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For a variation of a unit-speed geodesic, let $T$ be its tangent, $J$ its variation vector field and $A={\nabla }_{s}J$ its variation acceleration. The second variation of Riemannian arc length has the displayed endpoint term in addition to the Riemannian index form of $J^{\perp }=J-⟨J,T⟩T$. Differentiating the speed twice gives $|D_{t}J{|}^2-⟨D_{t}J,T{⟩}^2+⟨D_s{D}_{t}J,T⟩$. Commuting derivatives and integrating $D_{t}A$ by parts gives the formula. The term cannot be discarded unless the endpoints are fixed or an appropriate endpoint condition makes it zero.

Let $\gamma :[0,\ell ]\to M$ be a unit-speed geodesic, let $F(s,t)$ be a smooth variation with fixed endpoints, and letbe its variation vector field. Then $V(0)=V(\ell )=0$. If $V^{\perp }=V-⟨V,T⟩T$ is its component normal to $\gamma$, the second variation of Riemannian arc length isHere $D_t={\nabla }_T$ is the covariant derivative along $\gamma$, $R$ is the Riemann curvature tensor, and $I$ is the Riemannian index form. Fixed endpoints remove the boundary term. The normal projection removes a tangential change of parametrization, which does not change length to second order.

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.