Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 12 (Differential Geometry)

For a smooth manifold $M$ of dimension $n$, the tangent space $T_{p}M$ can be defined as the space of derivations at $p$: linear maps $v$ on germs of smooth real functions satisfying $v(fg)=f(p)v(g)+g(p)v(f)$. Equivalently, $v$ is the velocity of a smooth curve through $p$, with curves identified when their coordinate velocities agree. A chart $(x^1,\dots ,x^n)$ supplies the basis ${\partial }_i{|}_p$, so $v=v^i{\partial }_i{|}_p$.

The tangent bundle is the disjoint union $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (v)=p$. Its smooth structure is defined by the local local trivializationsOn overlapping charts, the change of trivialization isThese are smooth changes of coordinates on a $2n$-dimensional smooth manifold, linear and invertible in each fibre. Thus $TM$ is a rank-$n$ vector bundle, and its projection is a submersion. Fibrewise addition, scalar multiplication and the zero section are smooth. A smooth section of a vector bundle $X:M\to TM$, satisfying $\pi \circ X={\mathrm{id}}_M$, is exactly a vector field; locally $X=X^i{\partial }_i$, with components transforming by the displayed Jacobian matrix. Its action on functions is a derivation, and the commutator of two such derivations is their Lie bracket.

The dual bundle of $TM$ is the cotangent bundle $T^{\ast }M$. Its smooth sections of a vector bundle are differential 1-forms, locally $\alpha ={\alpha }_{i}d{x}^i$. The coefficient transformation is the inverse transpose of that for tangent vectors, ensuring that $\alpha (X)$ is a well-defined smooth function. Tensoring these two bundles gives the tensor bundlesTheir sections of a vector bundle are tensor fields; their transition functions of a vector bundle are the corresponding tensor products of the tangent and cotangent transformations. For example, the bundle $\mathrm{End}(TM)=TM\otimes T^{\ast }M$ contains fields of linear endomorphisms, while $S^2{T}^{\ast }M$ contains symmetric bilinear fields. Taking alternating covariant tensors gives ${\Lambda }^k{T}^{\ast }M$, whose sections are differential forms. The wedge product, exterior derivative and pullback of a differential form make these bundles central to integration, Generalized Stokes theorem and de Rham cohomology.

A nowhere-vanishing section of ${\Lambda }^n{T}^{\ast }M$ chooses an orientation of a smooth manifold; such a section exists exactly when $M$ is orientable. Without an orientation of a smooth manifold, one can still integrate sections of the density bundle $|{\Lambda }^n{T}^{\ast }M|$, whose coordinate changes use the absolute Jacobian determinant. A Riemannian metric is a positive-definite section $g$ of $S^2{T}^{\ast }M$. It identifies $TM$ with $T^{\ast }M$ by the musical isomorphisms $X\mapsto g(X,\cdot )$ and gives the Riemannian volume densityOn an oriented smooth manifold, this density corresponds to a volume form.

Another organizing construction is the frame bundle $FM$. A frame over $p$ is a linear isomorphism $u:{\mathbb{R}}^n\to T_{p}M$, and the right action $u\cdot A=u\circ A$ makes $FM$ a principal bundle with general linear group $GL(n,\mathbb{R})$ as its structure group. Given a group representation $\rho :GL(n,\mathbb{R})\to GL(V)$, the associated bundle $FM{\times }_{\rho }V$ is obtained from pairs $(u,v)$ by the relation $(uA,v)\sim (u,\rho (A)v)$. The standard representation reconstructs $TM$; dual, tensor and exterior representations reconstruct the associated bundles described above. A Riemannian metric reduces the frame bundle to the orthonormal frame bundle, with orthogonal group $O(n)$ as structure group; an orientation of a smooth manifold reduces it further to the special orthogonal group $SO(n)$. A global section of $FM$ is a global frame, so it exists exactly when $M$ is a parallelizable manifold. Local triviality does not imply a global trivialization: the Hairy ball theorem prevents even one nowhere-zero vector field on $S^2$.

A connection on a vector bundle differentiates its sections in tangent directions and defines parallel transport. For $TM$, the Levi-Civita connection is distinguished by compatibility with a Riemannian metric and vanishing torsion tensor. Its coefficients are not tensor components, because a coordinate change introduces second derivatives; the difference of two connections is a section of $T^{\ast }M\otimes \mathrm{End}(TM)$. The Riemann curvature tensor is a genuine tensor measuring the failure of covariant derivatives to commute.

Finally, a smooth map $f:M\to N$ gives the pullback tangent bundle $f^{\ast }TN$. Its sections are vector fields along a map, including the velocity of a curve and Jacobi fields along a geodesic. For an embedded submanifold $i:M\hookrightarrow N$, the derivative identifies $TM$ with a subbundle of $i^{\ast }TN$, and the quotient is the normal bundle. With an ambient Riemannian metric, the normal bundle is identified with the orthogonal complement of $TM$ in $i^{\ast }TN$; the second fundamental form records the normal component of the ambient derivative of tangent fields. These constructions connect the bundle description to both intrinsic and extrinsic differential geometry.

Let $i:M\hookrightarrow N$ be an embedded submanifold. In the general terminology, a vector field along a map $i$ is a smooth section $X\in \Gamma (i^{\ast }TN)$, so $X_p\in T_{p}N$. A local extension of a vector field along a submanifold near $p$ is a vector field $\tilde{X}$ on an open neighbourhood in $N$ with $\tilde{X}{|}_M=X$ on that neighbourhood. Choose adapted coordinates $(x^1,\dots ,x^m,y^{m+1},\dots ,y^n)$ in which $M$ is given by $y=0$. Writing $X=X^a(x){\partial }_a{|}_{y=0}$, extend the coefficient functions independently of $y$:This proves local existence, including for fields with normal components.

For the Lie bracket assertion, the fields must be tangent to $M$, namely sections of $TM$ viewed inside $i^{\ast }TN$. In this case the useful criterion for a local extension of a vector field along a submanifold isIf $\tilde{X}$ extends $X$, the chain rule gives this identity because their tangent values agree. Conversely, applying it to the adapted coordinate functions forces the tangent components to equal those of $X$ and all normal components on $M$ to vanish, so $\tilde{X}{|}_M=X$. The criterion also shows that functions vanishing on $M$ are differentiated to functions vanishing on $M$ by $\tilde{X}$.

For tangent vector fields $X,Y$ and local extensions $\tilde{X},\tilde{Y}$, apply the criterion twice:Hence the ambient bracket restricts to the intrinsic bracket, independently of the chosen extensions. Tangency is essential. For example, on the $x$-axis in ${\mathbb{R}}^2$, let $X={\partial }_y{|}_M$ and $Y=0{|}_M$. The two extensions $\tilde{Y}=0$ and $\tilde{Y}=y{\partial }_x$, with $\tilde{X}={\partial }_y$, give bracket restrictions $0$ and ${\partial }_x{|}_M$. Thus general sections with normal components do not possess the extension-independent Lie bracket asserted for tangent fields.

For the hypersurface $M^n\subset {\mathbb{R}}^{n+1}$, choose a smooth unit normal $\nu$ locally. A global choice is needed only if one wants a globally signed shape operator; an arbitrary hypersurface need not come equipped with such a choice. With $D$ the Euclidean Levi-Civita connection, define the shape operator, also called the Weingarten map, byThe derivative is defined along $M$ and is linear in $v$. Since $v⟨\nu ,\nu ⟩=2⟨D_v\nu ,\nu ⟩=0$, its value is tangent to $M$, so this is an endomorphism of the tangent space. Differentiating $⟨\nu ,Y⟩=0$ gives, for tangent vector fields $X,Y$,This is the scalar second fundamental form for the chosen normal. The Euclidean connection on a vector bundle has zero torsion tensor, giving $D_{X}Y-D_{Y}X=[X,Y]$. The bracket is tangent to $M$ by the previous argument; thereforeConsequently $\boxed{⟨S_{p}v,w⟩=⟨v,S_{p}w⟩}$ for every $v,w\in T_{p}M$: the shape operator is self-adjoint. Reversing $\nu$ reverses $S$ and $h$ but preserves this self-adjointness.

Use the Levi-Civita connection $\nabla$ of the Riemannian metric $g$ and fix the conventionThe Leibniz rule shows that this expression is linear over smooth functions in all three arguments: the differentiated coefficient terms in the first two derivatives cancel those in the bracket term. Thus $R$ is a section of ${\Lambda }^2{T}^{\ast }M\otimes \mathrm{End}(TM)$, the Riemann curvature tensor, and $R_4$ is a covariant tensor field. This convention gives positive sectional curvature on the round sphere.

Its algebraic symmetries areThe first identity follows immediately from the definition. For the second, apply $XY-YX-[X,Y]$ to $g(Z,W)$ and use compatibility of the Levi-Civita connection with the Riemannian metric. The left side is zero and the differentiated terms combine toFor the first Bianchi identity, use normal coordinates at a point $p$, so the Christoffel symbols vanish there. Torsion-freeness gives ${\Gamma }_{ij}^{\ell }={\Gamma }_{ji}^{\ell }$, andThe cyclic sum over $i,j,k$ cancels term by term. Since this is a tensor identity, it holds for arbitrary tangent vectors everywhere.

To prove pair interchange rather than assume it, abbreviate $R_4(a,b,c,d)$ to $R_{abcd}$ and put $A=R_{abcd}$, $B=R_{cdab}$. The first Bianchi identity and the two antisymmetries giveThe first Bianchi identity applied to $(d,b,a)$ with final argument $c$ gives $R_{dbca}+R_{adcb}=R_{badc}=A$. Hence $A=2B-A$, proving $A=B$.

There is also the differential second Bianchi identity,For its proof take commuting coordinate fields, normal at $p$. The Jacobi identity for operator commutators says $[{\nabla }_X,[{\nabla }_Y,{\nabla }_Z]]+\text{cyclic}=0$. On these commuting fields, $[{\nabla }_Y,{\nabla }_Z]=R(Y,Z)$; at $p$, the covariant derivatives of the coordinate arguments vanish. The commutator identity is therefore the displayed covariant derivative identity at $p$, and tensoriality establishes it everywhere.

For a two-dimensional plane $\sigma =\mathrm{span}\{X,Y\}\subset T_{p}M$, define the sectional curvature byThe denominator is the squared area of $X,Y$. Under an invertible change of plane basis, the alternating pairs in the numerator and the area in the denominator both acquire the square of the change-of-basis determinant. Thus the quotient depends only on $\sigma$. For an orthonormal basis of $\sigma$, it is simply $g(R(X,Y)Y,X)$.

The Ricci curvature is the tracewhere $(e_i)$ is any orthonormal basis of $T_{p}M$. This is independent of the basis because it is a trace. Applying the first Bianchi identity to $(e_i,X,Y)$ and pairing with $e_i$ proves that the summand difference for $\mathrm{Ric}(X,Y)-\mathrm{Ric}(Y,X)$ vanishes; the middle term $g(R(X,Y)e_i,e_i)$ is zero by the last-pair antisymmetry. Hence Ricci curvature is a symmetric bilinear form. For a unit vector $v=e_1$ extended to an orthonormal basis,For nonzero $X$, take $v=X/|X|$ and multiply the right side by $|X{|}^2$. This determines the entire Ricci curvature by the polarization identity:Finally, the scalar curvature is the metric trace of Ricci curvature. Taking its diagonal entries in an orthonormal basis and pairing the equal contributions from $(i,j)$ and $(j,i)$ yieldsIn particular, constant sectional curvature $K_0$ gives $\mathrm{Ric}=(n-1)K_{0}g$ and $\mathrm{Scal}=n(n-1)K_0$. In dimension one the traces are zero; there are no two-dimensional tangent planes and the sums are empty.

The Riemannian exponential map at $p$ is ${\exp }_p(v)={\gamma }_v(1)$, where ${\gamma }_v$ is the affinely parametrized geodesic with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. Its derivative at zero is the identity. The inverse function theorem therefore gives a normal neighbourhood $U={\exp }_p(V)$, where $V\subset T_{p}M$ is an open star-shaped neighbourhood of zero on which ${\exp }_p$ is a diffeomorphism. Star-shaped means $tv\in V$ whenever $v\in V$ and $0\le t\le 1$; it ensures the radial geodesic remains in $U$. This is distinct from a convex normal neighbourhood, which imposes a unique short geodesic between every pair of its points.

The radial distance in a normal neighbourhood and the unit radial vector field in a normal neighbourhood are defined, respectively, byThe function $r$ is continuous at $p$ and smooth away from $p$, but is not differentiable at $p$ in positive dimension. Some conventions use the scaled radial field $W_q=(d{\exp }_p{)}_{v}v$ and the smooth radial function $r^2/2$ instead; the associated gradient identities will distinguish them.

The Gauss lemma states that for $v\in V$ and $w\in T_{p}M$,To prove it, consider the geodesic variation $F(s,t)={\exp }_p(t(v+sw))$ for $0\le t\le 1$ and small $s$. Set $T={\partial }_{t}F$ and $J={\partial }_{s}F$. The Levi-Civita connection has zero torsion tensor, so $D_{t}J=D_{s}T$. Each $t$-curve is a geodesic, hence $D_{t}T=0$, and its squared speed is the constant $|v+sw{|}^2$. At $s=0$,Since $J(0)=0$, integration gives $g(T,J)=t{g}_p(v,w)$. At $t=1$, $T=(d{\exp }_p{)}_{v}v$ and $J=(d{\exp }_p{)}_{v}w$, proving the Gauss lemma. Taking $w=v$ shows $|E|=1$; taking $w\perp v$ shows that $E$ is orthogonal to the images of tangent vectors to the tangent-space sphere, hence to the geodesic spheres in $U$.

For $u\in T_{p}M$ of unit length, write the unit-speed radial geodesic as ${\gamma }_u(t)={\exp }_p(tu)$. Differentiating this expression and substituting $v=tu$ givesTo identify the gradient, write an arbitrary tangent vector at $q={\exp }_p(v)$ as $\eta =(d{\exp }_p{)}_{v}w$. Differentiating $r({\exp }_p(v))=|v|$ and using the Gauss lemma,By the defining property of the gradient,The latter field extends smoothly across $p$ by $W_p=0$, because $r^2/2$ is the smooth quadratic function $|v{|}^2/2$ in normal coordinates.

The same calculation also proves local radial minimization: for a piecewise smooth curve in the normal neighbourhood, $|dr(\dot{\alpha })|\le |\dot{\alpha }|$, so its length from $p$ to $q$ is at least $r(q)$, attained by the radial geodesic. Take a small tangent ball with closure inside a larger normal ball. Any path leaving the small ball before reaching a point inside it has already accumulated length at least its radius. Therefore, for points in the small ball, $r(q)=d(p,q)$ even when competitors may leave the neighbourhood. This local argument does not assert that $r$ equals global Riemannian distance throughout every arbitrarily large star-shaped normal neighbourhood.

Work on a connected, finite-dimensional Riemannian manifold without boundary. It is geodesically complete if every maximal affinely parametrized geodesic has parameter domain $\mathbb{R}$. Equivalently, for every $p$ the Riemannian exponential map ${\exp }_p$ is defined on all of $T_{p}M$: any finite affine time can be rescaled to time one, and negative times correspond to reversing the initial velocity.

The Hopf-Rinow lemma in its minimizing-geodesic form says: if ${\exp }_p$ is defined on all of $T_{p}M$ for one point $p$, then every $q$ can be joined to $p$ by a geodesic of length $d(p,q)$. We first establish the local ingredient, the distance splitting through a small geodesic sphere. If $D=d(p,q)>0$, choose $0<\epsilon <D$ sufficiently small that the closed normal ball about $p$ is compact and radial lengths give Riemannian distance, as proved using the Gauss lemma. Take paths from $p$ to $q$ of lengths tending to $D$. Their first intersections $x_j$ with its boundary sphere have a subsequence converging to some $x$ on the sphere. Their lengths are at least $\epsilon +d(x_j,q)$, giving $D\ge \epsilon +d(x,q)$. The triangle inequality gives the reverse inequality. ThereforeThis splitting uses only a compact local geodesic sphere, not completeness or the existence of a globally minimizing path.

We also need that a minimizing broken geodesic has no corner. To see this directly, let the incoming and outgoing unit tangents at a junction be $u$ and $v$. Choose one nearby point on each segment so that they and the junction lie in a common convex normal neighbourhood. Move the junction with velocity $z$, joining it to those two fixed points by the unique short geodesics. The Gauss lemma, or the first variation formula, gives the derivative of the sum of their lengths as $⟨u-v,z⟩$. If $u\ne v$, setting $z=v-u$ makes this derivative $-|u-v{|}^2<0$, contradicting minimality. Thus the tangent vectors agree, and uniqueness for the geodesic equation makes the broken path one smooth geodesic.

Now choose the unit-speed radial geodesic $\gamma$ through the point $x$ obtained by the distance splitting, with $\gamma (0)=p$ and $\gamma (\epsilon )=x$. By the hypothesis on ${\exp }_p$, it is defined at least up to time $D$. DefineThis is a nonempty closed set. If $t\in A$, the triangle inequality and the length of $\gamma {|}_{[0,t]}$ giveso $d(p,\gamma (t))=t$: that geodesic segment minimizes. Let $t_{\ast }=\max A$. If $t_{\ast }<D$, apply the distance splitting through a small geodesic sphere at $\gamma (t_{\ast })$ with radius $0<\delta <D-t_{\ast }$. There is a point $y$, joined to $\gamma (t_{\ast })$ by a short minimizing radial geodesic $\sigma$, for whichThe broken path $\gamma {|}_{[0,t_{\ast }]}\ast \sigma$ has length $t_{\ast }+\delta$. Moreover,Consequently that broken path minimizes. It has no corner, so $\sigma$ is the continuation of $\gamma$ and $y=\gamma (t_{\ast }+\delta )$. The last distance equality puts $t_{\ast }+\delta$ in $A$, a contradiction. Hence $D\in A$ and $\gamma (D)=q$. This proves the Hopf-Rinow lemma, with the case $p=q$ supplied by the constant geodesic.

The Hopf-Rinow theorem asserts the equivalence of

These conditions imply the existence of a minimizing geodesic between any two points. Here is the deduction, including the implication back to geodesic completeness. If the manifold is geodesically complete, the Hopf-Rinow lemma applies at each point. For every $R\ge 0$,One inclusion follows from the length of the radial geodesic; the other follows by taking an initial velocity of a minimizing one. The tangent-space ball is compact, so the metric ball is compact as its continuous image. Every closed bounded set is a closed subset of such a ball and is compact. A Cauchy sequence is bounded and therefore has a convergent subsequence in a compact ball; the Cauchy property makes the whole sequence converge. This proves completeness of the Riemannian distance.

Conversely, assume metric completeness and let $\gamma :[0,b)\to M$ be a geodesic with $b<\infty$. Its speed $c$ is constant, so $d(\gamma (s),\gamma (t))\le c|s-t|$. Completeness therefore gives a limit $q$ as $t\uparrow b$. Eventually $\gamma$ lies in a coordinate neighbourhood with compact closure on which the Riemannian metric is uniformly comparable to the Euclidean metric. Its coordinate velocity is bounded by the constant-speed estimate. The Christoffel symbols are bounded there, so the geodesic equationmakes its coordinate acceleration bounded. Thus the coordinate velocity also has a limit $v$ at time $b$. The local existence and uniqueness theorem for smooth ordinary differential equations extends the solution from initial data $(q,v)$ past $b$, contradicting maximality. Reversing time treats a finite left endpoint. Hence metric completeness implies geodesic completeness, and the preceding argument supplies compactness of closed bounded sets. Compactness of closed bounded sets already implies metric completeness by the Cauchy-sequence argument, completing all equivalences.

In fact, the one-point hypothesis of the Hopf-Rinow lemma suffices for the theorem: the same closed-ball image argument at that point puts every Cauchy sequence in a compact ball, giving metric completeness and then geodesic completeness at every point. By contrast, the existence of a minimizing geodesic for every pair alone does not imply completeness: an open Euclidean ball has minimizing straight segments between all its points but has Cauchy sequences converging to its missing boundary.

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.