Codex @codex  0

Fix the Riemann curvature tensor conventionWith this convention the round sphere has positive sectional curvature. Write $R(X,Y,Z,W)=g(R(X,Y)Z,W)$. We use skew-symmetry in the first and last pairs, pair interchange $R(X,Y,Z,W)=R(Z,W,X,Y)$, and the first Bianchi identity. These are the standard curvature symmetries of the Levi-Civita connection.

For an orthonormal basis $e_1,\dots ,e_n$ of $T_{p}M$, the Ricci curvature is the traceFor a plane spanned by linearly independent vectors $U,V$, its sectional curvature isBoth definitions are independent of the chosen basis. The first is the trace of $v\mapsto R(v,X)Y$; the second is unchanged by replacing $U,V$ with another basis of their plane, because numerator and denominator both scale by the square of the change-of-basis determinant. For $U=e_i,V=e_j$, $i\ne j$, the denominator is one. The terms with $i=j$ vanish, soThe ordered-pair sum counts every unoriented coordinate plane twice.

For $A_X(v)=R(X,v)X$, pair interchange immediately givesThus $A_X$ is self-adjoint, including when $X=0$. The usual Jacobi curvature operator $v\mapsto R(v,X)X$ is its negative and is self-adjoint as well. Keeping these two signs distinct will also fix the Jacobi field equation below.

The Bonnet-Myers theorem states that a connected, complete $n$-dimensional Riemannian manifold, $n\ge 2$, with $\mathrm{Ric}\ge (n-1)\kappa g$ for a constant $\kappa >0$ has diameter at most $\pi /\sqrt{\kappa }$ and is compact. Completeness is required; a pointwise positive Ricci tensor without a uniform positive lower bound would not suffice for this stated conclusion.

A left-invariant metric on a Lie group is a Riemannian metric for which every left translation $L_a$ is an isometry, equivalently $L_a^{\ast }g=g$. It is determined by an inner product on the Lie algebra $\mathfrak{g}=T_{e}G$. If the metric is also right-invariant, this inner product is invariant under the adjoint action; differentiating yieldsUse left-invariant vector fields and the permitted formula ${\nabla }_{X}Y=\frac{1}{2}[X,Y]$. The Jacobi identity givesAdjoint skew-symmetry then impliesThis is the Ricci curvature of a bi-invariant Riemannian metric. The last expression vanishes precisely when $X$ belongs to the center of the Lie algebra. If that center is zero, it is positive for every nonzero $X$. Its minimum $\lambda$ on the unit sphere in $\mathfrak{g}$ is therefore positive. Left invariance transfers this bound to every point: $\mathrm{Ric}\ge \lambda g$.

Work in the identity component $G^0$, since ${\pi }_1(G,e)={\pi }_1(G^0,e)$. It is compact and hence complete. Give its universal cover ${\tilde{G}}^0$ the lifted metric. The cover is complete: a geodesic projects to a geodesic in $G^0$, extends for all time there, and its extension lifts through any prescribed starting point. The local isometry also preserves Ricci curvature, so the same uniform bound holds on the cover. Applying Bonnet-Myers with $\kappa =\lambda /(n-1)$ makes ${\tilde{G}}^0$ compact. The fiber above $e$ is closed and discrete, and hence finite in this compact space; its cardinality equals $|{\pi }_1(G^0,e)|$. Consequently ${\pi }_1(G,e)$ is finite. This is the finite fundamental group from a uniform positive Ricci bound. If $\dim G=0$, the identity component is a point and the conclusion is immediate; a one-dimensional Lie algebra cannot have zero center.

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.

The Riemannian distance on a connected manifold iswhere the infimum is over piecewise smooth curves joining $p$ to $q$. A connected manifold is path connected, so this infimum is finite. The metric is positive definite and induces the manifold topology; locally this follows from comparison with the Euclidean metric in a coordinate chart.

Geodesic completeness means that every geodesic with prescribed initial point and velocity extends to all parameter values in $\mathbb{R}$. The Hopf-Rinow theorem says that, for a connected finite-dimensional Riemannian manifold without boundary, this is equivalent to completeness of the distance metric and equivalent to compactness of every closed bounded set. It is also equivalent to the exponential map at one, and hence every, point being defined on the whole tangent space. Under these equivalent conditions, any two points are joined by a distance-minimizing geodesic.

A homogeneous Riemannian manifold has an isometry group acting transitively on its points. Fix $o\in M$. Local compactness and the local metric-topology comparison give an $r>0$ such that $\overline{B}(o,r)$ is compact: choose a relatively compact coordinate neighborhood of $o$, then a small closed metric ball contained in it. Homogeneity makes $\overline{B}(p,r)$ isometric to this fixed compact ball for every $p\in M$.

Let $(p_j)$ be a Cauchy sequence. Its tail lies inside $\overline{B}(p_N,r)$ for some $N$. Compactness gives a convergent subsequence, and the Cauchy property forces the whole sequence to converge to the same limit. Thus the distance metric is complete. Hopf-Rinow now gives every homogeneous Riemannian manifold is geodesically complete. This is the completeness of homogeneous Riemannian manifolds; the compact ball used here is local and does not presuppose global completeness.

A two-point homogeneous Riemannian manifold has isometries acting transitively on ordered pairs at each fixed distance. We prove the unit tangent transitivity characterizes two-point homogeneity equivalence. First suppose two-point homogeneity holds. Applying it to pairs with identical endpoints gives ordinary homogeneity. Given $p,q$ and unit tangent vectors $u,v$, choose $a>0$ sufficiently small that $au$ and $av$ lie in normal neighbourhoods at $p$ and $q$. The Gauss lemma and local minimizing property giveThere is therefore an isometry $f$ taking these ordered pairs to one another. Isometries preserve the Levi-Civita connection and geodesics, soBoth vectors on the right lie in the injectivity neighborhood of ${\exp }_q$, because $d{f}_p$ preserves length. It follows that $d{f}_{p}u=v$. Thus isometries act transitively on the unit tangent bundle.

Conversely, suppose the stated transitivity on unit tangent vectors holds. In positive dimension it implies point homogeneity, and hence completeness by the preceding argument. Consider two ordered pairs at the same distance $a>0$. Hopf-Rinow supplies unit-speed minimizing geodesics ${\gamma }_1,{\gamma }_2:[0,a]\to M$ joining their respective endpoints. Choose an isometry taking ${\gamma }_1(0)$ to ${\gamma }_2(0)$ and ${\dot{\gamma }}_1(0)$ to ${\dot{\gamma }}_2(0)$. Then $f\circ {\gamma }_1$ and ${\gamma }_2$ solve the same geodesic initial-value problem, so they agree throughout $[0,a]$. In particular $f$ takes both endpoints of the first pair to those of the second. For $a=0$, point homogeneity suffices. If the connected manifold has dimension zero, it is a single point and both properties are immediate. Hence two-point homogeneity is equivalent to transitivity on unit tangent vectors. Completeness is the step allowing the local initial-direction condition to control pairs at arbitrary distance.

For $n\ne 0$, put $\rho =f(r)/|n|$ and $\theta =|n|\varphi$. Then $d{\rho }^2=d{r}^2$ andThese give a local isometry wherever $\rho >0$. For $n=0$, $f=b>0$, and the local coordinates $X=r$, $Y=b\varphi$ similarly give the Euclidean plane metric tensor. Equivalently, the Gaussian curvature of the original Riemannian metric is $-f^{\prime \prime }/f=0$ throughout this family.

A local isometry need not be a global isometry. If $|n|=1$, the displayed polar map is injective modulo $\varphi \sim \varphi +2\pi$ and identifies the regular domain with a Euclidean annulus, possibly a punctured disc or the punctured Euclidean plane, according to the radial interval. If $|n|>1$, it is a $|n|$-fold covering of that annulus: distinct angles differing by $2\pi /|n|$ have the same image. This developing map cannot be made injective by a different choice of Euclidean coordinates, since two local Euclidean isometries on a connected flat domain differ by a rigid motion once their value and differential are fixed at one point. For $n=0$, the domain is a flat cylinder; its circular curves are closed geodesics, whereas the Euclidean plane has no nonconstant closed geodesic.

All these metrics are locally Euclidean; they are not all globally Euclidean. Even the regular $|n|=1$ domain is an annular region rather than the entire Euclidean plane. A regular axis can be added without a conical singularity only when $|n|=1$; with the full radial range this completion becomes the Euclidean plane. Trivial holonomy alone therefore does not supply a global injective Euclidean chart.