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?   
 A: 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.
A: 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.
A: 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.
A: 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. 
