 Solution

ID: past-exam-of-the-geometry-course-of-the-university-of-oxford/2007/b/8/vi/solution

Apply Riemann-Roch at

D = K

and use

ℓ (K) = g

and

ℓ (0) = 1

g - 1 = de g K + 1 - g .

Rearranging gives $de g K = 2 g - 2$ . This holds for every canonical divisor, since any two differ by a principal divisor of degree zero.

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