Diameter of universal cover

Question

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 $\pi_1(M)$)?

Comments:

• 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?

Examples:

• 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 and also this question). It is much more than $\log m$, but still $k=o(\log^2 m)$.

Answer 1

As both Anton and Greg pointed out, it is enough to look at the Cayley graph of a presented finite group $G$ with only quadratic and cubic relators and study how large $diam(G)$ can be in terms of $\vert G \vert$. Note that the property that all relators are quadratic or cubic implies that $G$ is the $1$-skeleton of a simply connected $2$-dimensional simplicial complex.

It can be shown that $diam(G) \leq \sqrt{ 4 \vert G \vert +1}-2$, which implies the effective bound $diam(\tilde{M} )/ diam(M) \leq 4 \sqrt{\vert \pi_1(M) \vert } $. For the strictly asymptotic behavior, the main result of this paper implies that $diam (G) = o (\vert G \vert ^p) $ for any $p>0$, implying that $diam(\tilde{M} ) / diam(M) = o ( \vert \pi_1 (M) \vert ^p)$ for any $p>0$.

An sketch of the proof of the first inequality follows. Take $h \in G$ with $d(h,e)= diam(G)$ and a path $e=g_0, g_1 , \ldots, g_{diam(G)}=h$. Set $S_i$ to be the induced subgraph by the set $ \{ g \in G \mid d(g,e)= i \}$ for each $i \in \mathbb{N}$ and $T_i $ the connected component of $g_i$ in $S_i$. Fix $1 \leq i < diam(G)$. Removing $T_i$ disconnects $G$, otherwise there would be a loop in $G$ which cannot be contracted in any simplicial complex with $G$ as its $1$-dimensional skeleton.

Denote by $C_1$ and $C_2$ the connected components of $e$ and $h$, respectively in $G \backslash T_i$. Since the left multiplications are graph isomorphisms, removing $ g_i^{-1}T_i$ disconnects $G$ in two components $C_1^{\prime}$ and $C_2^{\prime}$ isomorphic to $C_1$ and $C_2$ respectively. If $g_i^{-1} T_i$ does not intersect $T_i$, then one of $C_1^{\prime}$ or $C_2^{\prime}$ properly contains $C_2$, but since $\vert C_2^{\prime} \vert = \vert C_2 \vert$, we have $C_1^{\prime } \subsetneq C_2$ and $\vert C_1 \vert > \vert C_2 \vert$. The same way, if $hg^{-1}_iT_i$ does not intersect $T_i$, then $\vert C_2 \vert > \vert C_1 \vert$.

Therefore $T_i$ has to intersect either $g_i^{-1}T_i$ or $hg_i^{-1}T_i$, so $\vert T_i \vert \geq diam(T_i) \geq \min \{ i, diam(G) -i \} +1$. Since the $T_i$'s are disjoint, summing over all $i$'s yield

$$\vert G \vert \geq \frac{(diam(G)+2)^2 +1}{4}.$$

Which is the desired inequality.

Answer 2

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 $\log |G|$.

Answer 3

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.

Answer 4

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 $\pm 2^s$)...

Answer 5

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

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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 $$\mathop{diam}\tilde M\leqslant \mathop{Const}\cdot(\mathop{diam} N_2+1).$$