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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.

Let $p_i:M_1\times M_2\to M_i$ be the projections. The derivative of the two projections identifies the tangent bundle withTake the direct sum of the two pullback connections. This is the product affine connection. In product coordinates $(x^a,y^{\alpha })$, its only nonzero Christoffel symbols are the two blocks inherited from the factors; all mixed coefficients are zero.

To make the connection on arbitrary fields explicit, write $V=V^a{\partial }_{x^a}+V^{\alpha }{\partial }_{y^{\alpha }}$ and $W=W^a{\partial }_{x^a}+W^{\alpha }{\partial }_{y^{\alpha }}$, allowing all coefficients to depend on both variables. ThenThe terms $V(W^a)$ and $V(W^{\alpha })$ include derivatives in both factors. Omitting these cross derivatives for fields with variable coefficients would not define a connection.

The formula is real-bilinear, linear over smooth functions in $V$, and obeys the Leibniz rule in $W$. Under a change of product coordinates each block has the factor's connection transformation law, so the local formulas agree; equivalently the pullback connection construction already guarantees this gluing.

A field $X_i$ on one factor has a canonical lift to the product. Opposite-factor lifts commute, and their mixed covariant derivatives vanish. For such lifted fields the coefficient derivatives reduce to their factor derivatives, givingThese values determine the connection uniquely: lifted coordinate frames span locally, and the connection's product rules determine its action on all their smooth linear combinations.

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.

The product of Riemannian manifolds has metricThe tangent summands are orthogonal, and positivity follows because a nonzero product tangent vector has a nonzero component. Its Levi-Civita connection is the product affine connection of the two factor Levi-Civita connections: mixed torsion is zero and metric compatibility follows from independence of the two coordinate sets.