I'm reading the classic paper "Harmonic maps into singular spaces and p-adic superrigidity for lattices in groups of rank one" by Gromov-Schoen. In Section 6, they define the notion of F-connectedness as follows:

We say that a nonpositively curved complex $\mathrm{X}^k$ is F-connected if any two adjacent simplices are contained in a totally geodesic subcomplex $\mathrm{X}_0$ which is isometric to a subset of the Euclidean space $\mathbb{R}^k$.

Here, $k$ is the maximum dimension of the simplices in $\mathrm{X}^k$. I understand the notion of nonpositive curvature (in the Alexandrov sense), but I can't understand how F-connectedness places any further restriction on the geometry of the complex. For the entire paper, the standing assumption has been made that the complex is (locally) isometrically embedded in $\mathbb{R}^n$. It seems to me (and I'm sure this is wrong, but I'm not sure why) that an isometry can be cooked up between the union of any two adjacent simplices and some subset of $\mathbb{R}^k$--since the simplices are of dimension at most k--and that this union should itself be a totally geodesic subcomplex.

In particular, what would be an example of a non-positively curved complex that is not F-connected?

I feel like I'm missing something obvious but important.

  • $\begingroup$ What definition of "adjacent" is being used in the context of this definition? $\endgroup$ – Lee Mosher Aug 1 '15 at 14:26
  • $\begingroup$ For example, if "adjacent" simply means "nonempty intersection", then a pair of 2-simplices identified at a vertex forms a nonpositively curved complex which is not F-connected. $\endgroup$ – Lee Mosher Aug 1 '15 at 14:29
  • $\begingroup$ It is not exactly explained in the paper what they mean by "adjacent," but I believe nonempty intersection is correct. $\endgroup$ – guest Aug 1 '15 at 17:41
  • $\begingroup$ @LeeMosher, Regarding your example, does F-connectedness fail because the inclusion map into R^2 is not an isometry? (By a rotation, we can assume the two simplices are embedded in R^2.) If so, why not? It seems that allowing the isometry to be to any subset of Euclidean space is very unrestrictive. $\endgroup$ – guest Aug 1 '15 at 17:46
  • $\begingroup$ $F$-connectedness of this example would fail because there is no neighborhood of the common vertex $V$ which is isometric to a subset of Euclidean space: for all sufficiently small $r>0$, the "sphere of radius $r$" around $V$ is a union of two closed intervals $I$, $J$ such that the distance function $d(x,y)=2r$ for $x \in I$, $y \in J$ is constant. This does not happen for any subsets of Euclidean space with the Euclidean metric. $\endgroup$ – Lee Mosher Aug 1 '15 at 18:36

Here's a write-up of the comments, giving an example as asked for.

We assume that for two simplices in $X$ to be "adjacent" means that their intersection is nonempty.

Let $X$ be the quotient of a pair of 2-simplices by identifying a vertex on one with a vertex on the other to a single vertex $V$. Each 2-simplex is given the metric of the standard 2-simplex (an equilateral triangle of side length 1). This is extended to a geodesic metric on $X$: the distance between a pair of points $x$ in one 2-simplex and $y$ in the other is the sum of the distances from $x$ and $y$ to $V$.

It follows that for sufficiently small $r$, the sphere around $V$ of radius $r$ contains two components $I$ and $J$, each an interval, with the property that for each $x \in I$ and $y \in J$ we have $d(x,y)=r$. This does not happen for any subsets of Euclidean space with the Euclidean metric.


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.