Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-15/2/b/solution

The expression is in general a vector field along a map, rather than a vector field on . Its intended meaning in the required formula is pointwise evaluation of the vector-bundle-valued differential form :
using the canonical identification . This interpretation works for every smooth map, including a constant map or a map with self-intersections.
Choose a frame of a vector bundle over , written as a row . Define its connection one-form by
for component columns . The pulled-back frame is a frame of a vector bundle over . For any component column of smooth functions on that open set, define
Here the matrix entries of are ordinary pullbacks of a differential form. The displayed operator is linear and satisfies the Leibniz rule, hence is locally a connection on a vector bundle.
To prove these local operators glue, change the original frame of a vector bundle to . Expanding gives the change of frame of a vector-bundle connection
The pullback of a differential form and the chain rule give
This is precisely the required transformation law for the pulled-back frame . Thus the definitions agree and give a global pullback connection.
For , substitute in the defining formula and use . This proves the required identity. Conversely that identity fixes ; the Leibniz rule then fixes the derivative of every , proving uniqueness. If the identity is initially imposed only on global sections of a vector bundle, multiply each local frame section by a smooth bump function equal to one near the point in question and supported in , and extend by zero. These global sections agree with the frame locally, so the same uniqueness argument applies. The pullback connection exists and is unique.

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