Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am reading do Carmo's Riemannian Geometry Chapter 7, and I want to do some exercises. I think that I need some hints to solve the following:


  1. How do I construct a counterexample that a local diffeomorphism does not presreve completeness?

  2. I want to show that any homogeneous manifold is complete, but I have no idea about the property "homogeneous".

Please give me an answer or some hints.

share|cite|improve this question
In 1., could you please state precisely what properties you need your example to satisfy? In 2., when you say you have no idea about the property homogeneous, do you mean that you do not know the what the definition is? If so, can you look up the definition? – Jonas Meyer Jan 7 '12 at 5:57
Sorry! I will ask question more precisely next time. Thank you very much :) – Peter Hu Jan 7 '12 at 8:34
up vote 1 down vote accepted

Your question is incomplete. Please be more precise next time. The problem is:

Let $f: M_1\rightarrow M_2$ be a local diffeomorphism of a manifold $M_1$ onto a Riemannian manifold $M_2$. Introduce on $M_1$ a Riemannian metric such that $f$ is a local isometry. Show by an example that if $M_2$ is complete, $M_1$ need not be complete.

Take $M_2$ to be the circle, $S^1$ in $\mathbb{R}^2$ with the usual Euclidean norm. Take $M_1 = (0,2)$. Wrap $(0,2)$ around $S^1$ twice (your usual covering map: $f: x\mapsto e^{2\pi i x}$), and pull back the metric from $S^1$ onto $(0,2)$. This will give you a local isometry, but $(0,2)$ with respect to this metric is not complete. The trick here is that $f$ is not one-to-one, so even though $(0,2)$ is not complete, there are enough points to cover $S^1$ and do so locally isometrically.

In the second question, a Riemannian manifold $M$ is homogeneous provided that for any pair $p, q\in M$, there exists an isometry of $M$ mapping $p$ to $q$.

Let me give you a hint here: try using the Hopf-Rinow theorem, which is covered in do Carmo, I think in the same chapter.

share|cite|improve this answer
Sorry! I will write down my question clear next time. And I will try to using Hopf-Rinow theorem. Thank you very much :) – Peter Hu Jan 7 '12 at 8:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.