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The Bonnet-Myers theorem states that if a complete connected $n$-dimensional Riemannian manifold satisfiesfor some $k>0$, thenIn 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 $E_1,\dots ,E_{n-1}$ normal to $T=\dot{\gamma }$ and setThe 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 givesUsing the Ricci curvature bound and integrating ${\sin }^2$ and ${\cos }^2$ yieldsso $\ell \le \pi /\sqrt{k}$. 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.

Give $S^2\times \mathbb{R}$ the Riemannian product of the unit round metric and the Euclidean metric. It is complete and has infinite diameter. The round sphere has scalar curvature $2$, while the line has scalar curvature $0$; scalar curvature is additive under Riemannian products, soThus this manifold has a strictly positive uniform lower bound on scalar curvature but violates the conclusion of the Bonnet-Myers theorem. Its Ricci curvature vanishes in the $\mathbb{R}$ direction, showing precisely why a scalar-curvature bound is insufficient.