Could any one tell me about the orientability of riemann surfaces? well, Holomorphic maps between two open sets of complex plane preserves orientation of the plane,I mean conformal property of holomorphic map implies that the local angles are preserved and in particular right angles are preserved, therefore the local notions of "clockwise" and "anticlockwise" for small circles are preserved. can we transform this idea to abstract riemann surfaces? thank you for your comments
orientability of riemann surface
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riemann-surfaces
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0I think a Riemann manifold of dimension $n$ is orientable if you can define a continuous $n$-form on it which doesn't vanish anywhere. A comment, not an answer, because I'm not sure it's really true. – 2012-07-23
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2@celtschk: Note that a "Riemannian manifold" is not the same thing as a "Riemann surface". – 2012-07-23
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0@HenningMakholm: Ah, I didn't know that. – 2012-07-23
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0@celtschk But it's indeed the case, for the reason you state, that complex manifolds are canonically orientable. – 2012-07-23
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0@HenningMakholm: So at least I was not completely off. :-) – 2012-07-23
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1@DylanMoreland: it is a general fact about $n$-dimensional manifolds that they are orientable iff they admit a nowhere-vanishing $n$-form. This has nothing to do with a Riemannian metric or complex structure, so I don't understand your comment? – 2012-07-23
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0@wildildildlife I read the comment a bit quickly; I assumed he was getting at the fact that the orientation induced by $(i/2)^n(dz_1 \wedge d\bar z_1 \wedge \cdots dz_n \wedge d\bar z_n)$ on $\mathbb C^n$ is preserved by holomorphic transition maps. – 2012-07-23