12
$\begingroup$

A Riemannian manifold $(M,g)$ is said to have flat ends if the curvature tensor of $g$ vanishes outside a compact set $K$. I was wondering if such manifolds are of bounded geometry. Recall that a manifold is of bounded geometry if

  1. The curvature tensor and all its covariant derivatives are uniformly bounded.
  2. The injectivity radius has a uniform positive lower bound.

It is obvious that a manifold with flat ends satisfies the first property, but it is not clear to me that a manifold with flat ends satisfies the second property. Two simple counterexamples come to mind.

  1. The manifold $M=\mathbb{R}^n-\{0\}$ has flat ends, but has no uniform lower bound on the injectivity radius.
  2. A cylinder $\mathbb{R}\times S_r^1$ of radius $r$ has injectivity radius $\pi r$. Take a countable union of such cylinders with decreasing radius $$M=\coprod_{n=1}^\infty \mathbb{R}\times S_{\frac 1n}^1.$$ This manifold is flat, but does not admit a uniform positive lower bound on the injectivity radius.

Counterexample one might be excluded by assuming that $g$ is complete, and counterexample 2 might be excluded by demanding that $M$ is connected. Therefore my question is:

Is any connected and complete Riemannian manifold $(M,g)$ with flat ends of bounded geometry?
$\endgroup$
7
  • $\begingroup$ The question is interesting. My intuition screams yes but I don't have any better reason than the fact that all the examples I can think off do have positive injectivity radius. I think it would be enough to prove that : 1) there's a finite number of ends, 2) up to a finite cover, these ends are isometric to open sets either in $\mathbb{S}^{n}\times\mathbb{R}$ or in a flat metric cone over $\mathbb{S}^n$ (minus the origin). But I wouldn't be surprised if there is a simpler way. $\endgroup$ Jan 15, 2013 at 12:11
  • $\begingroup$ @Thomas Richard, This paper: math.sciences.univ-nantes.fr/~carron/flat_end.pdf claims that the the number of ends is finite (I think they implicitly assume that $M$ is connected). $\endgroup$
    – Thomas Rot
    Jan 15, 2013 at 13:28
  • 4
    $\begingroup$ @Thomas Rot: If you use the isometric classification of flat ends quoted in that paper, I think you can answer your own question (affirmatively). The classification is due to Eschenburg and Schroeder. $\endgroup$
    – Misha
    Jan 15, 2013 at 13:53
  • $\begingroup$ If the answer to the displayed question is no, then there is a sequence of asymptotically flat n-manifolds that collapses to an Alexandrov space of dimension $<n$ and nonnegative curvature. Analysis of this collapse can in principle lead to a contradiction (or counterexamples) but I do not know how to do this. For partial answers check survey of Greene library.msri.org/books/Book30/files/greene.pdf (see page 120-121), and a paper by Petrunin-Tuscmann mis.mpg.de/publications/preprints/1999/prepr1999-47.html. $\endgroup$ Jan 15, 2013 at 14:01
  • $\begingroup$ Actually, as Misha says Eschenburg-Schroeder seems to do the job. $\endgroup$ Jan 15, 2013 at 14:06

1 Answer 1

14
$\begingroup$

Here is a self-contained proof not using any classification. With some effort, it can be made to work under weaker assumptions: the curvature is nonnegative outside a compact set and is bounded from above. It is a variation of the proof of the Soul Theorem via Sharafutdinov's retraction.

Fix a reference point $o\in M$ and define a Busemann function $b:M\to\mathbb R$ by $$ b(x) = \limsup_{y\in M,\ d(o,y)\to\infty} ( d(o,y)-d(x,y)) $$ where $d$ is the Riemannian distance (note the non-standard sign convention). The flat ends assumption implies the following key features:

  • Sublevel sets of $b$ are compact.

  • $b$ is (locally) convex outside a compact set.

To prove this, first observe some general properties of $b$. First, $b$ is 1-Lipschitz. Second, $b$ goes to $+\infty$ along any geodesic ray (essentially by the triangle inequality). Third, for any $x\in M$ there exists a geodesic ray starting at $x$ such that $b$ grows at unit speed along this ray (take a limit of geodesic segments $[xy_i]$ where a sequence $y_i$ realizes the $\limsup$). Let me call such rays "calibrating". Finally, for every compact set $K$ there is a compact set $K'$ such that no calibrating ray starting outside $K'$ intersects $K$. Indeed, otherwise a sequence of such rays would converge to a geodesic line, and $b$ would going to $-\infty$ in one of the directions along this line.

Now convexity of $b$ outside a compact set follows easily. Let $x_0\in M$ and $\gamma$ be a calibrating ray starting from $x_0$. Then $b(\gamma(t))=b(x_0)+t$ for all $t\ge 0$. Since $b$ is 1-Lipschitz, it follows that $$ b(x) \ge b(x_0) + (t-d(\gamma(t),x)) $$ for every $y\in M$, with equality for $x=x_0$. So $b$ is supported from below at $x_0$ by a function of the form $x\mapsto const - d(\gamma(t),x)$. If $x_0$ is sufficiently far from $o$, the ray $\gamma$ is contained in the flat part. Hence the distance to $\gamma(t)$ is bounded from above by a similar Euclidean distance function which is nearly linear near $x_0$ if $t$ is large. Existence of such lower bounds for $b$ implies that $b$ is convex.

Now let me show that sublevels of $b$ are compact. Suppose the contrary, then there is a sequence $x_i$ with $d(o,x_i)\to\infty$ but $b(x_i)\le const$. Choose a subsequence such that geodesic segments $[ox_i]$ converge to a geodesic ray $\gamma$. On this ray, mark a point $y$ where it leaves a ball containing the non-flat part of $M$. Once a segment $[ox_i]$ contains a point $y_i$ near $y$ and almost the same direction as $\gamma$ there, one observes that the derivative of $b$ along this segment at $y$ is greater than, say, 1/2. This and convexity of $b$ yield a linear lower bound for $b(x_i)$, contrary to the assumption $b(x_i)\le const$.

This proves the two key properties of $b$. Let $B_r$ denote the sublevel set $\{x:b(x)\le r\}$. Let $r_0$ be such that $b$ is convex outside $B_{r_0}$. Then, for every $R\ge r\ge r_0$, there is a distance non-increasing retraction (homotopic to identity) $f:B_R\to B_r$. Indeed, let $\varepsilon$ be smaller than the minimum injectivity radius on $B_R$. Then one retracts $B_R$ to $B_{R-\varepsilon}$ via a nearest-point projection, Locally it is just a nearest-point projection to a convex set in a Euclidean space, so it is well-defined and does not increases distances. Iterating this construction yields a retraction from $B_R$ to $B_r$. Moreover the complement of $B_r$ is mapped to the boundary of $B_r$.

Now let us turn to the injectivity radius. Let $r$ be such that $B_{r/2}$ contains the not-flat part of $M$ and $\rho_0$ is the minimum of $r/10$ and the injectivity radius on $B_{2r}$. I claim that the injectivity radius is no less than $\rho_0$ everywhere. Suppose the contrary. Then, somewhere outside $B_r$ there is a geodesic loop of length $2\rho<2\rho_0$. This loop is not contractible in the flat part. Apply the above retraction $f:B_R\to B_r$ where $R$ is so large that the loop is contained in $B_R$. The image is a loop in the boundary of $B_r$ with length $\le 2\rho$ and still non-contractible in the flat part. Hence the injectivity radius at $f(x)$ is at most $\rho$, a contradiction.

$\endgroup$
0

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.