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Motivation:

This is related to a different question I asked in April. 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 $H_{2N}$.

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    $\begingroup$ You should consider the extreme points as, if $C$ is a convex body and $x\in EP(C)$ (its extreme points) $C\setminus \{x\}$ is convex, therefore $EP(C)\subset S$. $\endgroup$ – Duchamp Gérard H. E. Jul 5 at 8:34
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    $\begingroup$ Insight : It seems to me that, if $C$ is intersection of closed half-spaces and bounded, then $S=EP(C)$, but I didn't check this formally. $\endgroup$ – Duchamp Gérard H. E. Jul 5 at 9:14
  • $\begingroup$ @DuchampGérardH.E. Thank you for sharing these insights. I suspect that your last point is true and I'm currently thinking about a proof. $\endgroup$ – Aidan Rocke Jul 5 at 9:56
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    $\begingroup$ Good Luck ! Do not hesitate to interact anyway. $\endgroup$ – Duchamp Gérard H. E. Jul 5 at 10:16
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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 $x \in V_{2N}.$

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