Tensor product of flat connections is flat

Statement

Suppose $M$ is a differential manifold and $E,E^{\prime }$ are vector bundles over $M$. Suppose $\mathrm{\nabla },{\mathrm{\nabla }}^{\prime }$ are connections on $E,E^{\prime }$ respectively. Suppose, further, that the Riemann curvature tensor of $\mathrm{\nabla }$ as well as of ${\mathrm{\nabla }}^{\prime }$ is zero. In other words, both $\mathrm{\nabla }$ and ${\mathrm{\nabla }}^{\prime }$ are flat connections. Then, the tensor product of connections $\mathrm{\nabla }\otimes {\mathrm{\nabla }}^{\prime }$ is also a flat connection.

Facts used

1. Formula for curvature of tensor product of connections: This states that:

$R_{\mathrm{\nabla }\otimes {\mathrm{\nabla }}^{\prime }}(X,Y)(s\otimes s^{\prime })=R_{\mathrm{\nabla }}(X,Y)(s)\otimes s^{\prime }+s\otimes {R_{{\mathrm{\nabla }}^{\prime }}}^{\prime }(X,Y)(s^{\prime })$.

Proof

Proof using the formula for curvature of tensor product

By the formula for tensor product of connections, we have:

$R_{\mathrm{\nabla }\otimes {\mathrm{\nabla }}^{\prime }}(X,Y)(s\otimes s^{\prime })=R_{\mathrm{\nabla }}(X,Y)(s)\otimes s^{\prime }+s\otimes {R_{{\mathrm{\nabla }}^{\prime }}}^{\prime }(X,Y)(s^{\prime })$.

The right side is zero. Thus, $R_{\mathrm{\nabla }\otimes {\mathrm{\nabla }}^{\prime }}(X,Y)$ is identically zero on all pure tensors. Further, since $R$ is a tensor, $R_{\mathrm{\nabla }\otimes {\mathrm{\nabla }}^{\prime }}(X,Y)(t)$ at a point depends only on the value of $t$ at that point. Thus, at every point, we have shown that $R_{\mathrm{\nabla }\otimes {\mathrm{\nabla }}^{\prime }}(X,Y)$ is identically zero on all pure tensors. Since every tensor is a sum of pure tensors and $R_{\mathrm{\nabla }\otimes {\mathrm{\nabla }}^{\prime }}(X,Y)$ is linear, $R_{\mathrm{\nabla }\otimes {\mathrm{\nabla }}^{\prime }}(X,Y)$ is identically zero at each point, and hence identically zero as a tensor. Thus, $\mathrm{\nabla }\otimes {\mathrm{\nabla }}^{\prime }$ is flat.