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 ?
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!