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Use the curvature convention $\mathcal{R}$ of Solution 1, and put $T=\dot{\gamma }$. On continuous piecewise smooth perpendicular vector fields with zero endpoint values, the Riemannian index form is the symmetric bilinear formA Jacobi field is a smooth field $J$ along $\gamma$ satisfying $D_t^{2}J+\mathcal{R}(J,T)T=0$. The point $\gamma (t_0)$ is a conjugate point to $\gamma (0)$ along this geodesic when there is a nonzero Jacobi field with $J(0)=J(t_0)=0$. Such a field is perpendicular to $T$, because $⟨J,T⟩$ is affine and vanishes at both endpoints.

The conjugate-point criterion for the Riemannian index form, stated without proof, isAt a first conjugate endpoint the form is nonnegative but has a nonzero nullspace, consisting of endpoint-vanishing Jacobi fields. A conjugate point strictly inside gives a negative direction. This explicitly distinguishes the non-strict and strict senses of positivity in the question; in terminology where “positive definite” already means strict positivity, the second line is the positive-definiteness condition.

Choose $n-1$ perpendicular orthonormal fields $E_1,\dots ,E_{n-1}$ by parallel transport along $\gamma$, and put $V_i(t)=\sin (\pi t/L)E_i(t)$. They are admissible andThe tangent has unit length. The assumed Ricci curvature bound and the two elementary sine-square and cosine-square integrals give the sine index-form bound for positive Ricci curvature:Since $n\ge 2$, at least one of the admissible $V_i$ therefore has $I(V_i,V_i)<0$.

If the Riemannian manifold is complete, the Hopf-Rinow theorem supplies a minimizing geodesic between any two points. Its Riemannian index form is nonnegative: any smooth fixed-endpoint variation has nonnegative second variation of geodesic energy, since a length minimizer of constant speed also minimizes energy on a fixed parameter interval by the Cauchy-Schwarz inequality. The same conclusion for piecewise smooth fields follows by smooth approximation, or by piecewise variations. The negative direction above excludes a minimizing segment longer than $\pi /\sqrt{\kappa }$. Taking the supremum over pairs yields the diameter conclusion of the Bonnet-Myers theorem:

The completeness assumption is essential. An explicit incomplete positively curved strip with infinite diameter isThe map $(u,v)\mapsto (\cos u\cos v,\cos u\sin v,\sin u)$ is a local isometry to the unit sphere, so $K=1$ and, in dimension two, $\mathrm{Ric}=g$. Thus the required Ricci curvature bound holds with $n=2$, $\kappa =1$. The unit-speed meridian $t\mapsto (t,0)$ reaches the missing boundary $u=\pi /4$ in finite time, so the strip is geodesically incomplete. On the other hand every curve joining $(0,0)$ to $(0,A)$ hassince $\cos u\ge 1/\sqrt{2}$ throughout the strip. Therefore its intrinsic diameter is infinite, in particular larger than $\pi$. This is the unwrapped strip with $v\in \mathbb{R}$, not a strip with longitude identified modulo $2\pi$; that distinction prevents spherical shortcuts from invalidating the example.

The Bonnet-Myers theorem says that a connected geodesically complete $n$-dimensional Riemannian manifold, with $n\ge 2$ and $\mathrm{Ric}\ge (n-1)\kappa g$ for a constant $\kappa >0$, hasThe dimension and positive lower bound are essential: in dimension one the Ricci condition is vacuous, and a zero lower bound does not imply bounded diameter.

By Hopf-Rinow, any two distinct points have a unit-speed minimizing geodesic $\gamma :[0,L]\to M$. Choose parallel orthonormal fields $E_1,\dots ,E_{n-1}$ perpendicular to its tangent $T$. For the endpoint-vanishing fields $V_i(t)=\sin (\pi t/L)E_i(t)$, the second variation of geodesic energy gives nonnegative index formsSumming and using the Ricci lower bound produces the sine index-form bound for positive Ricci curvature:If $L>\pi /\sqrt{\kappa }$, the last expression is negative, a contradiction. This proves the diameter bound. Hopf-Rinow makes closed bounded sets compact, so the entire manifold is compact.

Give the universal cover the pullback metric. Local isometry preserves its Ricci bound, and lifting complete base geodesics proves completeness of the cover. The same diameter and compactness argument applies there. A fiber of the covering is closed and discrete, hence finite in this compact cover; its cardinality is that of the fundamental group. This proves the final assertion.

For a counterexample that also breaks the diameter conclusion, use the incomplete positively curved strip with infinite diameterThe map $\Phi (u,v)=(\cos u\cos v,\cos u\sin v,\sin u)$ is a local isometry to the round unit sphere: its coordinate derivatives are orthogonal, with squared lengths one and ${\cos }^{2}u$. Thus $K=1$ and $\mathrm{Ric}=g$, satisfying the required lower bound with $n=2$, $\kappa =1$.

The meridian $(u,v)=(t,0)$ is a unit-speed geodesic and reaches the excluded boundary at $t=\pi /4$, so this metric is incomplete. Every curve joining $(0,0)$ to $(0,L)$ has length at least $\int \cos u|v^{\prime }|dt\ge |L|/\sqrt{2}$, because $\cos u\ge 1/\sqrt{2}$ on the strip. Therefore $\boxed{\mathrm{diam}M=\infty }$ despite the positive Ricci bound. Completeness cannot be omitted.