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Put $p=\gamma (0)$, $v=\dot{\gamma }(0)$, $a=J(0)$ and $b=D_{t}J(0)$. Choose the short curve $c(s)={\exp }_p(sa)$ and let $P_s:T_{p}M\to T_{c(s)}M$ be parallel transport along $c$. The tangent vectorssatisfy $V(0)=v$ and ${\nabla }_{s}V(0)=b$. Define a geodesic variation byFor each fixed $s$ this is an affinely parametrized geodesic with initial point $c(s)$ and initial velocity $V(s)$. Although the Riemannian manifold need not be complete, smooth dependence for the geodesic equation on a neighborhood of the compact reference segment provides a single $\epsilon >0$ on which $F$ exists for all $|s|<\epsilon$ and $0\le t\le 1$. Thus completeness is unnecessary.

Set $X(t)={\partial }_{s}F(0,t)$. The torsion-free Levi-Civita connection gives ${\nabla }_s{\partial }_{t}F={\nabla }_t{\partial }_{s}F$. Differentiating ${\nabla }_t{\partial }_{t}F=0$ with respect to $s$, and commuting the two covariant derivatives, yieldsAlso $X(0)=c^{\prime }(0)=a$ and $D_{t}X(0)={\nabla }_{s}V(0)=b$. Uniqueness for the Jacobi field equation now gives $X(t)=J(t)$ on the whole segment. This is realization of Jacobi fields by geodesic variations.

The printed formula switches the order of its two variables. The consistent notation here is $F(0,t)=\gamma (t)$ and $J(t)={\partial }_{s}F(0,t)$. A parametrized surface here means a smooth variation map; it cannot be required to be an immersion, since the allowed Jacobi field $J=0$ would already violate that condition along the reference curve.

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.

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.