## Diameter of m-fold cover

Let $M$ be a closed Riemannian manifold. Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$. Is it true that $$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$

• A complete solution is given by S. Ivanov [it can not be marked as accepted due to software limitations].

• This is a modification of a problem of A. Nabutovsky. Here is yet related question about universal covers.

• You can reformulate it for compact length metric space --- no difference.

• The answer is YES if the cover is regular (but that is not as easy as one might think).

• The estimate $\mathop{diam}\tilde M\le 2{\cdot}(m-1){\cdot} \mathop{diam} M$ for $m>1$ is trivial.

• We have equality in $(*)$ for covers of $S^1$ and for some covers of figure-eight.

-
In your inequality, the manifold on the right hand side should be M, not $\tilde{M}$, I think. – Jason DeVito Dec 4 2009 at 2:41
What do you mean by "for many covers of figure-eight"? It seems clear that it's true for all finite covers of the wedge of two circles. (The number of edges in the maximal tree is m-1.) – HW Dec 4 2009 at 2:53
An equilateral triangle with a loop at every vertex is a 3 fold cover. Right? – Andrey Gogolev Dec 4 2009 at 3:45
Oh, you're talking about when you get equality. Sorry, I misread the previous version of the question. – HW Dec 4 2009 at 4:05
To clarify, a regular cover is one you get from a free (group action), not a (free group) action. Also known as a Galois cover. See en.wikipedia.org/wiki/… – David Speyer Dec 11 2009 at 17:46

Here's a proposed sketch of an approach. I hope it actually works... [EDIT: it doesn't, as it stands. I guess the main take-away from the rough outline below is that whatever the answer is for graphs should carry over to manifolds].

First, we can prove an appropriate analog in the category of graphs. Let $G$ be a base graph and $\tilde{G}$ a connected $m$-cover of $G$ in the combinatorial sense (the mapping takes vertices to vertices and edges to edges, and preserves local neighborhoods). It's useful to visualize $\tilde{G}$ this as a set of discrete fibers over the vertices of $G$, the vertices of which can be aribtrarily numbered ${1,\ldots, m}$. Now the edge-fibers correspond to permutations in $S_m$. Also notice that we may relabel the vertex fibers in order to make certain edge fibers "flat", meaning the corresponding permutation is the identity. This can simultaneously be done for a set of edges of $G$ which contain no cycle, such as a path (or a tree).

Given two vertices $\tilde{x}, \tilde{y}$ in $\tilde{G}$, there's a path $P$ of length at most $d$ between their projections $x,y$ in $G$. We may assume that the permutations over the edges in $P$ are trivial. A path from $\tilde{x}$ to $\tilde{y}$ can now be formed by navigating across the floors (at most $d$ steps in each trip [EDIT: could be worse, since as you move to a new floor you're not guaranteed to land on the path]) and among the floors (at most $m$ steps overall), yielding $md+m$ steps in total. Sorry this is so vague but it's really quite simple if you draw a picture.

Now $m(d+1)$ is a bit too large (we want $md$) but this can't be helped in the category of graphs: for example, the hexagon (diameter 3) is a 2-cover of the triangle (diameter 1). But this is just because the triangle misrepresents the true diameter of the underlying geometry, which is really $3/2$. To resolve this nuisance, apply the procedure above to a fine subdivision of $G$ (and $\tilde{G}$), which make $d \to \infty$ and the ratio is brought back to the desired $m$.

Next, consider simplicial complexes of higher dimension. It seems to me that if $X$ is a sufficiently nice topological space triangluated by a simplicial complex $K$, then the diameter of $X$ can be well approximated by the diameter of the 1-skeleton of a sufficiently fine subdivision of $K$. Is this true? Given two points in $X$ and a long path between them, if the path is close to a PL one than this should be the case. I hope that if $X$ is not too pathological, its diameter is represented by a tame path.

Finally, I would hope that a general Riemannian manifold (or some other kind of space for which we need to prove this) can be effectively triangulated, although this extends beyond my off-the-top-of-my-head knowledge.

Can something like this work?

-
 I guess that "floor" means your tree? Then it might be up to $2d$ in each floor --- not $d$... – Anton Petrunin Dec 5 2009 at 21:05 That's right, and I modified my post to reflect that. However, I guess the point is that whatever the correct ratio is for graphs (up to subdivision, as outlined) should be the correct ratio for manifolds, right? If we can actually find a graph of diameter $d$ (say, a bipartite one to avoid the odd-cycle-diameter issue), and a connected $m$ cover with diamater $\alpha$md for $\alpha>1$, does this not also mean that the manifold case cannot be better? – Alon Amit Dec 5 2009 at 22:21 That is right --- it is sufficient to make it for graphs. In fact any metric space (in particular Riemannian manifold) can be approximated by a graph, say in Gromov--Hausdorff sense. – Anton Petrunin Dec 5 2009 at 22:42 It seems that on this way, the best you can get is $2(m-1)d$... – Anton Petrunin Dec 6 2009 at 3:11

I can show that $diam(\widetilde{M})\leq (m+1)diam(M)$. It follows from the fact that the fundamental group of $M$ is generated by "short" loops of length at most $2diam(M)$ (this is proved in Gromov's book "Metric structures ..."). Lets show that if $p$ and $q$ in $\widetilde{M}$ have the same projection $x$ in $M$, then $dist(p,q)\leq [m/2]\cdot diam(M)$. Consider the following graph. Vertices are $m$ preimages of $x$, edges correspond to short loops in $M$. This graph is 2-connected, therefore the diameter of the graph is at most $[m/2]$.

To improve the estimate it is sufficient to show, that for every two points $x,y\in M$ short loops with based point $x$ through $y$ generates the fundamental group.

-
 Well it works only for regular coverings. Look at the coverings of figure-eight its graph can be a tree... BTW, for regular coverings you can do $m\cdot\mathop{diam}M$. – Anton Petrunin Dec 9 2009 at 16:54 You forget to multiply by 2 in $dist(pq)[m/2]diam(M)$. – Anton Petrunin Dec 9 2009 at 20:11

I think I can prove that $diam(\tilde M)\le m\cdot diam(M)$ for any covering. Let $\tilde p,\tilde q\in\tilde M$ and $\tilde\gamma$ be a shortest path from $\tilde p$ to $\tilde q$. Denote by $p,q,\gamma$ their projections to $M$. I want to prove that $L(\gamma)\le m\cdot diam(M)$. Suppose the contrary.

Split $\gamma$ into $m$ arcs $a_1,\dots,a_n$ of equal length: $\gamma=a_1a_2\dots a_m$, $L(a_i)=L(\gamma)/m>diam(M)$. Let $b_i$ be a shortest path in $M$ connecting the endpoints of $a_i$. Note that $L(b_i)\le diam(M)< L(a_i)$. I want to replace some of the components $a_i$ of the path $\gamma$ by their "shortcuts" $b_i$ so that the lift of the resulting path starting at $\tilde p$ still ends at $\tilde q$. This will show that $\tilde\gamma$ is not a shortest path from $\tilde p$ to $\tilde q$, a contradiction.

To switch from $a_i$ to $b_i$, you left-multiply $\gamma$ by a loop $l_i:=a_1a_2\dots a_{i-1}b_i(a_1a_2\dots a_i)^{-1}$. More precisely, if you replace the arcs $a_{i_1},a_{i_2},\dots,a_{i_k}$, where $i_1< i_2<\dots< i_k$, by their shortcuts, the resulting path is homotopic to the product $l_{i_1}l_{i_2}\dots l_{i_k}\gamma$. So it suffices to find a product $l_{i_1}l_{i_2}\dots l_{i_k}$ whose lift starting from $\tilde p$ closes up in $\tilde M$. Let $H$ denote the subgroup of $\pi_1(M,p)$ consisting of loops whose lifts starting at $\tilde p$ close up. The index of this subgroup is $m$ since its right cosets are in 1-to-1 correspondence with the pre-images of $p$. While left cosets may be different from right cosets, the number of left cosets is the same $m$.

Now consider the following $m+1$ elements of $\pi_1(M,p)$: $s_0=e$, $s_1=l_1$, $s_2=l_1l_2$, $s_3=l_1l_2l_3$, ..., $s_m=l_1l_2\dots l_m$. Two of them, say $s_i$ and $s_j$ where $i< j$, are in the same left coset. Then $s_i^{-1}s_j=l_{i+1}l_{i+2}\dots l_j\in H$ and we are done.

-
I must be missing something - why do you say that L(bi) < L(ai) in the second paragraph? Why is that a strict inequality? – Alon Amit Mar 3 2010 at 0:36
@Alon. Otherwise the distance is $\le m\cdot \mathrm{diam}$ – Anton Petrunin Mar 3 2010 at 1:01
THAT IS GREAT!!! THANK YOU VERY MUCH [this stupid software does not let me to mark this answer as accepted] – Anton Petrunin Mar 3 2010 at 1:25
@Alon: I've edited the second paragraph to make it more clear. – Sergei Ivanov Mar 3 2010 at 9:56