5
$\begingroup$

Is it possible for SOME positive $c$, $c<1$ 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?

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 $1$.

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).

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.

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.

Of course this is not a real proof, but I am 100% it can be made real.

$\endgroup$
2
  • 1
    $\begingroup$ You should probably add hypotheses to your question, otherwise there is the trivial example mapping $\mathbb{H}^3 \to \mathbb{H}^3$ where in exponential coordinates, one takes every point closer to the origin by some factor. $\endgroup$
    – Ian Agol
    Dec 14, 2009 at 19:00
  • $\begingroup$ Thanks! I corrected the question, speaking about hyperbolic manifolds I had in mind "compact" manifold $\endgroup$ Dec 14, 2009 at 19:20

2 Answers 2

5
$\begingroup$

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.

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 [Edit: actually I'm not sure about this now, but in the example below there exists a contracting map]. There are finitely many tetrahedra in $\mathbb{H}^3$ which give rise to fundamental domains for discrete reflection groups (see Ratcliffe). 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.

$\endgroup$
4
  • $\begingroup$ Thanks a lot for the answer! I added a proof of 2-dimensional case, but of course Schwarz lemma do the job!! I will think about your 3-dimensional example! $\endgroup$ Dec 14, 2009 at 20:33
  • $\begingroup$ After reading this I looked at the Dirichlet domains of the (n,0) fillings of the figure eight knot, using SnapPea. It looks like (a) the domains are all combinatorially identical, as n grows, and (b) they are nested. So, if my eyes are correct, your argument will also apply to branched covers over the figure eight. $\endgroup$
    – Sam Nead
    Dec 14, 2009 at 20:47
  • $\begingroup$ @ Sam: As for the polyhedra, you might be interested in ams.org/mathscinet-getitem?mr=1382519. I'm not sure that there's a length-decreasing map for general polyhedra (the argument I had in mind works for a simplex). For example, there may be a very short edge near the core geodesic which increases in length in your examples as n decreases. So the natural combinatorial map of Dirichlet domains can't be length decreasing from the (nm,0) filling to the (m,0) filling. $\endgroup$
    – Ian Agol
    Dec 15, 2009 at 0:17
  • $\begingroup$ "there may be a very short edge near the core geodesic which increases in length in your examples as n decreases" Ok, I see this now. My mistake. Thanks for the link to the paper. I'll take a look. $\endgroup$
    – Sam Nead
    Dec 15, 2009 at 17:35
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Sam, I included a "proof" of 2-dimensional case $\endgroup$ Dec 14, 2009 at 20:19
  • $\begingroup$ Ok, that seems more reasonable... $\endgroup$
    – Sam Nead
    Dec 14, 2009 at 21:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.