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For an affinely parametrized geodesic $\gamma$, put $T=\dot{\gamma }$ and $D_t={\nabla }_T$. A Jacobi field is a smooth vector field $J$ along $\gamma$ satisfyingThis is a linear second-order equation, so $J$ is uniquely determined by $J(0)$ and $D_{t}J(0)$. The defining sign uses the curvature convention in the preceding solution.

To prove the realization of Jacobi fields by geodesic variations, write $p=\gamma (0)$, $V=T(0)$, $A=J(0)$ and $B=D_{t}J(0)$. Choose a small smooth curve $p(s)$ with $p(0)=p$ and $p^{\prime }(0)=A$, for instance $p(s)={\exp }_p(sA)$. Let $P_s:T_{p}M\to T_{p(s)}M$ be parallel transport along this curve and set $V(s)=P_s(V+sB)$. Then $V(0)=V$ and ${\nabla }_{s}V(s){|}_{s=0}=B$. DefineFor sufficiently small $\epsilon >0$, this is defined and smooth for every $(t,s)\in [0,1]\times (-\epsilon ,\epsilon )$. No completeness assumption is needed: smooth dependence for the geodesic ordinary differential equation gives an open set of initial data whose solutions exist throughout the compact interval $[0,1]$ around the given solution. The given geodesic also extends slightly beyond its endpoints by local existence.

Every $t$-curve of $f$ is a geodesic and $f(t,0)=\gamma (t)$. If $\hat{J}={\partial }_{s}f{|}_{s=0}$, torsion-freeness gives ${\nabla }_s{\partial }_{t}f={\nabla }_t{\partial }_{s}f$. Since ${\nabla }_t{\partial }_{t}f=0$, commuting covariant derivatives yieldsThus $\hat{J}$ satisfies the Jacobi equation. Moreover $\hat{J}(0)=A$ and $D_t\hat{J}(0)=B$. Uniqueness of the linear equation proves ${\partial }_{s}f(t,0)=J(t)$.

If $J(0)=0$, keep the initial point fixed and choose initial velocities $V+sB$. Differentiating $f(t,s)={\exp }_p(tV+stB)$ gives the general formula for Jacobi fields vanishing at their initial point:Here $T_{tV}(T_{p}M)$ is identified with $T_{p}M$. Every choice of $B$ gives a Jacobi field with zero initial value, and uniqueness shows that these are all such fields. The factor $t$ is part of the formula.

Points $\gamma (0)$ and $\gamma (L)$, with $L>0$, are conjugate along this geodesic if there is a nonzero Jacobi field with $J(0)=J(L)=0$. Equivalently, $(d{\exp }_p{)}_{LV}$ has a nontrivial kernel: for a field with zero initial value, $J$ is nonzero exactly when $B\ne 0$. This definition refers to the particular geodesic segment, not merely to the two endpoints.

Now suppose the sectional curvatures along the segment are nonpositive. Let $h(t)=|J(t){|}^2$. Metric compatibility and the Jacobi equation giveWhen $J,T$ are independent, the curvature term is $K(J,T)(|J{|}^2|T{|}^2-g(J,T{)}^2)\le 0$; it is zero when they are dependent. Therefore $h^{\prime \prime }\ge 0$. If $J$ vanishes at both endpoints, convexity gives $h(t)\le (1-t/L)h(0)+(t/L)h(L)=0$, whereas $h(t)\ge 0$ by definition. Hence $h=0$ and $J=0$. There are no conjugate points along the segment. This proves that nonpositive sectional curvature excludes conjugate points; it does not require the manifold to be complete or simply connected.

The divergence of a Riemannian vector field $X$ is the trace of the covariant derivative map $Y\mapsto {\nabla }_{Y}X$:It is characterized by ${\mathcal{L}}_X{\omega }_g=(\mathrm{div}X){\omega }_g$ in the oriented case. The metric-dual vector field $X_{\theta }={\theta }^{♯}$ is defined by $g(X_{\theta },Y)=\theta (Y)$ for every $Y$, using the musical isomorphism.

On a compact oriented manifold without boundary, the permitted exactness assertion and Stokes theorem imply ${\int }_M\mathrm{div}Y{\omega }_g=0$ for every smooth vector field $Y$. For a smooth function $a$, the product rule givesSince $⟨da,\theta ⟩=X_{\theta }(a)$, the formal-adjoint identity yieldsThis holds for all $a$; taking $a=\delta \theta +\mathrm{div}{X}_{\theta }$ proves the pointwise formulaIt agrees with $\delta \theta =-\ast d\ast \theta$, since $\ast \theta ={\iota }_{X_{\theta }}{\omega }_g$ and $d{\iota }_{X_{\theta }}{\omega }_g={\mathcal{L}}_{X_{\theta }}{\omega }_g$.

The Bochner formula for one-forms, with the positive Hodge Laplacian, isHere $\Delta =d\delta +\delta d$ acts on one-forms. The covariant derivative $\nabla \theta$ is a section of $T^{\ast }M\otimes T^{\ast }M$, and its formal-adjoint composition, the rough Laplacian, isThis expression is independent of the local orthonormal frame. The Ricci term is the zero-order bundle endomorphism specified byThus the identity separates the differential energy of the form from the curvature contribution; the sign of the Ricci term matches positive sectional curvature on the sphere.

Let $\theta$ be a harmonic one-form on a compact connected manifold with nonnegative Ricci curvature. Pair the formula with $\theta$ and integrate:Both integrands are nonnegative and continuous, so each vanishes everywhere. In particular $\nabla \theta =0$, proving that harmonic one-forms are parallel under nonnegative Ricci curvature. This integration uses the Riemannian volume density and formal adjoints, so orientation is not necessary for the final conclusion.

Fix $p\in M$. The evaluation mapis injective: a parallel form with zero value at $p$ stays zero under parallel transport along every path, and connectedness makes every point reachable from $p$. ThereforeThis is the dimension bound for harmonic one-forms under nonnegative Ricci curvature. The bound is sharp: on a flat $n$-torus the $n$ constant coordinate one-forms are parallel and harmonic. If Ricci curvature is positive definite at even one point, the same integral and parallelism argument force every harmonic one-form to vanish.