This is a problem from Riemannian Geometry by Do Carmo, namely Ch. 7, Sec. 3, Problem 7 on pg. 153.

Let $M, N$ be Riemannian manifolds with $N$ complete, and $f: M \to N$ a diffeomorphism for which there exists a $c > 0$ such that $|v| \geq c |df_p(v)|$ $\forall p \in M$ $\forall v \in T_p M$.

I figure I need to apply the Hopf-Rinow theorem or just go from the definition and show $exp_p$ is defined on all of $T_p M$, i.e. that any geodesic $\gamma(t)$ starting from $p$ is defined $\forall t \in \mathbb{R}$. I don't really know how to go about this, though.


It is easier to apply the following Hopf Rinow theorem:

$(N, h)$ is geodesically complete if and only if the following metric $$ d_N (r, s) := \inf\left\{ \int_0^1 |\dot \gamma(t)|_h dt \bigg| \gamma :[0,1] \to N, \gamma(0)= r, \gamma(1) = s\right\}$$ is a complete metric on $N$.

Together with your condition, one has (check!)

$$\tag{1} d_N (f(p), f(q)) \le \frac 1c d_M(p, q),\ \ \ \forall p,q\in M.$$

Now we can claim:

$(M, d_M)$ is complete.

Let $\{p_n\}$ be a Cauchy sequence in $M$. By $(1)$, $\{f(p_n)\}$ is Cauchy in $N$ and so $f(p_n) \to Q \in N$ by completeness of $N$. Note that there is $C >0$ and a closed ball $B_Q$ around $Q$ (since $f$ is a diffeomorphism) so that

$$ |v| \le C | df_q v|,\ \ \ \forall f(q) \in B_Q.$$

Thus for large $n$ we have $f(p_n) \in B_Q$ and so

$$ d_M (p_n, f^{-1}(Q))\le Cd_N(f(p_n), Q)$$

Thus $p_n \to f^{-1}Q$ and so $(M, d_M)$ is complete.

  • $\begingroup$ I think the constant in your formula (1) is 1/c. $\endgroup$ – Yilong Zhang Mar 21 '16 at 23:06
  • $\begingroup$ @YilongZhang : Yes, thanks and edited. $\endgroup$ – user99914 Mar 21 '16 at 23:08
  • $\begingroup$ Where did you get this Hopf-Rinow theorem? Do Carmo has a different version, and I wanted to restrict myself to his machinery. Is there a way to do it differently? Hmm, upon further inspection, this would seem to follow from his Proposition 2.6, right? $\endgroup$ – Ryker Mar 22 '16 at 1:31
  • $\begingroup$ That's Theorem 2.8 (c) (second printing 1993 version) in the book. @Ryker $\endgroup$ – user99914 Mar 22 '16 at 1:35
  • $\begingroup$ Ah, so the statement that $M$ is complete as a metric space? $\endgroup$ – Ryker Mar 22 '16 at 1:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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