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Is there a quadrangle $Q \subset \Bbb CP^2$, namely $Q$ is a set of four points, such that every permutation of $Q$ can be realizad by an isometric projectivity of $\Bbb CP^2$? Clearly the analogous question for a triangle in $\Bbb CP^1 \cong S^2$ has affirmative answer: it is a maximal equilateral triangle.

More generally, is there a subset of $n+2$ points in $\Bbb CP^n$ such that all of its permutations are restrictions of some isometric projectivity?

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That is true, since this is so for $\mathbb RP^n$ - take $n+2$ vertices of a "regular simplex" in it -- i.e. take the regular simplex in $S^n$ and project its vertices to $\mathbb RP^n$.

In coordinates, take points $(1,0,\ldots,0)$, ... $(0,\ldots,0,1)$ in $\mathbb R^{n+2}$, take the hyperplane $\sum_i x_i=1$ and take the $S^n$ in it that passes through these points. Then project this to $\mathbb RP^n$ by the central symmetry of $S^n$

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  • $\begingroup$ I don't understand why transpositions are realizable by isometries $\endgroup$
    – Daniele Zuddas
    Mar 18, 2018 at 19:32
  • $\begingroup$ Because this is so for $S^n$ $\endgroup$
    – Dmitri Panov
    Mar 18, 2018 at 19:33
  • $\begingroup$ but a simplex in $\Bbb RP^n$ has at most $n+1$ vertices, depending on its dimension $\endgroup$
    – Daniele Zuddas
    Mar 18, 2018 at 19:35
  • $\begingroup$ THE regular simplex in $S^n$ has $n+2$ vertices. The example that I propose is just the generalization of yours $\endgroup$
    – Dmitri Panov
    Mar 18, 2018 at 19:36
  • $\begingroup$ sorry, I misunderstood. So you take a regular simplex in $\Bbb R^n$ all of whose vertices belong to $S^n$, isn't? $\endgroup$
    – Daniele Zuddas
    Mar 18, 2018 at 19:43

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