Does $\pi_k(M)\neq 0$ implies $\operatorname{ind}(\gamma) < k$?

Question

Cross post from MSE. and sorry if this is an obvious question.

Here is a line of proof of Theorem 1.15 from

Brendle, Simon, Ricci flow and the sphere theorem, Graduate Studies in Mathematics 111. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4938-5/hbk). vii, 176 p. (2010). ZBL1196.53001.

Let us fix two points $p, q \in M$ such that $d(p, q) = \operatorname{diam}(M, g) > \frac{\pi}{2}$. Since $\pi_k(M)\neq 0$, there exists a geodesic $\gamma : [0,1] \to M$ such that $\gamma(0) = \gamma(1) = p$ and $\operatorname{ind}(\gamma) < k$.

Q: Why $\pi_k(M)\neq 0 \implies \operatorname{ind}(\gamma) < k$? Is this a general fact?

Note: $\operatorname{ind}(\gamma):=$ Morse index of $\gamma$.

Answer 1

Very roughly speaking: Geodesics $\gamma:[0,1]\to M$ with $\gamma(0)=\gamma(1)=p$ and $\operatorname{ind}(\gamma)<k$ correspond to critical points of Morse index less than $k$ of the energy functional $E:\Omega_p(M)\to \mathbb{R}$, where $\Omega_p(M)$ is the space of loops based at $p$.

The assumption that $\pi_k(M)\cong\pi_{k-1}(\Omega_p(M))\neq 0$ implies that $H_i(\Omega_p(M))\neq 0$ for some $i\leq k-1$, by the Hurewicz Theorem. Then Morse homology implies there must be a critical point of index less than $k$.