Diameter of m-fold cover - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T13:23:38Z http://mathoverflow.net/feeds/question/7732 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7732/diameter-of-m-fold-cover Diameter of m-fold cover Anton Petrunin 2009-12-04T00:28:20Z 2010-10-29T03:57:07Z <p>Let $M$ be a closed Riemannian manifold. Assume $\tilde M$ is a connected Riemannian $m$-fold cover of $M$. Is it true that $$\mathop{diam}\tilde M\le m\cdot \mathop{diam} M\ ?\ \ \ \ \ \ \ (*)$$</p> <p><strong>Comments:</strong></p> <ul> <li><p>A <strong>complete solution</strong> is given by S. Ivanov [it can not be marked as accepted due to software limitations].</p></li> <li><p>This is a modification of a problem of A. Nabutovsky. <a href="http://mathoverflow.net/questions/8534/diameter-of-universal-cover" rel="nofollow">Here</a> is yet related question about universal covers.</p></li> <li><p>You can reformulate it for compact length metric space --- no difference. </p></li> <li><p>The answer is YES if the cover is <a href="http://en.wikipedia.org/wiki/Covering_space#Deck_transformation_group.2C_regular_covers" rel="nofollow"><strong>regular</strong></a> (but that is not as easy as one might think).</p></li> <li><p>The estimate $\mathop{diam}\tilde M\le 2{\cdot}(m-1){\cdot} \mathop{diam} M$ for $m>1$ is trivial.</p></li> <li><p>We have equality in $(*)$ for covers of $S^1$ and for some covers of figure-eight.</p></li> </ul> http://mathoverflow.net/questions/7732/diameter-of-m-fold-cover/7880#7880 Answer by Alon Amit for Diameter of m-fold cover Alon Amit 2009-12-05T18:12:40Z 2009-12-05T22:18:57Z <p>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].</p> <p>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). </p> <p>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.</p> <p>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 $m$.</p> <p>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. </p> <p>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.</p> <p>Can something like this work?</p> http://mathoverflow.net/questions/7732/diameter-of-m-fold-cover/8357#8357 Answer by Oleg Eroshkin for Diameter of m-fold cover Oleg Eroshkin 2009-12-09T15:12:33Z 2009-12-09T15:12:33Z <p>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 $[m/2]$. </p> <p>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.</p> http://mathoverflow.net/questions/7732/diameter-of-m-fold-cover/16939#16939 Answer by Sergei Ivanov for Diameter of m-fold cover Sergei Ivanov 2010-03-03T00:08:11Z 2010-03-03T09:54:55Z <p>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.</p> <p>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 $L(b_i)\le diam(M)&lt; L(a_i)$. 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.</p> <p>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 $i_1&lt; i_2&lt;\dots&lt; i_k$, 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 $m$.</p> <p>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 $i&lt; j$, 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.</p>