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On an oriented even-dimensional Riemannian manifold of strictly positive sectional curvature, a nonconstant closed geodesic has a length-decreasing smooth variation. Its parallel transport fixes the tangent and acts on the odd-dimensional normal space by a special orthogonal group element. An odd-dimensional special orthogonal transformation has a fixed vector, so there is a nonzero periodic parallel normal field $V$. The second variation of geodesic energy is then strictly negative. The Cauchy-Schwarz inequality converts lower energy into strictly lower length because the original geodesic has constant speed. For an embedded geodesic the small variation is a smooth isotopy; otherwise it is a deformation through immersed loops.

Let a nonzero Jacobi field $J$ vanish at times zero and $a$. On a longer geodesic segment ending at $b>a$, extend it by zero to a continuous piecewise smooth field $Z$. Then $I(Z,Z)=0$. Choose an endpoint-vanishing field $P$ with $P(a)=-D_{t}J(a)$; integration by parts gives $I(Z,P)=-|D_{t}J(a){|}^2<0$. Thus $I(Z+\epsilon P,Z+\epsilon P)<0$ for small positive $\epsilon$. Smooth approximation preserves negativity. The second variation of geodesic energy supplies a shorter path, so a minimizing geodesic cannot extend beyond its first conjugate point.

// Target: differential-geometry.bigb

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.

Let $\tau >0$ be the first conjugate time and choose a nonzero Jacobi field $J$ with $J(0)=J(\tau )=0$. It is normal by part 1(b), and $D_{t}J(\tau )\ne 0$, since zero value and derivative there would force $J=0$. Fix any $T>\tau$ and extend $J$ by zero to a continuous piecewise smooth vector field $V$ on $[0,T]$.

For endpoint-vanishing fields, the Riemannian index form isIntegration by parts on $[0,\tau ]$ and the Jacobi field equation give $I(V,V)=0$ andChoose a smooth normal field $W$, zero at $0,T$, with $W(\tau )=-D_{t}J(\tau )$. Thenfor sufficiently small positive $\epsilon$. The corner in $V$ causes no difficulty: smooth endpoint-vanishing fields approximate $V+\epsilon W$ in the Sobolev space $H^1$, and the Riemannian index form is continuous in this norm on the compact segment. A smooth approximant therefore still has negative index form.

Realize this smooth field as a fixed-endpoint curve variation using the Riemannian exponential map. The allowed second variation of geodesic energy gives $E^{\prime }(0)=0$ and $E^{\prime \prime }(0)<0$, so some nearby curve has energy less than $T/2$, the energy of the unit-speed reference geodesic. The Cauchy-Schwarz inequality bounds its length byIt is shorter than the reference segment. Thus a geodesic cannot minimize beyond its first conjugate time. This argument deliberately uses $T>\tau$; it does not rule out minimization up to the conjugate point itself.

There is a normalization issue in the printed constant. Let ${\mathrm{Ric}}_{\mathrm{tr}}$ denote the trace convention in the catalog's Ricci curvature, and write $\overline{\mathrm{Ric}}={\mathrm{Ric}}_{\mathrm{tr}}/(n-1)$ for normalized Ricci curvature, with $n\ge 2$. The claimed $\pi /\sqrt{k}$ bound corresponds to $\overline{\mathrm{Ric}}\ge kg$, or equivalently ${\mathrm{Ric}}_{\mathrm{tr}}\ge (n-1)kg$. We prove the estimate in both conventions.

Take a unit-speed minimizing geodesic of length $L>0$ and perpendicular parallel orthonormal fields $E_1,\dots ,E_{n-1}$. Set $V_i(t)=\sin (\pi t/L)E_i(t)$. The fields vanish at the endpoints. The allowed second variation of geodesic energy makes each Riemannian index form nonnegative. Summing them givesunder the normalized lower bound. Thus $L\le \pi /\sqrt{k}$. Completeness and the Hopf-Rinow theorem supply a minimizing geodesic between any two points, so this bounds the diameter. A closed ball of that radius about any point contains the whole manifold and is compact, again by the Hopf-Rinow theorem.

Hence the intended Bonnet-Myers theorem conclusion isIf the printed $\mathrm{Ric}\ge k$ means trace Ricci curvature instead, the same integral has lower curvature term $kL/2$ and yieldsThe smaller bound is false under that convention when $n>2$: the round unit $S^n$ has trace Ricci tensor $(n-1)g$ and diameter $\pi$. Taking $k=n-1$ contradicts the printed $\pi /\sqrt{k}$ inequality. The compactness assertion remains true.

Parametrize the nonconstant closed geodesic at unit speed on $[0,L]$. Its parallel transport around the loop fixes the tangent $\dot{\gamma }(0)$. Since the manifold is orientable, it preserves orientation on the full tangent space, and hence has determinant one on the normal space. This normal space has odd dimension because the manifold has even dimension.

An odd-dimensional special orthogonal transformation has a fixed vector. Indeed, nonreal eigenvalues of a real orthogonal transformation occur in conjugate pairs with product one; the real eigenvalues are $\pm 1$, and determinant one makes the number of $-1$ eigenvalues even. Odd dimension therefore requires a $+1$ eigenvalue. Transport a unit fixed normal vector around the geodesic to obtain a periodic parallel field $V$, with $D_{t}V=0$ and $V\perp \dot{\gamma }$.

The variation ${\gamma }_s(t)={\exp }_{\gamma (t)}(sV(t))$ is defined uniformly for small $s$, because the image of the original loop is compact. Periodicity gives closed curves and a homotopy through loops. With the standard positive-sectional-curvature Riemannian index form, the second variation of geodesic energy isThe first variation vanishes for a closed geodesic, since the endpoint terms cancel. Therefore $E({\gamma }_s)<E({\gamma }_0)=L/2$ for sufficiently small nonzero $s$. The Cauchy-Schwarz inequality givesThus the closed geodesic is homotopic to a strictly shorter closed curve. This proves the instability of a closed geodesic in positive even-dimensional curvature; completeness of the manifold is unnecessary for this compact-loop variation. Constant loops are excluded by the usual nonconstant meaning of a closed geodesic in this assertion.

Use the curvature convention $\mathcal{R}$ of Solution 1, and put $T=\dot{\gamma }$. On continuous piecewise smooth perpendicular vector fields with zero endpoint values, the Riemannian index form is the symmetric bilinear formA Jacobi field is a smooth field $J$ along $\gamma$ satisfying $D_t^{2}J+\mathcal{R}(J,T)T=0$. The point $\gamma (t_0)$ is a conjugate point to $\gamma (0)$ along this geodesic when there is a nonzero Jacobi field with $J(0)=J(t_0)=0$. Such a field is perpendicular to $T$, because $⟨J,T⟩$ is affine and vanishes at both endpoints.

The conjugate-point criterion for the Riemannian index form, stated without proof, isAt a first conjugate endpoint the form is nonnegative but has a nonzero nullspace, consisting of endpoint-vanishing Jacobi fields. A conjugate point strictly inside gives a negative direction. This explicitly distinguishes the non-strict and strict senses of positivity in the question; in terminology where “positive definite” already means strict positivity, the second line is the positive-definiteness condition.

Choose $n-1$ perpendicular orthonormal fields $E_1,\dots ,E_{n-1}$ by parallel transport along $\gamma$, and put $V_i(t)=\sin (\pi t/L)E_i(t)$. They are admissible andThe tangent has unit length. The assumed Ricci curvature bound and the two elementary sine-square and cosine-square integrals give the sine index-form bound for positive Ricci curvature:Since $n\ge 2$, at least one of the admissible $V_i$ therefore has $I(V_i,V_i)<0$.

If the Riemannian manifold is complete, the Hopf-Rinow theorem supplies a minimizing geodesic between any two points. Its Riemannian index form is nonnegative: any smooth fixed-endpoint variation has nonnegative second variation of geodesic energy, since a length minimizer of constant speed also minimizes energy on a fixed parameter interval by the Cauchy-Schwarz inequality. The same conclusion for piecewise smooth fields follows by smooth approximation, or by piecewise variations. The negative direction above excludes a minimizing segment longer than $\pi /\sqrt{\kappa }$. Taking the supremum over pairs yields the diameter conclusion of the Bonnet-Myers theorem:

The completeness assumption is essential. An explicit incomplete positively curved strip with infinite diameter isThe map $(u,v)\mapsto (\cos u\cos v,\cos u\sin v,\sin u)$ is a local isometry to the unit sphere, so $K=1$ and, in dimension two, $\mathrm{Ric}=g$. Thus the required Ricci curvature bound holds with $n=2$, $\kappa =1$. The unit-speed meridian $t\mapsto (t,0)$ reaches the missing boundary $u=\pi /4$ in finite time, so the strip is geodesically incomplete. On the other hand every curve joining $(0,0)$ to $(0,A)$ hassince $\cos u\ge 1/\sqrt{2}$ throughout the strip. Therefore its intrinsic diameter is infinite, in particular larger than $\pi$. This is the unwrapped strip with $v\in \mathbb{R}$, not a strip with longitude identified modulo $2\pi$; that distinction prevents spherical shortcuts from invalidating the example.

The Bonnet-Myers theorem says that a connected geodesically complete $n$-dimensional Riemannian manifold, with $n\ge 2$ and $\mathrm{Ric}\ge (n-1)\kappa g$ for a constant $\kappa >0$, hasThe dimension and positive lower bound are essential: in dimension one the Ricci condition is vacuous, and a zero lower bound does not imply bounded diameter.

By Hopf-Rinow, any two distinct points have a unit-speed minimizing geodesic $\gamma :[0,L]\to M$. Choose parallel orthonormal fields $E_1,\dots ,E_{n-1}$ perpendicular to its tangent $T$. For the endpoint-vanishing fields $V_i(t)=\sin (\pi t/L)E_i(t)$, the second variation of geodesic energy gives nonnegative index formsSumming and using the Ricci lower bound produces the sine index-form bound for positive Ricci curvature:If $L>\pi /\sqrt{\kappa }$, the last expression is negative, a contradiction. This proves the diameter bound. Hopf-Rinow makes closed bounded sets compact, so the entire manifold is compact.

Give the universal cover the pullback metric. Local isometry preserves its Ricci bound, and lifting complete base geodesics proves completeness of the cover. The same diameter and compactness argument applies there. A fiber of the covering is closed and discrete, hence finite in this compact cover; its cardinality is that of the fundamental group. This proves the final assertion.

For a counterexample that also breaks the diameter conclusion, use the incomplete positively curved strip with infinite diameterThe map $\Phi (u,v)=(\cos u\cos v,\cos u\sin v,\sin u)$ is a local isometry to the round unit sphere: its coordinate derivatives are orthogonal, with squared lengths one and ${\cos }^{2}u$. Thus $K=1$ and $\mathrm{Ric}=g$, satisfying the required lower bound with $n=2$, $\kappa =1$.

The meridian $(u,v)=(t,0)$ is a unit-speed geodesic and reaches the excluded boundary at $t=\pi /4$, so this metric is incomplete. Every curve joining $(0,0)$ to $(0,L)$ has length at least $\int \cos u|v^{\prime }|dt\ge |L|/\sqrt{2}$, because $\cos u\ge 1/\sqrt{2}$ on the strip. Therefore $\boxed{\mathrm{diam}M=\infty }$ despite the positive Ricci bound. Completeness cannot be omitted.

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.

Write $T=\dot{\gamma }$ and $D_t={\nabla }_T$. A Jacobi field is a vector field along the geodesic satisfyingThe curvature convention is that used in Question 1. A geodesic variation is a smooth map $F:(-ϵ,ϵ)\times [0,1]\to M$ with $F(0,t)=\gamma (t)$ whose $t$-curves are affinely parametrized geodesics. Torsion-freeness gives $D_s{\partial }_{t}F=D_t{\partial }_{s}F$. Differentiating $D_t{\partial }_{t}F=0$ and commuting covariant derivatives gives the Jacobi equation for ${\partial }_{s}F{|}_{s=0}$.

For the converse, let $a=J(0)$ and $b=D_{t}J(0)$. Choose a curve $p(s)$ with $p(0)=\gamma (0)$ and $p^{\prime }(0)=a$. parallel transport along it identifies its tangent spaces with $T_{\gamma (0)}M$. Set $v(s)=P_s(T(0)+sb)$, so $v(0)=T(0)$ and $D_{s}v(0)=b$. The geodesics with initial data $(p(s),v(s))$ give a variation $F(s,t)$. Smooth dependence on initial conditions and compactness of the original time interval ensure that this family is defined for all $t\in [0,1]$ after shrinking $ϵ$, even if $M$ is incomplete. Its variation field has the initial values $a,b$, so uniqueness for the linear Jacobi equation identifies it with $J$. This proves the realization of Jacobi fields by geodesic variations.

The endpoint-vanishing pointwise normal fields form a vector space, since their conditions and equation are linear. If $\gamma$ is nonconstant, the map $J\mapsto D_{t}J(0)$ is injective because $J(0)=0$ and zero initial derivative force the zero solution. Differentiating $⟨J,T⟩=0$ gives $D_{t}J(0)\perp T(0)$. Hence the endpoint-vanishing normal Jacobi fields have dimension at most n-1 bound isFor a constant geodesic, $D_t^{2}J=0$, and both zero endpoint values force $J=0$, so the bound still holds.

On the unit round $S^n$, let $\gamma (t)=\cos (\pi t)e_0+\sin (\pi t)e_1$. Its speed is $\pi$. The constant ambient vectors orthogonal to $e_0,e_1$ give $n-1$ independent parallel normal fields $E_j(t)$. Since $R(J,T)T={\pi }^{2}J$ on normal fields, the fields $J_j(t)=\sin (\pi t)E_j(t)$ satisfy the Jacobi equation and vanish at both antipodal endpoints. They attain dimension $n-1$.

For the last clause, interpret a closed geodesic as a nonconstant smoothly periodic geodesic. A constant loop has length zero and cannot be shortened. Parametrize ${\gamma }_0$ on $[0,1]$ at constant nonzero speed. parallel transport once around it fixes $T(0)$. Because the manifold is orientable, this transport preserves orientation; on the normal space of dimension $n-1$, which is odd, it lies in $SO(n-1)$. An odd-dimensional special orthogonal transformation has a fixed vector: nonreal eigenvalues pair with their conjugates, real eigenvalues are $\pm 1$, and determinant one in odd dimension forces an eigenvalue $+1$.

Transport such a nonzero normal fixed vector around the loop. It gives a periodic parallel normal field $V$, with $D_{t}V=0$ and $V(1)=V(0)$. Its jets also agree at the seam, so it is smooth as a field on the parametrizing circle. For small $s$, $F(s,t)={\exp }_{{\gamma }_0(t)}(sV(t))$ is a smooth variation through closed curves, providing their homotopy to ${\gamma }_0$.

For energy $E(s)=\frac{1}{2}{\int }_0^1|{\partial }_{t}F{|}^{2}dt$, the first variation vanishes at the closed geodesic. The permitted second variation of geodesic energy has no endpoint term for this periodic variation, soThus $E(s)<E(0)$ for small nonzero $s$. The Cauchy-Schwarz inequality gives $L(F_s{)}^2\le 2E(s)$, while constant speed gives $L({\gamma }_0{)}^2=2E(0)$. Consequently the instability of a closed geodesic in positive even-dimensional curvature yieldsThe deformation need not remain a geodesic or an embedded curve.

Use the energy of a curve normalizationFor a variation $F(s,t)$ put $T={\partial }_{t}F$ and $V={\partial }_{s}F$. The Levi-Civita connection is torsion free, so ${\nabla }_{s}T={\nabla }_{t}V$. Differentiating energy once gives $E^{\prime }=[⟨V,T⟩{]}_a^b-\int ⟨V,{\nabla }_{t}T⟩$. Along a geodesic, ${\nabla }_{t}T=0$. Differentiating again, commuting covariant derivatives, and integrating by parts gives the second variation of geodesic energy:The curvature convention is $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, consistent with the specified positive sectional curvature. Indeed ${\nabla }_s{\nabla }_{t}V={\nabla }_t{\nabla }_{s}V+R(V,T)V$, and $⟨R(V,T)V,T⟩=-⟨R(V,T)T,V⟩$. The endpoint term vanishes for fixed endpoints or for periodic variations of a closed geodesic. The integral is the Riemannian index form $I(V,V)$.

Let $\dim M=2m$. Parallel transport around the closed geodesic preserves the metric and orientation, so lies in the special orthogonal group. It fixes the nonzero tangent $T$, and its restriction to $T^{\perp }$ is an orientation-preserving orthogonal map of odd dimension $2m-1$. An odd-dimensional special orthogonal transformation has a fixed vector: nonreal eigenvalues occur in conjugate pairs, while an odd-dimensional real orthogonal map with determinant one must have an eigenvalue $+1$. Choose a nonzero fixed vector $v$ normal to $T$ and parallel-transport it along the curve. It gives a nonzero smooth periodic normal field $V$ with ${\nabla }_{t}V=0$.

For the exponential variation ${\gamma }_s(t)={\exp }_{\gamma (t)}(sV(t))$, the endpoints match periodically. Strictly positive sectional curvature givesThus $E({\gamma }_s)<E(\gamma )$ for small nonzero $s$. By the Cauchy-Schwarz inequality,where the final equality uses the geodesic's constant speed. This proves the instability of a closed geodesic in positive even-dimensional curvature.

For an embedded closed geodesic, sufficiently small variations remain embeddings, hence give a smooth isotopy with strictly shorter curves. For a nonembedded closed geodesic the construction gives a smooth deformation through immersions; an isotopy class of embeddings is not literally defined for such a curve. The stated isotopy conclusion therefore uses the usual embedded-curve interpretation, while the shorter-loop variation holds without it.

Along a unit-speed minimizing geodesic of positive length $L$, take perpendicular parallel orthonormal fields $E_i$ and set $V_i=\sin (\pi t/L)E_i$. The second variation of geodesic energy makes each index form nonnegative, while $\mathrm{Ric}\ge (n-1)\kappa g$ bounds their sum as displayed. For $n\ge 2$ and $\kappa >0$, lengths greater than $\pi /\sqrt{\kappa }$ are impossible. Completeness is what supplies a minimizing geodesic between arbitrary points in the Bonnet-Myers theorem.