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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.

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.

For the Levi-Civita connection, the curvature form of a connection is the endomorphism-valued two-formIt has tensoriality in all three arguments. Its first Bianchi identity follows from torsion-freeness: the cyclic sum is ${\sum }_{\mathrm{cyc}}({\nabla }_X[Y,Z]-{\nabla }_{[Y,Z]}X)={\sum }_{\mathrm{cyc}}[X,[Y,Z]]=0$ by the Jacobi identity.