Difference between revisions of "Curvature of direct sum of connections equals direct sum of curvatures"

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Revision as of 22:19, 24 July 2009

Statement

Suppose E,E' are vector bundles over a differential manifold M. Suppose \nabla,\nabla' are connections on E,E' respectively. Let R_\nabla denote the Riemann curvature tensor of \nabla and R_\nabla' denote that Riemann curvature tensor of \nabla'. Then if E \oplus E' is the direct sum of vector bundles, \nabla \oplus \nabla' is the direct sum of connections, and R_{\nabla \oplus \nabla'} is its Riemann curvature tensor, we have:

R_{\nabla \oplus \nabla'}(X,Y)(s ,s') = \left(R_\nabla(X,Y)(s),R_\nabla(X,Y)s'\right).