Connection is splitting of first-order symbol sequence

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Statement

Suppose E is a vector bundle over a differential manifold M. Denote by \mathcal{E} the sheaf of sections of E. Consider the first-order symbol sequence for E, given by:

0 \to \operatorname{Hom}(\mathcal{E},\mathcal{A}) \to \mathcal{D}^1(\mathcal{E},\mathcal{A}) \to \operatorname{Hom}(\mathcal{E},\mathcal{T}) \to 0

Here \mathcal{A} is the sheaf of infinitely differentiable functions on M, \mathcal{D}^1 denotes the space of first-order differential operators from \mathcal{E} to \mathcal{A}, and \mathcal{T} denotes the sheaf of derivations of M.

A connection on E is equivalent to a choice of splitting for this sequence.

Proof

A splitting of the above sequence is equivalent to a section map:

\operatorname{Hom}(\mathcal{E},\mathcal{T}) \to \mathcal{D}^1(\mathcal{E},A)

which is equivalent to a map (satisfying some additional conditions):

\mathcal{T} \to \mathcal{E} \otimes \mathcal{D}^1(\mathcal{E},\mathcal{A})

The right side is equivalent to \mathcal{D}^1(\mathcal{E},\mathcal{E}), so a splitting of the sequence is equivalent to a map:

\mathcal{T} \to \mathcal{D}^1(\mathcal{E},\mathcal{E})

satisfying some additional conditions. Clearly, a connection is also a map of the above form, so it remains to check that the additional condition that comes from it being a splitting, is equivalent to the Leibniz rule for the connection.

References

Textbook references