In Federer's book "Geometric measure theory", he says in section 1.8.3 (where $\|\cdot\|$ is the comass norm):

Very little appears to be known about the structure of the convex sets $\wedge^m(\mathbb{R}^n)^\ast\cap\{\phi:\|\phi\|\leq 1\}$. What are their extreme points?

In section 1.8.4 he says:

Suppose $S$ and $T$ are mutually orthogonal subspaces of an inner product space $V$, $s:S\to V$ and $t:T\to V$ are the inclusion maps, $\xi\in\operatorname{im}\wedge_ps$ and $\eta\in\operatorname{im}\wedge_qt$. The equation $\|\xi\wedge\eta\|=\|\xi\|\cdot\|\eta\|$ holds if either $\xi$ or $\eta$ is simple. [...] I do not know whether the above equation holds in case neither $\xi$ nor $\eta$ is simple.

Are these matters understood by now?

edit 1, for completeness: given a finite-dimensional real inner product space $V$, the comass of $\phi\in\wedge^mV^\ast$ is defined by $$\|\phi\|=\sup\Big\{\phi(\xi):\text{decomposable }\xi\in \wedge^mV\text{ with }|\xi|\leq 1\Big\}$$ and the mass of $\xi\in\wedge^mV$ is defined by $$\|\xi\|=\sup\Big\{\phi(\xi):\phi\in\wedge^mV^\ast\text{ with }\|\phi\|\leq 1\Big\}.$$ Here $|\cdot|$ is the standard norm on $\wedge^mV$, defined by an orthonormal basis $e_{i_1}\wedge\cdots\wedge e_{i_k}$. It seems to me that the mass and comass are essentially analogous to (and perhaps the inspiration for) the Gromov norm on homology.

edit 2 : As pointed out in the answers to the question Pietro Majer links to, the question of 1.8.4 is addressed in Frank Morgan's article "The exterior algebra $\Lambda^k({\mathbb R}^n)$ and area minimization" (Linear Algebra and its Applications Volume 66, April 1985, Pages 1–28) where certain cases (by dimension) are proved.

  • $\begingroup$ Some references to recent results on Federer's problems are listed here : mathoverflow.net/questions/176544/… $\endgroup$ – Pietro Majer Jul 3 '20 at 9:27
  • $\begingroup$ Thanks, I'd seen that when posting but hadn't realized that the second question there is equivalent to the second question here $\endgroup$ – Quarto Bendir Jul 3 '20 at 16:04
  • $\begingroup$ I have to admit it's hard for me to understand, from a high-level perspective, why the comass is hard to understand. As I understand it, it's just the minimization of a linear function with a quadratic constraint. It feels like it should be more tractable! $\endgroup$ – Quarto Bendir Jul 3 '20 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.