4
$\begingroup$

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!

$\endgroup$

1 Answer 1

5
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.