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I have troubles trying to prove almost complex two-dimensional manifold is orientable.

Let I is complex structure on two-dimensional manifold M. Fix a basic $X_1,IX_1$ in each $T_xM$.

Easy to see any two such bases differ by a linear transformation with positive determinant.

To fix an orientation on M we consider the family of all coordinate systems $x_1,x_2$ of M such that in each coordinate neighborhood, the coordinate basics $\frac{\partial}{\partial x_1},\frac{\partial}{\partial x_2}$ of $T_xM$ at x differ from the chosen basis $X_1,IX_1$ by a linear transformation of positive determinant. I think These coordinate systems determine a complete oriented atlas for M but i don't prove it.

Here I am stuck. Could somebody show me how to prove it ?

Jonas Adler
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lam
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1 Answers1

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The observation here is that the Cauchy-Riemann equations actually give us orientability! Here's how to see this. Every manifold is locally orientable. What is important is that the transition map between the charts have positive determinant, so that we can patch together the local orientations into a consistent global orientation. A negative determinant will cause the local orientations to flip on the overlap, which is what we want to avoid.

Let's write the transition map as $f= u + iv$ in local coordinates. The derivative of $f$, written in terms of $x$ and $y$ (where $y$ is the imaginary component) is

$$df = \begin{pmatrix} u_{x} & u_{y}\\ v_{x} & v_{y} \end{pmatrix}$$

The determinant is $u_{x}v_{y} - u_{y}v_{x}$. And the Cauchy-Riemann equations tell us that

$$u_{x} = v_{y}$$ $$u_{y} = - v_{x}$$

Making these substitutions into the determinant turns it into

$$u_{x}^2 + u_{y}^2 > 0$$

and hence the determinant must be positive!

Elchanan Solomon
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  • Thank you very much . But i want to prove it without using complex analysis. Can you help me see how to build oriented atlas is right or wrong. – lam Apr 12 '16 at 03:42
  • In fact, an almost complex manifold in general does not have a holomorphic atlas, so this does not help. – Mariano Suárez-Álvarez Aug 23 '17 at 16:05
  • @MarianoSuárez-Álvarez I think that in 2-dimensions having an almost-complex structure is equivalent to having a holomorphic atlas. Theorem 3.11.1 in Jost's Compact Riemann Surfaces states that if $M$ is an oriented surface (2-manifold) with a (smooth) Riemannian metric then $M$ admits a conformal structure. – Yousuf Soliman Aug 23 '17 at 20:43
  • Well, you need orientability, which is precisely what this question is after, no? – Mariano Suárez-Álvarez Aug 23 '17 at 20:45
  • Yes, I'm not saying we can use that to help solve this problem. The remark was just to say that after solving this problem we learn that an almost complex surface does have a holomorphic atlas (unless I'm misunderstanding something in the statement of the theorem). – Yousuf Soliman Aug 23 '17 at 20:47
  • But the almost complex structure is irrelevant for that, really. Any orientable surface has a complex structure. – Mariano Suárez-Álvarez Aug 23 '17 at 22:26