Prove the theorem of Hurwitz that a curve \(X\) of genus \(g \geq 2\) over a field of characteristic 0 has at most \(84(g-1)\) automorphisms.
We will see later (Ex. 5.2) or (V, Ex. 1.11) that the group \(G=\Aut X\) is finite. So let \(G\) have order \(n\). Then \(G\) acts on the function field \(K(X)\). Let \(L\) be the fixed field. Then the field extension \(L \subseteq K(X)\) corresponds to a finite morphism of curves \(f: X \to Y\) of degree \(n\).
If \(P \in X\) is a ramification point, and \(e_P=r\), show that \(f^{-1} f(P)\) consists of exactly \(n / r\) points, each having ramification index \(r\). Let \(P_1, \ldots, P_s\) be a maximal set of ramification points of \(X\) lying over distinct points of \(Y\), and let \(e_{P_i}=r_i\). Then show that Hurwitz’s theorem implies that \[ {2 g-2 \over n } =2 g(Y)-2+\sum_{i=1}^s\left(1- {1\over r_i}\right) \]
Since \(g \geq 2\), the left hand side of the equation is \(>0\). Show that if \(g(Y) \geq 0\), \(s \geq 0\), \(r_i \geq 2\), \(i=1, \ldots, s\) are integers such that \[ 2g(Y)-2+\sum_{i=1}^s\left(1- {1\over r_i}\right)>0, \] then the minimum value of this expression is \(1/42\). Conclude that \(n \leq 84(g-1)\).
See (Ex. 5.7) for an example where this maximum is achieved. It is known that this maximum is achieved for infinitely many values of \(g\) (Macbeath). Over a field of characteristic \(p>0\), the same bound holds, provided \(p>g+1\), with one exception, namely the hyperelliptic curve \(y^2=x^p-x\), which has \(p=2g+1\) and \(2p(p^2-1)\) automorphisms (Roquette).
Solution
Let \[ G=\Aut X, \qquad n=\size G. \]
1. It is enough to prove the bound after replacing the ground field byits algebraic closure.
Base change to an algebraic closure preserves the genus. Every \(k\)-automorphism of \(X\) induces an automorphism of the base-changed curve, and this map on automorphism groups is injective. Thus a bound for the geometric automorphism group also bounds \(\Aut_kX\).
We may therefore assume from now on that \(k\) is algebraically closed. In particular every closed point of the curves below has residue field \(k\).
2. If\[ f:X\longrightarrow Y=X/G \]is the quotient map and \(P\in X\) has ramification index \(e_P=r\), then\[ f^{-1}(f(P)) \]has exactly \(n/r\) points, and every one of them has ramification index \(r\).
The extension of function fields \[ k(Y)=k(X)^G\subseteq k(X) \] is Galois with group \(G\). Hence \(G\) acts transitively on the points of \(X\) lying over a fixed point of \(Y\).
Let \[ G_P=\{\sigma\in G:\sigma(P)=P\} \] be the stabilizer. For a Galois extension of function fields, \(G_P\) is the decomposition group at \(P\). Because the residue field extension is trivial over the algebraically closed field \(k\), the decomposition group equals the inertia group. Its order is therefore the ramification index: \[ \size G_P=e_P=r. \]
Orbit–stabilizer now gives \[ \size(G\cdot P)=\frac nr. \] This orbit is the whole fibre. Stabilizers of points in the same orbit are conjugate, so they all have order \(r\) and hence all points of the fibre have ramification index \(r\).
3. If \(P_1,\ldots,P_s\) lie over the distinct branch points of \(Y\) and\[ r_i=e_{P_i}, \]then\[ \boxed{ \frac{2g-2}{n} = 2g(Y)-2 + \sum_{i=1}^s\left(1-\frac1{r_i}\right). } \]
Characteristic \(0\) ramification is tame, so Riemann–Hurwitz for the degree-\(n\) morphism \(f\) gives \[ 2g-2 = n\bigl(2g(Y)-2\bigr) + \sum_{P\in X}(e_P-1). \]
By step 2, the fibre over the branch point represented by \(P_i\) contains \(n/r_i\) points, each with ramification index \(r_i\). Its total contribution to the ramification sum is therefore \[ \frac n{r_i}(r_i-1) = n\left(1-\frac1{r_i}\right). \] Summing over all branch fibres and dividing by \(n\) gives the displayed formula.
4. Put\[ E = 2g(Y)-2 + \sum_{i=1}^s\left(1-\frac1{r_i}\right), \qquad r_i\ge2. \]If \(E>0\) and \(g(Y)\ge1\), then\[ E\ge\frac12. \]
If \(g(Y)\ge2\), then already \[ 2g(Y)-2\ge2. \]
If \(g(Y)=1\), the genus term is \(0\). Positivity then forces at least one branch point, and every summand satisfies \[ 1-\frac1{r_i}\ge\frac12. \] Thus \(E\ge1/2\) in the only remaining case.
5. Suppose \(g(Y)=0\). If \(E>0\), then \(s\ge3\). For \(s\ge4\), one has\[ E\ge\frac16. \]
For \(g(Y)=0\), \[ E = -2+s-\sum_{i=1}^s\frac1{r_i}. \] If \(s\le2\), this is negative, so \(s\ge3\).
If \(s\ge5\), then \(1-1/r_i\ge1/2\) gives \[ E\ge-2+\frac s2\ge\frac12. \]
If \(s=4\), then \[ E=2-\sum_{i=1}^4\frac1{r_i}. \] The choice \((2,2,2,2)\) gives \(E=0\). Subject to \(E>0\), the reciprocal sum is maximized by changing just one denominator from \(2\) to the next possible integer \(3\). Hence \[ \sum_{i=1}^4\frac1{r_i} \le \frac12+\frac12+\frac12+\frac13 = \frac{11}{6}, \] and therefore \[ E\ge\frac16. \]
6. Suppose \(g(Y)=0\) and \(s=3\). Then the smallest positive value of \(E\)is\[ \frac1{42}, \]attained for\[ (r_1,r_2,r_3)=(2,3,7) \]up to permutation.
Reorder the indices so that \[ 2\le r_1\le r_2\le r_3. \] Now \[ E = 1-left(\frac1{r_1}+\frac1{r_2}+\frac1{r_3}\right), \] so we must maximize the reciprocal sum subject to it being strictly less than \(1\).
If \(r_1\ge4\), then the sum is at most \(3/4\).
Suppose \(r_1=3\). If \(r_2=3\), positivity excludes \(r_3=3\), so \(r_3\ge4\) and the reciprocal sum is at most \[ \frac13+\frac13+\frac14 = \frac{11}{12}. \] If \(r_2\ge4\), it is at most \[ \frac13+\frac14+\frac14 = \frac56. \]
It remains to take \(r_1=2\). The case \(r_2=2\) cannot give a positive \(E\). If \(r_2=3\), positivity requires \[ \frac12+\frac13+\frac1{r_3}<1, \] so \(r_3\ge7\). The largest possible reciprocal sum is therefore \[ \frac12+\frac13+\frac17 = \frac{41}{42}. \] If \(r_2=4\), positivity excludes \(r_3=4\), so \(r_3\ge5\) and the sum is at most \[ \frac12+\frac14+\frac15 = \frac{19}{20}. \] Finally, if \(r_2\ge5\), the sum is at most \[ \frac12+\frac15+\frac15 = \frac9{10}. \]
Among all cases the largest reciprocal sum strictly below \(1\) is therefore \(41/42\), attained at \((2,3,7)\). Hence the least positive value of \(E\) is \[ 1-\frac{41}{42}=\frac1{42}. \]
7. One has\[ \boxed{\size\Aut X\le84(g-1).} \]
Since \(g\ge2\), step 3 gives \[ E=\frac{2g-2}{n}>0. \] Steps <1>4–<1>6 show that every positive value of \(E\) is at least \(1/42\). Therefore \[ \frac{2g-2}{n}\ge\frac1{42}, \] so \[ n\le42(2g-2)=84(g-1). \] This is Hurwitz’s bound.
8. Q.E.D.
Steps <1>2–<1>3 prove part (a), steps <1>4–<1>6 prove the numerical minimum in part (b), and step 7 gives the required automorphism bound.