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.



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


