I'm working on problem from Do Carmo's "Riemannian Geometry" (Ex. 1, chapter 10). The problem, called Klingenberg's Lemma, puts a lower bound on the length of some curve in a homotopy, provided the homotopy is between distinct geodesics in a complete manifold with bounded sectional curvature.

This question already deals with some aspects of the same problem, so I'll quote some of it. My thanks to @Colescu for the monumental typing effort. Here's the problem, plus the provided hints:

(Klingenberg's Lemma). Let M be a complete Riemannian manifold with sectional curvature $K<K_0$, where $K_0$ is a positive constant.
Let $p,q \in M$ and let $\gamma_0,\gamma_1$ be two distinct geodesics joining p to q with $\ell[\gamma_0]\leq \ell[\gamma_1]$.
Let $\alpha_t$ be a homotopy from $\gamma_0$ to $\gamma_1$.
Prove that there exists $t_0\in[0,1]$ such that $$ \ell[\gamma_0]+\ell[\alpha_{t_0}]\geq \frac{2\pi}{\sqrt{K}} $$
Hint: Assume $\ell(\gamma_0)<\pi/\sqrt{K_0}$ (otherwise, we have nothing to prove). From Rauch's Theorem, $\exp_p:TpM\to M$ has no critical point in the open ball $B$ of radius $\pi/\sqrt{K_0}$, centered at $p$. For $t$ small, it is possible to lift the curve at to the tangent space $T_pM$, i.e., there exists a curve $\widetilde{\alpha}_t$ in $T_pM$, joining $\exp_p^{-1}(0)=0$ to $\exp_p^{-1}(q)=\widetilde{q}$, such that $\exp_p\circ\widetilde{\alpha}_t=\alpha_t$. It is clear that it is not possible to do the same for every $t\in[0,1]$, since $\gamma_1$ cannot be lifted keeping the endpoints fixed.

We conclude that for all $\varepsilon>0$ there exists a $t(\varepsilon)$ such that $\alpha_{t(\varepsilon)}$ can be lifted to $\tilde{\alpha}_{t(\varepsilon)}$ and $\tilde{\alpha}_{t(\varepsilon)}$ contains points with distance $<\varepsilon$ from the boundary $\partial B$ of $B$. In the contrary case, for some $\varepsilon>0$, all lifts $\tilde{\alpha}_t$ are at the distance $\geq\varepsilon$ from $\partial B$; the set of $t$'s for which it is possible to lift $\alpha_t$ will then be open and closed and $\alpha_1$ could be lifted, which is a contradiction.

So far, I've managed to cope - some adjustments to standard proofs of the path-lifting and homotopy lifting properties of cover maps generalize for a non-singular $\exp_p$, under some constraints.

The next part of the hint still bugs me:

Therefore, for all $\varepsilon>0$, we have $$\ell(\gamma_0)+\ell(\alpha_{t(\varepsilon)})\geq\frac{2\pi}{\sqrt{K_0}}-\varepsilon.$$ Now choose a sequence $\{\varepsilon_n\}\to0$, and consider a convergent subsequence of $\{t(\varepsilon_n)\}\to t_0$. Then there exists a curve $\alpha_{t_0}$ with $$\ell(\gamma_0)+\ell(\alpha_{t_0})\geq\frac{2\pi}{\sqrt{K_0}}.$$

I can see why, since $\tilde{\alpha}_{t(\varepsilon)}$ contains points with distance $<\varepsilon$ from the boundary $\partial B$ of $B$, we have: $$ \ell[\alpha_t]\geq \frac{\pi}{\sqrt{K_0}}-\varepsilon $$ But since we assume $\ell(\gamma_0)<\pi/\sqrt{K_0}$, I can't derive the inequality: $$\ell(\gamma_0)+\ell(\alpha_{t_0})\geq\frac{2\pi}{\sqrt{K_0}}-\varepsilon$$

I've tried to sharpen the bound on $\ell[\alpha_t]$, e.g. by proving it goes out for $\pi/\sqrt{K_0}$ and goes back by $\pi/\sqrt{K_0}$ but I don't see why would that be correct.

What am I missing?


1 Answer 1


Turns out it was simpler than I thought. Since there exists $b\in Im(\alpha_{t})$ such that: $$d(p,b)\geq \frac{\pi}{\sqrt{K}}-\frac{\epsilon}{2}$$ And since both $\alpha_{t}$ and $\gamma_{0}$ are paths from $p$ to $q$, we have, from the triangle inequality: $$ \ell[\alpha_{t}]+\ell[\gamma_0] \geq d(p,b)+d(b,q)+d(p,q)\geq d(p,b)+d(p,b)\geq \frac{2\pi}{\sqrt{K}}-\epsilon $$


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