Curves of genus $2$ correspond to six branch points on $\PP^1$ modulo $\Sigma_6$

By Exercise IV.1.7, the canonical system gives a finite morphism \[ f:X\longrightarrow\PP^1 \] of degree \(2\). Since \(\operatorname{char}k\ne2\), the induced degree-two extension of function fields is separable and any ramification is tame.

Riemann–Hurwitz gives \[ 2g(X)-2 = (\deg f)\bigl(2g(\PP^1)-2\bigr) + \sum_{P\in X}(e_P-1). \] Thus \[ 2 = 2(-2)+\sum_P(e_P-1), \] so \[ \sum_P(e_P-1)=6. \] For a degree-two map, every ramification index is at most \(2\); at a ramification point it is therefore exactly \(2\) and contributes \(1\). Hence there are exactly six ramification points.

Their images in \(\PP^1\) are distinct: a ramification point already has multiplicity \(2\) in its fibre, which exhausts the degree of \(f\). Thus the map has exactly six branch points as well.

Let \[ D=h^*(q) \] for a point \(q\in\PP^1\). Then \(\deg D=2\), and pullback of the two independent sections of \(\mco_{\PP^1}(1)\) gives \[ \ell(D)\ge2. \] Riemann–Roch on the genus-\(2\) curve gives \[ \ell(D)-\ell(K-D)=1. \] Hence \(\ell(K-D)\ge1\). But \[ \deg(K-D)=0. \] A degree-zero line bundle with a nonzero section is trivial: the zero divisor of such a section is effective of degree zero, hence zero. Therefore \[ K\sim D. \] Moreover \(\ell(D)=\ell(K)=2\), so the pencil defining \(h\) is the complete canonical pencil. Choosing a different basis of its two-dimensional section space changes the target coordinates by an element of \(\PGL_2(k)\). This proves the uniqueness assertion in part (a).

Put \[ h(x)=\prod_{i=1}^6(x-\alpha_i). \] At \(x=\alpha_i\), the function \(h\) has valuation \(1\). If \(v\) is a valuation of \(K(X)\) above the corresponding valuation of \(k(x)\), then \[ 2v(z) = v(h) = e_v. \] Thus \(e_v\) is even. Since the extension has degree \(2\), one must have \[ e_v=2. \] So each \(\alpha_i\) is a simple branch point.

At a finite point \(a\ne\alpha_i\), one has \(h(a)\ne0\). On the affine double cover \[ z^2=h(x), \] the derivative with respect to \(z\) is \(2z\), which is nonzero above such a point because \(z^2=h(a)\ne0\). Hence the cover is étale there.

It remains to inspect infinity. Put \[ u=x^{-1}, \qquad w=zu^3. \] Then near \(u=0\) the equation becomes \[ w^2 = \prod_{i=1}^6(1-\alpha_i u). \] At \(u=0\) this gives \[ w^2=1. \] Since \(\operatorname{char}k\ne2\), the two points \(w=\pm1\) are nonsingular and the derivative \(2w\) is nonzero there. Thus the cover is unramified at infinity. Hence the six \(\alpha_i\) are exactly the branch points.

The extension is separable because its degree is \(2\) and \(\operatorname{char}k\ne2\). By step 3 there are exactly six tame ramification points, each with \(e_P=2\). Riemann–Hurwitz gives \[ 2g(X)-2 = 2(-2)+6, \] so \[ 2g(X)-2=2 \] and therefore \[ \boxed{g(X)=2}. \]

Step 4 gives \(g(X)=2\), and \(f\) has degree \(2\). Step 2 says that every degree-two morphism from a genus-\(2\) curve to \(\PP^1\) is the canonical morphism up to an automorphism of the target. This proves the remaining assertion of part (b).

By Exercise I.6.6, \[ \Aut\PP^1=\PGL_2(k). \] Choose a fractional linear transformation sending \(P_1\) to \(0\) and \(P_3\) to \(\infty\). Its value at \(P_2\) is a nonzero finite scalar; multiply the target coordinate by its inverse to send \(P_2\) to \(1\). This proves existence.

For uniqueness, if \(\varphi\) and \(\psi\) both have the required values, then \[ \psi\circ\varphi^{-1} \] fixes \(0\), \(1\), and \(\infty\). A fractional linear transformation fixing \(0\) and \(\infty\) has the form \(x\mapsto cx\), and fixing \(1\) forces \(c=1\). Hence \(\psi=\varphi\).

Part (a) gives six distinct branch points on \(\PP^1\). Choose an ordering and apply step 6 to the first three. Since the six points are distinct, the remaining three normalized points are distinct from each other and from \(0\), \(1\), and \(\infty\). Thus they are finite elements of \(k\), different from \(0\) and \(1\).

Associate to a point of \(U\) the ordered six-tuple \[ (0,1,\infty,\beta_1,\beta_2,\beta_3). \] Given \(\sigma\in\Sigma_6\), reorder this six-tuple by \(\sigma\). Its first three entries are still distinct, so step 6 gives a unique projective automorphism sending them to \(0\), \(1\), and \(\infty\). Apply that automorphism to the remaining three entries; because all six entries were distinct, the resulting last three lie in \(U\).

The identity permutation clearly acts trivially. For two permutations, performing the first reordering and normalization and then the second produces the same normalized ordered six-tuple as performing the combined reordering once: both projective automorphisms send the same first three ordered points to \(0\), \(1\), and \(\infty\), and step 6 makes that normalizing automorphism unique. Hence the rule satisfies the group law and defines the stated action.

Let \[ F=k(\PP^1)=k(x), \] and let \(L_1/F\) and \(L_2/F\) be the function-field extensions of two such covers with the same branch locus. Since \(\operatorname{char}k\ne2\), each quadratic extension has Kummer form \[ L_i=F(\sqrt{d_i}) \] for some \(d_i\in F^\times\), determined modulo \(F^{\times 2}\).

For a place \(P\) of \(F\), the quadratic extension \(F(\sqrt{d_i})/F\) is ramified at \(P\) exactly when \(v_P(d_i)\) is odd. Equality of the branch loci therefore implies that \[ v_P(d_1/d_2) \] is even for every \(P\in\PP^1\). Hence \[ \operatorname{div}(d_1/d_2)=2D \] for an integral divisor \(D\) on \(\PP^1\). This divisor has degree \(0\). Since every degree-zero divisor on \(\PP^1\) is principal, there is \(g\in F^\times\) with \[ D=\operatorname{div}(g). \] Consequently \[ \operatorname{div}\!\left(\frac{d_1}{d_2g^2}\right)=0, \] so \[ \frac{d_1}{d_2g^2}=c \] for some \(c\in k^\times\). Because \(k\) is algebraically closed, \(c=a^2\) for some \(a\in k^\times\). Thus \[ d_1=d_2(ag)^2, \] and therefore \[ F(\sqrt{d_1})=F(\sqrt{d_2}). \] The corresponding nonsingular projective curves are consequently isomorphic over \(\PP^1\).

An isomorphism \(X\cong Y\) identifies \(H^0(Y,K_Y)\) with \(H^0(X,K_X)\). Hence the two canonical maps differ only by an automorphism of their targets, so isomorphic curves determine the same orbit.

Conversely, suppose two genus-\(2\) curves determine the same orbit. After choosing orderings and normalizations, their canonical maps have the same six-point branch locus. Step 9 then identifies the two quadratic function-field extensions of \(k(\PP^1)\), and hence identifies the two nonsingular projective curves. Thus distinct isomorphism classes cannot determine the same orbit.

Finally, let \[ (\beta_1,\beta_2,\beta_3)\in U. \] Set \[ B=\{0,1,\infty,\beta_1,\beta_2,\beta_3\}\subset\PP^1. \] Choose a point \(q\in\PP^1\setminus B\) and an automorphism \(\tau\in\Aut\PP^1\) with \(\tau(q)=\infty\). Then \(\tau(B)\) consists of six finite points. Part (b), proved in steps <1>3–<1>5, constructs a genus-\(2\) curve whose canonical map is branched exactly over \(\tau(B)\). Composing that map with \(\tau^{-1}\) gives branch locus \(B\), so the resulting curve maps to the prescribed orbit. Therefore the correspondence is surjective as well as injective.

Explicitly, \[ U = \AA_k^3 \setminus \left( \bigcup_{i=1}^3\{\beta_i=0\} \cup \bigcup_{i=1}^3\{\beta_i=1\} \cup \bigcup_{1\le i<j\le3}\{\beta_i=\beta_j\} \right), \] so \(U\) is open in \(\AA_k^3\). The group \(\Sigma_6\) is finite. This is the three-parameter description asserted in part (e).

Moreover an algebraically closed field is infinite. Fixing two admissible distinct values of \(\beta_1\) and \(\beta_2\) leaves infinitely many choices of \(\beta_3\), while every \(\Sigma_6\)-orbit is finite. Hence \(U/\Sigma_6\) is infinite, so there are infinitely many pairwise non-isomorphic curves of genus \(2\) over \(k\).

Steps <1>1–<1>2 prove part (a), steps <1>3–<1>5 prove part (b), steps <1>6–<1>7 prove part (c), step 8 proves part (d), and steps <1>9–<1>11 prove part (e).