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](https://mathoverflow.net/questions/7732/diameter-of-m-fold-cover), 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](https://mathoverflow.net/q/102737)). It is much more than $\log m$, but still $k=o(\log^2 m)$.



Diameter of universal cover
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](https://arxiv.org/abs/1807.08827) 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](https://arxiv.org/abs/1203.5624) 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.


