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I am in the process of proving that a complex manifold is orientable. Consider the case $m=1$ so that in some chart, the usual coordinates of $p\in M$ are $(x,y)$. In some overlapping chart, let the coordinates of $p$ be denoted $(u,v)$. The manifold by assumption is analytic and so the Cauchy Riemann equations hold for the transition maps. For this case the proof is straightforward (as explained in one of the related posts). One need only show that the determinant of the Jacobian matrix comes out positive. And for this case, it does indeed (by use of Cauchy Riemann equations):$$\begin{vmatrix}u_x&u_y\\v_x&v_y \end{vmatrix}=u_x^2+v_x^2>0.$$ The problem arises when I try to extend this argument. Consider the case $m=2$. Then the coordinates of $p\in M$ in the overlapping charts are $(x^1, y^1,x^2,y^2)$ and $(u^1, v^1,u^2,v^2)$. The Cauchy Riemann equations read: $$u_{x^1}^1=v_{y^1}^1,v_{x^1}^1=-u_{y^1}^1,u_{x^1}^2=v_{y^1}^2,v_{x^1}^2=-u_{y^1}^2,u_{x^2}^1=v_{y^2}^1,v_{x^2}^1=-u_{y^2}^1,u_{x^2}^2=v_{y^2}^2,v_{x^2}^2=-u_{y^2}^2.$$ Putting these to use I work out the Jacobian determinant as:$$\begin{vmatrix}u_{x^1}^1&u_{y^1}^1&u_{x^2}^1&u_{y^2}^1\\v_{x^1}^1&v_{y^1}^1&v_{x^2}^1&v_{y^2}^1\\u_{x^1}^2&u_{y^1}^2&u_{x^2}^2&u_{y^2}^2\\v_{x^1}^2&v_{y^1}^2&v_{x^2}^2&v_{y^2}^2\end{vmatrix}$$$$=(u^1_{x^1}u^2_{x^2}+v^1_{x^2}v^2_{x^1})^2+(u^1_{x^1}v^2_{x^2}-u^2_{x^1}v^1_{x^2})^2+(u^1_{x^2}u^2_{x^1}+v^1_{x^1}v^2_{x^2})^2+(u^2_{x^2}v^1_{x^1}-u^1_{x^2}v^2_{x^1})^2-2u^1_{x^1}u^1_{x^2}v^2_{x^1}v^2_{x^2}-2u^1_{x^1}u^1_{x^2}u^2_{x^1}u^2_{x^2}-2u^2_{x^1}u^2_{x^2}v^1_{x^1}v^1_{x^2}-2v^1_{x^1}v^1_{x^2}v^2_{x^1}v^2_{x^2}.$$ I am not being able to make progress beyond this point. The squared terms are obviously not the problem, rather it is the last four terms that are not allowing me to deduce that the determinant be positive. Kindly guide me.

Kong
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  • https://math.stackexchange.com/questions/1817959/almost-complex-manifolds-are-orientable – xsnl Mar 27 '18 at 19:48
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    Searching before posting a question could save you a lot of time. – xsnl Mar 27 '18 at 19:49
  • Another way of seeing that the transition map preserves orientations in the one-dimensional case (real dimension 2) is that an analytic map is conformal away from critical points (preserves orientation and angles for tangent vectors and scales lengths by $|f'(z)|$, from which it follows that the Jacobian determinant is $|f'(z)|^2$). Not familiar enough with multiple complex variables to tell how that will generalize, but I have the definite feeling that will greatly simplify the proof in that case too. – Daniel Schepler Mar 27 '18 at 19:56
  • I did see the related posts, in fact the proof for the case m=1 comes from one of them, don't know how to include related posts in my post though. I preferably wanted to compute along the same lines. – Kong Mar 28 '18 at 03:10

1 Answers1

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The general question has been answered elsewhere, but I somehow share the desire of the OP to see an explicit "certificate" of positivity for the given determinant. To simplify notation let me rewrite the matrix of derivatives as

$$M = \begin{pmatrix} a & b & c & d \\ -b & a & -d & c\\ e & f & g& h\\ -f & e & h & -g \end{pmatrix} $$ Then we have \begin{align*} \operatorname{det} M &= a^2 g^2+a^2 h^2 - 2 aceg - 2acfh - 2adeh + 2adfg +b^2g^2 +b^2h^2\\ & \quad\quad + 2bceh - 2bcfg - 2bdeg - 2bdfh + c^2e^2 + c^2f^2 + d^2e^2 +d^2f^2\\ &=(ah+bg-cf-de)^2+(ag-bh-ce+df)^2. \end{align*}

Lazzaro Campeotti
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