4
$\begingroup$

Let $M$ be a compact Riemannian manifold with geodesic distance function $d$, of (normalised) diameter $1$.

Some of my favourite manifolds $M$ have the property that there exists an integer $k$ such that for any $k-1$ points $p_1,p_2,..,p_{k-1}$ of pairwise distance $1$ in $M$, there exists a unique $p_k$ with $d(p_k,p_i)=1$ for $i\leq k-1$.

For example, spheres have this property with $k=1$, and less trivially, the Grassmannian of $m$ planes in $mn$ dimensional real or complex space has this property with $k=n-1$ (all with their usual homogenous metrics).

My question is what do Riemannian manifolds with this property look like? Is there a classification/reason for classification to be hopeless? This feels like a natural condition to consider, but as I am not any kind of differential geometer, any pointers to literature would be very welcome.

$\endgroup$
3
  • $\begingroup$ Is this property related to the q-extent of a metric space X which is the maximum average distance between q points in X? i.e. $xt_q(X):=\max_{x_1,\dots,x_q}xt_q(x_1 ,\dots ,x_q)$ where $xt_q(x_1 ,\dots,x_q) = {q\choose 2}^{-1}\sum_{i<j} \operatorname{dist}(x_i , x_j) .$ $\endgroup$
    – C.F.G
    Commented Jun 7, 2023 at 16:55
  • 1
    $\begingroup$ Is it possible to construct Riemannian manifolds of diameter $1$ where there is exactly one $k$-element set of points having pairwise distance $1$, maybe already surfaces? If yes, I would expect that these manifolds can be deformed a bit in some open subset not hitting any of the connecting arcs without losing that property. $\endgroup$ Commented Jun 9, 2023 at 7:46
  • $\begingroup$ Yea having very few possibilities for maximal distance $k$ tuples would probably be more flexible. The examples I had in mind also had a kind of nondegeneracy, where one can complete $i$ tuples to $k$ tuples, but I’d be interested in either case. Even for say, graphs I’d be curious what this property resembles. $\endgroup$
    – Chris H
    Commented Jun 9, 2023 at 10:56

0

You must log in to answer this question.