Hurwitz's automorphisms theorem

Remarks

Finiteness is a separate step; a standard proof uses the finite set of Weierstrass points of $X$, which every automorphism permutes. For the bound, let $G = \Aut(X)$ and apply the Riemann–Hurwitz formula to the quotient map $X \to X / G$, of degree $\abs{G}$. Here $X/G$ has genus $h$, and the branch values have ramification indices $m_1, \ldots, m_k$. It gives $2g - 2 = \abs{G} \left( 2h - 2 + \sum_{i=1}^{k} (1 - 1/m_i) \right)$. The bracket is positive, and its smallest positive value is $1/42$, reached for $h = 0$ and $(m_1, m_2, m_3) = (2, 3, 7)$; this gives $\abs{G} \le 84(g - 1)$. The hypothesis $g \ge 2$ is needed: a torus $\C / \Lambda$ of genus $1$ has infinitely many automorphisms, the translations, and so does the Riemann sphere. Example: the Klein quartic $x^3 y + y^3 z + z^3 x = 0$ has genus $3$ and automorphism group of order $168 = 84 \cdot 2$, so the bound is attained. The bound is not attained for every genus; for $g = 2$ the largest automorphism group has order $48$.

Going deeper

For a student

The theorem is a typical application of the Riemann–Hurwitz formula to a quotient by a group action. The surfaces that reach the bound are called Hurwitz surfaces, and their automorphism groups are the finite quotients of the $(2, 3, 7)$ triangle group, called Hurwitz groups. The same bound holds for smooth projective curves over an algebraically closed field of characteristic $0$, but in positive characteristic it can fail because of wild ramification.