$\def\RR{\mathbb{R}}$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):

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 $-1 \leq \lambda_1, \lambda_2, \ldots, \lambda_n \leq 1$.

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 $(A^k)^2 = - \text{Id}$.

$\omega_1 \wedge \omega_2 \cdots \wedge \omega_n$ 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 $\pm \sqrt{-1}$.

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.

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.

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