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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

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I 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. –  celtschk Jul 23 '12 at 18:06
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@celtschk: Note that a "Riemannian manifold" is not the same thing as a "Riemann surface". –  Henning Makholm Jul 23 '12 at 18:51
    
@HenningMakholm: Ah, I didn't know that. –  celtschk Jul 23 '12 at 19:37
    
@celtschk But it's indeed the case, for the reason you state, that complex manifolds are canonically orientable. –  Dylan Moreland Jul 23 '12 at 19:39
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@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? –  wildildildlife Jul 23 '12 at 20:01

4 Answers 4

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.

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+1 for a thorough answer: you provide all linear algebraic details which I assumed as known in my answer :) –  wildildildlife Jul 23 '12 at 20:06
    
@wildildildlife thanks :) –  Olivier Bégassat Jul 23 '12 at 20:07
    
Very nice answer Olivier. To clarify a point: I have heard the argument "since the diagonalizable matrices are dense in $M_n(\mathbb{C})$ we can work with diagonalisable matrices" before but I have never seen how it actually works. Does it involve some kind of limiting procedure: you start with a sequence of diagonalisable matrices which converge to your (not diagonalisable) matrix, apply $\rho$ and then take the limit? –  GFR Jan 16 at 11:26
    
@GFR thanks! Yes: if $A$ is a complex matrix, there is a sequence $(A_n)$ of diagonalisable matrices converging to $A$. By continuity of the determinant, we have $$\det(A)=\lim_{n\to\infty}\det(A_n)$$ Actually, my argument is redundant if you know that every complex matrix is similar to an upper triangular matrix. This upper triangular matrix translates to an upper triangular real matrix of $2\times 2$ blocs, and the determinant of an upper triangular bloc matrix being the product of the determinants of the diagonal blocs gives the answer again. So no limits are required after all. –  Olivier Bégassat Jan 16 at 12:48

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 $M$.

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A simple closed contour $\Gamma$ in a patch $U$ of the surface is positively oriented if $\int_\Gamma f(z)\ dz = 0$ for all analytic functions on $U$. That provides an orientation.

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I believe this defines an orientation on the patch $U$, but the questioner already knew that (an open in) $\mathbb{C}$ is orientable. I think the question was about the globalization of this fact, i.e. why this yields a well-defined orientation on the whole manifold. This follows from the definition of a holomorphic atlas, as other answers explained. –  wildildildlife Jul 23 '12 at 19:50
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@wildildildlife I don't think this does anything really, for instance, if $U$ is the open disc in $\mathbb C$, then for any loop and any holomorphic function on $U$ the integral is $=0$. –  Olivier Bégassat Jul 23 '12 at 20:00

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)

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