MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 17 down vote accepted

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).

share|cite|improve this answer
Many thanks for the surprising answer. – Alexander Lytchak Dec 15 '10 at 11:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.