= Solution
The <Bonnet-Myers theorem> states that if a complete connected $n$-dimensional <Riemannian manifold> satisfies
$$
\operatorname{Ric}\geq(n-1)k g
$$
for some $k>0$, then
$$
\operatorname{diam}(M)\leq\frac{\pi}{\sqrt k}.
$$
In particular, $M$ is compact and has finite <fundamental group>.
By the <Hopf-Rinow theorem>, points $p,q\in M$ are joined by a unit-speed length-minimizing <geodesic> $\gamma:[0,\ell]\to M$. Choose a parallel orthonormal frame $E_1,\ldots,E_{n-1}$ normal to $T=\dot\gamma$ and set
$$
V_i(t)=\sin\left(\frac{\pi t}{\ell}\right)E_i(t).
$$
The endpoint-vanishing fields $V_i$ arise from fixed-endpoint variations. Since $\gamma$ minimizes length, its <Riemannian index form> is nonnegative on each $V_i$. Summing the <second variation of Riemannian arc length> gives
$$
0\leq\sum_{i=1}^{n-1}I(V_i,V_i)
=\int_0^\ell\left[
(n-1)\frac{\pi^2}{\ell^2}\cos^2\left(\frac{\pi t}{\ell}\right)
-\operatorname{Ric}(T,T)\sin^2\left(\frac{\pi t}{\ell}\right)
\right]dt.
$$
Using the <Ricci curvature> bound and integrating $\sin^2$ and $\cos^2$ yields
$$
0\leq\frac{(n-1)\ell}{2}\left(\frac{\pi^2}{\ell^2}-k\right),
$$
so $\ell\leq\pi/\sqrt k$. Taking the <supremum> over $p,q$ proves the diameter bound. Hopf-Rinow now makes the closed bounded space $M$ compact. Finally, the same bound applies to the complete <universal cover>; a compact universal cover has finite fibres over $M$, so $\pi_1(M)$ is finite.
Solved by gpt-5.6-sol high.
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