How to find extreme points of a set related to Minkowski's Theorem?

Question

Let $S^{n-1}$ be the unit sphere in $\mathbb{R}^n$. For $m>n$, we can define $\Lambda$ to be the set $$\{(\lambda_1, ..., \lambda_m):\sum_{i=1}^m \lambda_i=1, \lambda_i\ge0, and \mbox{ there exist}\, v_i \in S^{n-1}, i=1, ..., m \, \mbox{such that}\, \sum_{i=1}^m \lambda_i v_i=0 \}.$$

The motivation of the set $\Lambda$ is that $\lambda_i$ can serve as the surface area of a $m$-face polytope in $\mathbb{R}^n$ by Minkowski's Theorem.

My question is how to find extreme points of $\Lambda$. My guess is that $m-n$ component of the extreme points in $\Lambda$ must be zero, but I've no idea how to prove this. Can anyone give me some hint? Thanks in advance.

Answer 1

I'll assume $m, n \ge 2$. I claim that

$$\Lambda = \left\{\lambda \in \mathbb R^m: \; \sum_{i=1}^m \lambda_i = 1,\; 0 \le \lambda_i \le \frac{1}{2} \ \text{for all}\ i\right \}$$

The necessity of the condition $\lambda_i \le 1/2$ comes from the fact that for $v_j \in S^{n-1}$, $$\left\|\sum_{j=1}^m \lambda_j v_j \right\| \ge \lambda_i - \sum_{j \ne i} \lambda_j = 2 \lambda_i - 1$$

Conversely, suppose $\lambda \in \mathbb R^{m}$ with $\sum_i \lambda_i = 1$ and $0 \le \lambda_i \le 1/2$. WLOG we have $0 \le \lambda_1 \le \ldots \le \lambda_m$. Let $$\eqalign{U &= \left\{ \sum_{j=1}^{m-1} \lambda_j v_j: v_j \in S^{n-1}\right\}\cr L &= \lambda_{m-1} - \lambda_{m-2} + \ldots \pm \lambda_1 = \sum_{j=1}^{m-1} (-1)^{m-1-j} \lambda_j\cr R &= \lambda_{m-1} + \lambda_{m-2} + \ldots + \lambda_1 = \sum_{j=1}^{m-1} \lambda_j = 1 - \lambda_m\cr}$$ We have $0 \le L \le \lambda_{m-1} \le \lambda_m \le R$. $U$ contains vectors of norm $L$, e.g. $\sum_{j=1}^{m-1} \lambda_j (-1)^{m-1-j} \lambda_j v$ for any $v \in S^{n-1}$, and vectors of norm $R$, e.g. $\sum_{j=1}^{m-1} \lambda_j v$ for any $v \in S^{n-1}$. Since $S^{n-1}$ is path-connected, so is $U$, so it contains an element of norm $\lambda_m$, say $\lambda_m u = \sum_{j=1}^{m-1} \lambda_j v_j$ with $u \in S^{n-1}$, and then $ 0 = \lambda_m u - \sum_{j=1}^{m-1} \lambda_j v_j$. This proves the claim.

Now I claim the extreme points of $\Lambda$ are the vectors with two coordinates $1/2$ and the others $0$. It's easy to see that any vector of this form is an extreme point of $\Lambda$. Conversely, if $\lambda \in \Lambda$ is not of this form, there are at least two coordinates, say $i$ and $j$, where $0 < \lambda_i, \lambda_j < 1/2$. Then for $\epsilon > 0$ sufficiently small, $\lambda$ is the average of $\lambda + \epsilon (e_i - e_j) \in \Lambda$ and $\lambda - \epsilon (e_i - e_j) \in \Lambda$ (where $e_i$ and $e_j$ are standard unit vectors), so $\lambda$ is not an extreme point.