$\def\RR{\mathbb{R}}$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](https://mathoverflow.net/questions/448599/how-big-can-a-wedge-of-2-forms-be) and [math.SE](https://math.stackexchange.com/questions/4713029/how-big-can-a-wedge-of-2-forms-be); 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](https://math.stackexchange.com/posts/comments/9985908). 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](https://mathoverflow.net/posts/comments/1159176).
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](https://doi.org/10.1007/s00022-023-00685-3).



How big can a wedge of $n$ 2-forms in $\mathbb{R}^{2n}$ be?
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
$$
|(\ast \beta)(v\_1,\dots,v\_q)|= |\beta(w\_1,\dots ,w\_p)|=|\beta|.$$


