orientability of riemann surface 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

Holomorphic maps not only preserve nonoriented angles but oriented angles as well. Note that the map $z\mapsto\bar z$ preserves nonoriented angles, in particular nonoriented right angles, but it does not preserve the clockwise or anticlockwise orientation of small circles.
Now a Riemann surface $S$ is covered by local coordinate patches $(U_\alpha, z_\alpha)_{\alpha\in I}\ $, and the local coordinate functions $z_\alpha$ are related in the intersections $U_\alpha\cap U_\beta$ by means of holomorphic transition functions $\phi_{\alpha\beta}$: One has $z_\alpha=\phi_{\alpha\beta}\circ z_\beta$ where the Jacobian $|\phi'_{\alpha\beta}(z)|^2$ is everywhere $>0$. It follows that the positive ("anticlockwise") orientation of infinitesimal circles on $S$ is well defined througout $S$. In all, a Riemann surface $S$ is oriented to begin with.
For better understandng consider a Moebius band $M:=\{z\in {\mathbb C}\ | \ -1<{\rm Im}\, z <1\}/\sim\ ,$ where points $x+iy$ and $x+1-iy$ $\ (x\in{\mathbb R})$ are identified. In this case it is not possible to choose coordinate patches $(U_\alpha, z_\alpha)$ covering all of $M$ such that the transition functions are properly holomorphic. As a consequence this $M$ is not a Riemann surface, even though nonoriented angles make sense on $M$.
