Lecture notes from Karl Christ’s Algebraic Curves course, UT Austin, Spring 2024. Live-TeXed during lecture; any errors are mine.
Continuing Examples
We’re going to continue a series of examples we began last class. As a reminder, we’re looking to compute the dimension of \(\mathcal M_g\) for low genus.
Example: Take \(g = 4\). The locus of hyperelliptic curves has dimension \(7\) and is parameterized by varying the right number of branch points on \(\mathbb P^1\) (as before).
If \(X\) is not hyperelliptic, \(K_X\) is very ample and realizes \(X\) as a curve of degree \(6\) in \(\mathbb P^3\). Let’s first study what kind of curve we end up with in \(\mathbb P^3\).
Consider the space \(H^0(\mathcal O_{\mathbb P^3}(2))\) of degree \(2\) polynomials over \(\mathbb P^3\) together with the restriction map \(H^0(\mathcal O_{\mathbb P^3}(2)) \to H^0(\mathcal O_{X}(2))\).
- \(\dim H^0(\mathcal O_{\mathbb P^r}(d)) = \binom{r + d}{r},\) so \(H^0(\mathcal O_{\mathbb P^3}(2)) = \binom{5}{2} = 10\)
- The divisor corresponding to \(\mathcal O_{\mathbb X}(2)\) has degree \(12\), so by Riemann-Roch \(\dim H^0(\mathcal O_{\mathbb X}(2)) = 12 - 4 + 1 = 9\). From this we can conclude the restriction map is surjective.
Thus there must exist a quadric \(Q \subset \mathbb P^3\) such that \(X \subset Q\). If \(X\) were to lie on a second quadric it would follow that \(\deg(X) \leq 4\). But \(\deg(X) = 6\). Thus there is a unique quadric \(Q\) which contains \(X\) (it is a general fact that the intersection of a degree \(d_1\) hypersurface and a degree \(d_2\) hypersurface in \(\mathbb P^3\) results in a degree \(d_1 \cdot d_2\) curve in \(\mathbb P^3\), as long as the hypersurfaces intersect transversally).
Consider then the map \(H^0(\mathcal O_{\mathbb P^3}(3) \to H^0(\mathcal O_X(3))\).
- Then \(\dim H^0(\mathcal O_{\mathbb P^3}(3) = \binom{20}{3} = 20\) and \(\dim H^0(\mathcal O_X(3)) = 18 - 4 + 1 = 15\) by similar arguments as before.
There is thus at least a \(4\) dimensional family of cubics containing \(X\). Of this family, a 3-dimensional family consists of cubics of the form \(Q\cup H\). This implies there is an irreducible cubic \(C\) containing \(X\). Then \(C\cap Q\) is a degree 6 curve containing \(X\) and hence must be equal to \(X\).
Conversely, if \(X\) is a complete intersection of a quadric and a cubic, then by adjunction
\begin{align*} K_X = (\mathcal O_{\mathbb P^3}(5 - 4))|_X = (\mathcal O_{\mathbb P^3}(1))|_X. \end{align*}The intersection of a quadric and a cubic is a canonically embedded genus 4 curve (if smooth).
Then
\begin{align*} 3g - 3 = (10 - 1) + (20 - 4) - 1 - 15 = 9. \end{align*}Remark
Why \(3g - 3\)? If \(X\) is a curve and \(L\) a line bundle with \(\deg(L) = g + 1\) then Riemann-Roch tells us \(h^0(L) \geq 2\). “The general line bundle of degree \(g + 1\) satisfies \(h^0(L) = 2\). There is a \(g\)-dimensional space of line bundles of degree \(g+1\)”. That \(h^0(L) = 2\) means I get a cover \(X\to \mathbb P^1\), and by Riemann-Hurwitz \(2g - 2 = -2(g + 1) + b\) where \(b\) is the number of branch points. This means \(4g = b\). We expect to have a \(4g - 3\) dimensional space of covers of \(\mathbb P^1\) by genus \(g\) curves of degree \(g + 1\). Thus we would expect that there is a \(3g - 3\) dimensional space of genus \(g\) curves.
Maps to \(\mathbb P^3\) and \(\mathbb P^2\)
We saw already that any curve can be embedded in \(\mathbb P^{g+1}\). This is because line bundles become very ample once \(d \geq 2g + 1\), in which case Riemann-Roch tells us the global sections of a line bundle \(L\) have dimension \(h^0(L) = g + 2\). The claim is that we can do better.
Interlude on Grassmanians
As a set, \(\mathbb{Gr}(k,V)\) is the set of \(k\)-dimensional subspaces of a vector space \(V\). When \(k = 1\) or \(k = \dim V - 1\), this is just projective space \(\mathbb P(V)\). (See Harris: Basic algebraic geometry for a good reference constructing this variety in terms of coordinates). Consider the Plücker embedding
\begin{align*} \mathbb{Gr}(k,V) &\to \mathbb P(\bigwedge^k V) \\ W &\mapsto v_1\wedge ... \wedge v_k \end{align*}where \(v_1,...,v_k\) is a basis of \(W\). The image is the indecomposable elements of \(\bigwedge^k V\). It’s also an injective (hence is an embedding). Indeed, given an indecomposable \(\lambda \in \mathbb P(\bigwedge^k V)\) then the vector space
\begin{align*} W_\lambda = \{\omega \in V ~\mid~ \lambda \wedge \omega = 0\} \end{align*}is the unique subspace of \(V\) mapped to \(\lambda\).
Now, fix a basis \(e_1,...,e_n\) of \(V\) and a basis \(v_1,...,v_k\) of \(W\). Then the Plücker coordinates are the \(k\times k\) minors of the \(k\times n\) matrix with columns the \(v_1\wedge...\wedge v_n\).