I'm looking for references on various inequalities involving the wedge product and exterior forms. The only references I could hunt down are references on Hadamard-Schwarz Inequality. The article of Iwaniec-Kauhanen-Kravetz-Scott may be quoted as an example.

More specifically, I'm interested in the validity of Cauchy-Schwarz type inequality with respect to the wedge product. As a very special case of what I have in mind, I'm interested the necessary and sufficient condition (or some reasonable sufficient condition) for the existence of $b\in\Lambda^4(\mathbb{R}^n),b\neq 0$ satisfying $$ \langle b;\omega_1\wedge\omega_2\rangle^2< \langle b;\omega_1^2\rangle \langle b;\omega_2^2\rangle,\ \langle b;\omega_1^2\rangle>0, \langle b;\omega_2^2\rangle>0, $$ where $\omega_1,\omega_2\in\Lambda^2(\mathbb{R}^n)$ is given.

Thank you.

  • $\begingroup$ What does $\langle b ; a \rangle$ mean? $\endgroup$ – Qiaochu Yuan Nov 17 '12 at 9:32
  • $\begingroup$ It means inner product. $\endgroup$ – tatin Nov 17 '12 at 9:39

Probably, the most reasonable sufficient condition is that ${\omega_1}^2$, $\omega_1\wedge \omega_2$, and ${\omega_2}^2$ be linearly independent in $\Lambda^4(\mathbb{R}^n)$, for this is generic (if $n>4$) and guarantees a solution. In fact, you can find such a $b$ in the $3$-dimensional span $L$ of these $4$-vectors if this linear independence holds. To see this, suppose that they are linearly independent and let $x$, $y$, and $z$ be the linear functions on $L$ defined by $x(\alpha) = \langle \alpha; {\omega_1}^2\rangle$, $y(\alpha) = \langle \alpha; {\omega_2}^2\rangle$, and $z(\alpha) = \langle \alpha; \omega_1{\wedge}\omega_2\rangle$. Then $x$, $y$, and $z$ are coordinates on $L$, and you are asking that the open set defined by the inequalities $x>0$, $y>0$, and $xy >z^2$ be nonempty, which it clearly is.

In the case that there is a linear relation among ${\omega_1}^2$, $\omega_1\wedge \omega_2$, and ${\omega_2}^2$, whether there is a nonempty open set defined by your conditions will depend on the nature of that linear relation (or relations, if there is more than one). However, these are special cases that are easily worked out.

  • $\begingroup$ Unfortunately, we do not have the linear independence of $\omega_1^2, \omega_2^2$ and $\omega_1\wedge \omega_2^2$. What we have is that $$ (\alpha\omega_1+\beta\omega_2)^2\neq 0,\text{ for all }\alpha,\beta\neq 0. $$ In this case, is there any way to guaranty that the open set passes through the region? $\endgroup$ – tatin Nov 20 '12 at 16:01

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