Riemannian curvature tensor of product manifolds

Question

Let $(M_{1},g_{1})$ and $(M_{2},g_{2})$ be two Riemannian manifolds. Let $% R_{1}$ and $R_{2}$ be the (1,3)-type Riemannian curvature tensors of $M_{1}$ and $M_{2}$, respectively. Finally, let $R$ be the Riemannian curvature tensor of the product manifold $M=M_{1}\times M_{2}$. Can I write a relation between $R$, $R_{1}$ and $R_{2}$? I know that if we select $f=1$ in a warped product manifold, we have a product manifold. So my guess is: if $% X_{1},Y_{1}\in \Gamma (M_{1}),$ $X_{2},Y_{2}\in \Gamma (M_{2}),$ $% X=(X_{1},X_{2})\in \Gamma (M),$ $Y=(Y_{1},Y_{2})\in \Gamma (M)$; then $% R(X,Y)=R_{1}(X_{1},Y_{1})+R_{2}(X_{2},Y_{2})$. Is this correct? If yes, how can I prove it? If not, what is the relation?

Answer 1

Yes, it's correct, assuming to be precise that what you mean by $``R_1 + R_2"$ is $$ R(X,Y)Z = (R_1(X_1, Y_1) Z_1, R_2(X_2, Y_2)Z_2). $$

I imagine there's some elegant way to prove this, but it certainly works to just do a bunch of silly calculations; in particular, you can show that for product vector fields $(X_1, X_2)$ and $(Y_1, Y_2)$, the connection of $M$ is given by

$$ \nabla_{(X_1, X_2)} (Y_1, Y_2) = (\nabla^{M_1}_{X_1} Y_1, \nabla^{M_2}_{X_2} Y_2) $$

using, for instance, the Koszul formula (I think you can also just check directly that the right side satisfies the properties of a Levi-Civita connection on $M$); then you can just use the standard defining formula for $R$

$$ R(X,Y) = \nabla_X\nabla_Y - \nabla_Y \nabla_X - \nabla_{[X,Y]} $$

and it seems to just pop out.