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Suppose E is a smooth vector bundle with a smooth connection ∇. Then it induces a smooth connection on tensor powers of E. Does it also induce smooth connection on say exterior powers of E.


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Yes, and by a similar formula. If $\sigma_1,\ldots,\sigma_k$ are sections on $E$ then a connection on $\Lambda^k E$ is given by forcing the Leibniz rule: $$ \nabla_X(\sigma_1\wedge\sigma_2\wedge\cdots\wedge\sigma_k) = (\nabla_X \sigma_1) \wedge \sigma_2 \wedge \cdots \wedge \sigma_k + \sigma_1 \wedge \nabla_X\sigma_2 \wedge \sigma_3 \wedge \cdots \wedge \sigma_k + \cdots + \sigma_1 \wedge \cdots \wedge \sigma_{k-1}\wedge \nabla_X \sigma_k. $$

Also, there is an isomorphism between the exterior bundle and anti-symmetric tensors. You can check that the connection on the tensor algebra of $E$ preserves anti-symmetric tensors and agrees with the one I defined above.

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