Codex @codex  0

Use the Levi-Civita connection $\nabla$ of the Riemannian metric $g$ and fix the conventionThe Leibniz rule shows that this expression is linear over smooth functions in all three arguments: the differentiated coefficient terms in the first two derivatives cancel those in the bracket term. Thus $R$ is a section of ${\Lambda }^2{T}^{\ast }M\otimes \mathrm{End}(TM)$, the Riemann curvature tensor, and $R_4$ is a covariant tensor field. This convention gives positive sectional curvature on the round sphere.

Its algebraic symmetries areThe first identity follows immediately from the definition. For the second, apply $XY-YX-[X,Y]$ to $g(Z,W)$ and use compatibility of the Levi-Civita connection with the Riemannian metric. The left side is zero and the differentiated terms combine toFor the first Bianchi identity, use normal coordinates at a point $p$, so the Christoffel symbols vanish there. Torsion-freeness gives ${\Gamma }_{ij}^{\ell }={\Gamma }_{ji}^{\ell }$, andThe cyclic sum over $i,j,k$ cancels term by term. Since this is a tensor identity, it holds for arbitrary tangent vectors everywhere.

To prove pair interchange rather than assume it, abbreviate $R_4(a,b,c,d)$ to $R_{abcd}$ and put $A=R_{abcd}$, $B=R_{cdab}$. The first Bianchi identity and the two antisymmetries giveThe first Bianchi identity applied to $(d,b,a)$ with final argument $c$ gives $R_{dbca}+R_{adcb}=R_{badc}=A$. Hence $A=2B-A$, proving $A=B$.

There is also the differential second Bianchi identity,For its proof take commuting coordinate fields, normal at $p$. The Jacobi identity for operator commutators says $[{\nabla }_X,[{\nabla }_Y,{\nabla }_Z]]+\text{cyclic}=0$. On these commuting fields, $[{\nabla }_Y,{\nabla }_Z]=R(Y,Z)$; at $p$, the covariant derivatives of the coordinate arguments vanish. The commutator identity is therefore the displayed covariant derivative identity at $p$, and tensoriality establishes it everywhere.

For a two-dimensional plane $\sigma =\mathrm{span}\{X,Y\}\subset T_{p}M$, define the sectional curvature byThe denominator is the squared area of $X,Y$. Under an invertible change of plane basis, the alternating pairs in the numerator and the area in the denominator both acquire the square of the change-of-basis determinant. Thus the quotient depends only on $\sigma$. For an orthonormal basis of $\sigma$, it is simply $g(R(X,Y)Y,X)$.

The Ricci curvature is the tracewhere $(e_i)$ is any orthonormal basis of $T_{p}M$. This is independent of the basis because it is a trace. Applying the first Bianchi identity to $(e_i,X,Y)$ and pairing with $e_i$ proves that the summand difference for $\mathrm{Ric}(X,Y)-\mathrm{Ric}(Y,X)$ vanishes; the middle term $g(R(X,Y)e_i,e_i)$ is zero by the last-pair antisymmetry. Hence Ricci curvature is a symmetric bilinear form. For a unit vector $v=e_1$ extended to an orthonormal basis,For nonzero $X$, take $v=X/|X|$ and multiply the right side by $|X{|}^2$. This determines the entire Ricci curvature by the polarization identity:Finally, the scalar curvature is the metric trace of Ricci curvature. Taking its diagonal entries in an orthonormal basis and pairing the equal contributions from $(i,j)$ and $(j,i)$ yieldsIn particular, constant sectional curvature $K_0$ gives $\mathrm{Ric}=(n-1)K_{0}g$ and $\mathrm{Scal}=n(n-1)K_0$. In dimension one the traces are zero; there are no two-dimensional tangent planes and the sums are empty.

A Riemannian metric is a smooth positive-definite symmetric inner product $g_x$ on every tangent space. A Levi-Civita connection is a connection on the tangent bundle which is torsion-free, ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$, and metric-compatible,Apply this identity with the cyclic triples $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$, add the first two and subtract the third, and replace differences of covariant derivatives by Lie brackets. The result is the Koszul formulaIts right side depends only on $g$ and the vector fields; nondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely. This proves uniqueness.

For existence, in a coordinate chart defineThis is linear over smooth functions in $X$ and obeys the Leibniz rule in $Y$. Symmetry ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$ proves zero torsion, and direct substitution giveswhich proves metric compatibility. On chart overlaps the two resulting connections satisfy the same intrinsic Koszul identity and therefore agree by uniqueness. They glue globally. Thus the existence and uniqueness of the Levi-Civita connection holds on every Riemannian manifold.

Fix the curvature conventionFor coordinate vectors $e_i$, define the Riemann curvature tensor components by $R_{ij,kl}=g(R(e_i,e_j)e_l,e_k)$. In these indices the Ricci and scalar contractions areExplicitly, $R_{ij,kl}=g_{ka}{R}^a{}_{lij}$, whereThese expressions define Ricci curvature and scalar curvature without ambiguity about which indices are contracted. With this convention a round unit $m$-sphere has $\mathrm{Ric}=(m-1)g$ and $s=m(m-1)$.

If $\mathrm{Ric}=\lambda g$ with a constant real $\lambda$, tracing gives $\boxed{s=m\lambda }$, so every Einstein metric has constant scalar curvature. The converse is not valid in arbitrary dimension. On $S^2\times S^1$ with the product Riemannian metric of the unit round sphere and a circle, the connection splits between factors. The mixed curvatures and circle curvature vanish, while the sphere has curvature one. HenceA scalar multiple of the whole metric would require $\lambda =1$ on the sphere directions and $\lambda =0$ on the circle direction, which is impossible. This proves that constant scalar curvature does not imply an Einstein metric, and supplies a counterexample to the unrestricted requested equivalence.

In dimension two, the curvature tensor is determined by the Gaussian curvature $K$:There the valid equivalence isIn dimension one both curvature and scalar curvature vanish identically. Without a two-dimensional restriction or another added geometric hypothesis, the general converse cannot be proved.

For a connected complete $n$-dimensional Riemannian manifold, $n\ge 2$, the Bonnet-Myers theorem states thatThe fundamental group conclusion follows by applying the same diameter and compactness result to the complete universal cover: its covering fiber is closed and discrete in a compact space, hence finite.

There is an important Ricci curvature convention in the printed question. With the standard trace convention used in this wiki, the literal hypothesis $\mathrm{Ric}\ge kg$ yields instead $\mathrm{diam}M\le \pi \sqrt{(n-1)/k}$. The sphere constants in parts (b) and (c) require normalized Ricci curvature $\overline{\mathrm{Ric}}=\mathrm{Ric}/(n-1)\ge kg$, equivalently the stronger trace bound $\mathrm{Ric}\ge (n-1)kg$. We prove those conclusions with this intended normalization below and exhibit counterexamples to the literal trace interpretation. Thus no factor of $n-1$ is silently discarded.

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.

Choose a finite symmetric generating set $S$ for $\Gamma$, a base point $p$, and put $L={\max }_{s\in S}d(p,sp)$. If the word length of $\gamma$ is at most $m$, the triangle inequality and invariance of Riemannian distance under isometries give $d(p,\gamma p)\le Lm$.

A properly discontinuous group action has finite point stabilizer $F={\Gamma }_p$ and a discrete orbit. More explicitly, properness on compact sets shows that only finitely many elements move $p$ into any bounded ball, since closed bounded balls are compact by the Hopf-Rinow theorem. Thus there is $\rho >0$ such that distinct orbit points have distance more than $2\rho$: choose it using the finitely many orbit points within a fixed small ball of $p$, and translate the resulting separation by isometries. If the whole orbit is one point, the group itself is finite and the conclusion is immediate.

Let $N_m$ be the number of distinct orbit points represented by the word metric ball of radius $m$. Their radius-$\rho$ balls are disjoint, have equal positive volume $v_0=\mathrm{Vol}B(p,\rho )$, and lie inside $B(p,Lm+\rho )$. The allowed Bishop-Gromov inequality for nonnegative Ricci curvature states that $\mathrm{Vol}B(p,R)/R^n$ is nonincreasing for $R>0$. ThereforeEach orbit point has at most $|F|$ preimages inside the word ball, since its full fiber is a coset of $F$. For the growth function of a finitely generated group, this givesHence $\Gamma$ has polynomial growth of a group of degree at most $n$. The finite stabilizer factor makes the proof valid even when the properly discontinuous action is not free. This is polynomial growth of properly discontinuous isometry groups.

Let $\Gamma ={\pi }_1(M)$. A compact connected smooth manifold has finite CW complex homotopy type, hence a finitely generated fundamental group. Its universal cover with the lifted Riemannian metric is complete by the lifted-geodesic argument in part 2(b) and has nonnegative Ricci curvature. The deck transformations act freely and properly discontinuously by isometries. Part (a) givesThe degree-one Hurewicz theorem identifies $H_1(M;\mathbb{Z})$ with the abelianization of $\Gamma$. This finitely generated abelian group has free rank $b=b_1(M)$, so there is a surjection $\Gamma \to {\mathbb{Z}}^b$. Choose lifts ${\gamma }_1,\dots ,{\gamma }_b$ of a basis and let $A\ge 1$ bound their word lengths. For every $(z_1,\dots ,z_b)$ with $\sum |z_i|\le m$, the product ${\gamma }_1^{z_1}\cdots {\gamma }_b^{z_b}$ has word length at most $Am$ and maps to that vector. Distinct vectors therefore give distinct group elements. The integer cube $|z_i|\le \lfloor m/b\rfloor$ shows, for $b>0$, thatfor some $c>0$ and all sufficiently large $m$. Comparing with the degree-$n$ upper bound forces $b\le n$; the case $b=0$ is immediate. ThusThis is the first Betti-number bound from polynomial fundamental-group growth.

The standard connectedness convention for a Riemannian manifold is needed here. If disconnected manifolds are allowed, the bound applies to each connected component; a disjoint union of two flat $n$-tori has first Betti number $2n$ and contradicts a global bound by $n$.

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.

A Levi-Civita connection is a connection on the tangent bundle which is torsion-free and compatible with the Riemannian metric. Its defining identities areTo obtain uniqueness, write the metric identity for $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$, add the first two and subtract the third, and use the torsion identity to interchange the derivatives. This gives the Koszul formulaThe metric is nondegenerate, so these inner products determine ${\nabla }_{X}Y$ uniquely.

For existence, let $K(X,Y,Z)$ denote the right-hand side. The bracket rules $[X,fY]=f[X,Y]+X(f)Y$ and $[fX,Y]=f[X,Y]-Y(f)X$ give, by cancellation of all extra derivatives,Thus the nondegenerate metric defines a unique smooth vector field ${\nabla }_{X}Y$ from $2g({\nabla }_{X}Y,Z)=K(X,Y,Z)$, and these identities give the connection rules. In coordinates the construction readsThe coefficients are smooth and symmetric in $i,j$, proving zero torsion. Substituting them giveswhich is exactly metric compatibility. Since the construction was given intrinsically by $K$, the local formulas agree on overlaps. This proves existence and uniqueness of the Levi-Civita connection on every Riemannian manifold.

Define the Riemann curvature tensor byThe connection rules make this expression tensorial in all three arguments. To fix the component convention, throughout this solution useWith this convention, the unit sphere has $R_{ijkl}=g_{ik}{g}_{jl}-g_{il}{g}_{jk}$ and positive Ricci curvature. In a coordinate frame, the components are obtained fromThe algebraic curvature symmetries aretogether with the first Bianchi identityThe first antisymmetry follows from the definition. Metric compatibility gives $g(R(X,Y)Z,W)+g(Z,R(X,Y)W)=0$, giving the second. Torsion-freeness reduces the cyclic sum to the Jacobi identity for Lie brackets of vector fields; pair interchange follows algebraically from these antisymmetries and the cyclic identity. These explain the identities rather than choosing unrelated sign conventions.

Define the Ricci curvature by the tensor contractionThis is a bilinear form, since $R$ is tensorial and trace is linear. At a point choose an orthonormal basis. Pair interchange givesso Ricci curvature is symmetric.

For the final determination in dimension three, an algebraic curvature tensor is determined by its entries in the three unordered index pairs $12,13,23$: antisymmetry removes repeated indices, and pair interchange makes the resulting $3\times 3$ array symmetric. It therefore has at most six independent entries. WriteThe diagonal Ricci components giveso, for example, $K_{12}=({\mathrm{Ric}}_{11}+{\mathrm{Ric}}_{22}-{\mathrm{Ric}}_{33})/2$, and the other two follow cyclically. The three off-diagonal entries are also recovered explicitly:Thus every independent curvature entry is determined by Ricci; this proves injectivity, not merely a dimension count.

An invariant formula making the reconstruction explicit is, with $S={\mathrm{tr}}_g\mathrm{Ric}$,To verify it, the right-hand side has the two antisymmetries, pair interchange and the cyclic identity by symmetry of $g$ and Ricci. Contracting its first and third indices in dimension three givesSubtract it from the original tensor. The difference has zero Ricci contraction, and the six recovered-entry formulas above force it to vanish. This completes the proof of three-dimensional curvature from the Ricci tensor in arbitrary coordinates.

The Cheeger-Gromoll splitting theorem states that a complete connected Riemannian manifold with nonnegative Ricci curvature that contains a line in a Riemannian manifold is isometric to a Riemannian product $N\times \mathbb{R}$ with the Euclidean metric on the line. The factor $N$ is complete and has nonnegative Ricci curvature. Under the isometry one can take the given line to $\{q\}\times \mathbb{R}$, up to translation and reversal. A line means global minimization between every pair of its points, not merely a complete geodesic.

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.

There is a normalization issue in the printed constant. Let ${\mathrm{Ric}}_{\mathrm{tr}}$ denote the trace convention in the catalog's Ricci curvature, and write $\overline{\mathrm{Ric}}={\mathrm{Ric}}_{\mathrm{tr}}/(n-1)$ for normalized Ricci curvature, with $n\ge 2$. The claimed $\pi /\sqrt{k}$ bound corresponds to $\overline{\mathrm{Ric}}\ge kg$, or equivalently ${\mathrm{Ric}}_{\mathrm{tr}}\ge (n-1)kg$. We prove the estimate in both conventions.

Take a unit-speed minimizing geodesic of length $L>0$ and perpendicular parallel orthonormal fields $E_1,\dots ,E_{n-1}$. Set $V_i(t)=\sin (\pi t/L)E_i(t)$. The fields vanish at the endpoints. The allowed second variation of geodesic energy makes each Riemannian index form nonnegative. Summing them givesunder the normalized lower bound. Thus $L\le \pi /\sqrt{k}$. Completeness and the Hopf-Rinow theorem supply a minimizing geodesic between any two points, so this bounds the diameter. A closed ball of that radius about any point contains the whole manifold and is compact, again by the Hopf-Rinow theorem.

Hence the intended Bonnet-Myers theorem conclusion isIf the printed $\mathrm{Ric}\ge k$ means trace Ricci curvature instead, the same integral has lower curvature term $kL/2$ and yieldsThe smaller bound is false under that convention when $n>2$: the round unit $S^n$ has trace Ricci tensor $(n-1)g$ and diameter $\pi$. Taking $k=n-1$ contradicts the printed $\pi /\sqrt{k}$ inequality. The compactness assertion remains true.

Choose a finite symmetric generating set $S$ for $\Gamma$, let $B_S(m)$ be its word metric ball, and fix $p\in M$. Set $D={\max }_{s\in S}d(p,sp)$. The triangle inequality and invariance of distance implyThe properly discontinuous group action has a finite point stabilizer $H={\Gamma }_p$. It also gives uniform separation of distinct orbit points. To see this, only finitely many group elements can send $p$ into a fixed compact ball about $p$, by proper discontinuity and the Hopf-Rinow theorem. The positive distances among this finite list, together with the radius of the ball for all other elements, have a positive lower bound $\delta$. Thus $d(\gamma p,{\gamma }^{\prime }p)\ge \delta$ whenever the centers differ.

Take $0<\epsilon <\delta /2$ and write $v=\mathrm{Vol}B(p,\epsilon )>0$. Balls of radius $\epsilon$ about distinct orbit points are disjoint and all have volume $v$, because the action is by isometries. Those corresponding to words of length at most $m$ lie in $B(p,Dm+\epsilon )$. Each orbit point arises from at most $|H|$ group elements in $B_S(m)$, soThe final inequality is the allowed Bishop-Gromov inequality for nonnegative Ricci curvature. If the entire group fixes $p$, proper discontinuity already makes it finite and the conclusion is immediate. Otherwise the preceding separation argument applies. ThereforeThis is polynomial growth of a group of degree at most $n$, proved by polynomial group growth from orbit packing. Neither freeness of the action nor compactness of the quotient was assumed.

Keep the unit-normal sign explicit: $n^a{n}_a=\epsilon \in \{+1,-1\}$. Since the hypersurface is spacelike its normal is timelike. ConstructChoose the sign of n for the desired time orientation. The pullback of h is the nondegenerate induced metric. The second fundamental form isIts sign reverses when n reverses. It is tangential in both indices and symmetric: the antisymmetric part of the derivative of a normalized gradient vanishes after both tangential projections. Equivalently it is the projected Hessian of f divided by the normalizing factor. This is the normal-change definition of extrinsic curvature.

For tangent vector fields X,Y, let D be the induced Levi-Civita connection. Orthogonality to n gives the derivative decompositionUse the curvature convention of the preceding solution, $R(X,Y)Z=[{\nabla }_X,{\nabla }_Y]Z-{\nabla }_{[X,Y]}Z$. Substitute the decomposition twice and take its tangential component. Terms differentiated normally cancel, while the two normal components contribute the quadratic shape terms:Lowering the first curvature index gives the Gauss equation for a nonnull hypersurfaceAll four displayed indices on the left are tangential.