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Diagonalize the self-adjoint Jacobi curvature operator at the initial point and transport its orthonormal eigenbasis parallel along the geodesic. The Jacobi field equation reduces to $j_i^{\prime \prime }+{\lambda }_i{j}_i=0$. For zero initial position and initial derivative $b_i$, the solution is $b_i\sin (\sqrt{{\lambda }_i}t)/\sqrt{{\lambda }_i}$ when ${\lambda }_i>0$, $b_{i}t$ when ${\lambda }_i=0$, and $b_i\sinh (\sqrt{-{\lambda }_i}t)/\sqrt{-{\lambda }_i}$ when ${\lambda }_i<0$. Only positive eigenvalues contribute conjugate points, at $t=k\pi /\sqrt{{\lambda }_i}$. At a common zero the multiplicity is the sum of the contributing eigenspace dimensions. The displayed convention uses the curvature operator positive on a positively curved plane; reversing the tensor convention also reverses its tensor formula.

For a Jacobi field along a geodesic with nonpositive sectional curvature, the curvature convention with positive round-sphere curvature givesThus $|J{|}^2$ is nonnegative and convex. If it vanishes at both ends of a segment, convexity forces it to vanish everywhere. No nonzero Jacobi field can vanish at both ends, so the segment has no conjugate points. Only curvature of planes containing the geodesic tangent is needed, and completeness is unnecessary.

Use the curvature convention $R(X,Y)={\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}$ of the earlier curvature calculation. A geodesic variation of $\gamma :[a,b]\to M$ is a smooth map $F:(-\epsilon ,\epsilon )\times [a,b]\to M$ with $F(0,t)=\gamma (t)$ such that every $t\mapsto F(s,t)$ is an affinely parametrized geodesic. Its variation vector field $J={\partial }_{s}F{|}_{s=0}$ satisfies the Jacobi equationIndeed $D_{s}T=D_{t}J$ by vanishing torsion tensor; differentiating $D_{t}T=0$ in $s$ and commuting the covariant derivatives gives this equation. A Jacobi field is a field satisfying this equation. Conversely, any prescribed pair $J(a),D_{t}J(a)$ is obtained by smoothly varying the initial point and velocity of a geodesic; smooth dependence of the geodesic equation then constructs a geodesic variation on the compact parameter interval, whose variation vector field is the prescribed Jacobi field by uniqueness of this linear equation.

For continuous piecewise smooth vector fields along a curve $V,W$ along $\gamma$, the Riemannian index form is the symmetric bilinear formContinuity at the breakpoints is part of the admissibility condition; without it an arbitrary jump could destroy the claimed index inequality. The curvature symmetries make $V\mapsto R(V,T)T$ self-adjoint. For fields with zero endpoints, this form is the second variation of geodesic energy $E=\frac{1}{2}{\int }_a^b|\dot{\gamma }{|}^{2}dt$. More explicitly, twice differentiating energy, commuting $D_s,D_t$ and integrating the term $⟨D_t(D_s{\partial }_{s}F),T⟩$ by parts leaves the displayed integral; $D_{t}T=0$ and fixed endpoints remove the remaining terms.

The endpoints $p=\gamma (a)$ and $q=\gamma (b)$ are conjugate points along $\gamma$ when there is a nonzero Jacobi field $J$ with $J(a)=J(b)=0$. Equivalently, the Riemannian exponential map at the initial velocity for this segment has singular differential: variations of initial velocity give precisely the Jacobi fields with zero initial value. For a unit-speed geodesic starting at $p$, its cut time isIf finite and the geodesic is defined there, its endpoint $\gamma (c(v))$ is the Riemannian cut point along that geodesic. Before the cut time the segment minimizes, at that time it still minimizes by continuity, and every positive extension ceases to minimize. This is the Riemannian notion, independent of whether deleting the endpoint disconnects the space.

Let ${\mathcal{H}}_0$ denote the continuous piecewise smooth vector fields along a curve with zero endpoints. The precise characterization at conjugate endpoints isThus Jacobi fields form the radical of the fixed-endpoint index form. For a smooth field $J$, integration by parts givesA Jacobi field makes both terms vanish. Conversely, if $I(J,W)=0$ for all tests, first take tests supported inside each smooth subinterval. The fundamental lemma gives the Jacobi equation on each piece. Integrating by parts piecewise then leaves, at an interior breakpoint $t_i$, the term $⟨D_{t}J(t_i^{-})-D_{t}J(t_i^{+}),W(t_i)⟩$. The values of the tests there are arbitrary, so all derivative jumps vanish. The continuous field and its continuous first covariant derivative solve the smooth Jacobi equation globally, making $J$ smooth by uniqueness of its initial-value problem. In particular, conjugacy means that this radical is nonzero. Merely imposing $I(J,J)=0$ is insufficient when the Riemannian index form is indefinite; a zero value of an indefinite quadratic form need not mean membership in its radical.

Now suppose there are no conjugate points on $(0,1]$. In a parallel orthonormal frame along $\gamma$, write the Jacobi equation as $j^{\prime \prime }+K(t)j=0$, where the symmetric matrix $K$ represents $Z\mapsto R(Z,T)T$. LetEvery Jacobi field vanishing at zero has components $j(t)=A(t)c$. The nonconjugacy assumption means $A(t)$ is invertible for every $0<t\le 1$, and hence the unique Jacobi field with the requested endpoint values isPut $U=W-V$, so $U(0)=U(1)=0$. Since $V$ solves the Jacobi equation, integration by parts gives $I(V,U)=0$ and thereforeTo establish the needed strict positivity, use the Riccati factorization of the Riemannian index form. Differentiation givesand the initial data make this conserved quantity zero. Consequently $S=A^{\prime }{A}^{-1}$ is symmetric and obeys the matrix Riccati equation $S^{\prime }+S^2+K=0$. For the components $u$ of $U$,Near zero, the differential equation gives $A(t)=t{I}_n+O(t^3)$ and $S(t)=t^{-1}{I}_n+O(t)$. Since $U$ is piecewise smooth and $U(0)=0$, $u(t)=O(t)$, so $u^{T}Su\to 0$. At time one the same boundary term vanishes because $u(1)=0$. Integration, including cancellation at all interior breakpoints by continuity, therefore givesIf equality holds, $u^{\prime }=Su$ on each piece, so $(A^{-1}u{)}^{\prime }=0$ on $(0,1]$. Continuity across the breakpoints gives $u=Ac$ with one constant vector $c$; $u(1)=0$ and invertibility of $A(1)$ force $c=0$. HenceThe nonzero endpoint condition is compatible with this proof, but the minimizing property itself holds for arbitrary specified final value.

For the relation to the Riemannian cut point, let $a>0$ be the first conjugate time along a unit-speed geodesic, and let $J$ be a nonzero Jacobi field on $[0,a]$ vanishing at both endpoints. Necessarily $D_{t}J(a)\ne 0$, otherwise uniqueness of the Jacobi equation would give $J=0$. Fix any later time $b>a$ for which the geodesic exists. Extend $J$ by zero to a continuous piecewise smooth field $Z$ on $[0,b]$. Then $I(Z,Z)=0$, while for any endpoint-vanishing smooth field $P$,Choose $P(a)=-D_{t}J(a)$, using a smooth bump times parallel fields, so this equals $-|D_{t}J(a){|}^2<0$. For small $\epsilon >0$,If a smooth variation is desired, smooth the single corner of this field while preserving the endpoints. The field and its derivative converge in the integral defining the Riemannian index form, so strict negativity persists. The second variation of geodesic energy then gives a fixed-endpoint path of smaller energy. By the Cauchy-Schwarz inequality, $L^2\le 2bE$ for paths on $[0,b]$, with equality for the original unit-speed geodesic; the new path is therefore shorter. This proves loss of geodesic minimality beyond a conjugate point, and henceA missing conjugate time is interpreted as infinity. If a finite conjugate time exists and the geodesic continues past it, every later segment fails to minimize, so its first loss of minimality occurs no later than that conjugate time.

Finally, the cut-point dichotomy for complete Riemannian manifolds is the following characterization, stated without proof. On a complete connected Riemannian manifold, the cut time is the first time at which either the endpoint is conjugate to $p$ along the given geodesic, or a distinct minimizing geodesic from $p$ reaches the same endpoint with the same length. Thus a finite Riemannian cut point is either the first conjugate point or the meeting point of at least two distinct minimizing geodesics from the initial point; both alternatives may occur together. Completeness is the hypothesis guaranteeing that competing minimizing geodesics exist at a nonconjugate cut endpoint. It is not needed for the preceding local index-form argument on an existing segment.

Expand a Jacobi field in the parallel orthonormal eigenbasis: $J(t)={\sum }_i{j}_i(t)e_i(t)$. Part (b) reduces its equation to the scalar equations $j_i^{\prime \prime }+{\lambda }_i{j}_i=0$. DefineFor arbitrary initial values $J(0)={\sum }_i{a}_i{e}_i$ and $D_{t}J(0)={\sum }_i{b}_i{e}_i$, the complete solution isA point at parameter $T>0$ is a conjugate point of $p$ precisely when a nonzero Jacobi field vanishes at both $0$ and $T$. The first condition makes every $a_i=0$. The second requires $b_i{S}_{{\lambda }_i}(T)=0$ for every $i$, with at least one nonzero $b_i$. For $T>0$, the factors for zero and negative eigenvalues never vanish; positive eigenvalues give exactly the sine zeros. ThereforeThe corresponding points are $\gamma (T)$. If several positive eigenspaces have a zero at the same parameter, the conjugate multiplicity is the sum of their dimensions. This also shows that a locally symmetric Riemannian manifold has no conjugate points along a given geodesic when that geodesic's Jacobi curvature operator has no positive eigenvalues.

A simple Riemannian manifold has strictly convex boundary, no conjugate points, and a unique connecting geodesic depending smoothly on its endpoints. In particular its unit-speed geodesics are nontrapping. Write $\nu$ for the inward unit normal, and identify inward and outward parts of its boundary unit tangent bundle byThe signs are a convention; here $+$ means inward. If $(x,v)$ is strictly inward, let $\tau (x,v)>0$ be its first exit time. The geodesic scattering relation isThus it records both the exit point and the outward unit velocity, not just the exit point. On strictly outward vectors extend $\alpha$ by following the same geodesic flow backwards to its previous entry point, retaining the velocity orientation. On tangential vectors set $\alpha (x,v)=(x,v)$. The extended scattering relation is an involution on the full boundary unit bundle, since its forward and backward constructions undo one another. On a simple Riemannian manifold, the travel time equals the boundary distance function between the two endpoints; travel time is often recorded separately as lens data.

Take the normalizationThe first variation of geodesic energy says that the critical points on the fixed-endpoint path space are exactly the affinely parametrized geodesics. The kernel of the second variation of geodesic energy, or Riemannian index form, is the space of Jacobi fields vanishing at both endpoints. Hence a geodesic is a nondegenerate critical point exactly when that space is zero, or equivalently when its terminal endpoint is not a conjugate point to its initial endpoint along the given geodesic.

The Morse index theorem states that the index is the sum of the multiplicities of all conjugate points strictly inside the parameter interval:The multiplicity at the terminal endpoint gives nullity rather than an additional contribution to the index. These statements use the usual Sobolev $H^1$ completion of the path space; smooth paths have the same homotopy type.

First suppose $p\ne q$ as well as $p\ne -q$, and put $\theta =d(p,q)\in (0,\pi )$. There is a unique great circle through these two points. Every critical path runs along it with constant speed, possibly passing around it extra times. Ifthen a complete enumeration without repetitions isIts length is $L=|\ell |$. Equivalently the two positive length lists areThere are no other critical paths, because a nonconstant geodesic of the round sphere is a constant-speed great circle and its plane must contain $p,q$.

To compute conjugate points and indices of round-sphere geodesics, the round unit sphere has constant sectional curvature $1$. Along a geodesic of speed $L$, a normal Jacobi field in a parallel direction satisfies $y^{\prime \prime }+L^{2}y=0$. A field with $J(0)=0$ is therefore a constant multiple of $\sin (Lt)$ in each of the $n-1$ normal directions. Its tangential component satisfies $y^{\prime \prime }=0$ and contributes no endpoint-vanishing field. Consequently the conjugate points occur at $Lt=j\pi$, each with multiplicity $n-1$. Since neither length list contains an integer multiple of $\pi$, all the displayed critical paths are nondegenerate, and the Morse index theorem gives

The permitted case $p=q$ requires separate treatment. There is a constant geodesic, which has index and nullity zero. All other critical paths areFor each $r$, this is a family parametrized by $S^{n-1}$; both directions of traversal are included through $v$ and $-v$. The interior conjugate points number $2r-1$, and the endpoint is also conjugate. ThusThese are Morse-Bott critical manifolds for $n>1$, rather than nondegenerate critical points. For $n=1$ the same formulas give zero index and nullity, with the two isolated directions comprising $S^0$.

For the requested homology calculation, choose distinct nonantipodal endpoints. Concatenation with a fixed path back to the basepoint gives a homotopy equivalence between this fixed-endpoint path space and the based loop space $\Omega S^n$. We use the Morse cell-attachment theorem for geodesic energy: on a complete compact Riemannian manifold, the fixed-endpoint energy, when all its critical points are nondegenerate, gives a CW complex of the same homotopy type with one cell of dimension equal to the index of each critical point. One can obtain this theorem from finite-dimensional broken-geodesic approximations and ordinary Morse theory; critical energy values tend to infinity here.

The two index lists interleave to give exactly one cell in every dimension $j(n-1)$, $j\ge 0$. If $n>2$, these dimensions are separated by at least two. The cellular homology groups therefore have zero boundary maps, since no occupied cell dimension has an occupied dimension one lower. With integer coefficients,This determines the graded abelian groups. The argument does not need an identification of the multiplication on loop-space homology.

Finally apply the Freudenthal suspension theorem: for an $(n-1)$-connected based CW complex $X$, $n\ge 2$, the suspension homomorphismis an isomorphism for $k\le 2n-2$ and a surjection for $k=2n-1$. A sphere $S^n$ is $(n-1)$-connected, and its reduced suspension of a topological space is homeomorphic to $S^{n+1}$. HenceThe isomorphism is the suspension map. Its stable range is also reflected by the Morse theory cell structure of $\Omega S^{n+1}$: after its bottom $n$-cell, the next positive-dimensional cell has dimension $2n$. For the stated homology problem $n>2$, all the connectivity hypotheses apply.

In a parallel orthonormal frame along a geodesic, write the Jacobi equation as $A^{\prime \prime }+KA=0$, with $A(0)=0$, $A^{\prime }(0)=I$ and symmetric $K$. In the absence of conjugate points on $(0,b]$, $A$ is invertible there. The conserved quantity $A^T{A}^{\prime }-A^{\prime T}A=0$ makes $S=A^{\prime }{A}^{-1}$ symmetric, and $S^{\prime }+S^2+K=0$. For a field $u$ vanishing at both endpoints,Indeed the difference of the integrands is $(u^{T}Su{)}^{\prime }$; its boundary value at zero vanishes because $S=t^{-1}I+O(t)$ and $u=O(t)$. Equality forces $u=Ac$ and then $A(b)c=0$, so $u=0$. This proves strict positivity and the minimizing property of the Jacobi field with prescribed endpoint values.

// Target: riemannian-geometry.bigb