Lecture notes from Karl Christ’s Algebraic Curves course, UT Austin, Spring 2024. Live-TeXed during lecture; any errors are mine.
More Riemann-Roch
Recall that for a divisor \(D\) on a curve \(X\), if we write
- \(\ell(D) = h^0(X,\mathcal O_X(D))\)
- \(i(D) = h^0(X,K_X\otimes \mathcal O_X(D)^{-1}) = h^1(X,\mathcal O_X(D))\) (here \(h^0(X,K_X\otimes \mathcal O_X(D)^{-1})\) is called the “index of speciality”)
Then Riemann-Roch says \(\ell(D) - i(D) = d - g + 1\) where \(d\) is the degree of the line bundle. Today we’ll collect some easy Corollaries of Riemann Roch.
Definition
\(\mathcal O_X(D)/D\) is special if \(\ell(D) > 0\) and \(i(D) > 0\).
Corollary
Let \(K_X\) be the canonical sheaf. Then \(h^0(K_X) = g\) and \(\deg(K_X) = 2g - 2\).
Proof
Apply Riemann-Roch to \(\mathcal O_X\). Then \(h^0(\mathcal O_X) = 1\) since the only global regular functions on \(X\) are constant, \(h^0(K_X\otimes \mathcal O_X^{-1}) = h^0(K_X)\) and \(d = 0\) since the degree of \(\mathcal O_X\) is \(0\). Then
\begin{align*} h^0(K_X) = g. \end{align*}Now applying Riemann-Roch to \(K_X\) gives us \(\deg(K_X) = 2g - 2\).
Corollary
If \(\deg(L) < 0\) or \(\deg(L) > 2g- 2\) then \(L\) is not special and
\begin{align*} h^0(L) = \begin{cases} 0 & \text{if} \deg(L) < 0 \\ d - g + 1 &\text{if} \deg(L) > 2g - 2 \end{cases} \end{align*}Proof
The first case where \(\deg(L) < 0\) was already established, so assume \(\deg(L) > 2g - 2\). Then \(h^0(L) - h^0(K_X - L) = d - g + 1\).
Karl says that \(h^0(K_X - L) = 0\), I’m not sure why, but this gives you the result.
Corollary
The Hilbert Polynomial of a degree d subcurve of \(\mathbb P^n\) \(h_X(m) = d\cdot m - g + 1\).
Proof
By definition the Hilbert function is the function
\begin{align*} f_X(m) = h^0(X,\mathcal O_X(m)). \end{align*}Applying Riemman-Roch gives us
\begin{align*} h^0(X,\mathcal O_X(m)) - h^1(X,\mathcal O_X(m)) = dm - g + 1. \end{align*}For \(m \gg 0\), \(h^1(X,\mathcal O_X(m)) = 0\).
Corollary
\(X\) is rational (i.e. isomorphic to \(\mathcal P^1\)) if and only if \(g = 0\). That is, there is only one genus \(0\) curve up to isomorphism.
Proof
Suppose \(g=0\). We now want to construct a map to \(\mathbb P^1\), Pick \(p\in X\) and consider \(\mathcal O_X(p) = L\). By Riemann-Roch, \(h^0(X,L) \geq d - g + 1 = 2\) since \(g = 0\) and \(d = \deg(\mathcal O_X(p) = 1\). This gives us a degree 1 map \(X \dashrightarrow \mathcal P^1\), which implies \(X\cong \mathbb P^1\).
Now suppose that \(X\cong \mathbb P^1\). Then \(h_X(m) = h^0(\mathcal O_{\mathbb P^1}(m)) = m + 1.\) Then the previous corollary concerning the Hilbert polynomial means \(d\cdot m - g + 1 = m + 1\), and since \(d = 1\) (viewing \(X\) as a curve with a degree 1 embedding in \(\mathbb P^1\)) we have \(g = 0\).
Corollary
For any line bundle \(L\), \(h^0(L) \leq \deg(L) + 1\). With equality if and only if \(X\) is rational.
Proof
Suppose \(h^0(L) > \deg(L) + 1\). Let \(D\) be an effective divisor of degree \(d + 1\). Then \(h^0(L(-D))\geq 0\). But \(0 > \deg(L(-D))\), which is a contradiction.
If \(X\) is rational, \(L\) has degree d. Then \(h^0(L) \geq d + 1\). Then \(h^0(L)\geq d + 1\) since \(h^1(L) = 0\). Conversely, suppose there is a line bundle \(L\) of degree \(d > 0\) and space of global sections \(h^0(L) = d + 1\).
Now let \(D\) be an effective divisor of degree \(d - 1\). Then \(h^0(L(-D)) \geq 2\) and the degree of \(L(-D)\) is \(1\) (since \(L\) is degree \(d\) and \(-D\) is of degree \(1 - d\)). This again gives a map \(X\dashrightarrow \mathbb P^1\) of degree \(1\) and hence \(X\cong \mathbb P^1\).
Now for the most important corollary:
Corollary
- Any line bundle of degree \(2g\) is base point free
- Any line bundle of degree \(2g + 1\) is very ample.
Proof
If \(\deg(L) \geq 2g\), then \(h^0(L) = d - g + 1\) and \(h^0(L(-p)) = d - 1\) for all \(p\). This implies there are no base points.
If \(\deg(L)\geq 2g + 1\) then \(h^0(L) = h^0(L(-p-q)) + 2\) for all \(p,q\).