# Is a manifold with flat ends of bounded geometry?

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?
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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. – Thomas Richard Jan 15 '13 at 12:11
@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). – Thomas Rot Jan 15 '13 at 13:28
@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. – Misha Jan 15 '13 at 13:53
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. – Igor Belegradek Jan 15 '13 at 14:01
Actually, as Misha says Eschenburg-Schroeder seems to do the job. – Igor Belegradek Jan 15 '13 at 14:06

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.

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Thank you for your very informative answer. – Thomas Rot Jan 17 '13 at 12:40