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Pointwise positive Ricci curvature does not imply compactness or a finite diameter. The complete positively curved paraboloid $z=x^2+y^2$ has induced metricand Gaussian curvatureFor example, this follows from the graph formula $K=\det D^{2}z/(1+|Dz{|}^2{)}^2$. The metric dominates the Euclidean plane metric, so it is complete by upward stability of Riemannian completeness. Its diameter is infinite since its distance dominates Euclidean distance. On a surface the Ricci tensor is $Kg$, so it is strictly positive everywhere, but has no uniform positive lower bound.

There is no positive-Ricci metric on $S^2\times S^1$. If one existed, compactness of the unit tangent bundle would give ${\mathrm{Ric}}_{\mathrm{tr}}\ge cg$ for some $c>0$. The lifted metric on the universal cover is complete by Q2 and satisfies the same positive lower bound. Part (a) makes this cover compact. A covering fiber is closed and discrete and therefore finite, implying a finite fundamental group. Butwhich is infinite. This is the finite fundamental group from a uniform positive Ricci bound obstruction; it rules out every possible metric, not just the standard product metric.

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.