Lecture notes from Karl Christ’s Algebraic Curves course, UT Austin, Spring 2024. Live-TeXed during lecture; any errors are mine.
Last week, we saw
- there exists a unique \(g^1_2\) on any hyperelliptic curve with \(g \geq 2\).
- \(K_X\) is very ample if and only if \(X\) is not hyperelliptic.
Proposition
If \(C\subseteq \mathbb P^n\) is a complete intersection curve and is smooth of genus at least 2, then \(C\) is not hyperelliptic.
Proof
\(N_{C/\mathbb P^n} = \bigoplus \mathcal O_{\mathbb P^n}(d)\), \(\bigwedge^{n-1} N_{C/\mathbb P^1} = \mathcal O_{\mathbb P^n}(\sum d_i)|_C\). The canonical bundle of \(C\) is given by
\begin{align*} K_C = \left(K_{\mathbb P^n} \otimes \bigwedge^{n-1} N_{C/\mathbb P^n}\right)|_C = \mathcal O_{\mathbb P^n}(\sum d_i - n - 1)|_C, \end{align*}where we’ve used adjunction and the fact that \(K_{\mathbb P^n} = \mathcal O_{\mathbb P^n}(-n-1)\). Since \(g(C) \geq 2\), we need to have \(\sum d_i - n - 1 > 0\).
Through any pair of points \(p\) and \(q\) on \(C\), we can find a hypersurface of degree \(\sum d_i- n - 1\) which passes through \(p\) but not \(q\) (this is true as long as \(\sum d_i -n - 1 > 0\)). This works for double points as well – we can find such a hypersurface which vanishes to order 1 at a specified point. Since \(K_C\) is cut out by hypersurfaces, this means \(K_C\) is very ample, hence \(C\) is not hyperelliptic curve by (2) above.
Riemann existence theorem
Question: If given a \(\mathbb P^1\) with specified branch points \(p_1,...,p_b\in \mathbb P^1\), can you find a corresponding cover \(X\to \mathbb P^1\) whose branch points are precisely these \(p_i\)?
First attempt at constructing this: You want a cover of \(\mathbb P^1\) that is unramified away from the \(p_i\). So consider first \(\mathbb P^1 \setminus \{p_1,...,p_b\}\) topologically. The fundamental group of this is generated by loops \(\gamma_i\) based at \(p_0\) and going around \(p_i\). They are subject to the relation \(\prod \gamma_i = \id\). This means that interchanging any of these two generators corresponds to permuting the sheets of the cover.
For any cover \(X\xrightarrow{d:1} \mathbb P^1\) branched at the \(p_i\), then we have \(d\) sheets and therefore we get a group homomorphism
\begin{align*} \pi_1(\mathbb P^1\setminus \{p_1,...,p_b\}, p_0) \to S_d. \end{align*}Thus we get permutations \(\tau_i\in S_d\) such that
\begin{align*} \prod_{i=1}^d\tau_i = \id. \end{align*}These permutations additionally must generate a transitive subgroup of \(S_d\), if we want \(X\) to be connected.
Note 1 at this point: We haven’t prescribed ramificaiton indices to the points \(p_i\), but they are encoded in this picture as well. As constructed, the ramification index of \(p_i\) corresponds to the “number of sheets” in the cover of \(\mathbb P^1\set minus \{p_1,...,p_b\}\) which meet at \(p_i\), and this number is equal to the number of cycles in the cycle decomposition of \(\tau_i\). To see this, consider the analytic picture: locally near \(p_i\), the map will be given \(z \mapsto z^{r_i}\) where \(r_i\) is the ramification index of \(p_i\).
Note 2 at this point: We’re trying to construct \(X\) and a map to \(\mathbb P^1\) by first constructing a topological cover \(U\to \mathbb P^1\set minus \{p_1,...,p_b\}\) and then gluing our branch points back in. One way to construct this cover is to take \(U = \widetilde U \times \{p_1,...,p_b\}/G\), where \(\widetilde{U}\) is the universal cover and \(G = \pi_1(\mathbb P^1\set minus \{p_1,...,p_b\}, p_0)\) acts diagonally by deck transformation.
This leads us to a theorem that says constructing a cover in this way is always possible.
Theorem
(Riemann Existence). There is a bijection between degree d covers \(X\to \mathbb P^1\) simply branched over \(p_1,...,p_b\) up to isomorphism and transpositions \(\tau_i\in S_d\) (\(1\leq i\leq b\)) such that \(\prod \tau_i = \id\) and the group generated by the \(\tau_i\) acts transitively on \(\{1,...,d\}\) up to simultaneous conjugation by an element in \(S_d\).
Example: For any even number \(b\) of branch points there is a unique hyperelliptic curve \(X\) with \(X\rightarrow{2:1} \mathbb P^1\) is branched over the \(p_i\). Riemann-Hurwitz says
\begin{align*} [2g - 2 = d_i (2h - 2) + \deg(R)] \end{align*}so
\begin{align*} 2g - 2 = -4 + b \iff b = 2g + 2. \end{align*}The dimension of the space of hyperelliptic curves of genus \(g\) is \(2g + 2 - 3 = 2g - 1\).
Note that this theorem only works when the base field is \(\mathbb C\). Moving from the topological construction of the cover \(U\xrightarrow{d:1} \mathbb P^1\setminus \{p_1,...,p_b\}\) as we sketched above to an algebraic cover \(X\to \mathbb P^1\) requires that there is a unique complex structure we can impose on \(X\). The analytic picture is therefore quite essential in this case.
Example: (Degree 3 simply ramified covers.) The relevant permutation group is \(S_3\). Then \(2g - 2 = -6 + b \iff b = 2g +4\). If I choose \(2g + 3\) transpositions, their composition is again a transposition. Then \(\frac{3^{2g + 3} - 3}{6}\) is the number of simpy ramified degree 3 covers \(X\to \mathbb P^1\) with \(g(X) = g\) and branched over fixed points \(p_1,...,p_b\) up to isomorphism.
Example: For a fixed ramification profile, there does not need to exist a cover of that profile, even if the numerics of Riemann-Hurwitz are satisfied: for \(d = 4\), \(\{(2)(2), (2)(2), (3)\}\). Then \(2g - 2 = -8 + 6 = -2\) which implies \(g = 0\). But there are no elements \(\tau_1,\tau_2,\tau_3 \in S_4\) of types \((2)(2), (2)(2)\) and \((3)\) such that \(\prod \tau_i = \id\).
It’s a very open question to determine which covers are allowed and which are not.
Low genus examples
Genus 2: Any genus 2 curve is hyperelliptic, and the unique \(g^1_2\) is the canonical sheaf. The number of branch points allowed by a map \(X\to \mathbb P^1\) is \(2 = -4 + b \iff b = 6\). Morally, genus \(2\) curves are paramterized by \((\mathbb P^1)^3\). We have an action of \(S_6\) on \(\{(0,1,\infty)\}\times (\mathbb P^1)^3\) given on \((0,1,\infty, p_1,p_2,p_3)\) given by applying the permutation naturally on the 6-tuple and then applying an automorphism of \(\mathbb P^1\) to each coordinate so that the first three coordinates are again \(0\), \(1\) and \(\infty\). Removing \(0,1\) and \(\infty\) from each copy of \(\mathbb P^1\) as the diagonal from the product \((\mathbb P^1)^3\) makes this action free, allowing us to take a quotient giving us the moduli space \(M_2\):
\begin{align*} M_2 = (\mathbb A^1 \setminus \{0,1,\}^3 \setminus \Delta / S^6, \hspace{1cm} \dim = 3 = 2g - 1 = 3g - 3. \end{align*}Genus \(3\): If \(X\) is not hyperelliptic then \(\varphi_{K_X}\) realizes \(X\) as a degree \(2g - 2 = 4\) curve in \(\mathbb P^{g - 1} = \mathbb P^2\). Conversely, any smooth plane quartic is a canonically embedded genus \(2\) curve. Indeed, by adjunction,
\begin{align*} K_C = (\mathcal O_{\mathbb P^2}(-3 + 4))|_C = \mathcal O_{\mathbb P^2}(1)|_C. \end{align*}Suppose \(C\) is hyperelliptic with \(g^1_2\) with \(g^1_2 = \mathcal O_C(p+q)\). Then \(2 = h^0(\mathcal O_C(p+q)\) . Then
\begin{align*} 2 = h^0(\mathcal O_C(p+q)) = 2 - 3 + 1 + h^0(K_C - p - q). \end{align*}But for geometric reasons \(h^0(K_c - p - q) = 1\). The dimension of the space of non-hyperelliptic genus 3 curves thus is \(h^0(\mathcal O_{\mathbb P^2}(4)) - 1 - 8 = \binom{6}{2} - g = 15 - g = 6\). As before, the space of hyperelliptic curves has dimension \(2g - 1 = 5\).