Motivation:
-----------


This is [related to a different question I asked in April](https://mathoverflow.net/questions/328558/understanding-the-geometry-of-h-n-vecx-in-n-nn-sum-i-1n-x-i-0). It occurred to me while thinking about the sums of uniform random variables and it stuck in my mind because it's the special case of a more general problem:



> 
> Given the hyperplane cut of $[-1,1]^N$,$H\_{N}=\{\vec{x} \in [-1,1]^N:\sum\_{i=1}^N x\_i = 0\}$, what is the vertex set of $H\_N$?
> 
> 
> 


which reduces to the original question when $N \in 2\mathbb{N}$.


 My conjecture for this problem:
--------------------------------


I conjecture that:


\begin{equation}
V\_{2N}=\{\vec{x} \in \{-1,+1\}^{2N}:\sum\_{i=1}^{2N} x\_i = 0\} \tag{1}
\end{equation}


is the smallest set such that $\mathrm{conv}(V\_{2N})=H\_{2N}$ where we note that:


\begin{equation}
\lvert V\_{2N} \rvert= {2N \choose N} \tag{2}
\end{equation}


and it's relatively easy to show that:


\begin{equation}
\mathrm{conv}(V\_{2N}) \subset H\_{2N} \tag{3}
\end{equation}


So far I haven't managed to show that $\mathrm{conv}(V\_{2N}) = H\_{2N}$ but I have managed to show that if we define:


\begin{equation}
\forall k < N, V\_{2k}=\{\vec{x} \in H\_{2N}:\sum\_{i \in \Gamma} x\_i = \sum\_{j \notin \Gamma} \lvert x\_j \rvert = 0 \land \{x\_i\}\_{i \in \Gamma} \in \{-1,+1\}^{2k} \} \tag{4}
\end{equation}


then we have:


\begin{equation}
\forall k < N-1, \mathrm{conv}(V\_{2k}) \subset \mathrm{conv}(V\_{2k+2}) \tag{5}
\end{equation}


I have spent some time thinking about this problem and at this point I'm not sure how to proceed. The challenge is basically to show that $V\_{2N}$ is the vertex set of $H\_{2N}$.



Given $H_{N}=\{\vec{x} \in [-1,1]^N:\sum_{i=1}^N x_i = 0\}$, what is the smallest subset $S \subset H_{2N}$ such that $\mathrm{conv}(S)=H_{2N}$
It suffices to verify (see the comment by Duchamp Gérard H. E. and (<https://en.wikipedia.org/wiki/Extreme_point>)) that the set $V\_{2N}$ is precisely the set of extreme points ${\bf EP}(H\_{2N})$ of $H\_{2N}$. Clearly 
$V\_{2N} \subset {\bf EP}(H\_{2N})$. To see the converse, suppose $x \in {\bf EP}(H\_{2N})$.
Let $\{e\_j\}\_{j=1}^{2N}$ denote the standard basis vectors in ${\bf R}^{2N}$. If there were two coordinates $1 \le i<j \le 2N$ such that $|x\_i|<1$ and $|x\_j| <1$, then the representation
$$x=\frac{(x+\delta (e\_i-e\_j))+(x-\delta (e\_i-e\_j))}{2}$$
for small enough $\delta$, would show that $x$ is not an extreme point of $H\_{2N}$.
If precisely one index $i \le 2N$ satisfies $|x\_i|<1$, then the sum 
$\sum\_{j \le 2N \,: \, j \ne i}x\_j$ is $\pm 1$, so $x$ could not be in $H\_{2N}$.
The only remaining possibility is that $x \in V\_{2N}.$


