Index and length of closed geodesics

Question

Consider the round metric on $S^n$. The geodesics are (multiples of) great circles, and one can verify that this metric is of Morse-Bott type. The Morse indices of the n-covered great circles are (if I recall correctly) $(n-1), 3(n-1), 5(n-1), \dots, (2k-1)(n-1), \dots$.

Observe in particular that the Morse index of geodesics grows linearly with the length. i.e. there is a constant $C>0$ (depending only on the metric) such that for any geodesic $\gamma$, we have $$\operatorname{ind}(\gamma) > C \operatorname{length}(\gamma).$$

Question: let $g$ be a small, suitably generic perturbation of the round metric with the property that all geodesics are non-degenerate (such a metric is called "bumpy", I believe). Is it still the case that the Morse index of the geodesics grows linearly with the length?

I am also interested in the weaker question: does there exist a bumpy metric on $S^n$ with the property that the Morse index of the geodesics grows linearly with the length?

Answer 1

This follows from Bonnet-Myers. For a metric near the round metric (in the $C^\infty$ topology), the sectional curvature will be pinched below by $k > 0$, where $k\thickapprox 1$. Hence a segment of length $>\pi/\sqrt{k}$ of any geodesic will be unstable by Myers' theorem. Hence the index of a geodesic of length $L$ should be at least $L\sqrt{k}/\pi$: divide it up into segments of length $\geq \pi/\sqrt{k}$ which each have index at least 1. In fact, using a comparison theorem, one could probably do better and get an extra constant factor depending on the dimension (Myers theorem is using the weaker input of a lower bound on Ricci curvature, so a lower bound on sectional curvature will give a higher index estimate). You may be familiar with it, but Milnor's book on Morse theory is a good introduction to these topics (see Part III). For comparison theorems I like Gallot-Hulin-Lafontaine.