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On the unit round sphere $S^n$, a geodesic of speed $L$ on $[0,1]$ has normal Jacobi equation $y^{\prime \prime }+L^{2}y=0$. Its conjugate parameters are $j\pi /L$, each with multiplicity $n-1$. If $L/\pi$ is not an integer, its index is $(n-1)\lfloor L/\pi \rfloor$. For $L=m\pi >0$, its index is $(m-1)(n-1)$ and its nullity is $n-1$.

On continuous piecewise smooth vector field along a curve with zero endpoints, the radical of a bilinear form of the Riemannian index form is exactly the space of endpoint-vanishing Jacobi fields. Integration by parts against tests supported inside each smooth piece gives the Jacobi equation; tests with arbitrary values at breakpoints force continuity of the covariant derivative, hence a global smooth Jacobi field. This characterization uses $I(V,W)=0$ for every test field $W$, not merely $I(V,V)=0$, which is insufficient for an indefinite form.

// Target: riemannian-geometry.bigb

Assume the Strengthened Legendre condition $P>0$ and let $q(\phi )={\int }_a^b(P{\phi }^{\prime 2}+Q{\phi }^2)$. Then $q$ is strictly positive on all nonzero fixed-endpoint Lipschitz variations exactly when the normalized Jacobi equation solution with $w(a)=0$, $P(a)w^{\prime }(a)=1$ has no zero in $(a,b]$. A zero at $c$ gives a nonzero variation equal to $w$ up to $c$ and zero thereafter, with $q=0$. Conversely, if no zero occurs, the Jacobi identity for the second variation writes $q$ as a nonnegative square. Equality forces $\phi =Cw$, and the endpoint at $b$ forces $C=0$.

A parallel frame along a curve consists of a basis of vector fields $E_a$ satisfying $D_t{E}_a=0$ for its covariant derivative along a curve. Transporting a basis at the initial point constructs the frame. If the connection preserves the metric, an initially orthonormal basis stays orthonormal. In this frame the Jacobi equation becomes a linear matrix ordinary differential equation.

For fixed-endpoint variations, the second variation iswith $p=f_{vv}$, $r=f_{uv}$, $s=f_{uu}$ evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$. The Jacobi equation, written without differentiating $r$ separately, isWhen $r$ is differentiable this is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$.

Positive definiteness means $q(\phi )>0$ for every nonzero admissible variation. Uniform or coercive positivity of a second variation is the stronger assertion that $q(\phi )\ge c\int {\phi }^{\prime 2}$ for some fixed $c>0$. On a fixed-endpoint interval this is equivalent to a positive lower bound in the $W^{1,2}$ norm.

The strict Legendre implication uses the uniform interpretation of positivity. If $q\ge c\int {\phi }^{\prime 2}$, localized rapidly oscillating variations show $p(x)\ge c$: otherwise support them in an interval where $p-c<0$, so the derivative term of $q-c\int {\phi }^{\prime 2}$ is negative of order frequency squared, overwhelming its mixed and zero-order terms. Continuity extends the bound to the endpoints, giving $p>0$ on $[a,b]$. Under mere positive definiteness this strict conclusion is false: $f=u^2$ at $\bar{u}=0$ gives $q=2\int {\phi }^2>0$ but $p=0$. In that weaker convention only the ordinary Legendre necessary condition $p\ge 0$ follows. This distinction makes the implication precise.

Under the stated regular condition $p>0$, a conjugate point $c\in (a,b]$ to $a$ is an endpoint admitting a nonzero Jacobi solution with $z(a)=z(c)=0$. The disconjugacy form of the Jacobi positivity criterion says that no such point in $(a,b]$ exists if and only if there is a Jacobi solution strictly positive on all of $[a,b]$.

For the required identity, let that positive solution be $z$ and write $\phi =zh$. Set $A=pz{z}^{\prime }+r{z}^2$. The Jacobi equation givesExpanding the second-variation integrand therefore yieldsFor compactly supported $\phi$, the boundary term vanishes, provingThis is the Jacobi identity for the second variation. The divergence form is also appropriate under the original $C^1$ regularity; with $p>0$, the Euler-Lagrange equation gives the usual further regularity along the extremal.

For a fixed-endpoint variation $\phi \in W_0^{1,\infty }(a,b)$, the second variation iswhere all coefficients are evaluated at $(x,\bar{u}(x),{\bar{u}}^{\prime }(x))$. The notation ${\delta }^{2}I(\bar{u})>0$ means this is strictly positive for every nonzero admissible variation. This is pointwise positive definiteness of the quadratic form, not an assertion of a prescribed uniform lower bound.

PutAfter integration by parts, the form is $q(\phi )={\int }_a^b(P{\phi }^{\prime 2}+Q{\phi }^2)dx$. Its Jacobi equation isIn the regular case $P>0$ on $[a,b]$, a conjugate point $c\in (a,b]$ to $a$ is a point for which a nonzero solution satisfies $w(a)=w(c)=0$. Equivalently it is a zero of the normalized solution with $w(a)=0$, $P(a)w^{\prime }(a)=1$.

Under this regular Strengthened Legendre condition, the Jacobi positivity criterion isThe endpoint $b$ must be excluded as a conjugate point too. To prove necessity, suppose a Jacobi solution vanishes at $a$ and $c$. Define $\phi =w$ on $[a,c]$ and zero on $[c,b]$. It is a nonzero Lipschitz admissible variation, anda contradiction. For completeness, the converse follows from the Jacobi identity for the second variation: when the normalized $w$ has no further zero,The boundary contribution $[P(w^{\prime }/w){\phi }^2{]}_a^b$ vanishes, including the limit at $a$ since $w$ has a simple zero and $\phi =O(x-a)$. Equality forces $\phi =Cw$, and $\phi (b)=0$ with $w(b)\ne 0$ forces $C=0$.

Independently of regularity, positivity necessarily implies the ordinary Legendre necessary condition $P\ge 0$. If $P<0$ on an interval, localized rapidly oscillating test functions have a negative derivative contribution of order frequency squared, dominating the bounded zero-order term. This contradicts positivity.

The assumptions stated here do not themselves impose $P>0$. That strengthened condition is a regularity hypothesis for the familiar Jacobi criterion, and is not necessary for strict positivity: the integrand $f=u^2$ gives $q(\phi )=2\int {\phi }^2>0$ with $P=0$. For a criterion covering the full possibly degenerate class, use Regularized Jacobi criterion for a degenerate second variation. The following conditions are necessary and sufficient:

These conditions are necessary because $q>0$ gives $q_{\eta }(\phi )=q(\phi )+\eta \int {\phi }^{\prime 2}>0$, so the regular criterion just proved applies; a weak Jacobi solution tested against itself would give $q=0$. They are sufficient as well: the regular criterion gives $q_{\eta }>0$ for every $\eta >0$, hence $q\ge 0$ by letting $\eta \downarrow 0$. If $q(\phi )=0$, nonnegativity of $q(\phi +t\psi )$ for every real $t$ forces its mixed bilinear term to vanish for every test $\psi$. This is precisely the weak Jacobi equation. The third condition then forces $\phi =0$. Thus the criterion treats degeneracy without silently assuming the strengthened condition.

Here $P=2$, $Q=-2{\lambda }^2$ and the Jacobi equation is $w^{\prime \prime }+{\lambda }^{2}w=0$. With $w(0)=0$ and $w^{\prime }(0)=1$, its solution is $w(x)=\sin (|\lambda |x)/|\lambda |$ for $\lambda \ne 0$, and $w(x)=x$ for $\lambda =0$. Its first positive zero is $\pi /|\lambda |$ when $\lambda \ne 0$.

If ${\lambda }^2<{\pi }^2/b^2$, no conjugate point occurs in $(0,b]$. The Jacobi positivity criterion givesSince $I_{\lambda }$ itself is this quadratic form divided by two, $I_{\lambda }(u)>0=I_{\lambda }(0)$ for every nonzero admissible $u$. Thus zero is the unique global minimizer. This argument establishes the result before using the sharp Poincaré inequality.

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.

Use the Riemann curvature tensor convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$. A Jacobi field is a smooth vector field $J$ along $\gamma$ satisfyingwhere $D_t$ is the covariant derivative along a curve for the Levi-Civita connection. Differentiating a geodesic variation gives this equation, and conversely every Jacobi field arises from a geodesic variation by varying its initial point and velocity.

Choose a parallel frame $E_1(t),\dots ,E_n(t)$ along $\gamma$ and write $J={\sum }_a{y}_a{E}_a$. The Jacobi equation becomes the linear systemThe existence and uniqueness theorem for linear ordinary differential equations says that each pair $(y(0),y^{\prime }(0))\in {\mathbb{R}}^n\oplus {\mathbb{R}}^n$ determines exactly one solution on $[0,1]$. Addition and scalar multiplication preserve the equation. Thus evaluation of initial position and covariant velocity is a linear isomorphismand the dimension is

For the convexity assertion, use the convex-normal meaning of a geodesically convex open set: its points are joined by a unique geodesic within the set, and the joining geodesic depends smoothly on its endpoints. Equivalently, the appropriate star-shaped restriction of ${\exp }_p$ is a diffeomorphism onto the set. The convex-normal-neighbourhood theorem supplies such sets around every point. Mere existence of some minimizing geodesic, without this uniqueness and normality, would not imply the assertion.

Let $p=\gamma (0)$ and $v=\dot{\gamma }(0)$. For a Jacobi field with $J(0)=0$ and $D_{t}J(0)=w$, differentiating ${\exp }_p(t(v+sw))$ gives the standard differential-of-the-exponential map identityIn a convex normal neighbourhood, $(d{\exp }_p{)}_v$ is invertible. Therefore a Jacobi field vanishing at both endpoints has $w=0$ and hence is identically zero. The difference of two fields with the same endpoint values consequently vanishes. More precisely, the mapis an isomorphism: it is injective and both spaces have dimension $2n$. The constant-geodesic case follows directly from $D_t^{2}J=0$. This uses the convex-normal-neighbourhood theorem, the differential-of-the-exponential map identity, and linear ordinary differential equation uniqueness.

For the special unitary group example, use Jacobi fields from conjugation at a central endpoint. WriteThe endpoints of $\gamma (t)=\exp (tA)$ are $I$ and $zI$. The latter is central. For every $X\in \mathfrak{su}(3)$ considerConjugation is an isometry for a bi-invariant Riemannian metric, and geodesics of a bi-invariant metric are one-parameter subgroups. Thus $F_X$ is a geodesic variation. Its Jacobi field iswhich vanishes at $t=0$ and $t=1$ because both endpoints are central. Its initial covariant derivative is $[X,A]$.

The linear map $X\mapsto J_X$ has kernel equal to the centralizer of an element of a Lie algebra of $A$. Indeed, $J_X\equiv 0$ implies $[X,A]=0$ on differentiating at $t=0$, and that commutation conversely implies $J_X\equiv 0$. The repeated first two eigenvalues giveAs ${\dim }_{\mathbb{R}}\mathfrak{su}(3)=8$, rank-nullity theorem givesExplicit independent generators are the fields belonging to $E_{13}-E_{31}$, $i(E_{13}+E_{31})$, $E_{23}-E_{32}$ and $i(E_{23}+E_{32})$. Their initial derivatives $[X,A]$ are independent because the eigenvalue differences in the $13$ and $23$ entries are $2\pi i$. This also demonstrates directly why the endpoint-value conclusion fails along this geodesic.

False with the $\mathrm{limsup}$ printed in the paper. Let the even-indexed surfaces be one fixed unit sphere and the odd-indexed surfaces one fixed ring torus. Thenbut infinitely many surfaces are tori. The two fixed surfaces lie in one ball and have uniformly bounded area. Their injectivity radii have a positive common lower bound, and compactness gives a common bound on $|\nabla K|$, hence on every radial derivative $|{\partial }_{r}K|$. Thus this alternating sequence satisfies all four additional conditions as well.

The intended statement becomes true if $\mathrm{limsup}$ is replaced byHere is the proof. Put ${\delta }_n=\max (0,-{\inf }_{S_n}{K}_n)$, so ${\delta }_n\to 0$. Choose $p_n$ maximizing distance from the centre of the fixed containing ball. The supporting-sphere curvature bound givesWrite $K_0=K_n(p_n)$ and $j(r,\theta )=\sqrt{G(r,\theta )}$ in geodesic polar coordinates. Choose a fixed sufficiently small $c>0$ and set ${\rho }_n=c{K}_0^{-1/2}$. Conditions (3) and (4) allow $c$ to be chosen so that ${\rho }_n\le {ϵ}_0$ andThe Jacobi equation $j_{rr}=-K_{n}j$, with $j(0)=0$ and $j_r(0)=1$, and the Sturm comparison theorem then give $j(r,\theta )\ge c_{1}r$ on this ball for a constant $c_1>0$ independent of $n$. ConsequentlySince $K_n\ge -{\delta }_n$ elsewhere and $\mathrm{Area}(S_n)\le M$,for all sufficiently large $n$. Gauss-Bonnet forces $\chi (S_n)>0$, and the classification theorem for surfaces makes each such $S_n$ a sphere.

Take a variation $x+\epsilon \xi$ with $\xi (a)=\xi (b)=0$. Expanding the action givesAn integration by parts and the fixed endpoint conditions give the first variationThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationand the second variation is

Linearizing the equation of motion about $x$ givesso the Jacobi equation isWhen $u$ has no zero on $[a,b]$,The total derivative integrates to zero because $\xi (a)=\xi (b)=0$, hence

For the simple harmonic oscillator, the Jacobi equation is $\ddot{u}+{\omega }^{2}u=0$. Put $t_0=(a+b)/2$ and chooseIf $b-a<\pi /\omega$, then $|\omega (t-t_0)|<\pi /2$ throughout $[a,b]$, so $u$ is positive there. The preceding square identity proves that the classical path is a local minimum of the action whenever the elapsed time is less than half an oscillation period.

Along the reference curve put $p=f_{vv}$, $r=f_{uv}$, $s=f_{uu}$. The second variation isStrict positivity means $q(\varphi )>0$ for every nonzero zero-trace variation. Under the given regular Legendre condition $p>0$, the no-conjugate-point criterion also supplies a uniform coercive lower bound. The Jacobi equation is $-(p{z}^{\prime }+rz{)}^{\prime }+r{z}^{\prime }+sz=0$. A point $c>a$ is conjugate to $a$ when a nonzero solution vanishes at both $a$ and $c$. In this regular setting, absence of conjugate points in $(a,b]$ is equivalent to existence of a Jacobi solution positive throughout $[a,b]$.

Choose such a positive solution $z$. The given square identity extends from smooth compactly supported variations to $H_0^1$ by density and boundedness of its coefficients. With $v=\varphi /z$ it reads $q(\varphi )=\int p{z}^2{v}^{\prime 2}$. The positive minima of $p,z$ bound this below by $c\Vert v^{\prime }{\Vert }_2^2$. A zero-trace function obeys $\Vert v{\Vert }_2\le C\Vert v^{\prime }{\Vert }_2$, while ${\varphi }^{\prime }=z{v}^{\prime }+z^{\prime }v$ and $\varphi =zv$ give $\Vert \varphi {\Vert }_{H^1}\le C_1\Vert v^{\prime }{\Vert }_2$. ConsequentlyIn particular the required second variation is positive for all nonzero Lipschitz zero-trace variations.

The Jacobi equation is $z^{\prime \prime }+{\lambda }^{2}z=0$. For $|\lambda |<\pi$ the function $z=\cos (\lambda (x-1/2))$ is positive on the closed interval. The preceding square identity giveswith $z=1$ if $\lambda =0$. As in part (a), its positive minimum and bounded derivative provide a constant $\gamma >0$ with $I(u)\ge \gamma \Vert u{\Vert }_{H^1}^2$. The second variation at zero is $2I$, so is positive, and zero is the unique global minimizer.

For a fixed zero-trace $u$, positivity of $\int u^{\prime 2}-{\lambda }^2\int u^2$ for all ${\lambda }^2<{\pi }^2$ gives, by letting ${\lambda }^2\uparrow {\pi }^2$,This derivation did not assume the sharp inequality in advance. For equality, the nonnegative quadratic form $Q(u)=\int (u^{\prime 2}-{\pi }^2{u}^2)$ has $Q(u)=0$. Expanding $Q(u+t\varphi )\ge 0$ for both signs of $t$ forces $\int u^{\prime }{\varphi }^{\prime }-{\pi }^{2}u\varphi =0$ for every test. Thus $u^{\prime \prime }+{\pi }^{2}u=0$, and the zero endpoint values give exactly $u(x)=C\sin (\pi x)$, including $C=0$. These functions do give equality.

With coefficients evaluated along $(x,\bar{u},{\bar{u}}^{\prime })$, put $p=f_{vv}$, $r=f_{uv}$ and $s=f_{uu}$. The second variation isThe Jacobi equation in divergence form is $-(p{z}^{\prime }+rz{)}^{\prime }+r{z}^{\prime }+sz=0$; if $r$ is differentiable it is $-(p{z}^{\prime }{)}^{\prime }+(s-r^{\prime })z=0$. A conjugate point $c>a$ admits a nonzero Jacobi solution vanishing at $a,c$.

There are two positivity conventions worth distinguishing. Strict pointwise positivity means $q(\varphi )>0$ for each nonzero fixed-endpoint variation. Coercive positivity means $q(\varphi )\ge c\int {\varphi }^{\prime 2}$ with one $c>0$ for all such variations; the endpoint Poincaré estimate makes this equivalent to a positive lower bound in the full $H^1$ norm. Under the regular Legendre hypothesis $p>0$, the necessary and sufficient Jacobi criterion is no conjugate point in $(a,b]$; in that regular case it gives both strict and coercive positivity. Equivalently the Jacobi equation has a solution positive on $[a,b]$, and the square identitygives positivity and a uniform derivative lower bound by the positive minimum of $z$ and boundedness of $z,z^{\prime }$. Conversely a conjugate Jacobi field, extended by zero beyond its second zero, is a nonzero Lipschitz variation with $q=0$, contradicting either positivity convention.

Coercivity itself forces $p>0$: local high-frequency variations make the leading coefficient at least the coercivity constant. Mere strict positivity need not force it, as $q(\varphi )=\int {\varphi }^2$ shows. If that weaker convention is intended without regularity, a complete criterion is $p\ge 0$, no conjugate point for every regularized form $q_{\eta }=q+\eta \int {\varphi }^{\prime 2}$ with $\eta >0$, and no nonzero zero-endpoint Lipschitz weak solution of the original Jacobi equation. The regularized criteria imply $q_{\eta }\ge 0$, hence $q\ge 0$ by passage to the limit. For a nonnegative quadratic form, $q(\varphi )=0$ forces all mixed terms with test variations to vanish, giving exactly the weak Jacobi kernel; excluding it gives strict positivity. Necessity follows in reverse. This specifies the degenerate case without imposing an unstated strict Legendre condition.

For weak local minimality, sufficient conditions are the Euler-Lagrange equation and a coercively positive second variation with the stated smoothness. Taylor's remainder is bounded by $\epsilon \Vert \varphi {\Vert }_{H^1}^2$ for small $W^{1,\infty }$ perturbations, so the positive quadratic term dominates it. These are the classical weak minimum sufficient conditions.

For continuous $P\ge 0,Q$, a scalar second variation $q=\int (P{\phi }^{\prime 2}+Q{\phi }^2)$ is strictly positive on nonzero fixed-endpoint Lipschitz variations if and only if every regularized form $q_{\eta }$ for $\eta >0$ has no conjugate point in $(a,b]$, and the original weak Jacobi boundary problem has no nonzero Lipschitz solution. The Jacobi positivity criterion applies to each regularized coefficient $P+\eta >0$. Letting $\eta$ decrease to zero gives $q\ge 0$. For a nonnegative quadratic form, $q(\phi )=0$ forces its mixed bilinear form with every test function to vanish, by considering $q(\phi +t\psi )$ for both signs of $t$. This is the weak Jacobi equation, so the kernel condition gives strict positivity. Necessity follows by the same argument in reverse. It distinguishes strict positivity from a uniform coercive estimate when the derivative coefficient degenerates.

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

Strict positivity of the highest-derivative coefficient on the closed parameter interval is the strengthened Legendre condition. Continuity then gives a uniform positive lower bound and makes the Jacobi equation a regular second-order equation. It permits the Jacobi positivity criterion. It is not necessary for mere strict positivity of a second variation: $\int {\phi }^2$ is strictly positive on nonzero variations even though its derivative coefficient vanishes.