Curvature of a Riemannian product

Use [F2] to define, at $(p,q)$, $k_{(p,q)}((u_1,u_2),(v_1,v_2)):=g_p(u_1,v_1)+h_q(u_2,v_2)\mathrm{.}$ In the product chart supplied by [F1], its matrix is $\mathrm{diag}((g_{ij}(x)),(h_{\alpha \beta }(y)))$. Its entries are smooth and it is symmetric; moreover $k((u_1,u_2),(u_1,u_2))=g(u_1,u_1)+h(u_2,u_2)>0$ unless both components vanish. Thus [F3] proves that $k={\pi }_M^{\ast }g+{\pi }_N^{\ast }h$ is a Riemannian metric. Its inverse matrix is $\mathrm{diag}((g^{ij}(x)),(h^{\alpha \beta }(y)))$; all mixed entries vanish, the first block has no $y$-dependence, and the second has no $x$-dependence.

Applying [F4] to the blocks in step 1.1 gives ${\mathrm{\Gamma }}^k{}_{ij}={\mathrm{\Gamma }}^k{}_{ij}(g)$ and ${\mathrm{\Gamma }}^{\gamma }{}_{\alpha \beta }={\mathrm{\Gamma }}^{\gamma }{}_{\alpha \beta }(h)$, while every symbol whose indices meet both blocks is zero. For instance, ${\mathrm{\Gamma }}^k{}_{i\beta }=\frac{1}{2}{g}^{k\mathrm{\ell }}({\mathrm{\partial }}_i{G}_{\beta \mathrm{\ell }}+{\mathrm{\partial }}_{\beta }{g}_{i\mathrm{\ell }}-{\mathrm{\partial }}_{\mathrm{\ell }}{G}_{i\beta })=0$ and ${\mathrm{\Gamma }}^k{}_{\alpha \beta }=-\frac{1}{2}{g}^{k\mathrm{\ell }}{\mathrm{\partial }}_{\mathrm{\ell }}{h}_{\alpha \beta }=0$; exchanging the factors covers an upper $N$-index.

A factor lift has coefficients depending only on that factor. Expanding covariant derivatives with the connection laws in [F4] and the symbols from step 2.1 gives ${\mathrm{\nabla }}_{\tilde{X}}\tilde{Y}=\widetilde{{\mathrm{\nabla }}_X^{M}Y}$ and ${\mathrm{\nabla }}_{\tilde{U}}\tilde{V}=\widetilde{{\mathrm{\nabla }}_U^{N}V}$. In a cross derivative, the differentiated lift's coefficients are constant in the differentiating factor and all relevant mixed symbols vanish, so ${\mathrm{\nabla }}_{\tilde{X}}\tilde{V}={\mathrm{\nabla }}_{\tilde{V}}\tilde{X}=0$.

In [F5], if the output and all three lower indices lie in the $M$ block, step 2.1 reproduces exactly the coordinate formula for $R^M$ because the symbols and their $x$-derivatives agree; all-$N$ indices similarly reproduce $R^N$. If the indices meet both blocks, each derivative term is either the derivative of a zero mixed symbol or a cross derivative of a factor-only symbol, and every quadratic term contains a zero mixed symbol. Hence every mixed curvature component is zero. Tensoriality and [F2] now give $R^{M\times N}((u_1,u_2),(v_1,v_2))(w_1,w_2)=(R^M(u_1,v_1)w_1,R^N(u_2,v_2)w_2)$.

For $a=(u,0)$ and $b=(0,v)$, step 3.2 gives $R^{M\times N}(a,b)b=0$, so [F6] makes the sectional-curvature numerator zero. By step 1.1, $a$ and $b$ are orthogonal with squared norms $g(u,u)>0$ and $h(v,v)>0$, so their Gram determinant is the positive product $g(u,u)h(v,v)$; therefore their plane has sectional curvature zero.

If a factor is empty, the product and every assertion about its points or mixed planes are vacuous. A zero-dimensional factor contributes an empty coordinate block and no nonzero vector for a mixed plane; one-dimensional factors are fully covered by the same formulas. Positive definiteness in step 1.1 excludes degenerate product metrics and step 4.1 checks the only denominator. There is no interval, scale endpoint, or manifold-boundary claim in this example. All charts and vectors are supplied locally and the Levi–Civita connection is unique, so no further family choice is made beyond the stated inherited assumption. No biconditional is asserted.