orientability of riemann surface

Question

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

Answer 1

Riemann surfaces are one dimensional complex manifolds. It is in fact true that any complex $n$-manifold is orientable as a real manifold. One possible way of showing this, is to calculate the determinant of the jacobians of the transition functions.

Suppose you have an $n$-dimensional complex manifold $M$. That is, $M$ is a (first countable, Hausdorff) space such that every point $p\in M$ has an open neighborhood $V$ homeomorphic to an open subset $\Omega\subset\mathbb C^n$, $$\phi:V\rightarrow\Omega~\mathrm{homeomorphism}$$ and when two such neighborhoods intersect, the transition functions are holomorphic, $$\psi\circ\phi^{-1}:\phi(V\cap V')\rightarrow\psi(V\cap V')~\mathrm{biholomorphism.}$$ By identifying $\mathbb C^n$ with $\mathbb R^n$ like so: $(z^1,\dots,z^n)\leftrightarrow(x^1,y^1,\dots,x^n,y^n)$ where $z^k=x^k+iy^k$ is the real part/imaginary part decomposition, we get a real manifold structure on $M$. We can calculate the jacobian matrices of the transition functions for both structures.

Let's set up some notation. The first coordinate chart will be $\phi:V\rightarrow\Omega, p\mapsto (z^1(p),\dots,z^n(p))$, and the second will be $\psi:V'\rightarrow\Omega', p\mapsto (Z^1(p),\dots,Z^n(p)).$ In the complex case, we have $$\mathrm{Jac}_{\phi(p)}(\psi\circ\phi^{-1})=\left(\frac{\partial~[\psi\circ\phi^{-1}]^k}{\partial~z^l}\right)_{1\leq k,l\leq n}=\left(\frac{\partial~Z^k(z^1,\dots,z^n)}{\partial~z^l}\right)_{1\leq k,l\leq n}\\=\left(c_l^k\right)_{1\leq k,l\leq n}\in\mathrm{GL}(n,\mathbb C),$$ and in the real case you'll find $$\mathrm{Jac}_{\phi(p)}(\psi^{\mathbb R}\circ(\phi^{\mathbb R})^{-1})= \left(\begin{array}{cc} \frac{\partial~X^k(x^1,y^1,\dots,x^n,y^n)}{\partial~x^l} & \frac{\partial~X^k(x^1,y^1,\dots,x^n,y^n)}{\partial~y^l} \\ \frac{\partial~Y^k(x^1,y^1,\dots,x^n,y^n)}{\partial~x^l} & \frac{\partial~Y^k(x^1,y^1,\dots,x^n,y^n)}{\partial~y^l} \end{array}\right)_{1\leq k,l\leq n}= \left(\begin{array}{cc} \mathrm{Re}(c_l^k) & -\mathrm{Im}(c_l^k) \\ \mathrm{Im}(c_l^k) & \mathrm{Re}(c_l^k) \end{array}\right)_{1\leq k,l\leq n}\in\mathrm{GL}(2n,\mathbb R),$$ using the Cauchy-Riemann equations. We will calculate the determinant of these matrices, show that it is always $>0$, which is equivalent to $M^{\mathbb R}$ being orientable.

We move on to calculating the determinants of these. Consider the $\mathbb R$-algebra homomorphism $$\rho:M_n(\mathbb C)\rightarrow M_{2n}(\mathbb R),~\left(c_l^k\right)_{1\leq k,l\leq n}\mapsto \left(\begin{array}{cc} \mathrm{Re}(c_l^k) & -\mathrm{Im}(c_l^k) \\ \mathrm{Im}(c_l^k) & \mathrm{Re}(c_l^k) \end{array}\right)_{1\leq k,l\leq n}$$ Since it is $\mathbb R$-linear and the spaces involved are finite dimensional, it is continuous. Also, being an algebra homomorphism, we have $\mathrm{det}~\rho(P^{-1}AP)=\mathrm{det}(\rho(P)^{-1}\rho(A)\rho(P))=\mathrm{det}(\rho(A))$. Finally, the diagonalizable matrices are dense in $M_n(\mathbb C)$, so we can restrict our calculations to diagonal matrices in $M_n(\mathbb C)$. For these, the calcuations are easy: $$\mathrm{det}\big(\rho(\mathrm{Diag}(c_1,\dots,c_n))\big)=\mathrm{det}~\mathrm{Diag}\left(\left(\begin{array}{cc} \mathrm{Re}(c_1) & -\mathrm{Im}(c_1) \\ \mathrm{Im}(c_1) & \mathrm{Re}(c_1) \end{array}\right),\dots,\left(\begin{array}{cc} \mathrm{Re}(c_n) & -\mathrm{Im}(c_n) \\ \mathrm{Im}(c_n) & \mathrm{Re}(c_n) \end{array}\right)\right)\\=\prod_{i=1}^n\mathrm{det}\left(\begin{array}{cc} \mathrm{Re}(c_i) & -\mathrm{Im}(c_i) \\ \mathrm{Im}(c_i) & \mathrm{Re}(c_i) \end{array}\right)=\prod_{i=1}^n|c_i|^2=\left|\mathrm{det}~\mathrm{Diag}(c_1,\dots,c_n)\right|^2,$$ so we conclude that $$\forall A\in M_n(\mathbb C),~\mathrm{det}~\rho(A)=|\mathrm{det}~A|^2$$

Finally, we can conclude that the transition functions for the charts $\phi^{\mathbb R}$ for $M^{\mathbb R}$ have positive determinants, thus the real underlying manifold $M^{\mathbb R}$ is orientable.

Answer 2

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 $M$.

Answer 3

Every coordinate-change of a holomorphic atlas is by definition a bi-holomorphism between opens of $\mathbb{C}^n$, i.e. its derivative (tangent map) is a $\mathbb{C}$-linear isomorphism of $\mathbb{R}^{2n}$, hence has positive determinant.

($n=1$ for a Riemann surface)