Is it true that $E^2$ is the only Euclidean space, in which the convex hull of $n+2$ points in convex configuration has two inner diagonals and in all other cases there is only one such diagonal?

In this context an inner diagonal shall be a line-segment that connects two corners of the convex hull and, for which each inner point can be expressed as the convex-combination of all corners and all weights in the open unit interval,
i.e. if the inner points of the segment connecting corners $p_i$ and $p_j$ can be expressed as $\ \alpha p_i+(1-\alpha)p_j\ $ and, as $\ \sum_{k=1}^{n+2}\beta_kp_k$, $\quad0\lt \alpha, \beta_k\lt 1$

In $E^1$ maximally $2$ points can be in convex configuration, but the linesegment connecting the two points complies with the definition of inner diagonals.


1 Answer 1


If $p_1,\dots,p_{n+2}$ are the vertices and $\sum \lambda_i p_i=0$, $\sum \lambda_i=0$, is their (let it be unique for a moment) affine dependence, the point $q=\alpha p_i+(1-\alpha) p_j$ on the diagonal between $p_i$ and $p_j$ has affine representations of the form $q=\alpha p_i+(1-\alpha) p_j+c\sum_k \lambda_k p_k$ for variable $c$. We see that if we need all coefficients being positive, the coefficients $\lambda_k,k\notin \{i,j\}$ must have the same sign. If $n>2$, this may happen for at most one pair $\{i,j\}$ by obvious reasons.

If the affine dependence between the vertices is not unique, there exist $n+1$ vertices belonging to the same hyperplane (and in convex position in this hyperplane) and it is seen that we have no inner diagonal.


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