Riemann curvature product metric

Question

Suppose that $M=M_1 \times M_2,$ with the product metric $g= g_1 \oplus g_2.$ Let $p\in M$ and suppose that $X \in T_pM_1$ and $Y\in T_pM_2.$

I want to show that $R(X,Y,Y,X)=0,$ at the point $p.$ I can show this using the coordinate formula for the curvature.

Surely there is a more "invariant" way?

Answer 1

Letting $X_{1}, X_{2}, \ldots, X_{l}$ be a local frame for $M_{1}$ and $Y_{1}, Y_{2}, \ldots, Y_{m}$ be a local frame for $M_{2}$, the Levi-Civita connection $\nabla$ of the product metric can be shown to satisfy the following:

1. $\nabla_{X_{i}}{X_{j}} =\tilde{\nabla}_{X_{i}}{X_{j}}$, where $\tilde{\nabla}$ is the Levi-Civita connection of $M_{1}$,
2. $\nabla_{Y_{i}}{Y_{j}} = \dot{\nabla}_{Y_{i}}{Y_{j}}$, where $\dot{\nabla}$ is the Levi-Civita connection on $M_{2}$, and
3. $\nabla_{X_{i}}{Y_{j}} = \nabla_{Y_{j}}{X_{i}} = 0$.

In fact, Extending the conditions from 1., 2., and 3. via the usual linearity properties, one can show that $\nabla$ so defined is in fact torsion free and symmetric (and thus is the Levi-Civita Connection on $M_{1} \times M_{2}$).

With the above in mind, working out that $R(X, Y, Y, X) =0$ should be straight forward.