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The product Riemannian metric of two round unit spheres has Ricci curvature equal to its metric, since the factor Ricci tensors equal their metrics and the product connection splits. Its directional Ricci curvatures are all one. A two-plane tangent to a single sphere has sectional curvature one, while a mixed plane has sectional curvature zero. This four-dimensional Einstein manifold therefore need not have constant sectional curvature.

The product Riemannian metric of the round unit two-sphere and a circle has Ricci curvature equal to $g_{S^2}$ on the sphere directions and zero on the circle direction. Its scalar curvature is constantly two, but no single scalar multiple of the whole metric equals that Ricci tensor. In two dimensions the identity $\mathrm{Ric}=\mathrm{Scal}g/2$ does make constant scalar curvature equivalent to an Einstein metric.

For a product affine connection, its curvature satisfiesExpand the defining derivative commutator on lifted vector fields. Opposite-factor fields commute and have zero mixed covariant derivatives, leaving exactly the two factor curvatures. Tensoriality gives the formula for arbitrary tangent vectors. Thus the product connection is flat if the two factors are flat; for a product Riemannian metric this applies to its Levi-Civita connection.

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.

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.

The product Riemannian metric has block matrixIn the given Christoffel symbols formula, every mixed coefficient vanishes. Off-diagonal metric entries are zero, ${\partial }_{y^{\alpha }}{g}_{ij}=0$, and ${\partial }_{x^i}{g}_{\alpha \beta }=0$. These facts give ${\Gamma }_{i\alpha }^A=0$ for every output index $A$, as well as ${\Gamma }_{ij}^{\alpha }=0$ and ${\Gamma }_{\alpha \beta }^i=0$. The remaining pure blocks are precisely the Christoffel symbols of the respective factor metrics.

For the lifted fields of (i), ${\partial }_{x^i}{Y}^{\alpha }=0$. ConsequentlyThus $\boxed{{\nabla }_{X}Y=0}$. The same calculation gives ${\nabla }_{Y}X=0$; differentiating two fields from one factor gives the lifted factor covariant derivative. This identifies the product Levi-Civita connection with the product affine connection and proves the curvature splitting used in the root solution.

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.

A Levi-Civita connection on a Riemannian manifold is a covariant derivative $D$ that is a torsion-free connection, $D_{X}Y-D_{Y}X=[X,Y]$, and a metric connection, $Xg(Y,Z)=g(D_{X}Y,Z)+g(Y,D_{X}Z)$. A covariant derivative is $C^{\infty }(M)$-linear in $X$, real linear in $Y$, and obeys $D_X(fY)=X(f)Y+f{D}_{X}Y$.

Combining metric compatibility for the cyclic triples and eliminating reversed derivatives with the torsion identity gives the Koszul formulaSince $g$ is nondegenerate, this determines $D_{X}Y$ uniquely. For existence, use the right-hand side to define $D_{X}Y$: expansion of the Lie bracket of vector fields shows it is $C^{\infty }(M)$-linear in $Z$, hence defines a smooth one-form, which the musical isomorphism converts to a smooth vector field. The same expansion shows $D_{fX}Y=f{D}_{X}Y$ and $D_X(fY)=X(f)Y+f{D}_{X}Y$. Subtracting the formula with $X,Y$ exchanged gives torsion zero; adding its versions for $Y,Z$ exchanged gives metric compatibility. Thus it is a Levi-Civita connection. Equivalently its coefficients in a manifold chart are the Christoffel symbolsThe intrinsically defined Koszul formula ensures these local expressions fit together. This proves the fundamental theorem of Riemannian geometry.

The product Riemannian metric on $M\times N$ isThe two factor tangent spaces are orthogonal, and the sum is positive definite. Lift $X$ from $M$ and $Y$ from $N$. We check $D_{X}Y$ against lifted local frame fields from both factors, which together span each product tangent space.

If $Z$ is lifted from $M$, then $g(Y,Z)=g(X,Y)=0$, while $g(Z,X)$ depends only on $m$, so $Yg(Z,X)=0$. Also $[Y,Z]=[X,Y]=0$ and $[Z,X]$ is horizontal, hence orthogonal to $Y$. Every term in the Koszul formula is zero. If $W$ is lifted from $N$, $g(W,X)=g(X,Y)=0$ and $g(Y,W)$ depends only on $n$, so $Xg(Y,W)=0$. Now $[W,X]=[X,Y]=0$ and $[Y,W]$ is vertical, hence orthogonal to $X$. Again every term is zero. Thus $D_{X}Y$ is orthogonal to both spanning frame families, and positive definiteness yieldsThis calculation uses the lifts' independence of the other factor coordinates; a vector field with varying coefficients in those coordinates can have a nonzero mixed derivative.

The product Riemannian metric isIt is smooth and symmetric. If $(u_1,u_2)\ne 0$, at least one component is nonzero and contributes a strictly positive squared norm; the other contributes a nonnegative term. Thus $g$ is positive definite.

Let the factor connections be Levi-Civita connections. The torsion form of the product affine connection vanishes on two lifted fields from one factor because it equals the factor torsion; on opposite-factor fields it vanishes because both covariant derivatives and the Lie bracket of vector fields vanish. By tensoriality, torsion vanishes for all fields.

Check metric compatibility on lifted fields as well. When the two paired fields are from one factor, their metric pairing depends only on that factor. Differentiation along that factor gives the usual factor compatibility identity, and differentiation along the other factor gives zero on both sides. When the paired fields are from different factors, their pairing is identically zero and their derivatives stay in their original summands, so both sides are again zero. Since $\nabla g$ has tensoriality, these checks imply $\nabla g=0$ globally.

The existence and uniqueness of the Levi-Civita connection therefore givesEquivalently, applying the Christoffel symbol formula to the block metric gives the factor coefficients and zero mixed coefficients, because each factor metric is independent of the other coordinates.

At $(p,q)$, choose neighborhoods on which the two factors admit local isometries to open subsets of Euclidean space. Their product is a local isometry of the product Riemannian metric to the standard Euclidean metric on ${\mathbb{R}}^{2n}$. In these product coordinates the metric matrix is constant, so every Christoffel symbol of the Levi-Civita connection is zero. Its Riemann curvature tensor therefore vanishes at every point of that neighborhood. Curvature has tensoriality, so this is independent of the chosen coordinates; the neighborhoods cover the product.

The splitting provides another useful direct check. Expanding the definition on lifted fields, using zero mixed derivatives and commuting opposite-factor fields, gives the curvature splitting for a product connectionTensoriality extends this identity from lifts to arbitrary tangent vectors at a point. Each factor curvature is zero under the stated local-isometry hypothesis. ConsequentlyThis is a local conclusion about the connection; no global identification of either factor or their product with Euclidean space is required.