Algebraic Curves Lecture 4: Riemann Roch

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Lecture notes from Karl Christ’s Algebraic Curves course, UT Austin, Spring 2024. Live-TeXed during lecture; any errors are mine.

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Last time we stopped with the following characterization of “very ampleness”:

Proposition

Let \(L\) be a line bundle on a curve \(X\). Then

  1. \(L\) is base point free if and only if \(h^0(X,L) = h^0(X,L(-p)) + 1\) for all \(p\in X\).
  2. \(L\) is very ample if \(h^0(X,L) = h^0(X,L(-p-q))+2\) for all \(p,q \in X\).

Remark

The same is true for linear series, and the definitions (of base point, of very ample) are the same as well.

Example

Take a map \(\mathbb P^1 \to \mathbb P^2\) given by \((s:t) \mapsto (s^3:st^2:t^3)\). Take the linear series \(V = \mathrm{span}\{s^3,st^2,t^3\} \subseteq H^0(\mathbb P^1, \mathcal O_{\mathbb P^1}(3))\). In the affine chart \(r\mapsto (r^2, r^3)\) this is given by the equation \(x^3 - y^2\); it’s the cuspoidal cubic. The linear series \(V\) is NOT very ample because at the singularity the differential is not injective: the sections in \(V\) that vanish at \(0\) are of the form \(a\cdot r^2 + b\cdot t^3 = r^2\cdot (a+b\cdot r)\), meaning

\begin{align*} \dim(V(-0)) = \dim(V(-2\cdot 0)). \end{align*}

Here \(0\) is the origin \((0,0)\), the divisor.

Here are other ways to see that \((0,0)\) is singular in the the image:

  1. \(k\mapsto (2r \cdot k, 3r^2 \cdot k)\)
  2. Jacobi’s criterion
  3. If you have a plane curve you can directly read off the equation of the Zariski tangent space at the origin – the equation of the Zariski tangent space at the origin is the linear term in the equation. In the case of \(x^3 - y^2 = 0\) we don’t have a linear term, hence singular at the origin.
  4. You can also use the definition of a singularity; show that the dimension of the Zariski tangent space is larger than the dimension of the ring. In our case,
\begin{align*} \dim_{\mathbb C}\left(\frac{(x,y)}{(x^2,xy,y^2)} = 2 \neq 1.\right) \end{align*}

Note that the equation for the tangent cone is given by the lowest order terms. The tangent cone to an affine variety \(X\) given by an ideal \(I\) at the origin is the Zariski closed subset corresponding to the ideal \(in(I)\), the ideal generated by all the lowest degree terms of elements in \(I\).

Riemann Roch

Lemma

  1. If \(\deg(L) < 0\) then \(h^0(X,L) = 0\).
  2. If \(\deg(L) = 0\) and \(h^0(X,L) > 0\) then \(L\cong \mathcal O_X\).
Proof
  1. Assume \(h^0(X,L) > 0\). We know that \(L\cong \mathcal O_X(D)\) for some \(D\in \textrm{Div}(X)\). If \(h^0(X,L) > 0\), then \(D\sim D'\) with \(D'\) effective which then implies that \(\deg(D') = \deg(L) \geq 0\).
  2. This follows since the only effective divisor of degree \(0\) is 0 (all coefficients equal to 0 in the divisor).

Definition

We denote by \(K_X\) the canonical sheaf on \(X\). Analytically, the sheaf of holomorphic one forms.

Theorem

(Serre Duality).

\begin{align*} H^1(X,L) \cong H^0(X,K_X\otimes L^{-1}). \end{align*}

Note that we call \(K_X\otimes L^{-1}\) the residual of \(\mathbf{L}\).

Theorem

(Riemann Roch).

\begin{align*} h^0(X,L) - h^0(X,K_X\otimes L^{-1}) = d - g + 1. \end{align*}
Proof

By Serre Duality, we need to show that

\begin{align*} h^0(X,L) - h^1(X,L) = d - g + 1. \end{align*}

We prove this by induction, and we write \(h^i(-)\) for \(h^i(X,-)\) to simplify notation a bit.

Base case: The claim is true for \(L = \mathbb O_X\):

\begin{align*} h^0(\mathcal O_X) = 1, ~h^1(\mathcal O_X) = g, ~d = 0. \end{align*}

We show that equality holds for \(L\) if and only if it holds for \(L(-p)\).

We have a short exact sequence

\begin{align*} 0\to L(-p)\to L\to L|_p \to 0 \end{align*}

where \(L|_p\) denotes the skyscraper sheaf of \(L\) at \(p\). I can then pass to the long exact sequence of cohomology to get

\begin{align*} 0 \to H^0(L(-p)) \to H^0(L) \xrightarrow{\mathrm{ev}} K \to H^1(L(-p)) \to H^1(L) \to 0 \end{align*}

noting that \(H^0(L|_p) = K\) and \(H^1(L|_p) = 0\). From this long exact sequence we have that

\begin{align*} h^0(L) = h^0(L(-p)) + \dim(\ker \mathrm{ev}) \end{align*}

and

\begin{align*} h^1(L(-p)) &= h^1(L) + \textrm{codim}(\ker \mathrm{ev}) \\ &= h^1(L) + 1 - \dim(\ker\textrm{ev}). \end{align*}

This implies

\begin{align*} h^0(L) - h^0(L(-p)) = h^1(L) + 1 - h^1(L(-p)) \end{align*}

and then

\begin{align*} h^0(L) - h^1(L) = h^0(L(-p)) - h^1(L(-p)) + 1. \end{align*}

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