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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.

In coordinates, write ${\nabla }_{{\partial }_i}{\partial }_{\ell }={\Gamma }_{i\ell }^k{\partial }_k$. The Riemannian curvature two-form is defined by the matriceswith repeated indices summed. It assigns to $X,Y$ the fiberwise endomorphism $R(X,Y)$ whose entries are $R^k{}_{\ell ij}{X}^i{Y}^j$. We can prove both coordinate independence and the required intrinsic identity as follows.

Define an operator on vector fields by $C(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$. It is alternating in $X,Y$. For a smooth function $f$, the identities $[fX,Y]=f[X,Y]-Y(f)X$ and ${\nabla }_Y(fZ)=Y(f)Z+f{\nabla }_{Y}Z$ givethe two terms involving $Y(f){\nabla }_{X}Z$ cancel. Alternation gives linearity over smooth functions in $Y$ as well. Expanding in the final argument yieldsThus $C$ is tensorial in all three arguments, defining a global endomorphism-valued two-form.

For coordinate fields ${\partial }_i,{\partial }_j$, their bracket is zero. Expand $[{\nabla }_{{\partial }_i},{\nabla }_{{\partial }_j}]Z$ using $Z=Z^{\ell }{\partial }_{\ell }$. The second derivatives of $Z^{\ell }$ cancel by equality of mixed partials. The terms containing first derivatives of $Z^{\ell }$ cancel in pairs, and the remaining coefficient of $Z^{\ell }{\partial }_k$ is exactly $R^k{}_{\ell ij}$ above. Hence the coordinate definition equals the global tensor $C$, establishing coordinate independence and provingThis convention for the Riemann curvature tensor is valid for any affine connection; in particular it applies to the Levi-Civita connection.

Use the sign conventionFor the Levi-Civita connection this is the Riemannian curvature two-form, an element of ${\Omega }^2(M;\mathrm{End}(TM))$: $R(X,Y)$ is an endomorphism of the tangent bundle, alternating in $X,Y$, and the expression has tensoriality in all arguments. It is the curvature form of a connection for the tangent-bundle connection.

For the first Bianchi identity, take the cyclic sum in $X,Y,Z$. Torsion-freeness says ${\nabla }_{Y}Z-{\nabla }_{Z}Y=[Y,Z]$, so the double-derivative terms combine to ${\sum }_{\mathrm{cyc}}{\nabla }_X[Y,Z]$. The remaining terms may be cyclically relabelled as $-{\sum }_{\mathrm{cyc}}{\nabla }_{[Y,Z]}X$. Using torsion-freeness once more, followed by the Jacobi identity for vector fields, givesThe Ricci curvature is the tracefor any orthonormal basis. The sectional curvature of the two-plane spanned by independent $u,v$ isThese conventions give positive curvature on a round sphere. In dimension three, put $K_{ij}=K(e_i,e_j)$ and $r_i=\mathrm{Ric}(e_i,e_i)$. Curvature symmetries giveSolving this linear system yields the sectional curvatures from Ricci curvature in dimension three formulaHere $\mathrm{Scal}=r_1+r_2+r_3$ is the scalar curvature. The chosen orthonormal basis need not diagonalize Ricci; its diagonal evaluations already determine these three sectional curvatures.