## 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

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

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

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