Tensor product of metric connections is metric

Statement

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

Proof

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

Let Failed to parse (unknown function "\Ga"): {\displaystyle 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