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Diagonalize the self-adjoint Jacobi curvature operator at the initial point and transport its orthonormal eigenbasis parallel along the geodesic. The Jacobi field equation reduces to $j_i^{\prime \prime }+{\lambda }_i{j}_i=0$. For zero initial position and initial derivative $b_i$, the solution is $b_i\sin (\sqrt{{\lambda }_i}t)/\sqrt{{\lambda }_i}$ when ${\lambda }_i>0$, $b_{i}t$ when ${\lambda }_i=0$, and $b_i\sinh (\sqrt{-{\lambda }_i}t)/\sqrt{-{\lambda }_i}$ when ${\lambda }_i<0$. Only positive eigenvalues contribute conjugate points, at $t=k\pi /\sqrt{{\lambda }_i}$. At a common zero the multiplicity is the sum of the contributing eigenspace dimensions. The displayed convention uses the curvature operator positive on a positively curved plane; reversing the tensor convention also reverses its tensor formula.

A Riemannian manifold is locally symmetric when its Riemann curvature tensor is parallel for the Levi-Civita connection. Along a geodesic, parallel transport then conjugates the Jacobi curvature operator at the initial point to the operator at every later point. Thus its eigenvalues stay constant. The condition is local and does not assert that global point reflections extend to isometries of the whole manifold.

Use the curvature sign convention implicit in part (c): the Jacobi field equation is $D_t^{2}J+K_{\dot{\gamma }}J=0$, so the operator in a positively curved normal direction has positive eigenvalue. With the alternative convention $R_{\mathrm{std}}(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$ used for the catalog's Jacobi curvature operator, the paper's tensor is $R=-R_{\mathrm{std}}$, and its operator is $K_v(x)=R_{\mathrm{std}}(x,v)v$. This fixes the sign throughout Q1.

Let $r(A,B,C,D)=⟨R(A,B)C,D⟩$. Pair interchange for the Riemann curvature tensor givesThus $K_v$ is self-adjoint with respect to the Riemannian metric. The curvature sign does not affect this conclusion. The spectral theorem for real symmetric matrices therefore supplies the real orthonormal eigenbasis used in part (b). Also $K_{v}v=0$, so the tangent direction is a zero-eigenvalue direction.

Expand a Jacobi field in the parallel orthonormal eigenbasis: $J(t)={\sum }_i{j}_i(t)e_i(t)$. Part (b) reduces its equation to the scalar equations $j_i^{\prime \prime }+{\lambda }_i{j}_i=0$. DefineFor arbitrary initial values $J(0)={\sum }_i{a}_i{e}_i$ and $D_{t}J(0)={\sum }_i{b}_i{e}_i$, the complete solution isA point at parameter $T>0$ is a conjugate point of $p$ precisely when a nonzero Jacobi field vanishes at both $0$ and $T$. The first condition makes every $a_i=0$. The second requires $b_i{S}_{{\lambda }_i}(T)=0$ for every $i$, with at least one nonzero $b_i$. For $T>0$, the factors for zero and negative eigenvalues never vanish; positive eigenvalues give exactly the sine zeros. ThereforeThe corresponding points are $\gamma (T)$. If several positive eigenspaces have a zero at the same parameter, the conjugate multiplicity is the sum of their dimensions. This also shows that a locally symmetric Riemannian manifold has no conjugate points along a given geodesic when that geodesic's Jacobi curvature operator has no positive eigenvalues.

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.