3
$\begingroup$

Let $G:=(E,V,W)$ be a weighted graph and let $d_G$ be its graph metric, defined by on any two edges $e_1,e_2\in E$ by $$ d_G(e_1,e_2)\triangleq \inf_{\gamma}\, \sum_{v\in \gamma} W(v),\qquad\tag{0}\label{0} $$ where the infimum is taken over all sequences of vertices $\gamma=(v_1,\dots,v_t)$ connecting $e_1$ to $e_2$; where $W:V\rightarrow \mathbb{R}$ is the weight function defined on unweighted graph $(E,V)$.

In Ballmann's book Lectures on spaces of nonpositive curvature conditions are given which guarantee that $d_G$ has non-positive curvature (in the sense of Alexandrov); namely, there is some $c>0$ satisfying: $$ \inf_{v}\, W(v)\geq c.\tag{1}\label{1} $$ My question is, when is $(E,d_G)$ a $\operatorname{CAT}(\kappa)$ space, for some $\kappa\in \mathbb{R}$?


My hypothesis is that there exist some partition $C_1,\dotsc,C_N$ of $E$ into open balls such that \eqref{1} holds; where the infimum in \eqref{0} is taken over all paths $\gamma$ contained in $C_i$; where $e_1,e_2\in C_i$.

$\endgroup$
1
  • 3
    $\begingroup$ My rough guess is that a metric graph with smallest cycle of length $r$ is CAT($\kappa$) iff the round 2-sphere with equator of size $r$ (i.e., radius $r/2\pi$) has curvature $\le\kappa$. But I'm really unsure. $\endgroup$
    – YCor
    Dec 20, 2021 at 17:05

1 Answer 1

3
$\begingroup$

The generalized Cartan--Hadamard theorem states that a length space is CAT(1) if it is locally CAT(1) + any closed curve of length $<2{\cdot}\pi$ is null-homotopic in class of curves of length $<2{\cdot}\pi$.

If the space is a graph, then the latter is equivalent to saying that any cycle has length at least $2{\cdot}\pi$.

$\endgroup$
2
  • $\begingroup$ I can't manage to track down this generalized Cartan-Hadamard theorem; where could I find it/does it have some other name? $\endgroup$ Dec 21, 2021 at 9:58
  • 1
    $\begingroup$ @Carl_Petterson See p. 119 here arxiv.org/abs/1903.08539v1. $\endgroup$ Dec 21, 2021 at 10:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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