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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.

Xipan Xiao
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  • One should remark that $\omega|W$ takes the form $\sum{i=1}^m\lambda_i u^i\wedge v^i$ for any skew-symmetric bilinear form $\omega|_W$ on any $W$, and even if $dim(W)$ is not even. In general, ${u_1,...,u_m,v_1,...,v_m}$ need not be a basis of $W$ but just a collection of orthonormal vectors, and can be completed to a basis by adding a basis of $ker(\omega|_W)(=ker (pr_W\circ J|_W))$. This is exactly the same construction that shows that, if $J$ is an isometry on $V$ (and hence non-degenerate), it can be represented by the canonical matrix with only $+1$ and $-1$ on the diagonal. – No-one May 18 '25 at 14:57

1 Answers1

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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.

Ted Shifrin
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  • 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 09 '16 at 20:50
  • And how do we prove that the dimension of $W_1^\perp$ is $m-2$? – Xipan Xiao Dec 09 '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 09 '16 at 21:05
  • Let's say when we do the Gram Schmidt process, then $u_2$ is replaced by $u_2-a u_1-b v_1$, however this changes the evaluation of $\omega$ since $\omega(u_2, u_1)=0$ while $\omega(u_2-a u_1-b v_1, u_1)=b\lambda_1$. – Xipan Xiao Dec 09 '16 at 21:46
  • Yes, you're right. My apologies. Although we can probably make this work by a more intelligent process of choosing vectors more carefully, I'm going to give up and do this the usual way. See my edit. – Ted Shifrin Dec 09 '16 at 22:05
  • Thanks. $A^2$ then has a real eigenvalue $-\lambda_1^2$ and an eigenspace of 2 dimension, which is what I was asking for. A minor correction: Hermitian matrices have real eigenvalues (not pure imaginary). – Xipan Xiao Dec 10 '16 at 07:07
  • Yeah, I half rewrote what I was doing and garbled it. I fixed it now. Thanks. – Ted Shifrin Dec 10 '16 at 07:15
  • @TedShifrin I need some help to finish the proof. Why do we get a an inequality for the forms in the end and not for only their absolute values? – klirk Jan 04 '18 at 16:17
  • @klirk: The point is that you get the product of all the $\lambda_j$ (times $m!$ times the volume form), and (with the correct orientations) all the $\lambda_j=1$. – Ted Shifrin Jan 04 '18 at 17:26
  • @TedShifrin hm...did I understand that right? The point is, that as the $\lambda_i$ are constants, the sign of $\omega^m|W$ is fixed. But the volume form is only fixed up to sign, depending on the orientation of the basis, so we can choose an orientation in which the signs of the volume form and $\omega^m|_W$ coincide. \

    How do you get the $\lambda_i=1$ though? By norming the length of the basis vectors accordingly, but then it isn't a orthonormal basis anymore (in general). Why would that be helpful?

     – klirk Jan 04 '18 at 18:07
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    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 04 '18 at 18:51
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    @TedShifrin, what is the usual linear algebra protocol you're talking about? I can't see how taking the real and imaginary parts of the complex eigenvectors will result in the diagonal blocks $\left(\begin{matrix}0 & \lambda_j\ -\lambda_j & 0\end{matrix}\right)$ – rmdmc89 Jun 18 '19 at 14:10
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    @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 '19 at 22:26
  • What confuses me is that $A$ is a $2m\times 2m$ matrix. If we replace $T=A$ in your last comment, we get $2m$ purely imaginary eigenvalues and a $4m\times 4m$ block-diagonal matrix. Right? – rmdmc89 Jun 19 '19 at 12:16
  • Oh, I think I get it. The eigenvalues of $A$ come in pairs $(i\lambda_k, -i\lambda_k)$, so we only need to consider one eigenvalue from each pair. – rmdmc89 Jun 19 '19 at 13:56
  • Why the obtained basis is orthogonal in real sense? – XT Chen Aug 19 '20 at 17:02
  • @XTChen: This is standard linear algebra. We get a unitary basis diagonalizing $iA$. Because $A$ is real, these basis vectors come in conjugate pairs. If $\langle x+iy,x-iy\rangle = 0$, we conclude that $|x|=|y|$ and $x\cdot y = 0$, so, renormalizing, we get a real orthonormal basis $x',y'$. – Ted Shifrin Aug 20 '20 at 01:39