12
$\begingroup$

Let $M$ be a compact simply connected R. manifold. Let $x$ be a base point and let $\gamma$ be a smooth loop in $M$ starting and ending at $x$.

Is there a base point preserving retraction of $\gamma$ to $x$ such that every point on $\gamma$ travels a distance at most, say, $2diam(M)$?

I think the answer is yes. The scheme I have in mind is the following

Consider the cut locus $CL(x)$ of $x$. By perturbing the metric and the curve we can assume that $CL(x)$ is triangulable and $\gamma$ intersects it at finitely many points. Then slide these points off the $CL(x)$. Then the new curve does not intersect $CL(x)$ and can be contracted to $x$ following the geodesics.

This is probably too complicated...

Is there a simple proof? A reference?

Remark: 1) I would be happy with any estimate (not necessarily $2diam(M)$) which is independent of $\gamma$. 2) If there is a good curve shortening procedure that is not base-point-preserving please share it as well.

$\endgroup$

3 Answers 3

11
$\begingroup$

Yes there is a bound independent of $\gamma$. Fix a fine triangulation of $M$ (say, with simplices 10 times smaller than the injectivity radius of the metric). For each vertex $q$ of the triangulation, fix a shortest path $s_q$ connecting $q$ to the marked point $p$. It is easy to deform any loop $\gamma$, via a short homotopy, into a path in the 1-skeleton of the triangulation. Then, for every vertex occuring on this path, "pull" a small piece of the loop to $p$ along the corresponding $s_q$. Now the loop is a product of "elementary" loops of the form $s_q^{-1}[qq']s_{q'}$ where $[qq']$ is an edge of the triangulation. Contract these "elementary" loops one-by-one. Since there are only finitely many different elementary loops, and the same elementary loops work for all $\gamma$'s, we have a uniform upper bound (depending on $M$) for the length of the homotopy.

On the other hand, there is no bound for the homotopy length in terms of the diameter. In addition to the great Bill Thurston's answer, let me mention another construction. For every $\varepsilon>0$ there is a metric on $S^3$ with sectional curvature $\le 1$ and diameter $\le\varepsilon$ (it was first constructed in Gromov "Almost flat manifolds, see also Buser and Gromoll "On the almost negatively curved $3$-sphere"). In this metric, take a geodesic starting from $p$ of length $2\varepsilon$ and connect its endpoint back to $p$ by a shortest path. Since the curvature is bounded above by 1, any contraction of this curve will have a trajectory longer than $\pi$ (I think this fact is called Klingenberg's lemma but I am not sure).

$\endgroup$
1
  • $\begingroup$ Spasibo! I had this idea with elementary loops in mind, but I didn't think of triangulating the manifold and couldn't make the last step (bound uniformly contraction of elementary loop). $\endgroup$ Sep 21, 2010 at 23:56
12
$\begingroup$

The rub is here: "slide the curve off the cut locus". The problem: the cut locus might be a humongous growth whose diameter is large compared to that of $M$. The diameter of $M$ can be kept small by embedding the growth in $M$ in a way that attaches it to a small base. Added: Or, in dimension > 2, as shown in the answers of Ivanov and Agol, the diameter of the cut locus might remain modest in size, but the puzzle of "sliding the curve off it" can become tedious and tricky, even computationally unsolvable in dimension > 3.

This concept is closely related to the filling length of a loop, which Gromov defined to be the infimal $L$ such that the loop can be contracted to a point via loops of length no longer than $L$. You can visualize the two definitions as asking for a homotopy where the leaves of a certain foliation are bounded by some estimate. In the definition above, these are radii; in the other defition they are concentric circles. By reparametrizing, one can be converted to the other with only bounded loss of efficiency.

In The Morse Landscape of a Riemannian Disk by Frankel and Katz, they prove that Riemannian metrics on a disk of bounded length can have arbitrarily large filling length, that is (my rephrasing):

(Theorem 2) for every constant $C$ there is a $g$ on the disk $D^2$, of boundary length $1$ with every point of $D$ is within distance $1$ of the boundary such that every homotopy of $S^1$ to a point in $(D,g)$ contains an intermediate curve of length bigger than $C$.

Tim Riley and I developed a related construction in The absence of efficient dual pairs of spanning trees in planar graphs where we constructed cell complexes of bounded geometry (the complex and dual both have valence $\le 6$) where the filling length is a quadratic function of diameter. We also showed that quadratic behavInior is sharp. We used this to show that for all spanning trees in the complex, the length of the tree plus the length of the dual tree is bounded below by a quadratic function of the diameter of the graph.
You can interpret this as saying that there is no way to sweep across the disk with arcs whose maximum length is less than a quadratic function of the diameter of the disk.

These examples also show that for homotopies of the boundary to a point, some point must travel an unbounded distance for arbitrary diameter 1 metrics on the disk, and a quadratic function of diameter, if there are curvature bounds. These examples work for all higher dimensions.

In both constructions, the cut locus is has a large diameter, considerably larger than the diameter of the metric itself. To construct the metrics, start with a tree that branches 3 ways at each vertex, and thicken it by replacing the vertices triangles at the vertices and rectangles for edges. Now glue on a strip, all the way around, cut out of the hyperbolic plane between two concentric horocycles with the long edge along the tree and the short edge on the outside. This achieves the metric of bounded diameter. Here are the relevant figures from our paper. The idea is that the lines of any homotopy end up having to cut across the tree numerous times, each time losing some efficiency. See the two cited papers for details.

alt text http://dl.dropbox.com/u/5390048/Tree%20pictures.jpg

Added: See the answers of Sergei Ivanov and Ian Agol for nice and more general methods that work in higher dimensions.

$\endgroup$
1
  • $\begingroup$ Very interesting, thank you! Luckily there's at least some bound. $\endgroup$ Sep 22, 2010 at 1:36
8
$\begingroup$

I don't think this is possible. In fact, I don't think you can bound the length of the tracks of the homotopy by any computable function. I don't have a precise argument, but heuristically, suppose this were true. Then one could easily solve the simple-connectivity problem in any compact Riemannian manifold $M$ since for any given loop, one could contract it to a point a bounded number of ways. If this didn't succeed, then the loop is not contractible. So one could tell if the manifold is simply connected. But there is no algorithm to tell if a PL 4-dimensional manifold is simply-connected. I imagine this argument could be made precise by approximating the smooth structure by a PL structure, and I think something akin to this may have been worked out by Schmuel Weinberger.

$\endgroup$
2
  • $\begingroup$ The bound on the length used to contract loops could exist and be a non-computable function of $M$. $\endgroup$ Sep 22, 2010 at 0:04
  • 1
    $\begingroup$ I believe the length of tracks is a noncomputable function even for triangulations of the 4-ball, by the same line of reasoning. Of course for any single triangulation of a simply-connected manifold it's bounded, as Ivanov noted in his answer, but shrinking length can be larger than any computable function for contractible curves in non-simply-connected manifolds. Note that for triangulated 3-manifolds, Haken developed methods that transform the question to integer linear programming, whose solutions have explicit bounds in terms of the number of simplices. $\endgroup$ Sep 22, 2010 at 11:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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