Linear Complex Structure and Kahler Angles

Question

I am trying to read Donaldson's paper on symplectic submanifolds

Link

and am getting a bit stuck on some simple linear algebra at the beginning. On the fourth page of the paper (p. 669) there is the following assertion: let $A: \mathbb{C}^n \rightarrow \mathbb{C}$ be an $\mathbb{R}$-linear transformation and write $A$ as the sum of its linear and antilinear parts $a'$ and $a''$. Then

$A$ has real rank 2 unless $\bar{a''} = e^{i\alpha}a'$ for some $\alpha$

and

the tangent of the Kahler angle is given by $2\sqrt{|a'|^2||a''|^2 - |\langle a', a'' \rangle|^2}/(|a'|^2 - |a''|^2)$.

Donaldson says this is evident upon "a little calculation", but it seems to be eluding me (sorry if the answer is simple).

If someone could give a bit more detailed derivation of these two statements that would be very helpful. Thanks!

Answer 1

For the first assertion, note that $$a'(z)=\frac{1}{2}(A(z)-iA(iz)),\quad a''(z)=\frac{1}{2}(A(z)+iA(iz))$$ If the linear map $A$ has rank $\leq 1$ then its image lies in a line $e^{i\phi}\mathbb{R}$ with $0\leq \phi<\pi$. So $\overline{A}=e^{-2i\phi}A$, from which one gets $\overline{a''}=e^{-2i\phi}a'$.

Checking the assertion about the Kaehler angle is not (to me) an easy problem. This is a quantification of Wirtinger's inequality for 2-forms. You can find a neat derivation in a paper by Cieliebak and Mohnke http://arxiv.org/abs/math/0702887 (go to Remark 8.4).