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.

References

Textbook references

• Book:Globalcalculus^{More info}, Page 125-127