Tensor product of flat connections is flat

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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.