Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-17/4/b/solution

Conversely, choose the stated normal holomorphic coordinates at an arbitrary point. The coefficient matrix equals , so all of its first derivatives at the center vanish. Since the coordinate differentials themselves are closed,
Every point can serve as the center, hence on all of . Combined with the preceding construction, this proves the equivalence of (a) and (b); it is a first-jet characterization rather than a claim that the metric is flat on a neighborhood.

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