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

$$\def\RR{\mathbb{R}}$$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 $$\sum_{i. I wonder whether the $$n!$$ bound might continue to be correct if all that we assume about $$A^1, A^2, \ldots, A^n$$ is that $$\sum_{i. (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.

• Would you mind if I edited this to include the background from the previous questions? While at it, I'd also really like to tweak the notation in two ways: (1) Ask about $n$ two-forms, so as not to suggest an infinite sequence of questions :) and (2) use the notation $\alpha_1$, $\alpha_2$, ..., $\alpha_n$ rather than $\alpha$, $\beta$, $\gamma$, ... since the alphabetic notation is a pain to engage with. (EG: Imagine writing "let the eigenvalues of $\alpha$ be $(\rho_1, \rho_2, \dots, \rho_{2n})$, the eigenvalues of $\beta$ be $(\sigma_1, \sigma_2, \dots, \sigma_{2n})$ etcetera ...") Commented Jun 13, 2023 at 13:59
• @DavidESpeyer I would much appreciate it, go ahead. Commented Jun 13, 2023 at 14:00
• Okay, edits made. As you can tell, I like this question! Commented Jun 13, 2023 at 14:18
• I'll make one more comment. Here is a variant of the problem which doesn't mention skew-symmetry and which might already be known. Let $D(A^1, A^2, \ldots, A^m)$ be the unique symmetric multilinear polynomial in $m$ many $m \times m$ matrices such that $D(A, A, \dots, A) = \det(A)$. Suppose that each $A^k$ has operator norm $\leq 1$. Can we conclude that $|D(A^1, A^2, \ldots, A^m)| \leq 1$? Commented Jun 13, 2023 at 14:59
• I will be utterly astonished if there is a larger maximum than the n! as in the question. Commented Jun 15, 2023 at 21:59

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 $$|(\ast \beta)(v_1,\dots,v_q)|= |\beta(w_1,\dots ,w_p)|=|\beta|.$$

• Nice answer, this one might go in THE BOOK! Commented Jun 18, 2023 at 15:18

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 $$|\omega_1\wedge\cdots\wedge\omega_n|\le n!.$$

• Can this be considered a strengthening of the inequality for 2-forms of comass 1, seeing that $\max \lambda_j\geq \frac{\|\omega\|}{\sqrt{n}}$ ? Commented Jun 19, 2023 at 8:26
• @MikhailKatz It seems so. Commented Jun 19, 2023 at 9:17

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.

• You can use "insert citation" button to find well-formatted citation. Commented Jun 18, 2023 at 16:17
• And according to Roos, the proof is due to Hörmander in his paper "On a theorem of Grace", where it appears to be Theorem 4. Commented Jun 18, 2023 at 16:27
• @DeaneYang In the real case, Roos points out that the proof is due already to Banach. Commented Jun 18, 2023 at 16:29