# Regular simplex in projective space

Is there a reference or a very short argument proving the following statement?

Let $C$ be a set consisting of $r$ points in the real projective space $\mathbb RP ^k$ with its usual round metric. Assume that the distances between all pairs of points in $C$ are the same. Assume further that $r>k+1$. Then $r$ must be equal to $k+2$ and $C$ must be the image of the set of vertices of a regular simplex inscribed in the unit sphere.

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Is it possible that you mean the sphere and not real projective space? – Deane Yang Dec 15 '10 at 2:10
No, I really meant the projective space and am very surprised by the negative answer to my question. – Alexander Lytchak Dec 15 '10 at 11:21

Equidistant points in projective space correspond to equiangular lines in Euclidean space. The maximum number of equiangular lines in $\mathbb R^n$, $f(n)$ is a function which might not be as nice as you think, some computed values are in OEIS. For example, your claim fails even in the case $k=2$ where you can take the six large diagonals of the icosahedron and get 6 equidistant points in $\mathbb RP^2$. An absolute upper bound is given by $f(n)\le \binom{n+1}{2}$ (attributed to Gerzon in "Equiangular lines" by Lemmens and Siedel, proved again later by Koornwinder in "A note on the absolute bound for systems of lines" in both real and complex case).