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.

For $p,q\in M$, define the Riemannian distance byA connected smooth manifold is locally path connected, so any two points are joined by a finite concatenation of coordinate paths. Thus the infimum is finite. It is nonnegative, $d_g(p,p)=0$, and reversing a curve proves symmetry. Concatenating two curves whose lengths approach their respective infima proves the triangle inequality.

For positivity and the topology, choose a coordinate ball about $p$ whose closure is contained in a chart. On its closure there are constants $0<c<C$ withA curve staying in the ball has length at least $c$ times its coordinate displacement. A curve leaving it must first reach its boundary, costing at least $c$ times the positive coordinate distance from $p$ to that boundary. Therefore a distinct $q$ has $d_g(p,q)>0$, whether it lies inside or outside the ball. For $q$ sufficiently close to $p$, the coordinate line segment gives $d_g(p,q)\le C|x(q)-x(p)|$, while the preceding lower bounds prevent a short curve from taking a shortcut outside the chart. These inequalities prove that the Riemannian distance induces the manifold topology. In particular it defines a genuine metric space.

A Riemannian manifold is geodesically complete if every maximal affinely parametrized geodesic exists for all real parameter values. We prove both implications of the Hopf-Rinow theorem rather than assuming global distance minimizers.

First assume completeness as a metric space. Let a constant-speed geodesic have a finite maximal right endpoint $b$. The bound $d_g(\gamma (t),\gamma (s))\le c|t-s|$ makes $\gamma (t)$ a Cauchy sequence as $t\uparrow b$, hence gives a limit $p\in M$. The distance topology just established puts the final portion inside a relatively compact coordinate ball about $p$. The positive lower bound on the Riemannian metric bounds all coordinate velocity components. The Christoffel symbols are bounded on the closure of that ball, and the geodesic equationthen bounds the coordinate acceleration. The velocity has a limit $v$ as $t\uparrow b$. Local existence and uniqueness for the geodesic equation with data $(p,v)$ continue $\gamma$ through $b$, contradicting maximality. The same argument applies at a finite left endpoint. Hence metric completeness implies geodesic completeness.

Conversely assume geodesic completeness. Fix $p,q$, put $r=d_g(p,q)>0$, and choose $0<ϵ<r$ so small that a closed radius-$2ϵ$ normal ball about $p$ lies inside a convex normal neighborhood. By the Gauss lemma, its radial geodesics minimize; a curve leaving the larger ball cannot improve on a radius-$ϵ$ radial segment. Thus the radius-$ϵ$ geodesic sphere $S_{ϵ}(p)$ is compact and consists of points at distance exactly $ϵ$ from $p$.

The distance splitting through a small geodesic sphere has an elementary proof here. Every path from $p$ to $q$ must cross $S_{ϵ}(p)$, and its initial portion has length at least $ϵ$. HenceThe triangle inequality proves the reverse bound. Continuity of $d_g(\cdot ,q)$ and compactness give a point $x$ achieving the minimum, so $d_g(x,q)=r-ϵ$. Let $\gamma (t)={\exp }_p(tv)$ be the complete unit-speed radial geodesic through $x$ at time $ϵ$.

Consider the closed subsetIt is nonempty since $ϵ\in A$, and has a largest member $t_{\ast }$. Suppose $t_{\ast }<r$. At $x_{\ast }=\gamma (t_{\ast })$, choose a small normal sphere of radius $\eta <r-t_{\ast }$ and repeat the splitting argument. It supplies a radial endpoint $y$ withThe triangle inequality gives $d_g(p,y)\ge t_{\ast }+\eta$, while the broken path following $\gamma$ to $x_{\ast }$ and then the short radial segment to $y$ has exactly that length. It is therefore minimizing. Its two velocities at the joining point must agree: if unit incoming and outgoing velocities are $u,w$, moving the join in direction $w-u$ changes the sum of lengths by $-|u-w{|}^2$ to first order, using the first variation of geodesic energy or the equivalent unit-speed length formula. A corner would strictly shorten it. Uniqueness of the geodesic equation consequently gives $y=\gamma (t_{\ast }+\eta )$, contradicting the maximality of $t_{\ast }$. Thus $t_{\ast }=r$, $\gamma (r)=q$, and $\gamma {|}_{[0,r]}$ is a minimizing geodesic.

This radial continuation proof of Hopf-Rinow also proves compactness of closed balls. For every $R\ge 0$,The inclusion from right to left follows from the length of a radial geodesic; the reverse inclusion uses the minimizing geodesic just constructed. The Riemannian exponential map is defined on all of $T_{p}M$ by geodesic completeness, and the closed tangent ball is compact in a finite-dimensional vector space. Its continuous image is compact. Every Cauchy sequence is bounded, hence has a convergent subsequence in such a ball, and the Cauchy property makes the whole sequence converge. Therefore geodesic completeness implies metric completeness, completing both directions.

For the hyperbolic plane, use the upper half-plane model with $g=(d{x}^2+d{y}^2)/y^2$, $y>0$. Its geodesic equation isEvery nonconstant unit-speed solution is either a vertical curve $x=a$, $y=b{e}^{\pm t}$ or, after translating and possibly reversing $t$,Substitution verifies both equations and $g(\dot{\gamma },\dot{\gamma })=1$. These curves cover every unit initial tangent: a nonvertical tangent at $(x,y)$ determines the circle center $a=x+y{y}^{\prime }/x^{\prime }$ and radius $R=\sqrt{(x-a{)}^2+y^2}$, and its orientation determines the sign of the parameter. Constant solutions and constant-speed rescalings cover arbitrary initial velocities. In every case $y(t)>0$ for all finite $t$, and the solution is defined for every real $t$. Thus the real hyperbolic plane is geodesically complete, directly, and the proved Hopf-Rinow theorem also makes its distance complete.

A Riemannian isometry $F:(M,g)\to (N,h)$ is a diffeomorphism satisfying $F^{\ast }h=g$. A local isometry is a local diffeomorphism with the same pullback identity; equivalently $d{F}_p$ is an isometric linear isomorphism for every $p$. An isometric immersion into a higher-dimensional manifold does not satisfy this local-diffeomorphism requirement.

Let ${\varphi }_t$ be the local flow of the generating vector field $K$. The identity ${\varphi }_t^{\ast }g=g$ differentiates to ${\mathcal{L}}_{K}g=0$, where $\mathcal{L}$ is the Lie derivative of a tensor field. For arbitrary vector fields $X,Y$, metric compatibility and the torsion-free connection giveThis proves the Killing equation. Conversely, a smooth Killing field has a local flow, andSince ${\varphi }_0$ is the identity, ${\varphi }_t^{\ast }g=g$ wherever the flow is defined. Its inverse is ${\varphi }_{-t}$ locally, so the local flow consists of local isometries. Neither direction assumes that $K$ is complete.

Put $H(X,Y)={\nabla }_X({\nabla }_{Y}K)-{\nabla }_{{\nabla }_{X}Y}K$ and $B(X,Y,Z)=⟨H(X,Y),Z⟩$. These expressions are tensorial in $X,Y,Z$. Take the covariant derivative in direction $X$ of the Killing equation, subtracting the terms from differentiating $Y$ and $Z$. It follows thatWith the curvature convention $\mathcal{R}$ of Solution 1, the torsion-free connection also givesDefine $C(X,Y,Z)=⟨\mathcal{R}(X,K)Y,Z⟩$. It is skew in $Y,Z$ by skew-adjointness of the Riemann curvature tensor. The first Bianchi identity givesso $C$ has the same antisymmetrization in $X,Y$ as $B$. Consequently $D=B-C$ is symmetric in its first two slots and skew in its last two. These symmetries force it to vanish:We have proved the second covariant derivative of a Killing vector identity, including its sign:The printed identity uses the opposite curvature-sign convention in Killing derivative identities. If $R=-\mathcal{R}$, namely $R(X,Y)Z={\nabla }_Y{\nabla }_{X}Z-{\nabla }_X{\nabla }_{Y}Z+{\nabla }_{[X,Y]}Z$, the same result reads ${\nabla }_{X,Y}^{2}K=R(K,X)Y$, exactly as printed. The convention must change with the sign; under $\mathcal{R}$, writing $\mathcal{R}(K,X)Y$ would be incorrect.

Finally let $H_0=K-\tilde{K}$ have zero value and zero first covariant derivative at $p$. Along any geodesic issuing from $p$, the identity just proved saysThus $H_0$ is a Jacobi field with zero initial value and derivative. Uniqueness for this linear ordinary differential equation implies $H_0=0$ on each such geodesic, hence on a convex normal neighborhood of $p$. Consider the set of points where both $H_0$ and $\nabla H_0$ vanish. It is closed by continuity and open by this same local argument; on the resulting neighborhood $H_0$ vanishes identically, so its derivative does too. It is nonempty, and $M$ is connected, therefore it is all of $M$. This proves $K=\tilde{K}$ everywhere and the general statement that a Killing field is determined by its value and first covariant derivative.

Use the Levi-Civita connection and the Riemann curvature tensor conventionfor orthonormal $T,E$. Thus the round sphere has positive sectional curvature. Work on a fixed compact parameter interval $[a,b]$ inside the given open interval; this keeps all differentiations and endpoint evaluations well-defined. The base geodesic is regular and has constant speed, so a constant rescaling of $t$ makes it unit-speed. Put $T={\partial }_t\varphi$, $J={\partial }_s\varphi$, $D_t={\nabla }_T$, $D_s={\nabla }_J$ and evaluate at $s=0$. The torsion-free connection gives $D_{s}T=D_{t}J$ because the two parameter derivatives commute. The geodesic equation is $D_{t}T=0$.

Differentiating the length of a curve once and twice, using metric compatibility, givesandLet $A=D_{s}J$. The curvature commutator gives $D_s{D}_{t}J=D_{t}A+\mathcal{R}(J,T)J$, and the skew-adjoint curvature symmetry gives $⟨\mathcal{R}(J,T)J,T⟩=-⟨\mathcal{R}(J,T)T,J⟩$. Integration by parts therefore proves the moving-endpoint second variation of Riemannian arc length:Equivalently, with $W=J-⟨J,T⟩T$, the integral is $I(W,W)$, the Riemannian index form. The endpoints of this geodesic variation generally move, so the displayed boundary term must be retained. For a base speed $c$ instead of one, the boundary and integral have an overall factor $1/c$, and the subtracted square inside the integral is $⟨D_{t}J,T{⟩}^2/c^2$. Returning to the original parameter, every ruling has constant speed, so $L_{[a,b]}(s)=(b-a)|{\partial }_t\varphi (0,s)|$ and the full open-interval length is $2\delta |{\partial }_t\varphi (0,s)|$. ConsequentlyThe preceding curvature formula, with the base-speed factor $1/c$ when necessary, therefore evaluates the original full length derivative using any compact subinterval. No extension to the missing endpoints is required.

Differentiating $D_{t}T=0$ in $s$ yields the Jacobi field equationThe same ruling curves are geodesics of the surface with its induced metric, because the tangential part of their ambient acceleration is zero. Write its second fundamental form as $II$ and its induced Levi-Civita connection as $D^N$. In particular $II(T,T)=0$. The scalar $f=⟨J,T⟩$ satisfies $f^{\prime \prime }=0$, by the Jacobi field equation, so $W=J-fT$ is a Jacobi field both intrinsically and in the ambient manifold. In a regular surface chart $h=|W|>0$, and $E=W/h$ is the perpendicular unit tangent. Along a two-dimensional geodesic, $D_t^{N}E=0$: its derivative is orthogonal to both $E$ and the parallel field $T$. The intrinsic Jacobi field equation consequently givesFor the ambient derivative, the Gauss formula gives $D_{t}W=h^{\prime }E+hII(T,E)$. Applying the same norm differentiation used above to $h=|W|$ givesComparing these identities proves the curvature comparison for a geodesically ruled surface:This is also the Gauss equation in a curved ambient manifold with $II(T,T)=0$. It applies locally to each regular patch of the surface.

For strict inequality, take the Euclidean embedded submanifold parametrized by $\varphi (t,s)=(t,s,ts)$. Each $t$-curve is a straight-line geodesic. Its first fundamental form has coefficients $E_0=1+s^2$, $F_0=st$, $G_0=1+t^2$; the normal $(-s,-t,1)/\sqrt{1+t^2+s^2}$ gives second fundamental form coefficients $e=0$, $f_0=1/\sqrt{1+t^2+s^2}$, $g_0=0$. HenceThus equality can fail everywhere, although every ruling is an ambient geodesic.

The length differentiation requires a nonconstant base ruling; a regular parametrized surface patch guarantees this. If “image of a smooth map” is read literally without a rank assumption, a degenerate member can fail even to have a first length derivative. For example, in the flat cylinder $\mathbb{R}\times S^1$, the map $\varphi (t,s)=(st,e^{is})$ for $|t|<\delta$, $|s|<4\pi$ has geodesic rulings and a two-dimensional open image: every nonzero-$s$ point has full-rank parametrization, and the sole zero-$s$ image point also has a full-rank preimage at $s=2\pi ,t=0$. Yet $L({\gamma }_s)=2\delta |s|$ is not differentiable at zero. The variation formula therefore applies to nonconstant members, as usual. The curvature inequality still holds on the whole image: regular values are dense by Sard theorem, and both curvatures extend continuously to the remaining points.

We prove both inequalities, including the BK inequality, for the inhomogeneous product measure in the question. Identify a configuration with its set of coordinates equal to one. An up-set is then an increasing event. For an increasing event $A$, a set $S$ of open coordinates is a witness if the configuration with precisely $S$ open belongs to $A$: upward closure ensures that those coordinates alone force $A$, whatever the remaining coordinates do. The event $A\square B$ means that there are disjoint open witnesses $S,T$ for $A,B$ respectively.

First prove the Harris' inequality by induction on the number of coordinates. For the last coordinate, of parameter $p$, write $A_0\subseteq A_1$ and $B_0\subseteq B_1$ for the two sections in the remaining coordinates, and put $a_i=\mathbb{P}(A_i)$, $b_i=\mathbb{P}(B_i)$. Induction applied to each section givesSubtract the product of the marginal probabilities from the right-hand side:The base case of zero coordinates is immediate. Thus

For disjoint occurrence, first assume every coordinate parameter is at most $1/2$, and again use induction. Put $c_{ij}=\mathbb{P}(A_i\square B_j)$ on the remaining coordinates. The zero-section of $A\square B$ is exactly $A_0\square B_0$. The one-section is exactlyIndeed, the last open coordinate can belong to at most one of two disjoint witnesses. Assigning it to the $A$ witness leaves a witness for $A_1$ and a witness for $B_0$; assigning it to the $B$ witness gives the other event. If neither witness uses it, both descriptions still hold. Conversely, witnesses for either displayed event lift to disjoint witnesses by adding the last coordinate only to the appropriate one.

The intersection of the two displayed events contains $A_0\square B_0$, because the sections are increasing. ThereforeThe coefficient $1-2p$ is nonnegative. Applying the induction hypothesis to each term givesThe product of the marginal probabilities is exactly this last expression plusThis proves the desired inequality whenever every parameter is at most $1/2$.

To remove that restriction, suppose first $p_i<1$ for every coordinate. Replace coordinate $i$ by the logical disjunction of $m_i$ independent Bernoulli bits, each with parameterChoose $m_i$ large enough that $q_i\le 1/2$. Different blocks are independent, and their OR values have exactly the original distribution. Let $\tilde{A},\tilde{B}$ be the inverse images of $A,B$ under this block-OR map; they are increasing events. If the OR configuration has disjoint witnesses $S,T$, choose one open bit from each block indexed by $S$ or $T$. The bits chosen for $S$ force $\tilde{A}$, those for $T$ force $\tilde{B}$, and the two bit sets are disjoint. Thus the inverse image of $A\square B$ is contained in $\tilde{A}\square \tilde{B}$. The small-parameter result provesParameters equal to one follow by a limit: probabilities of any fixed event on a finite cube are polynomials in the coordinate parameters, hence continuous. Zero parameters already cause no difficulty. We have obtained

For the graph application, take coordinates to be the edges of the complete graph. Let $A$ be the increasing event that $x$ is connected by a graph path to $U$, and $B$ the corresponding event for $W$. Their probabilities are $\alpha ,\beta$. A simple graph path from $U$ to $W$ through $x$ splits at $x$ into two graph paths with disjoint edge sets. Those sets witness $A$ and $B$, so this graph path event is contained in $A\square B$. The BK inequality givesOn the other hand, if both $A$ and $B$ occur, concatenate an $x$-to-$U$ graph path and an $x$-to-$W$ graph path to obtain a walk from $U$ to $W$. Removing loops yields a simple graph path, although it need not pass through $x$. Thus $A\cap B$ is contained in the unrestricted connection event, and the Harris' inequality givesThe distinction between a simple graph path through $x$ and an unrestricted connection is essential: a walk may revisit an edge, so a through-$x$ walk would not in general provide disjoint witnesses.

Colour every vertex independently red or blue, each with probability $1/2$. A fixed $r$-element hypergraph edge is monochromatic with probability $2^{1-r}$. If there are $m<2^{r-1}$ hypergraph edges, the expected number of monochromatic hypergraph edges is $m{2}^{1-r}<1$. Equivalently, the union bound says that the probability of even one monochromatic hypergraph edge is less than one. A proper two-colouring therefore exists.

For the converse, suppose $r\ge 2$, put $N=2r^2$, and sample $m$ independent uniformly random $r$-subsets of an $N$-vertex set. Initially this is a multihypergraph. For a fixed two-colouring with $a$ red vertices, a sampled hypergraph edge is monochromatic with probabilityThe numerator is minimized by $a=N/2=r^2$. To see this discretely, use $\left(\genfrac{}{}{0ex}{}{a+1}{r}\right)-\left(\genfrac{}{}{0ex}{}{a}{r}\right)=\left(\genfrac{}{}{0ex}{}{a}{r-1}\right)$: the successive difference of the numerator is $\left(\genfrac{}{}{0ex}{}{a}{r-1}\right)-\left(\genfrac{}{}{0ex}{}{N-a-1}{r-1}\right)$, negative up to the middle and positive afterwards. Binomial coefficients with an upper index below $r$ are understood as zero.

For this balanced colouring,The elementary inequality ${\prod }_j(1-u_j)\ge 1-{\sum }_j{u}_j$, valid for $0\le u_j\le 1$, givesThus any fixed colouring is proper for all the sampled hypergraph edges with probability at most $e^{-m{2}^{-r}}$. There are $2^N$ colourings, so chooseA union bound makes the probability that any proper colouring exists at most $2^N{e}^{-m{2}^{-r}}\le e^{-1}<1$. Some sampled multihypergraph is consequently not two-colourable. Remove repeated copies of its hypergraph edges: this changes neither which colourings are proper nor non-two-colourability, and leaves at most $m$ distinct hypergraph edges. We have provedFor $r=1$, one singleton hypergraph edge is already not two-colourable, so that endpoint causes no exception.