Do highly symmetric cones have "small" supporting hyperplanes? Let $C$ be a full-dimensional cone in $\mathbb{R}^{d}$, defined as the positive span of $c = {n \choose 3} \gg d$ vectors.  $C$ is highly symmetric in the following sense: each such vector is labelled with a distinct unordered triple from $[n]$, and the automorphism group of $C$ is basically $S_n$: for each permutation $\pi : [n] \to [n]$, there is a linear transformation of $\mathbb{R}^d$ that takes each vector labelled $(x, y, z)$ to the vector $(\pi(x), \pi(y), \pi(z))$.
My intuition is that any supporting hyperplane to $C$ can have only $O(d/n)$ of these vectors, roughly because the symmetry constraints prevent too many supporting vectors from being concentrated on one side of the cone.  Is this true?
 A: The prescribed group action allows to start from it and build the cones in question. It is well-known how the permutation representation of $S_n$ on 3-subsets decomposes into irreducibles - there will be just 4 of these, one of them trivial 1-dimensional one. With the other ones (more precisely, subsets of these) you have a graph embedding, affording the group action. You could start by figuring out which ones satisfy your $c\gg d$ condition, and basically classify all the possibilities for $C$.
Specifically, the irreducibles correspond to partitions of $n$ as
$n=(n-j)+j$, with $0\leq j\leq 3$, and dimensions are given by
$$
d_j:=\frac{n-2j+1}{n-j+1}\binom{n}{j},\qquad 0\leq j\leq 3,
$$
Now, $d_j$ is a degree $j$ polynomial in $n$, and I guess that this only leaves $j=1$ as the $d=d_1$ or $d=d_1+1=n$, satisfying $c\gg d$.

E.g. in the case $d=n$ you get the generators of $C$ given by indicator vectors of the 3-subsets, and one should be able to expicitly classify the facets of this cone.  In any event, the facets will contain at least $O(n)$ of the generators, and not $O(1)$.
So it seems that the question needs an adjustment to make sense of this case.
