# Do cotangent bundles have “bounded geometry”?

I have often heard the phrase "a manifold $M$ has bounded geometry" thrown around without ever seeing a precise definition of what this means. Apparent examples are compact manifolds and $\mathbb{R}^n$. What is the definition and can I therefore assume that cotangent bundles of compact manifolds have bounded geometry?

• Bounded geometry is a property of a metric space, so your question doesn't make sense. A Riemannian manifold has bounded geometry if and only if the curvature tensor and all of its covariant derivatives are uniformly bounded. – Paul Siegel Jul 31 '15 at 10:50
• OK, so does a cotangent bundle equipped with the standard metric (induced from a metric on the base) have bounded geomerty if the metric satifies these curvature properties? – ss78 Jul 31 '15 at 11:16
• One should note that there are several different versions of the definition of "bounded geometry", and they are not mutually equivalent to each other. Everytime an author uses this notion one has to check which of these definitions he is using (and since some authors do not write it down explicitly, one often has to guess the version which is used). – AlexE Oct 9 '17 at 8:04

A Riemannian manifold is said to have bounded geometry if the curvature tensor and all of its covariant derivatives are uniformly bounded (by a bound depending on the order of the derivative) AND the injectivity radius is bounded below by $1$. Clearly any compact manifold has a metric of bounded geometry. It is less obvious but still true that any manifold has a metric of bounded geometry. There are several ways to prove this. One of them can be found in http://arxiv.org/abs/1303.5957.

For the tangent (or equivalently cotangent) bundle of a Riemannian manifold there is a standard way to produce a metric of bounded geometry on a neighborhood of the zero section that comes from a bounded geometry metric on the base. The key point is that the exponential map is a local diffeomorphism on the ball about the origin whose radius is $\ge \pi/\sqrt{K}$, where $K$ is the upper curvature bound of the base, so the pullback metric by such local diffeomorphisms has a uniform lower injectivity radius bound of $\ge \pi/\sqrt{K}$. (I omit some details here). This contruction is used in collapsing theory.

Let me complement (and compliment?) Igor's answer by providing an explicit definition of bounded geometry from the work of Cheeger and Gromov.

A Riemannian manifold $(M,g)$ has $C^k$-bounded geometry if there is a uniform radius $r > 0$ so that

• for each $x \in M$ the exponential map $e_x: T_xM \to M$ is a diffeomorphism from $B_r(0)$ to $B_r(x)$,
• the pullback of the metric $g_{ij}$ along $e_x$ is bounded in the $C^k$ topology on $T_xM$, and
• the inverse $g^{ij}$ is bounded in sup-norm.

You can find this in Sec 3 of Finite Propagation Speed, Kernel Estimates for Functions of the Laplace Operator, and the Geometry of Complete Riemannian Manifolds in J Diff Geo 17 (1982) pp 15--53, available here.

No, the tangent (and co-tangent, I think) bundle of a bounded geometry manifold do not have bounded geometry, unless the original manifold was flat to begin with. The heuristic reason is that any curvature on the base manifold blows up along the fibers.

See for example Gudmundsson and Kappos, On the geometry of tangent bundles (2002), Theorem 7.8. This is for tangent bundles only, but I'd assume the equivalent result holds for the cotangent bundle.

By the way, a reference for the definition of bounded geometry for Riemannian manifolds is Eichhorn, The Banach manifold structure of the space of metrics on noncompact manifolds, (1991), page 255.

• The corollary 7.11 you mention (called Theorem 7.11) claims "Let (M, g) be a Riemannian manifold and the tangent bundle TM be equipped with the Sasaki metric g. Then (TM, g) has constant scalar curvature if and only if (M, g) is fiat." How you infer that TM admits no bounded geometry unless M is flat? – valeri Jul 31 '15 at 18:57
• @valeri: Sorry, I wasn't careful and reference the wrong thing. Now corrected it to Theorem 7.8. If $TM$ would have bounded curvature, then also its sectional curvature would be bounded, and Theorem 7.8 then implies that $TM$ is flat, which is equivalent to $M$ flat using Theorem 7.6. But can also look at the explicit formulas for the curvature of $TM$ expressed in terms of the curvature of $M$. – Jaap Eldering Jul 31 '15 at 19:57
• all these claims are about very special Sasaki metric on TM, and do not imply non existence of bounded geometry. Actually, as was already answered above - every manifold admits bounded geometry [Greene], which means that your answer is wrong. To "believe" in Greene's result you might just represent a manidold as a handlebody and endow every handle with bounded geometry which is product near the boundary - then any union of handles is provided with bounded geometry. – valeri Jul 31 '15 at 21:32
• @valeri: well, the OP clarified in the comments to the question that he was using the "standard metric induced from a metric on the base", which I interpreted as the natural Sasaki metric. So I think my answer as it stands correctly answers the question. – Jaap Eldering Aug 1 '15 at 1:50
• Yes, if "standard metric" means Sasaki. Although with this correction the question seems quite different. Btw, for TS^n the Cheeger-Gromoll metric for me much more "standard" - which I believe has bounded geometry. – valeri Aug 1 '15 at 9:28

Alexander Grigor'yan ("Heat Kernel and Analysis on Manifolds", 2009) gives the following definition:

a Riemannian manifold $(M, g)$ (of $\dim M = n$) has bounded geometry if and only if there exist the "bounds" $0<c<C$ and the radius $\varepsilon > 0$ such that for every $x \in M$ the geodesic ball $B(x, \varepsilon)$ is diffeomorphic to the standard Euclidean ball $B(0, \varepsilon) \subset \Bbb R^n$ under the diffeomorphism $\varphi _x : B(0, \varepsilon) \to B(x, \varepsilon)$ and $c g_0 \le \varphi _x ^* g \le C g_0$ (where $g_0$ is the usual Riemannian structure on $\Bbb R^n$).

(The definition is assembled from pieces found at pages 93 and 312.)

Notice that in general $T^*M$ has no natural Riemannian structure, therefore asking whether it has bounded geometry is meaningless, regardless of whether $M$ is compact or not.