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## Main Question or Discussion Point

If we consider a Riemannian surface as a one-dimensional complex manifold, what does that tell us about its intrinsic curvature? 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.

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.

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.