# Federer's questions on the mass and comass norms

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.

• Some references to recent results on Federer's problems are listed here : mathoverflow.net/questions/176544/… – Pietro Majer Jul 3 '20 at 9:27
• Thanks, I'd seen that when posting but hadn't realized that the second question there is equivalent to the second question here – Quarto Bendir Jul 3 '20 at 16:04
• 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! – Quarto Bendir Jul 3 '20 at 16:06