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Ben Niehoff
Oct31-11, 05:16 PM
Sci Advisor
P: 1,593
Quote Quote by TrickyDicky View Post
If we consider a Riemannian surface as a one-dimensional complex manifold, what does that tell us about its intrinsic curvature?
Having a complex structure means that the structure group of the manifold is reduced from [itex]GL(2n, \mathbb{R})[/itex] to [itex]GL(n, \mathbb{C})[/itex]. This means that the curvature 2-form is only allowed to take values within this subgroup.

On a Riemann surface, [itex]SO(2) \simeq U(1)[/itex], so the structure group reduction is tautologous. This is just another way to state the obvious fact that all Riemann surfaces have a complex structure.

However, since this is tautologous, the fact that there is a complex structure tells you nothing about the curvature. There is one independent curvature in 2 dimensions --- the Ricci scalar --- and it can be anything whatsoever.

I mean for one-dimensional curves we know they only have extrinsic curvature so it depends on the embedding space, this doesn't seem to be the case for one-dimensional complex manifolds, but I'd like to understand this better.
1-complex-dimensional curves have 2 real dimensions, hence they can have intrinsic curvature given by the Ricci scalar.

On the other hand, thinking of Riemannian surfaces as 2-dimensional real manifolds they inmediately have an intrinsic curvature that determines to a certain extent their topology (number of handles:g=0 positive curvature,g=1 flat, g>1 negative). Or said differently, in these manifolds the topology can be determined by their metric, I think this is a property of manifolds up to 3 real dimensions at least for the constant curvature cases but I'm not completely sure. Maybe someone could clarify. Certainly in the GR 4-manifold the metric does not give us the general topology.
The metric never tells you everything about the global topology. For example, a flat metric might correspond to an infinite plane, an infinite cylinder, or a torus. Each of these has different first homotopy group.

However, on a closed 2-manifold, you can calculate the Euler characteristic by integrating the Ricci scalar, yes. In higher dimensions, it is a very non-trivial task to find useful topological invariants that can be expressed as integrals of the curvature.

There are some topological invariants that cannot be expressed this way. For example, the property of having a complex structure in even dimensions > 2 is highly-nontrivial. It is still an open question whether some manifolds can be given complex structures, for example the 6-sphere.