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.
$\begingroup$
$\endgroup$
9
-
1$\begingroup$ What do you want holonomy to mean on a non-Riemannian manifold? Do you want it to be a manifold with some kind of non-Levi-Cevita connection? Any restrictions? Ehresmann minimally? $\endgroup$– Ryan BudneyCommented Apr 11, 2012 at 3:50
-
1$\begingroup$ Sorry if I'm being dense here but how would you define the holonomy group for a non-Riemannian (compact) manifold? In my limited understanding, the holonomy groups are associated to a bundle with a connection. For Riemannian manifolds, it is the holonomy group associated with the Levi-Civita connection on the tangent bundle which is of interest and classified by Berger. $\endgroup$– Somnath BasuCommented Apr 11, 2012 at 3:52
-
$\begingroup$ Assuming you have a connection and the holonomy is linear, I think it's compact if and only if its Levi-Cevita. $\endgroup$– Ryan BudneyCommented Apr 11, 2012 at 3:54
-
$\begingroup$ Maybe you mean a pseudo-Riemannian manifold. Then it is non-compact, of couse. $\endgroup$– MishaCommented Apr 11, 2012 at 4:02
-
1$\begingroup$ 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) is trivial, because the metric is flat, but the full holonomy can be a countable dense subgroup of $SO(2)$ if you choose correctly the angle of opening of the cone. $\endgroup$– Claudio GorodskiCommented Apr 11, 2012 at 16:32
|
Show 4 more comments
4 Answers
$\begingroup$
$\endgroup$
2
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), 223-257.
What is true is that, for a simply-connected 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.
answered Apr 11, 2012 at 13:27
-
$\begingroup$ So the compactness comes from the type of metric and the simply-connectedness of the manifold, rather than the compactness of the manifold... Is there any way to see this directly and intuitively? $\endgroup$– EarthliŋCommented Apr 12, 2012 at 13:27
-
$\begingroup$ @s.barmeier: No simple way to see it is known that doesn't use some structure theory, both from Lie groups and from de Rham's Theorem. See Besse's comments in Chapter 10, Section E, where the Borel-Lichnerowicz theorem is proved, which states that the identity component of the holonomy of a Riemannian metric is always a closed (and hence compact) subgroup of the orthogonal group. For the effect of the fundamental group, see Section 10.115, but note that this was written before the paper of Wilking appeared, so that particular question is now settled. $\endgroup$ Commented Apr 12, 2012 at 15:14
$\begingroup$
$\endgroup$
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 counter-clockwise 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 countably-infinite.
answered Apr 11, 2012 at 4:59
$\begingroup$
$\endgroup$
If a manifold equipped with a pseudo-Riemmanian (= 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.
answered Apr 11, 2012 at 7:49
$\begingroup$
$\endgroup$
1
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.
answered Apr 11, 2012 at 16:46
-
2$\begingroup$ @Claudio: There is an easy (and standard) way to convert your example to a compact one (affine flat structure on the 2-torus). Take a subgroup of $R_+\times SO(2)$ generated by a dilation $A$ and a rotation $B$ of infinite order. Then $A$ and $B$ match opposite edges of a "fundamental membrane" $D$ which is bounded by two concentric circular arcs and two segments contained in lines passing through the origin. Gluing the sides of $D$ via $A$ and $B$ yields a torus with a flat affine structure whose holonomy is nondiscrete (generated by $A$ and $B$). See more in Thurston's Notes or book. $\endgroup$– MishaCommented Apr 11, 2012 at 17:35