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Let $M$ be Riemannian manifold and $\tilde M$ be its universal cover (with induced metric). What is the upper bound for $k=\mathop{diam}\tilde M/\mathop{diam} M$ in terms of $m=|\pi_1(M)|$ (or $\pi_1(M)$)?


  • There is a similar answered question here, but there cover is NOT universal. So we get $k\leqslant m$, but $k\ll m$ is expected.

  • The question is open even in case $\pi_1=\mathbb Z_m$ (even asymptotics is not known).

  • Clearly, $\sup k$ for given finite group $\Gamma=\pi_1(M)$ is an invariant of $\Gamma$. Is it an interesting invariant?


  • For $\pi_1=\mathbb Z_{3\cdot 2^n}$ one can make $k\sim n$ or $k=O(\log m)$ (see my answer below).

  • For $\pi_1=S_n$, one can make $k$ of order $n^2$ or (see Greg's answer). It is much more than $\log m$, but still $k=o(\log^2 m)$.

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Do you know Rob Kirby's book of open problems in topology? It is very influential. You should make a catalogue of open problems in comparison geometry. Your questions are excellent. – Greg Kuperberg Dec 18 '09 at 16:12
See also this shorter list of open problems in quantum information. It is too short to have the same influence, but it is still nice. – Greg Kuperberg Dec 18 '09 at 16:15
There is an old but very nice collection of open problems in comparison geometry by P. Petersen. – Anton Petrunin Dec 18 '09 at 17:09
up vote 5 down vote accepted

Here is an example provoked by Anton's Cayley graph reformulation. I thought of this example before, but for some reason I miscounted. To review: You can obtain a family of examples for Anton's question by looking at the Cayley graph of a presented finite group $G$ with cubic relators. (And quadratic relators, if you want to make some of the elements involutions.) The question is how large $\text{diam}(G)$ can be as a function of $|G|$. Up to a constant factor, it does not matter whether the relators are cubic or bounded in length by any fixed $k$. In Anton's new posting, he argues that this Cayley graph construction is actually equivalent to all examples, up to a constant factor.

For example, suppose that $G = S_n$ in the Coxeter presentation. Then the relators all have length 2, 4, and 6. It is well known that the length of the longest word is $n(n-1)/2$. This violates the proposed upper bound $\log |G|$.

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I thought that cyclic group should be the worse, but maybe it is the best --- It is always nice when you intuition brakes :) – Anton Petrunin Dec 24 '09 at 15:20

Here is an example that says nothing by itself asymptotically, but is still somewhat interesting and suggests further examples. The Poincaré homology sphere has a 120-fold universal cover, and you can calculate that in the round metric the diameter increases by a factor of $\pi/(\arccos \phi^2/\sqrt{8}) \approx 8.09$, where $\phi$ is the golden ratio. This is a vaguely more favorable constant than in your general construction.

If $G$ is a compact, simple, simply connected Lie group and $\Gamma$ is a finite subgroup, it seems separately interesting to calculate the diameter of $G/\Gamma$. Since $G$ acts transitively on this coset space, the covering radius of the subset $\Gamma$ is the diameter. Moreover, $G$ has a unique bi-invariant Riemannian metric, although you could also consider a left-invariant metric. The covering radius problem has been widely studied when $G$ is Euclidean space and $\Gamma$ is a lattice. People have also thought about it at least some in, e.g., hyperbolic geometry. I never considered that it would also be very interesting when $G$ is compact.

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Here is my example of length space with $\pi_1(X)=\mathbb Z_{3\cdot 2^n}$ such that $\mathop{diam}\tilde X$ has order of $n\cdot \mathop{diam} X$ --- I can not do better. (One can easely make a 4-dimensional manifold out of this.)

Consider sequence $\tau_n$ of triangulations of disc, defined inductively on the following way:

  • $\tau_0$ is one triangle

  • $\tau_{n+1}$ obtained from $\tau_n$ by gluing a triangle to each side on the boundary.

Clearly the boundary of $\tau_n$ consist of $3\cdot 2^n$ edges. Let us take cyclic sequence of $3\cdot 2^n$ copies of $\tau_n$ and glue each to the next one along the boundary, rotating by one edge. On the obtained a 2-dimensional complex, change each triangle to a Reuleaux triangle of width 1, we obtaine space $\tilde X$.

On $\tilde X$, we have an free isometric action of $\mathbb Z_{3\cdot 2^n}$, it acts by shifting sequence of $\tau_n$'s. Take $X=\tilde X/\mathbb Z_{3\cdot 2^n}$. It straight forward to see that and $\mathop{daim} X\le 2$ and $\mathop{diam}\tilde X\ge \tfrac n2$ (the distance from vertex $0$ to vertex $\ell$ is the least number of terms in presetation of $\ell$ as a sum of $\pm 2^s$)...

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Here is an approach to get an upper bound. It is "community wiki" --- feel free to make changes.

Let $\mathop{diam} M=1$ and $p_1,p_2,\dots,p_m\in \tilde M$ be the preimages of a point in $M$. Then $B(1+\epsilon,p_i)$ is a cover of $\tilde M$. Consider 2-skeleton $N_2$ of nerve of this covering. Clearly $N_2$ is simply connected and it admits natural action of $\pi_1 M$ which is transitive on the set of vertices.

All this should give some bounds on $\mathop{diam} N_2$ (assuming all triangles are standard) and it is clear that $$\mathop{diam}\tilde M\leqslant \mathop{Const}\cdot(\mathop{diam} N_2+1).$$

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What you are saying is that it is equivalent, up to a constant factor, to look at the Cayley graph of a presented finite group with cubic relators. I agree that that is a good reformulation of the question. Among other merits, it helps explain your example. – Greg Kuperberg Dec 24 '09 at 6:08
I think that the model of making a 2-complex with equilateral triangles from a presentation with cubic relators is promising. It seems very possible that you could prove a logarithmic bound when $G$ is abelian, because the set of relators is just an integer matrix. But I did not manage to find an argument. – Greg Kuperberg Dec 26 '09 at 3:30

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