Contracting maps of hyperbolic manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T01:20:43Z http://mathoverflow.net/feeds/question/8882 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8882/contracting-maps-of-hyperbolic-manifolds Contracting maps of hyperbolic manifolds Dmitri 2009-12-14T15:51:38Z 2009-12-15T02:27:33Z <p>Is it possible for SOME positive $c$, $c&lt;1$ to find a pair of COMPACT hyperbolic manfiolds $M^3$ and $N^3$ with a positive degree map $$f: M^3 \to N^3,$$ such that $f$ is contacting with constant $c$? Are there may examples like this?</p> <p>One can ask the same question of Riemann surfaces, and it seems to me that this should be possible. For example we can take a double cover of Riemann surface with many points or ramification. Though I don't know a proof even in this case. Of course for non-ramified cover the best possible constant $c$ is $1$. </p> <p>ADDED. Following the answer of Sam Need, let me give an approximative "proof" of the fact that this works in dimesnion 2. Let us triangulate a hyperbolic surface $N^2$ in triangles of very small size, that have acute angles (this is always possible). We want to show that a double cover of $N^2$ with ramifications at vertices of the triangulation will do the job. For this we need a lemma (without a proof).</p> <p>Lemma. Suppose we have two hyperbolic trianlges, one very small and acute with angles $a$, $b$, $c$, and the over with angles a/2, b/2, c/2. Then there is a contacting map from the second triangle to the first one. The lemma is true, since the second trianlge will be large.</p> <p>Now on the double cover we can take a trangulation that comes from $N^2$ and glue it from these triangles with half angles. Half angles come from doble cover. Then we just need to "adjust" the map.</p> <p>Of course this is not a real proof, but I am 100% it can be made real.</p> http://mathoverflow.net/questions/8882/contracting-maps-of-hyperbolic-manifolds/8902#8902 Answer by Sam Nead for Contracting maps of hyperbolic manifolds Sam Nead 2009-12-14T19:50:27Z 2009-12-14T19:50:27Z <p>This question confuses me, even in dimension two. The non-trivial branched coverings $f$ I can think of are extremely contracting near the branch points. So much so that $f$ is actually expanding elsewhere to produce enough area. </p> http://mathoverflow.net/questions/8882/contracting-maps-of-hyperbolic-manifolds/8904#8904 Answer by Agol for Contracting maps of hyperbolic manifolds Agol 2009-12-14T20:09:58Z 2009-12-15T02:27:33Z <p>In general, for any non-zero degree map from one closed negatively curved manifold to another, there is a canonical map (due to Besson-Courtois-Gallot) called the "natural map". However, it's only known to be pointwise volume decreasing, not necessarily contracting. They call this the "real Schwarz-Lemma". Applying the Schwarz lemma for Riemann surfaces I think gives the contracting map in this case for branched covers. Think of the induced map on the universal cover, which is the unit disk, or $\mathbb{H}^2$. The Schwarz lemma says that any conformal map from the disk to the disk is contracting, unless it's an isometry. </p> <p>I thought of one (not very explicit) example in 3-D. Take two simplices in hyperbolic space. There is a canonical affine map (say in the Lorentzian model) taking one simplex to the other. This will be a contracting map for the hyperbolic metric if one simplex sits inside the other <strong>[Edit: actually I'm not sure about this now, but in the example below there exists a contracting map]</strong>. There are finitely many tetrahedra in $\mathbb{H}^3$ which give rise to fundamental domains for discrete reflection groups (see <a href="http://books.google.com/books?id=hwQmbllvFMUC&amp;lpg=PP1&amp;dq=ratcliffe%2520hyperbolic&amp;pg=PA296#v=onepage&amp;q=&amp;f=false" rel="nofollow">Ratcliffe</a>). Two of these have one dihedral angle $\pi/5$, with opposite edge angle $\pi/2$ and $\pi/4$, respectively, and all other angles the same. There is a 1-parameter family of polyhedra interpolating between these (basically, just "push" the two faces closer together along the dihedral angle $\pi/5$ edge) which decreases distances. Also, the orbifold fundamental group (i.e. reflection group) from the $\pi/4$ one maps to that of the $\pi/2$ one. So there's a distance decreasing map from one orbifold to the other. Using Selberg's lemma, one may find finite-sheeted manifold covers with the same property. </p>