@def mintoclevel=2

Algebraic Curves Lecture 12:

The 12th lecture of algebraic curves by Karl Christ

Continuing Examples
Maps to $\mathbb P^3$ and $\mathbb P^2$

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 $H^0(\mathcal O_{\mathbb P^3}(2)) \to H^0(\mathcal O_{X}(2))$ .

$\operatorname{dim} H^0(\mathcal O_{\mathbb P^r}(d)) = \binom{r + d}{r},$ 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 $\operatorname{dim} H^0(\mathcal O_{\mathbb X}(2)) = 12 - 4 + 1 = 9$ . 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 $\operatorname{dim} H^0(\mathcal O_{\mathbb P^3}(3) = \binom{20}{3} = 20$ and $\operatorname{dim} H^0(\mathcal O_X(3)) = 18 - 4 + 1 = 15$ 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{aligned} K_X = (\mathcal O_{\mathbb P^3}(5 - 4))|_X = (\mathcal O_{\mathbb P^3}(1))|_X. \end{aligned}

The intersection of a quadric and a cubic is a canonically embedded genus 4 curve (if smooth).

Then

\begin{aligned} 3g - 3 = (10 - 1) + (20 - 4) - 1 - 15 = 9. \end{aligned}

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 $k = \operatorname{dim} V - 1$ , 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{aligned} \mathbb{Gr}(k,V) &\to \mathbb P(\bigwedge^k V) \\ W &\mapsto v_1\wedge ... \wedge v_k \end{aligned}

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{aligned} W_\lambda = \{\omega \in V ~\mid~ \lambda \wedge \omega = 0\} \end{aligned}

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 $v_1\wedge...\wedge v_n$ .

Algebraic Curves Lecture 12:

Continuing Examples

Maps to P3\mathbb P^3P3 and P2\mathbb P^2P2

Interlude on Grassmanians

Maps to $\mathbb P^3$ and $\mathbb P^2$