Codex @codex  0

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 Riemannian volume form of an oriented $n$-dimensional Riemannian manifold is the unique positive $n$-form taking value one on every positively oriented orthonormal frame. In positive coordinates,The metric induces an inner product on exterior powers of the cotangent bundle, with wedges of distinct orthonormal covectors forming an orthonormal basis. The Hodge star is the pointwise linear map $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ specified byIt is an isometry and satisfies $\ast \ast \alpha =(-1{)}^{p(n-p)}\alpha$. With the positive-sign convention, the Hodge Laplacian on differential forms isHere $d$ is the exterior derivative and $\delta$ its formal adjoint; $\delta$ on functions is zero. On functions this is the Laplace-Beltrami operatorThe convention $\mathrm{div}\mathrm{grad}$ gives the negative of this operator. All subsequent Laplacians use the nonnegative convention above.

On a compact oriented manifold without boundary, the Hodge decomposition theorem gives an orthogonal decomposition of smooth real forms,The three component forms are uniquely determined, although the potentials in the last two summands need not be unique. The space ${\mathcal{H}}^p$ is finite-dimensional, and each de Rham cohomology class has a unique harmonic representative. Integration by parts gives $⟨\Delta \alpha ,\alpha {⟩}_{L^2}=\Vert d\alpha {\Vert }_{L^2}^2+\Vert \delta \alpha {\Vert }_{L^2}^2$, so harmonic forms are exactly the closed and coclosed forms. Set ${\Omega }^k=0$ outside $0\le k\le n$. Compactness without boundary is a hypothesis of this version of the theorem; orientation alone does not supply it.

A connection on $TM$ induces a dual connection on $T^{\ast }M$ and then connections on its tensor and exterior powers by the Leibniz rule. Explicitly, for a $p$-form $\alpha$,To prove the parallelism of the Riemannian volume form, use a local positively oriented orthonormal frame $E_i$, its dual coframe ${\theta }^i$, and write ${\nabla }_X{E}_j={\sum }_i{a}_{ij}(X)E_i$. Metric compatibility makes the matrix $a(X)$ skew-symmetric. Since ${\omega }_g={\theta }^1\wedge \cdots \wedge {\theta }^n$, differentiating the wedge yieldsOff-diagonal terms vanish because they repeat a coframe factor. Thus $\nabla {\omega }_g=0$; equivalently parallel transport preserves both the metric and the continuously chosen orientation.

Now let $M$ be compact, oriented and without boundary. For representing functionals on harmonic forms by wedge pairing, choose an $L^2$-orthonormal basis $h_1,\dots ,h_b$ of ${\mathcal{H}}^p$ and defineIf $\phi ={\sum }_a{c}_a{h}_a$, linearity givesThis also works when $b=0$, using the empty sum. Since the Hodge star commutes with the Hodge Laplacian, ${\psi }_0$ is harmonic of degree $n-p$.

For the full ambiguity of harmonic wedge-pairing representatives, write another representative as ${\psi }_0+\chi$. Because the star is an isometry,The forms $\ast \phi$ range over all of ${\mathcal{H}}^{n-p}$. Hence $\chi$ gives zero functional exactly when its harmonic projection is zero. Hodge decomposition proves that the complete answer isBoth summands in this ambiguity are permitted; it is not only an exact-form ambiguity. The exact and coexact component forms are orthogonal to every harmonic form, so every displayed choice works. Conversely a nonzero harmonic component would pair nontrivially with some $\ast \phi$, and therefore cannot occur. The harmonic representative $\ast h_f$ is unique; the arbitrary smooth representative generally is not. The formula also covers the endpoint degrees and any exceptional case where the ambiguity space is zero.

For a concrete coexact ambiguity, take the torus $(\mathbb{R}/2\pi \mathbb{Z}{)}^2$ with metric $d{x}^2+d{y}^2$. The form $\chi =\delta (\sin ydx\wedge dy)=\cos ydx$ annihilates every harmonic one-form by orthogonality, yet $d\chi =\sin ydx\wedge dy$ is nonzero. Thus an annihilating representative need not even be closed, let alone exact.