Background: Let $W$ be a finite reflection group of rank $n$, acting on $\mathbb{R}^n$. The reflecting hyperplanes of $W$ meet the unit sphere $S^{n-1}\subset\mathbb{R}^n$, inducing a simplicial complex structure on $S^{n-1}$ ("Coxeter complex of $W$"). $W$ acts freely and transitively on the facets of this triangulation ("the chambers"). Fixing a specific chamber, the $n$ reflections in the $(n-1)$-faces ("walls") of this chamber are "Coxeter generators" and one gets a metric on the chambers: given chambers $P,Q$, find the unique $w\in W$ with $wP=Q$ and define $d(P,Q)$ to be the minimum length of the word in the Coxeter generators required to represent $w$. The Cayley graph of $W$ with respect to the Coxeter generators can be identified with the graph with vertices and edges the chambers and walls of the Coxeter complex; then this metric is the graph metric.

One also gets a metric by picking a distinguished point of each chamber and then just taking the distance between chambers $P$ and $Q$ to be the Riemannian distance in the sphere $S^{n-1}$. To pin down this possibility, let us use the barycenter of each chamber as its distinguished point.

Question: How do these two metrics relate?

It seems to me that they should be pretty closely related. The Coxeter complex is a very regular object. For example, $S_4$ acting in its standard representation on $\mathbb{R}^3$ is such a Coxeter group; the barycenters of the coxeter complex form the vertices of a truncated octahedron, embedded in $S^2$. The Cayley-graph distance is the path-length of the shortest path between vertices. Vertices further away on the sphere $S^2$ also require longer paths.

But can this be made precise? E.g. are there universal positive constants $A,\varepsilon$ with $\varepsilon$ reasonably small so that $A(1-\varepsilon)d_{graph}(P,Q)\leq d_{Riemannian}(P,Q)\leq A(1+\varepsilon)d_{graph}(P,Q)$? Or are there such constants that depend on $n$ in an explicitly controlled way?

Motivation: The present question follows up an attempt I made to answer this question. I posted a partial answer with the uncertainty in my answer coming precisely from my uncertainty about the present question.

  • $\begingroup$ What is $S_4$? The symmetric group? $\endgroup$ – Claudio Gorodski Oct 4 '16 at 2:09
  • $\begingroup$ If the distinguished point $P$ in a chamber $C$ is close to a wall $r$ of $C$, the spherical distance to the reflected point $rP$ in $rC$ would be close to 0 (in fact, can be made arbitrarily close). Thus the choice of distinguished point is crucial to any sort of comparison result. $\endgroup$ – Victor Protsak Oct 4 '16 at 2:40
  • $\begingroup$ @ClaudioGorodski - yes. $\endgroup$ – benblumsmith Oct 4 '16 at 20:58
  • $\begingroup$ @VictorProtsak - you're right. I'm not actually sure what I was thinking. I think I was imagining that since the points are images of the initially chosen point under $W\subset O(n)$ that perturbing the initial choice would perturb the whole orbit by an element of $O(n)$, preserving distances, but retrospectively this is nonsense. I will remove that comment and canonize the barycenter for the sake of the comparison. $\endgroup$ – benblumsmith Oct 4 '16 at 21:02

Your suggestion that the Riemannian metric on an orbit and the word metric will be almost homothetic in some sense is very far from the reality.

We consider two metrics on the symmetric group $S_n$, the word metric $|\cdot|_w$ and the metric coming from the restriction of the Riemannian metric on the sphere to an orbit (under the standard representation), $|\cdot|_r$. Here, as is standard, I denote $|g|=d(g,e)$ for whatever metric. I suggest to consider the ratio $\frac{|g|_w|h|_r}{|g|_r|h|_w}$ for some choice of $g,h\in S_n$ and see how it varies in $n$. Below we will take $g$ to be the longest element and $h$ a generator.

For $i=1,\ldots,n-1$, let us denote by $s_i$ the transposition $(i,i+1)$. These are the generators of $S_n$. Let us also consider the element $w_0$ sending $i\to n+1-i$. The word length of $w_0$ is the number of positive roots, thus $|w_o|_w={n+1\choose 2}$. Clearly, $|s_i|_w=1$.

Let us discuss now the Riemannian metric. We consider the standard representation of $S_n$ on $\mathbb{R}^n$ and consider the invariant $n-1$ dimensional subspace $V=\{x\mid \sum x_i=0\}$. For simplicity of the presentation let me assume now $n=2k+1$. It is easy to check that the point $p=(-k,-k+1,\ldots,-1,0,1,\ldots,k-1,k)$ has the property that for all $i$, $\langle s_ip,p \rangle$ is the same. In fact, deboting $R=\sqrt{k(k+1)n/3}$, $p$ is on the $R$ sphere and this common value is $R^2-1$. Instead of working on the 1-sphere we will work on the $R$-sphere endowed with angle metric, which is basically the same. We will work with the orbit of $p$. Taking any other base point will be worse for one of the generators.

For the angle metric on the $R$-sphere it is clear that $|w_o|_r=d(w_0p,p)=\pi$ and $$R^2-1=\langle s_1p,p \rangle=R^2\cos |s_1|_r \geq R^2(1-|s_1|_r^2/2),$$ thus $|s_1|_r \geq \sqrt{2}/R$. We conclude that $$\frac{|w_o|_w|s_1|_r}{|w_0|_r|s_1|_w} \geq \frac{\sqrt{2}{n+1\choose 2}}{\pi\sqrt{k(k+1)n/3}} \to \infty.$$

  • $\begingroup$ Dear Uri - this is a very satisfying answer to the part of the question that asks if I could find a universal $A,\varepsilon$. But perhaps there can still be some controlled dependence on $n$, or some other way to precisely capture the relationship between the metrics? What I think I understand from your argument is that the graph path from the identity to the longest element may be forced to be significantly "folded up" since the graph distance can so greatly exceed the Riemannian distance. This makes it seem possible that there is a steady though nonlinear, and dependent on $n$ relationship. $\endgroup$ – benblumsmith Jan 4 '17 at 19:17
  • $\begingroup$ +1, by the way. Thank you for this. $\endgroup$ – benblumsmith Jan 4 '17 at 19:18
  • $\begingroup$ (Marking the question answered, although of course still interested in the broader form of the question.) $\endgroup$ – benblumsmith Feb 17 '20 at 0:39

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.