Tensor product of metric connections is metric

Statement

A differential manifold $M$. Two metric bundles $(E,g)$ and $(E^{\prime },g^{\prime })$ (i.e., $E,E^{\prime }$ are vector bundles and $g,g^{\prime }$ are Riemannian metrics or pseudo-Riemannian metrics on these). $\mathrm{\nabla },{\mathrm{\nabla }}^{\prime }$ are metric connections on $(E,g)$ and $(E^{\prime },g^{\prime })$ respectively. Then, the tensor product of connections $\mathrm{\nabla }\otimes {\mathrm{\nabla }}^{\prime }$ is a metric connection on $(E\otimes E^{\prime },g\otimes g^{\prime })$.

Related facts

• Tensor product of torsion-free connections is torsion-free
• Tensor product of flat connections is flat