Algebraic Curves Lecture 6: Two more corollaries of Riemann Roch

The 6th lecture of algebraic curves by Karl Christ

Correction from last time
1. Final two corollaries
Adjunction

Correction from last time

The dimension of a variety is the degree of a Hilbert polynomial.
The leading coefficient of the Hilbert polynomial is $\frac{d}{n!}$ .

Final two corollaries

We finished with the following observation, a corollary following from Riemann-Roch:

Cor:

Any line bundle of degree $\geq 2g$ is base point free
Any line bundle of degree $\geq 2g+1$ is very ample
Any curve can be embedded in $\mathbb P^{g+1}$ .
A line bundle is ample if and only if its degree is positive.

You get the dimension of the embedding above in (3) directly from Riemann Roch: RR says $h^0(L) = d - g + 1$ , so if $d = 2g + 1$ then $h^0(L) = g + 2$ .

The last corollary we'll mention, at least for now, is the following.

Cor: For any set of points

\{p_1, ..., p_k\}

X

X\setminus \{p_1, ..., p_k\}

is affine.

Proof. Set $D = \sum_i P_i\in \textrm{Div}(X)$ . Let $L = \mathcal O_X(D)^{\otimes \ell}$ be very ample.

Embedd $X$ in $\mathbb P^r$ via $\varphi_L$ . Then there is a hyperplane $H\subseteq \mathbb P^r$ such that $H|_{\varphi_L(X)} = \ell \cdot D$ . Thus $X\setminus D = X|_{\mathbb P^r \setminus H}$ is affine.

\square

Note that this is definitely not true in general for varieties, even if you replace "points" with "divisors". For example, $\mathbb P^2$ minus a point is not affine. If you take $\mathbb P^2$ blown up at a point and then remove the exceptional divisor, you get the same thing as $\mathbb P^2$ with a point removed; so it's still not affine.

The proof above works because, if you take any effective divisor $D$ , then you have a global section of $\mathcal O_X(D)$ . This is because you can find a regular function which vanishes on $D$ .

Example: (genus 1 curves). Let $X$ be a curve of genus 1, $p\in X$ , and consider $L_n = \mathcal O_X(n\cdot p)$ for $n\mathbb N$ . Riemann Roch says that $h^0(L_n) = d - g + 1 = d$ as soon as $d\geq 2g - 1 = 1$ . A corollary from last week says

\begin{aligned} \deg(K_X) = 2g - 2 = 0 \end{aligned}

and

\begin{aligned} h^0(K_X) = g = 1, \end{aligned}

hence

K_X \cong \mathcal O_X

If $n=1$ , then $h^0(L_n) = 1$ .

If $n=2$ then $h^0(L_2) = 2$ , and hence we get 2:1 a map $X\xrightarrow{2:1} \mathbb P^1$ . Thus $H^0(L_2) = textrm{span}\{1, x\}$ .

If $n=3$ , then $h^0(L_3) = 3$ . We get a new global section $y$ with a triple pole at $p$ . $L_3$ defines a map $\varphi_{L_3}:X\to \mathbb P^2$ which can be written as an affine chart $[z^3 : zx : y]$ , where $x = x/z^2$ and $y = y/z^3$ . Here $z$ is a local coordinate which vanishes to order $1$ at $p$ . The intersection of $z = 0$ with the image of the map $\varphi_{L_3}$ will be $[0:0:1]$ . In particular, we have a "globally defined" marked point on $X$ .

If $n=4$ , $h^0(L_4) = 4$ , the new section is $x^2$ .

If $n=5$ , $h^0(L_5) = 5$ , the new section is $x\cdot y$ .

If $n=6$ , $h^0(L_4) = 4$ , but two new sections $x^3$ and $y^2$ . This tells you that you must have a linear relation between all the sections now; i.e. we have an equation of the form

\begin{aligned} y^2 + a_1xy + a_3y = x^3 + a_2 x^2 + a_4x + a_6 \end{aligned}

where

x

and

y

have been rescaled so the coefficients of

x^3

and

y^2

are

1

. This is some standard form of an elliptic curve.

Remark: A genus $1$ curve + a marked point $p$ is called an elliptic curve. It has a group structure compatible with the algebraic structure, making it an Abelian variety.

Not all distinct choices of $a_1,...,a_5$ yield distinct isomorphism classes of curves, and some choices yield singular curves. Nonetheless, we get a map $\mathbb A^5\dashrightarrow \mathcal M_{1,1}$ .

Remark: This shows that

\mathcal M_{1,1}

(the moduli space of genus one curves with one marked point, i.e. the moduli space of elliptic curves) is unirational. It's not particularly interesting here because

\mathcal M_{1,1}

is just

\mathbb A^1

, but as the genus and number of marked points increases this becomes more interesting.

Adjunction

Suppose we have $Y\subseteq X$ a subvariety of a variety $X$ . On any smooth variety $X$ we can define the canonical sheaf $K_X = \bigwedge^n_{i=1}\Omega_{X}$ .

Example: $K_{\mathbb P^n} = \mathcal O_{\mathbb P^n}(-n-1).$

In an open standard chart $u_1,...,u_n$ of $\mathbb P^n$ , $K_{\mathbb P^n}$ is generated by $du_1 \wedge ... \wedge du_n$ . On a different chart with standard coordinates $w_1,...,w_n$ , the transition functions between the charts are something like $u_1 = \frac{1}{w_i}$ and $u_j = \frac{w_j}{w_i}$ . Over this $w_1,...,w_n$ chart $K_X$ is generated by $dw_1\wedge ... \wedge dw_n$ , and

\begin{aligned} du_1 = d\frac{1}{w_i} = -\frac{1}{w_i^2} dw_i \end{aligned}

and

\begin{aligned} du_i = d\frac{w_i}{w_j} = \frac{1}{w_i}dw_j - \frac{w_j}{w_i^2}dw_j. \end{aligned}

Checking $du_1\wedge ...\wedge du_n$ , applying bilinearity and canceling accordingly, you should end up with something like

\begin{aligned} du_1 \wedge ...\wedge du_n = -\frac{1}{w_1^{n+1}}\cdot \left(dw_1\wedge ...\wedge dw_n\right). \end{aligned}

The point of this calculation is that I end up with a pole of order $n+1$ .