# Proof of the Wirtinger inequality

This is an exercise from the book "Complex Geometry, An Introduction" by Huybrechts. The statement involves to prove that the restriction of the fundamental form $\omega$ of a vector space $V, I, \langle, \rangle$ (with real dimension $2n$ and an almost complex structure $I$ compatible with the inner product $\langle, \rangle$), to a sub-space $W$ of real dimension $2m$, is not greater than $m! vol$ where $vol$ is the volume form of $W$.

$$\omega^m|_W\le m!\cdot vol$$ The book has a hint: There exists an orthonomal basis $\{u_i, v_i:i=1,2,\cdots, m\}$ of $W$ such that $$\omega|_W=\sum_{i=1}^m \lambda_i u^i\wedge v^i$$ Where $u^i, v^i, i=1, 2, \cdots, m$ are the dual basis. From this one is easy to prove the conclusion.

I did find a proof without using the hint from the book "Complex Analytic Sets" by E.M. Chirka and I understood the proof well. My question is how to prove the expression in the hint.

Let $P: V\rightarrow W$ be the orthogonal projection and $J=P\circ I$ then I was led to the conclusion that the existence of the above expression depends on a fact that $J^2: W\rightarrow W$ has a real eigenvalue whose eigenspace is of dimension (at least) $2$. Then I don't know how to prove this.

Any hints are appreciated.

After reading Ted's answer I realized that $J$ is actually skew-symmetric for $\langle Jv, w\rangle=\langle Iv, w\rangle$ since $(Iv-Jv)\perp W$. Then $J$ has the desired block diagonal form.

This is just linear algebra (and nothing to do with projection). It's really just the process for obtaining the normal form for a skew-symmetric bilinear form.

First of all, if $\omega|_W$ is degenerate, then the result is obvious, as $\omega^m|_W = 0$. So we suppose $\omega|_W$ is nondegenerate. This means that $\omega|_W$ is represented by a nonsingular skew-symmetric real $2m\times 2m$ matrix $A$ (here I'm assuming that $\langle\cdot,\cdot\rangle$ is a real inner product on $V$). $\sqrt{-1}A$ is a hermitian matrix, which is unitarily diagonalizable by the Spectral Theorem, and so therefore is $A$, with pure imaginary eigenvalues. Following the usual linear algebra protocol (taking real and imaginary parts of the complex eigenvectors), we get a real normal form for $A$ with diagonal blocks of the form $\begin{bmatrix} 0 & -\lambda_j \\ \lambda_j & 0\end{bmatrix}$ with respect to an orthonormal basis. [Note that I don't need to assume nondegeneracy. Degeneracy will manifest itself by $\lambda_j=0$ for some $j$.]

That is, we end up with an orthonormal basis $\{u_1,v_1,\dots,u_m,v_m\}$ for $W$. Letting $\{u^1,v^1,\dots,u^m,v^m\}$ denote the dual basis, this means that $$\omega|_W = \sum_{j=1}^m \lambda_j u^j\wedge v^j$$ for some scalars $\lambda_j$.

Remember that $\omega(u,v) = \langle Iu,v\rangle$. Then $\omega(u_j,v_j) = \lambda_j = \langle Iu_j,v_j\rangle$. By Cauchy-Schwarz, $$|\lambda_j| = \left|\langle Iu_j,v_j\rangle\right| \le \|Iu_j\|\|v_j\| = \|u_j\|\|v_j\| = 1,$$ and you know how to finish it from here.

• Thanks Ted. I think what I tried was to decompose using the inner product instead of using $\omega$ itself, which did not work. Question: is it necessary that $v\perp W_1, \forall v\in W_1^\perp$? – Xipan Xiao Dec 9 '16 at 20:50
• And how do we prove that the dimension of $W_1^\perp$ is $m-2$? – Xipan Xiao Dec 9 '16 at 20:58
• What do you mean when you say $v\perp W_1$? You mean using the inner product? No, definitely not. That's why I said you had to do Gram-Schmidt a bunch again. ... Nondegeneracy is what controls dimensions. Without that, dimensions could be large. But the dimension count should just come directly from nullity-rank. – Ted Shifrin Dec 9 '16 at 21:05
• Sorry. I haven't thought about this in ages. We have the inequality $|\lambda_j|\le 1$, so that gives the desired inequality. (You're worrying about absolute values, but evaluate on a correctly-oriented orthonormal basis for $W$.) – Ted Shifrin Jan 4 '18 at 18:51
• @rmdmc89: If $T(x+iy) = (\alpha+i\beta)(x+iy)$, then taking real and imaginary parts, we get $T(x) = \alpha x - \beta y$ and $T(y) = \beta x +\alpha y$, which gives the matrix $\begin{bmatrix} \alpha & \beta \\ -\beta & \alpha\end{bmatrix}$ with respect to the basis $\{x,y\}$. – Ted Shifrin Jun 18 at 22:26