Decomposition which is invariant under the action of holonomy group

Question

Let M be a pseudo-Riemannian maniflod and H be the holonomy group of M at the point m. It is possible that M_m has an H-invariant subspace if M_m is indecomposable. Does it admit a decomposition M_m=M_1+M_2 which M_i is H-invariant?

Answer 1

Tom Krantz proved that if the holonomy group $H$ of an indecomposable pseudo-Riemannian manifold preserves a non-trivial decomposition $T_xM=V_1\oplus V_2$, then there exists also an $H$-invariant decomposition $T_xM=U_1\oplus U_2$ into the direct sum of two totally isotropic subspaces, in particular, the manifold must have netral signature. The papers can be found on http://www.mathematik.uni-dortmund.de/~tkrantz/Web/Publikationen.html

Answer 2

Even in 2-dimensions, this happens all the time. For a 2-dimensional, simply-connected pseudo-Riemannian manifold, the two null lines in each tangent space are preserved by the holonomy group, and they sum to the whole tangent space. In this case, as long as the curvature is nonzero, the manifold is indecomposable.

A similar phenomenon happens in all higher even dimensions: You can have the tangent space split as the sum of two maximal null subspaces that are invariant under the holonomy.

In 3-dimensions, this can't happen. If the holonomy preserves a null line, it preserves its orthogonal 2-plane, so it's enough to ask whether you can preserve two distinct null lines. However, if this happens, the 2-plane they span is nondegenerate and will be preserved, so the splitting theorem implies that the metric is locally a product.

Answer 3

Thanks firstly. By Wu's paper, there is a similar De Rham's theorem into indecomposable manifolds. But in the level of tangent space, is there a decomposition of tangent space which isnot necessary orthogonal for an indecomposable maniflod?