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 $(E\otimes E^{\prime },g\otimes g^{\prime })$.

Proof

Let $X,Y \in \Ga(E), X',Y' \in \Ga(E')$. Then $\til X \some{def} X \otimes X' \in \Ga(E \otimes E'), \til Y \some{def} Y \otimes Y' \in \Ga(E \otimes E')$. Let $Z \in \Ga(\TM)$. We just expand the two expressions which are needed to be shown equal:

Related facts

• Tensor product of flat connections is flat