The comass of a 2-form $\alpha$ is the maximal value of $\alpha(u,v)$ for a pair of unit vectors $u,v$. The symplectic form $\alpha$ on $\mathbb R^{2n}$ has the property that $|\alpha^{\wedge n}| = n!$. For example, in $\mathbb R^6$ one has $|\alpha^{\wedge 3}| = 6$. Are there 2-forms $a,b,c$ of unit comass on $\mathbb R^6$ such that $|a\wedge b\wedge c|>6$ ?
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asked Jun 10, 2023 at 20:26
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4$\begingroup$ Isn't the wedge product (with the normalization making $|\alpha^{\wedge n}|=n!$) taken to be $a_1\wedge a_2\wedge a_3=\sum_{\sigma\in S_3}(-1)^{|\sigma|}a_{\sigma(1)}\otimes a_{\sigma(2)}\otimes a_{\sigma(3)}$, so that the triple wedge is just an alternating sum of six terms, made up of the tensor product of unit comass forms? In which case, the inequality $|a\wedge b\wedge c|\leq 6$ should just be the triangle inequality. $\endgroup$– Ben JohnsrudeCommented Jun 10, 2023 at 20:58
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3$\begingroup$ @BenJohnsrude: The question is about 2-forms, not 1-forms, so the triple wedge is actually a symmetric product, not an alternating one. $\endgroup$– Robert BryantCommented Jun 11, 2023 at 0:28
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4$\begingroup$ The triple wedge is a sum of $6!/2^3$ terms, not $3!$ terms. $\endgroup$– David E SpeyerCommented Jun 11, 2023 at 0:30
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3$\begingroup$ We can write each $2$-form $\alpha_i$ as $\sum_j v_{ij} \wedge w_{ij}$ with $v_{i1}$, ..., $v_{in}$, $w_{i1}$, ..., $w_{in}$ orthogonal and length at most $1$. Then $\alpha_1 \wedge \cdots \wedge \alpha_n$ is a sum of $n^n$ wedge products, and the triangle inequality gives $|\alpha_1 \wedge \cdots \wedge \alpha_n| \leq n^n$. This beats $(2n)!/2^n$ from the more naive triangle bound, but it isn't as good as $n!$, which I suspect is the right bound. $\endgroup$– David E SpeyerCommented Jun 11, 2023 at 0:44
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2$\begingroup$ For everyone's convenience, link to the math.SE thread math.stackexchange.com/questions/4713029/… $\endgroup$– David E SpeyerCommented Jun 11, 2023 at 0:49
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I think I figured this one out in the end. It is easy to show that it suffices to prove that for the standard symplectic form $\alpha=e_1\wedge e_2 + e_3\wedge e_4 + e_5\wedge e_6$, one has $|\alpha\wedge b \wedge c|\leq6$ for all 2-forms $b,c$ of comass $1$. Consider the endomorphism $M_\alpha$ of $\bigwedge^2(\mathbb R^6)$ sending $b$ to $\ast(\alpha\wedge b)$. Then $\bigwedge^2(\mathbb R^6)$ decomposes into invariant blocks $V+W$ where $V$ is spanned by $e_1\wedge e_2$, $e_3\wedge e_4$, and $e_5\wedge e_6$, whereas $W$ is spanned by the remaining $e_i\wedge e_j$. The restriction of $M_\alpha$ to $W$ is given by a permutation matrix, whereas the restriction to $V$ is $\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}$, with eigenvalues $-1,-1,2$. Hence the spectral radius of $M$ is $2$. Finally, by Cauchy-Schwarz $|\alpha\wedge b\wedge c|= |\langle M_\alpha(b),z\rangle|\leq2|b|\,|c|\leq 6 \|b\|^\ast \|c\|^\ast$. This calculation has apparently eluded researchers in the field for 40 years, and leads to a natural extension of Gromov's optimal (i.e., tight) stable systolic inequality for the complex projective space.
answered Jun 12, 2023 at 6:54
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$\begingroup$ For the general case, see mathoverflow.net/q/448770/613 $\endgroup$ Commented Jun 17, 2023 at 16:15