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By part (b), $M=S^n/\Gamma$, where the group of deck transformations acts freely by round isometries. Every round isometry is the restriction of an orthogonal transformation of ${\mathbb{R}}^{n+1}$: it preserves the inner product $⟨x,y⟩=\cos d(x,y)$, so its images of an orthonormal basis determine its linear extension.

Suppose $n$ is even. Every $A\in SO(n+1)$ has eigenvalue one. Indeed, the nonreal eigenvalues come in conjugate pairs with product one, while the remaining real eigenvalues are $\pm 1$. Their number is odd and their product is one, forcing at least one $+1$. Such a transformation fixes a unit vector. Freeness of the deck transformation action consequently implies that the determinant map $\Gamma \to \{\pm 1\}$ has trivial kernel. Thus $|\Gamma |\le 2$.

If $\Gamma$ is nontrivial with generator $A$, then $A^2=I$. Its eigenvalues are $\pm 1$, and freeness excludes the eigenvalue $+1$. Hence $A=-I$, the antipodal involution. We obtainThis proves the even-dimensional part of the classification of complete positive constant-curvature manifolds.

The conclusion fails in odd dimensions. For $n=2m-1\ge 3$, identify $S^{2m-1}$ with the unit sphere in ${\mathbb{C}}^m$. Scalar multiplication by $\zeta =e^{2\pi i/r}$, with $r\ge 3$, generates a freely acting cyclic group of isometries: ${\zeta }^{j}z=z$ for nonzero $z$ forces ${\zeta }^j=1$. The quotient is complete with constant sectional curvature one and has fundamental group $C_r$, so it is neither the simply connected sphere nor Real projective space with fundamental group $C_2$. In particular $S^3/C_3$ is a lens space counterexample.

Take unit-curvature Real projective space ${\mathbb{RP}}^n=S^n/\{\pm 1\}$ with $n\ge 2$, and any point $p=[v]$. It is complete with positive sectional curvature. A unit-speed geodesic through $p$ lifts toSince the two representatives of an endpoint are antipodal, Riemannian distance in the quotient is the smaller of the two spherical distances:This equals $t$ for $0\le t\le \pi /2$ and is strictly smaller just afterwards. Thus the Riemannian cut point occurs at time $\pi /2$ in every direction.

Along a curvature-one geodesic, every normal Jacobi field with initial value zero is $J(t)=\sin t{P}_{t}a$, where $P_t$ is parallel transport and $a\perp \dot{\gamma }(0)$. Its first further zero is $t=\pi$. The quotient map is a local isometry, so the Jacobi field equation and conjugacy times are unchanged. HenceThis is cut and conjugate times in round real projective space.

Use the intended trace bound $\mathrm{Ric}\ge (n-1)kg$ and write $D=\pi /\sqrt{k}$. We state precisely the allowed Bishop volume comparison with a positive-curvature model. Setwhere ${\omega }_{n-1}$ is the area of the unit $(n-1)$-sphere. For $V_p(r)=\mathrm{Vol}B(p,r)$, the ratio $V_p(r)/v_k(r)$ is nonincreasing on $0<r<D$, tends to one at zero, and is at most one. Before the cut time, the radial volume density $J(r,\theta )$ in geodesic polar coordinates is at most $s_k(r{)}^{n-1}$.

By the Bonnet-Myers theorem, $\mathrm{diam}M\le D$. Therefore $V_p(D)=\mathrm{Vol}M$. If distance-$D$ endpoints exist, they do not affect this volume equality: the distance sphere is contained in the smooth image ${\exp }_p(D{S}_p^{n-1})$, which has zero $n$-dimensional Riemannian volume. Taking limits in the comparison ratio at $D$ is legitimate. The assumed equality of total volume gives $V_p(D)/v_k(D)=1$, so monotonicity forcesfor every $p$.

We now extract the equality geometry, rather than merely naming a rigidity theorem. On a small normal neighborhood of $p$, the continuous nonnegative difference $s_k(r{)}^{n-1}-J(r,\theta )$ has integral zero. It is therefore identically zero. Let $A(X)={\nabla }_X{\partial }_r$ on a distance sphere and $h=\mathrm{tr}A={\partial }_r\log J$. Equality of the density gives $h=(n-1)s_k^{\prime }/s_k$. The traced radial Riccati equation for distance spheres readsSince $s_k^{\prime \prime }=-k{s}_k$, substituting $h$ makes the sum of the following two nonnegative quantities zero:Equality in the Cauchy-Schwarz inequality for the eigenvalues of the self-adjoint operator $A$ forces $A=(s_k^{\prime }/s_k)I$. The angular metric $g_r$ in geodesic polar coordinates consequently satisfies ${\partial }_r{g}_r=2(s_k^{\prime }/s_k)g_r$. Smoothness at the center fixes its integration constant, since $r^{-2}{g}_r\to g_{S^{n-1}}$. Henceon the normal ball: the Riemannian metric is locally the round curvature-$k$ metric. As $p$ was arbitrary, $M$ has constant sectional curvature $k$.

The classification of complete positive constant-curvature manifolds now gives $M=S_k^n/\Gamma$ with $\Gamma$ a finite freely acting group of isometries. Covering degree gives $\mathrm{Vol}M=\mathrm{Vol}{S}_k^n/|\Gamma |$. Volume equality forces $|\Gamma |=1$, provingThis establishes spherical rigidity of maximal total volume under the correct normalization.

Under the literal trace bound, the assertion is false in dimension four. Take the Riemannian product $M=S^2(r)\times S^2(r)$ with $r^2=1/(\sqrt{6}k)$. Its Ricci curvature is $r^{-2}g=\sqrt{6}kg\ge kg$, andBut a mixed tangent two-plane has sectional curvature zero, whereas a factor plane has curvature $\sqrt{6}k$. Thus this manifold is not the round sphere despite having the literal hypotheses.

Again assume the intended trace Ricci curvature bound $\mathrm{Ric}\ge (n-1)kg$, and put $D=\pi /\sqrt{k}$. Compactness from the Bonnet-Myers theorem ensures that the diameter is attained: choose $p,q$ with $d(p,q)=D$. For $0<r<D$, the open balls $B(p,r)$ and $B(q,D-r)$ are disjoint, because a point in both would violate the triangle inequality.

The clearly stated Bishop volume comparison with a positive-curvature model from part (b), applied between radius $r$ and radius $D$, givesHere both radius-$D$ ball volumes equal the total volume, as shown in part (b). The symmetry $s_k(D-t)=s_k(t)$ givesThe two lower bounds therefore sum to $\mathrm{Vol}M$, while disjointness bounds that same sum above by $\mathrm{Vol}M$. Both inequalities must be equalities. In particularLetting $r\downarrow 0$, the local Euclidean volume asymptotic gives one on the left. Thus $\mathrm{Vol}M=v_k(D)$, and the detailed equality proof in part (b) applies. We obtainThis proves maximal diameter rigidity from disjoint comparison balls directly, including the required equality argument.

The literal trace interpretation again has a counterexample. On $M=S^2(r)\times S^2(r)$ with $r^2=1/(2k)$, the Ricci curvature equals $2kg\ge kg$. For the product Riemannian metric, distances satisfy $d^2=d_1^2+d_2^2$: every product curve has length at least $\sqrt{d_1^2+d_2^2}$ by the integral triangle inequality, and a pair of constant-speed minimizing factor geodesics attains it. ConsequentlyThe mixed sectional curvature is zero, so this manifold is not $S_k^4$. The stronger bound, or normalized Ricci curvature convention, is essential.

For a connected smooth Riemannian manifold $(M,g)$ without boundary, let $I(M)$ be the group of all global diffeomorphisms $f$ with $f^{\ast }g=g$, endowed with its compact-open topology. Equivalently these maps preserve the intrinsic distance; the Myers-Steenrod theorem supplies smoothness of distance-preserving bijections and makes $I(M)$ a finite-dimensional Lie group acting smoothly. The group is metric-dependent: a topological manifold can carry a very symmetric metric or one with no continuous symmetries. Connectedness matters here. With infinitely many identical disconnected components, permutations of those components can produce a group which is not a finite-dimensional Lie group.

An isometry is determined by its value and derivative at one point. To prove this, suppose $f(p)=p$ and $d{f}_p=I$. Preservation of the Levi-Civita connection giveswhere the exponential map is defined locally. Thus $f$ is the identity in a normal neighborhood of $p$. The set where its value and derivative agree with the identity is both closed and, by the same argument, open. Connectedness makes it all of $M$. No completeness is needed for this determination argument.

The stabilizer $I(M{)}_p$ consequently acts faithfully and orthogonally on $T_{p}M$, so embeds in $O(n)$. More generally an isometry takes a chosen orthonormal frame at $p$ to one at its image point, providing an injective map into the orthonormal frame bundle. It is a smooth embedding, and the full isometry action is proper. The orbit-stabilizer dimension formula therefore givesIsotropy is compact. If $M$ is compact then $I(M)$ itself is compact, by the same frame description or Arzela-Ascoli applied to isometries and their inverses. Properness also means that the quotient by isometries has well-controlled local orbit structure, even though orbit dimensions may vary.

The infinitesimal equation is the Killing equation. Differentiating $f_t^{\ast }g=g$ at zero givesConversely a vector field satisfying this equation has local flows preserving the metric. A global one-parameter subgroup requires the field to be complete. On a geodesically complete manifold every Killing field is complete: its length is constant along each of its own trajectories, so a finite-time trajectory has bounded speed and stays in a closed bounded ball. The Hopf-Rinow theorem makes that ball compact, and the ordinary differential equation can then be continued. The caveat is real: on the Euclidean interval $(0,1)$, the field ${\partial }_x$ is Killing but its translations are not globally defined for all time. This illustrates Killing fields on incomplete manifolds need not generate global isometries.

The Killing transport identities determine a Killing field from $X(p)$ and the skew-adjoint map $(\nabla X{)}_p$. Along a geodesic the field is a Jacobi field, so uniqueness for its second-order differential equation propagates these initial data. They have $n+n(n-1)/2$ components, reproducing the dimension bound infinitesimally. The important distinction on an incomplete manifold is between all local Killing solutions and the Lie algebra of complete global isometry flows.

In dimension three the bound is six, and rotational isotropy gives stronger information. The Lie subalgebras of $\mathfrak{so}(3)$ have dimensions zero, one or three: identifying the bracket with the cross product, a two-plane is not closed under brackets. Suppose $\dim I(M)=5$. Since an orbit has dimension at most three, its stabilizer would have dimension at least two, and hence exactly three. Its connected isotropy acts as all of $SO(3)$ on the tangent space. But the orbit tangent space is invariant under that action and would have dimension $5-3=2$, impossible because the standard rotational representation is irreducible. Therefore dimension five cannot occur.

If the dimension is six, each orbit has dimension three and each stabilizer dimension three. All orbits are open, so connectedness makes the action transitive. At each point the isotropy is transitive on tangent two-planes, forcing sectional curvature to be independent of the plane; transitivity makes its value independent of the point. Thus a maximally symmetric connected three-manifold has constant sectional curvature. Homogeneous Riemannian manifolds are complete, so its simply connected cover is the round sphere, Euclidean space or hyperbolic space after normalization. Conversely these three simply connected metrics attain dimension six:Here $O^{+}(3,1)$ preserves the chosen sheet of the hyperboloid; it includes orientation-reversing hyperbolic isometries. Constant curvature alone does not imply that a quotient retains six-dimensional global symmetry.

Dimension four likewise forces transitivity: stabilizer dimension three would give a one-dimensional invariant orbit tangent space, again impossible for full rotational isotropy. The stabilizer thus has dimension one and the orbits are three-dimensional. Typical four-dimensional isometry groups occur for $S^2\times \mathbb{R}$ and ${\mathbb{H}}^2\times \mathbb{R}$, where the factor groups have dimensions three and one. The usual Nil metric has three translation symmetries and one rotational isotropy symmetry. In centred Heisenberg coordinates it iswhich exhibits the horizontal rotations. Its distinct vertical and horizontal Ricci eigenvalues bound continuous isotropy by that rotation group, so its full isometry dimension is four. Standard metrics on $\widetilde{S{L}_2(\mathbb{R})}$ and nonround Berger spheres also have dimension four. The standard Sol geometry has a three-dimensional translation group and only finite isotropy, giving dimension three. These examples show how a preferred vertical direction reduces rotational symmetry.

The isometry dimensions in dimension three are precisely $0,1,2,3,4,6$, all realizable. A compact hyperbolic three-manifold gives zero; the product of a circle and a closed hyperbolic surface gives one. For dimension two take $T^2\times (0,1)$ withIts scalar curvature is the constant $-14$, but its Ricci eigenvalues are the distinct constants $-3,-6,-5$ in the two horizontal directions and the vertical direction. Hence connected isotropy is trivial and possible isometries preserve these line distributions. A local isometry preserving them has $t^{\prime }=t+c$ and would require constant dilations $e^{-c}$ and $e^{-2c}$ of the two circle coordinates. A global circle diffeomorphism with such a constant dilation has degree $\pm 1$, forcing $c=0$. The identity component is therefore exactly the two circle translations. A flat three-torus gives dimension three; $S^2\times \mathbb{R}$ gives four; a simply connected constant-curvature model gives six. This construction distinguishes local homogeneity from the smaller global isometry group.

For quotients, let a complete simply connected Riemannian manifold $X$ cover $M=X/\Gamma$. Every isometry of $M$ lifts to an isometry of $X$ which normalizes the deck transformation group. Conversely, every normalizing isometry descends. Two lifts differ by a deck transformation, giving the isometry group of a Riemannian quotient formulaThus one should not replace the isometry group of a quotient by that of its model geometry. For a flat torus, for example,The second factor is finite, whereas translations give the three-dimensional identity component.

There is a useful curvature test for compact manifolds. Contracting and differentiating the Killing equation gives ${\nabla }^{\ast }\nabla X=\mathrm{Ric}(X)$, with the nonnegative rough-Laplacian convention. Integration yields the integrated Bochner identity for Killing fieldsIf Ricci curvature is negative definite, no nonzero Killing field exists. The compact isometry group then has dimension zero and is finite. This explains why closed hyperbolic manifolds have finite global symmetry, despite the six-dimensional isometry group of their universal cover. Compactness is essential to this conclusion.

For a compact flat Riemannian manifold, the same identity makes every Killing field parallel. Parallel fields are exactly the vectors fixed by linear holonomy. ThereforeApplied to the six flat types of Question 1, this gives dimensions three for the torus, one for each of the four nontrivial cyclic types, and zero for the Hantzsche-Wendt manifold. These are identity-component dimensions; disconnected finite symmetries can still be present. Local curvature symmetry, global deck-group constraints, and completeness thus play different roles in determining the isometry group.

Let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. A real semisimple Lie algebra has nondegenerate $B$, and its invariance gives $B([X,Y],Z)=B(X,[Y,Z])$. Translating $B$ defines a bi-invariant pseudo-Riemannian metric $g=B$ on the Lie group.

For left-invariant vector fields, all derivatives of their pairwise inner products vanish. The Koszul formula reduces toNondegeneracy gives ${\nabla }_{X}Y=[X,Y]/2$. With $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, the Jacobi identity now givesTaking the trace of $X\mapsto R(X,Y)Z=-\frac{1}{4}\mathrm{ad}Z\mathrm{ad}Y(X)$ yields the Ricci tensorThus the Killing-form Einstein metric makes the group an Einstein manifold, allowing an indefinite metric. Its scalar curvature in this convention is $-\dim G/4$.

For a connected compact semisimple Lie group, $B$ is negative definite: the adjoint operators are skew-adjoint for an invariant positive inner product, so $B(X,X)=\mathrm{tr}((\mathrm{ad}X{)}^2)<0$ for $X\ne 0$. The conventional positive Riemannian metric is instead $g_{+}=-B$. Its Ricci tensor is $g_{+}/4$, and its sectional curvature isIt can vanish on commuting two-planes; positivity is not automatic in every direction. With the literal negative metric $B$, the corresponding sectional-curvature signs are reversed.

For a noncompact real semisimple group, a Cartan decomposition $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}$ has $B$ negative on $\mathfrak{k}$ and positive on $\mathfrak{p}$. The Killing metric is indefinite and cannot be made positive by an overall sign. Its Einstein constant remains $-1/4$ for $g=B$, but that does not assert a uniform sign for sectional curvature of all nondegenerate planes. For connected semisimple groups, negative-definite $B$ characterizes the compact case. Disconnected groups require a separate condition on the number of components before compactness follows.

Suppose, for a contradiction, that $X=M\times \mathbb{R}$ has a complete metric with zero Ricci curvature. Since $M$ is closed and connected, the complement of $M\times [-A,A]$ has two unbounded components in any complete metric on $X$: their closures are noncompact, while closed bounded sets are compact by the Hopf-Rinow theorem. Thus $X$ is disconnected at infinity, and part (a) supplies a line. The Cheeger-Gromoll splitting theorem gives an isometryThe Riemannian product curvature formula makes $N$ complete and Ricci-flat. A three-dimensional Ricci-flat Riemannian manifold is flat: for any orthonormal triple, the three trace equations are ${\mathrm{Ric}}_{11}=K_{12}+K_{13}$ and their cyclic companions. Solving gives $K_{ij}={\mathrm{Ric}}_{ii}+{\mathrm{Ric}}_{jj}-\mathrm{Scal}/2=0$. Every plane can occur in such a triple, so every sectional curvature vanishes, and hence so does the Riemann curvature tensor.

The complete simply connected universal cover of $N$ is therefore Euclidean ${\mathbb{R}}^3$, and that of $X$ is Euclidean ${\mathbb{R}}^4$. But the topological universal cover of $X$ is $\tilde{M}\times \mathbb{R}$. Its deformation retraction onto $\tilde{M}\times \{0\}$ shows that contractibility of this cover would imply contractibility of $\tilde{M}$, contrary to the hypothesis. Therefore no complete Ricci-flat metric exists on $M\times \mathbb{R}$. Notice that no product assumption was made about the hypothetical original metric.

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.

An orthonormal coframe writes a metric as $g={\eta }_{ab}{\theta }^a\otimes {\theta }^b$ with constant diagonal signature matrix $\eta$. In Cartan's first structure equation, solveThese are the torsion-free and metric-compatibility conditions and determine the Levi-Civita connection. Then Cartan's second structure equation givesContract $R^a{}_{bad}$ to obtain the Ricci tensor and then contract again for the scalar curvature. This fixes the curvature convention; reversing the definition of the Riemann curvature tensor reverses the resulting curvature signs.

On $z>0$, choose ${\theta }^0=dt/z$, ${\theta }^1=dx/z$, ${\theta }^2=dy/z$, ${\theta }^3=dz/z$, with $\eta =\mathrm{diag}(-1,1,1,1)$. For $a=0,1,2$,A torsion-free metric connection therefore hasFor example, ${\Omega }^a{}_3=-d{\theta }^a=-{\theta }^a\wedge {\theta }^3$, while for $a,b<3$,Metric antisymmetry supplies the remaining components, giving uniformlyThus the metric has constant sectional curvature $-1$. In four dimensions,It is an Einstein manifold, with Einstein constant $-3$ in the stated curvature convention.

These are Poincare coordinates on anti-de Sitter spacetime of unit radius. Constant curvature gives the maximal ten-dimensional local isometry algebra $\mathfrak{so}(3,2)$. This can also be checked directly: set $u=(t,x,y)$, ${\eta }_{ab}=\mathrm{diag}(-1,1,1)$ and $u_a={\eta }_{ab}{u}^b$. The ten independent Killing vector fields areTranslations and Lorentz transformations leave the numerator and $z$ unchanged. A dilation rescales numerator and denominator equally. For $K_a$, direct differentiation gives ${\mathcal{L}}_{K_a}{\eta }^{(4)}=4u_a{\eta }^{(4)}$ and $K_a(z)=2u_{a}z$, so the conformal factor cancels and ${\mathcal{L}}_{K_a}g=0$.

Hence the maximally extended Anti-de Sitter spacetime has connected isometry group locally $S{O}_0(3,2)$, with the appropriate covering group if one unwraps its time coordinate. The displayed coordinates cover only a patch. The ten local generators do not all give globally complete flows preserving that patch; for example special conformal flows can cross its horizon. The manifest complete patch symmetries include the boundary Poincare transformations and positive dilations. This distinguishes local maximal symmetry from the global isometries of a chosen coordinate domain.

Since $Y$ is lifted from $N$, its covariant derivative ${\nabla }_{Y}Y$ is also lifted from $N$. Part (ii) therefore gives ${\nabla }_X({\nabla }_{Y}Y)=0$. It also gives ${\nabla }_{X}Y=0$, and part (i) gives $[X,Y]=0$. Substituting into the Riemannian curvature two-form definition yieldsWith the four-slot convention fixed above, this provesFor nonzero lifted vectors from different factors they are orthogonal, so the sectional-curvature denominator is positive. Hence the corresponding mixed plane has sectional curvature zero, providing the mixed-plane value in the $S^2\times S^2$ example.

Use the Riemannian curvature two-form conventionand define its four-covariant Riemann curvature tensor by $R(X,Y,Z,W)=g(R(X,Y)W,Z)$. This makes $R(X,Y,X,Y)$ positive on an orthonormal two-plane of the unit sphere. The full algebraic symmetries areTogether with multilinearity, these are antisymmetry in each pair, interchange of pairs, and the first Bianchi identity.

Define the Ricci tensor by the basis-independent contractionfor any orthonormal basis $(e_a)$. The trace is invariant under change of basis, and contraction of the smooth curvature tensor gives a smooth covariant two-tensor. Pair interchange gives $R(e_a,Y,e_a,W)=R(e_a,W,e_a,Y)$, so $\mathrm{Ric}(Y,W)=\mathrm{Ric}(W,Y)$. Thus the Ricci tensor is a symmetric covariant 2-tensor.

The sectional curvature of the two-plane $\sigma =\mathrm{span}(u,v)$ isBoth numerator and denominator multiply by the square of the determinant under a change of basis of $\sigma$, so this is a plane invariant. The directional Ricci curvature of $u\ne 0$ is $\mathrm{Ric}(u,u)/g(u,u)$. For an orthonormal basis beginning with $u/|u|$, it equals the sum of the sectional curvatures of the planes containing that basis vector.

In dimension three, suppose this directional value is $c$ at the point under consideration. For any orthonormal triple $(e_1,e_2,e_3)$, put $K_{ij}=K(\mathrm{span}(e_i,e_j))$. Thenso all three are $c/2$. Every two-plane can be the span of the first two members of such a triple. ConsequentlyThis is a pointwise conclusion, as required; it uses sectional curvatures from Ricci curvature in dimension three and asserts no spatial constancy.

For the product, the canonical identification $T_{(p,q)}(M\times N)=T_{p}M\oplus T_{q}N$ defines the product Riemannian metricIt is smooth in product charts and positive definite, since a nonzero vector has a nonzero component in at least one factor. Those components are orthogonal. The coordinate connection computation in the subparts below givesTaking the trace in orthonormal bases of the two factors therefore gives ${\mathrm{Ric}}_{M\times N}={\mathrm{Ric}}_M\oplus {\mathrm{Ric}}_N$.

For the final example choose the two round unit spheres. Each has sectional curvature one and Ricci tensor equal to its metric. Hence the product has $\mathrm{Ric}=g$, so every directional Ricci curvature is one. A two-plane tangent to one factor has sectional curvature one, whereas a mixed two-plane has sectional curvature zero by (iii). This gives constant Ricci curvature without constant sectional curvature:Both kinds of plane occur at every point, so the sectional curvatures fail to be constant even at a single point.

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.

The divergence of a Riemannian vector field $X$ is the trace of the covariant derivative map $Y\mapsto {\nabla }_{Y}X$:It is characterized by ${\mathcal{L}}_X{\omega }_g=(\mathrm{div}X){\omega }_g$ in the oriented case. The metric-dual vector field $X_{\theta }={\theta }^{♯}$ is defined by $g(X_{\theta },Y)=\theta (Y)$ for every $Y$, using the musical isomorphism.

On a compact oriented manifold without boundary, the permitted exactness assertion and Stokes theorem imply ${\int }_M\mathrm{div}Y{\omega }_g=0$ for every smooth vector field $Y$. For a smooth function $a$, the product rule givesSince $⟨da,\theta ⟩=X_{\theta }(a)$, the formal-adjoint identity yieldsThis holds for all $a$; taking $a=\delta \theta +\mathrm{div}{X}_{\theta }$ proves the pointwise formulaIt agrees with $\delta \theta =-\ast d\ast \theta$, since $\ast \theta ={\iota }_{X_{\theta }}{\omega }_g$ and $d{\iota }_{X_{\theta }}{\omega }_g={\mathcal{L}}_{X_{\theta }}{\omega }_g$.