3
$\begingroup$

Let $M$ be a connected and simply connected compact Riemannian manifold (without boundaries). Fix a point $p\in M$. The diameter set $D_p$ of $p$ is the set of points that maximize the distance from $p$, i.e., $D_p=\lbrace q : d(p,q)=\max_r d(p,r)\rbrace$. The cut locus $C_p$ is the set of points at which geodesics (starting at $p$) stop to be length minimizing.

It is clear that $D_p \subset C_p$. What are conditions under which equality holds? Does it always hold or is it, for instance, true for symmetric spaces?

Improve this question
$\endgroup$
8
  • 3
    $\begingroup$ For the first question - 'Does it always hold [...]?' - you can take a long cylinder $C$ modelled on $S^1 \times (0,10)$ for example, capped off to make it compact. For a point on the equator $S^1 \times \{ 0 \}$, its cut locus includes points lying on a vertical line, on the 'opposite' side of the cylinder. Most of these are much closer than the maximizers. $\endgroup$
    – Leo Moos
    Commented Mar 13, 2023 at 14:16
  • $\begingroup$ It also doesn't hold on symmetric spaces in general. For example, on $T^2$ with its standard flat metric (by identifying sides of a square), if $p$ denotes the "center" point, then $D_p$ is the unique corner point while $C_p$ is the entire "boundary" square. I don't know what happens for irreducible symmetric spaces. $\endgroup$
    – Jason DeVito - on hiatus
    Commented Mar 13, 2023 at 18:22
  • $\begingroup$ @Leo Moos, Jason DeVito yes I think both points are true but both these counterexamples are not simply connected $\endgroup$
    – Lau
    Commented Mar 13, 2023 at 19:31
  • 1
    $\begingroup$ @Lauritz How is my example not simply connected - it's a sphere? In any case, the $Sp(n)$ example that Jason gives below is simply connected, so I think it answers your question. $\endgroup$
    – Leo Moos
    Commented Mar 13, 2023 at 19:45
  • 1
    $\begingroup$ Any compact connected Riemannian manifold with $C_p=D_p$ has the cohomology of a unique compact rank one symmetric space; see arxiv.org/abs/1309.1326 $\endgroup$
    – Ben McKay
    Commented Mar 14, 2023 at 13:43

1 Answer 1

Reset to default
5
$\begingroup$

The equality $D_p = C_p$ holds for compact symmetric spaces (CROSSes) of rank $1$, but not in general for higher rank symmetric spaces.

A rank $1$ CROSS is isometric to a round sphere or projective space with Fubini-Study metric. For a sphere, we of course have $D_p = C_p = \{-p\}$. On $\mathbb{K}P^n$ with $\mathbb{K}\in\{\mathbb{R},\mathbb{C},\mathbb{H},\mathbb{O}\}$, we have $D_p = C_p \cong \mathbb{K}P^{n-1}$.

For higher rank, consider $Sp(n)$ (the $n\times n$ quaternionic unitary matrices) with bi-invariant metric and set $p = I$, the identity matrix.

We claim that $D_p = \{-I\}$. Indeed, given $q\in D_p$, we may select a minimizing geodesic $\gamma$ from $p$ to $q$. This geodesic is a $1$-parameter subgroup, so can be conjugated to lie within the standard maximal torus $T^n\subseteq Sp(n)$. On the torus (which inherits its standard flat metric) clearly $-I$ is the farthest point from $I$. So, $q$ is conjugate to $-I$, which is central.

On the other hand $C_p$ contains infinitely many points. To see this, recall that $Sp(n)\setminus \{p\}$ deformation retracts onto $C_p$. This implies that in cohomology we have $H^3(C_p)\neq 0$ for $n\geq 2$

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Actually, not for any higher rank symmetric space, since it contains a totally geodesic torus. $\endgroup$
    – Ben McKay
    Commented Mar 14, 2023 at 13:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .