe this is true for Riemannian manifolds, according to Berger's classification.
Any insights would be appreciated.
Is the holonomy group for general (not necessarily Riemannian) compact manifolds compact? I believe this is true for Riemannian manifolds, according to Berger's classification. Any insights would be appreciated. 


You are misinterpreting Berger's theorem; it's not even true for compact Riemannian manifolds. See On compact Riemannian manifolds with noncompact holonomy groups, Burkhard Wilking. J. Differential Geom. Volume 52, Number 2 (1999), 223257. What is true is that, for a simplyconnected Riemannian manifold, the holonomy group is connected and compact. This is a consequence of Berger's theorem, but it also needs the fact that the holonomy in this case is the product of holonomy groups of locally irreducible Riemannian manifolds. See Besse's treatment and discussion in his book Einstein manifolds for details. 


The holonomy group need not be compact. For example, take $S^1$, trivialize its tangent bundle and let $\Gamma_{1,1}^1 = 1$, constant on $S^1$. If you parallel transport any vector around $S^1$, the holonomy is multiplication by some number, call it $k$, $k > 0$ and $k \neq 1$. If we use the standard counterclockwise and euclidean unit vector trivialization of $TS^1$, I suppose $k = e^{2\pi}$. So the holonomy $\pi_1 S^1 \to Hom(T_1 S^1)$ is the map that sends $n \in \mathbb Z \equiv \pi_1 S^1$ to multiplication by $e^{2\pi n}$ in $T_1 S^1$. So it's a discrete holonomy, but still countablyinfinite. 


If a manifold equipped with a pseudoRiemmanian (= nondegenerate but not necessary positively definite) metric contains a region with constant nonzero curvature tensor, then its holonomy group is the whole connected component of the orthogonal group which is not compact unless the metric is positively or negatively definite. 


A very simple example is a cone of revolution in $\mathbb R^3$ minus the vertex. The restricted holonomy (that is, that generated by parallel transport along loops homotopic to a point, which is also the connected component of the full holonomy group) is trivial because the metric is flat. On the other hand, by flatenning the cone on a plane one sees that going once around the vertex by parallel transport gives a rotation of angle $2\pi(1\sin\theta)$, where $\theta$ is the angle of opening of the cone. Since the fundamental group of the cone is $\mathbb Z$, the holonomy group is either a finite or a countable dense subgroup of $SO(2)$ according to whether $1\sin\theta$ is rational or not. Edit: The cone is not compact! But the connected component of the holonomy group of a Riemannian manifold is always compact. 

