# Tensor product of flat connections is flat

## Statement

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

## Facts used

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

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

## Proof

### Proof using the formula for curvature of tensor product

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

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

The right side is zero. Thus, $R_{\nabla \otimes \nabla'}(X,Y)$ is identically zero on all pure tensors. Further, since $R$ is a tensor, $R_{\nabla \otimes \nabla'}(X,Y)(t)$ at a point depends only on the value of $t$ at that point. Thus, at every point, we have shown that $R_{\nabla \otimes \nabla'}(X,Y)$ is identically zero on all pure tensors. Since every tensor is a sum of pure tensors and $R_{\nabla \otimes \nabla'}(X,Y)$ is linear, $R_{\nabla \otimes \nabla'}(X,Y)$ is identically zero at each point, and hence identically zero as a tensor. Thus, $\nabla \otimes \nabla'$ is flat.