Product metric on product manifold

Suppose we have two Riemannian manifolds (M^m,g) and (N^n,\tilde g), what happened to their product manifold with product metric?
(M\times N, g\times\tilde g).

We will use a,b,c,..=1..m and A,B,C,..=1..n. Then by definition g_{aA}=0. Therefore

\Gamma_{aA}^b=\frac 12g^{bc}(\partial_Ag_{ac}+\partial_ag_{Ac}-\partial_cg_{aA})=0

Similarly for all mixed indices on \Gamma. This implies

 \nabla_{\partial_a}\partial_A=\Gamma_{aA}^b\partial_b+\Gamma_{aA}^B\partial_B=0.

 Therefore the Riemannian curvature operator is

 R(\partial_a,\partial_A)\partial_B=R(\partial_a,\partial_A)\partial_b=0.

 and Ricci tensor

 R_{aA}=0.

When you have the warped product metric like \mathbb{R}\times N with metric dr^2+\phi^2(r)\tilde g, the curvature is given like the following

R(\partial _r,X,\partial_r,Y)=-{\phi''}\phi\tilde g(X,Y)

R(\partial_r,X,Y,Z)=0

R(X,Y,Z,W)=\phi^2\tilde R(X,Y,Z,W)-(\phi'\phi)^2(\tilde g(X,Z)\tilde g(Y,W)-\tilde g(X,W)\tilde g(Y,Z))

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