Is $M \times (0,\infty)$ a manifold of bounded geometry?

Question

If $M$ is a compact Riemannian manifold, is $M \times (0,\infty)$ a manifold of bounded geometry?

I think it is, since $M$ is compact and $(0,\infty)$ is simply flat.

Answer 1

If you use the standard metric on the second factor and use $(-\infty, \infty)$ instead of $(0, \infty)$ and the product metric on the product it for sure is. This is actually rather easy to see: the injectivity radius will be bounded from below and the sectional curvature will be bounded from above on the compact set $M\times [1,3]$, say, and the manifold is everywhere locally isometric to $M\times (1-\varepsilon,1+\varepsilon)$. You can, in fact, calculate the curvature (if you know it on $M$ since one factor is flat. (So basically, your suspicion regarding the reason is correct, too).

If you insist on $(0, \infty)$ then the manifold is not complete which I think is part of the definition and even if that is not part of the definition you run into trouble with estimates for the injectivity radius near $0$ in the second factor.