# Difference between revisions of "Connection is splitting of first-order symbol sequence"

## 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.