How big can a wedge of $n$ 2-forms in $\mathbb{R}^{2n}$ be?

Question

$\def\RR{\mathbb{R}}$Let $\omega$ be a $2$-form on $\RR^{2n}$, where $\RR^{2n}$ has the usual Euclidean norm. The comass of $\omega$ is defined to be $\max_{|u|, |v| \leq 1} \omega(u,v)$. Here is the main question:

If $\omega_1$, $\omega_2,\ldots,\omega_n$ are $2$-forms of comass $\leq 1$ on $\RR^{2n}$, what is the maximum possible value of $$|\omega_1 \wedge \omega_2 \wedge \cdots \wedge \omega_n|?$$ In particular, if $\omega_1 = \omega_2 = \cdots = \omega_n = e_1 \wedge e_2 + e_3 \wedge e_4 + \cdots + e_{2n-1} \wedge e_{2n}$, then $\omega_1 \wedge \omega_2 \wedge \cdots \wedge \omega_n = n! (e_1 \wedge \cdots \wedge e_{2n})$, is this $n!$ optimal?

This is a followup to previous posts on Mathoverflow and math.SE; here are observations from those posts (plus a few more new ones):

1. Define the skew-symmetric matrix $A^k$ by $A^k_{ij} = \omega_k(e_i, e_j)$. Then the comass of $\omega_k$ is the spectral radius (largest norm of an eigenvalue) of $A^k$. Recall that the eigenvalues of a skew symmetric matrix are of the form $\pm \lambda_1 \sqrt{-1}$, $\pm \lambda_2 \sqrt{-1}, \ldots, \pm \lambda_n \sqrt{-1}$, so the condition that the comass is $\leq 1$ is equivalent to $-1 \leq \lambda_1, \lambda_2, \ldots, \lambda_n \leq 1$.

2. In fact, we may assume that, for each $k$, the eigenvalues of $A^k$ are $\pm \sqrt{-1}$, see David Speyer's comment on the math.SE post. In this case, the condition on $A^k$ is equivalent to saying that $A^k$ is an orthogonal matrix with $(A^k)^2 = - \text{Id}$.

3. $\omega_1 \wedge \omega_2 \cdots \wedge \omega_n$ is multilinear in the coefficients $A^k_{ij}$ of the skew-symmetric matrices. Specifically, it is the unique symmetric multilinear polynomial $P(A^1, A^2, \ldots, A^n)$ in $n$ skew-symmetrix matrices such that $P(A, A, \ldots, A)$ is $n! \operatorname{Pfaffian}(A)$. If $A$ has eigenvalues $\pm \lambda_1 \sqrt{-1}, \ldots, \pm \lambda_n \sqrt{-1}$, then $\operatorname{Pfaffian}(A) = \pm \lambda_1 \lambda_2 \cdots \lambda_n$, so, in the case $A^1 = A^2 = \cdots = A^n$, equality holds exactly when all the eigenvalues are $\pm \sqrt{-1}$.

4. Previous posts have proved the $n!$ bound in the cases $n=2$ and $n=3$. There is also an easy $n^n$ bound; see David Speyer's comment on the previous MO post.

5. A speculation from David Speyer: If $A$ is a skew symmetric matrix with eigenvalues $\pm \lambda_1 \sqrt{-1}, \ldots, \pm \lambda_n \sqrt{-1}$, then $\sum_{i<j} A_{ij}^2 = \sum \lambda_i^2$. I wonder whether the $n!$ bound might continue to be correct if all that we assume about $A^1, A^2, \ldots, A^n$ is that $\sum_{i<j} (A^k_{ij})^2 \leq n$. (Note that $\sum_{i=1}^n \lambda_i^2 \leq n$ implies $\prod \lambda_i \leq 1$ by AM-GM, so the case $A^1 = A^2 = \cdots = A^n$ still works with this weaker hypothesis.) The case of $n=2$ and two distinct two-forms also works.

A related article appeared here.

Answer 1

Yes! Here is a proof.

Define the norm $|\omega|$, for a $p$-form on $\mathbb R^m$, to be the least upper bound of $|\omega(v_1,\dots ,v_p)|$ for vectors $v_i$ of length one. For any $p$ and $q$ we can ask for the smallest number $C$ such that for every $p$-form $\alpha$ and every $q$-form $\beta$ on $\mathbb R^m$ we have $|\alpha\wedge \beta|\le C|\alpha||\beta|$. Call it $C_{p,q}$. It does not depend on $m$, because if $C$ is valid for forms in $\mathbb R^{p+q}$ and if $v_1,\dots,v_{p+q}$ span $V\cong \mathbb R^{p+q}$ in $\mathbb R^m$ then $$ |(\alpha\wedge \beta)(v_1,\dots ,v_{p+q})|\le C|\alpha_V||\beta_V||v_1|\dots |v_{p+q}|\le C|\alpha||\beta||v_1|\dots |v_{p+q}|,$$ where $\alpha_V$ and $\beta_V$ are the restrictions to $V$.

I claim that $C_{2,2n-2}\le n$. This implies that for $2$-forms $\omega_1,\dots ,\omega_n$ we have $|\omega_1\wedge \dots \wedge \omega_n|\le n|\omega_1\wedge \dots \wedge \omega_{n-1}||\omega_n|$, and therefore by induction on $n$ we have $|\omega_1\wedge \dots \wedge \omega_n|\le n!|\omega_1|\dots |\omega_n|$.

To prove the claim, first observe that $C_{p,q}$ has this other interpretation: it is the smallest $C$ such that the inner product of two $p$-forms on $\mathbb R^{p+q}$ always satisfies $\alpha\cdot \beta\le C|\alpha||\beta|$. This is so because the norm satisfies $|\beta|=|\ast\beta|$ (Hodge star) and $|\alpha\wedge \ast\beta|=|\alpha\cdot\beta|$.

So a restatement of the claim to be proved is that for $2$-forms on $\mathbb R^{2n}$ we have $\alpha\cdot\beta\le n|\alpha||\beta|$. This can be seen by making a change of orthogonal basis so that $$\alpha = a_1e_1\wedge e_2+\dots a_ne_{2n-1}\wedge e_{2n}$$ for some $a_1,\dots ,a_n$. Write $$\beta=b_1e_1\wedge e_2+\dots b_ne_{2n-1}\wedge e_{2n}+\dots ,$$ where the other terms will not matter. Then $$|\alpha|=max|a_i|$$ $$|\beta|\ge max |b_i|$$ $$\alpha\cdot\beta=\sum_i a_ib_i\le n\ max|a_ib_i|\le n|\alpha||\beta|$$.

I did not say why this norm is invariant under Hodge star. Suppose that $\beta$ is a $p$-form in $\mathbb R^{p+q}$. The claim is that $|(\ast \beta)(v_1,\dots,v_q)|\le |\beta||v_1|\dots |v_q|$. Without loss of generality the vectors $v_i$ are orthogonal and of length one. Complete them to an orthonormal basis by vectors $w_1,\dots ,w_p$. Then $$ |(\ast \beta)(v_1,\dots,v_q)|= |\beta(w_1,\dots ,w_p)|=|\beta|.$$

Answer 2

This is an answer to the point/question 5, suggested by David Speyer (as well as a strenghtening of the main inequality, see the Edit below). Consider the norm $$\|\omega\|=\left(\sum_1^n\lambda_j^2\right)^{\frac12}=\sqrt{-{\rm Tr} A^2}.$$ This is a Euclidian norm over the space of skew-symmetric matrices. Then one applies the van der Korput-Schaake inequality (actually discovered by Szegö, and probably known to Banach, see the MO answer). It says that if $P$ is a symmetric $N$-linear form over a Euclidian space $X$, and if $$\forall x\in X,\qquad |P(x,\ldots,x)|\le c\|x\|^N,$$ then $$\forall x_1,\ldots,x_N\in X,\qquad |P(x_1,\ldots,x_N)|\le c\prod_1^N\|x_k\|.$$ In the present case, we have $$|\omega\wedge\cdots\wedge\omega|=n!|\lambda_1\cdots\lambda_n|\le c_n\|\omega\|^n,\qquad c_n=n!n^{-n/2}.$$ We deduce therefore $$|\omega_1\wedge\cdots\wedge\omega_n|\le n!n^{-n/2}\prod_1^n\|\omega_k\|.$$

Edit. As remarked by Mikhail, this can be combined with the obvious inequality $$\|\omega\|\le n^{1/2}\max_j|\lambda_j|,$$ to recover the main inequality that when the comass of each $\omega_k$ is $\le1$, then $$|\omega_1\wedge\cdots\wedge\omega_n|\le n!.$$

Answer 3

I was informed by a colleague that there is also a solution to the Pfaffian formulation of the problem, in Lemma 2.1 of the following article:

Roos, Bero, On Bobkov’s approximate de Finetti representation via approximation of permanents of complex rectangular matrices, Proc. Am. Math. Soc. 143, No. 4, 1785-1796 (2015). ZBL1345.28002.