Riemann-Roch 07: Genus One
Riemann-Roch notes / VII
The elliptic curve case
On an elliptic curve, the canonical divisor is trivial and the dimension count shifts by one from the genus zero case.
Let \(X\) be a genus one compact Riemann surface. Analytically,
\[X\cong \mathbb C/\Lambda.\]After choosing a base point, it is an elliptic curve. The invariant differential \(dz\) descends to the torus and has no zeros or poles, so
\[K\sim 0.\]Riemann-Roch becomes
\[\ell(D)-\ell(-D)=\deg(D).\]Positive effective divisors
If \(D\) is effective and \(\deg(D)>0\), then \(-D\) has negative degree. Hence \(\ell(-D)=0\), and therefore
\[\ell(D)=\deg(D).\]This is the basic genus one count.
A single pole is not enough
For one point \(P\),
\[\ell(P)=1.\]Thus a meromorphic function on an elliptic curve cannot have only one simple pole unless it is constant.
For two points,
\[\ell(P+Q)=2,\]so a nonconstant function exists with poles bounded by \(P+Q\).
On the concrete elliptic curve \(E:y^2=x^3-x\), this appears in the familiar functions
\[L(2O)=\operatorname{span}\{1,x\}, \qquad L(3O)=\operatorname{span}\{1,x,y\}.\]The first space has dimension \(2\) because \(x\) has a double pole at \(O\). The second has dimension \(3\) because \(y\) has a triple pole at \(O\). These two computations are the genus one analogue of polynomial bases on \(\mathbb P^1\).
Geometric reason
A nonconstant meromorphic function with one simple pole would define a degree one map \(X\to\mathbb P^1\). A degree one map between smooth compact curves is an isomorphism. That would force a genus one curve to be the projective line, which is impossible.
The equality \(\ell(P)=1\) is the dimension-count version of that geometric obstruction.
The shift from genus zero
On \(\mathbb P^1\), a divisor of degree \(n\) gives dimension \(n+1\). On an elliptic curve, a positive effective divisor of degree \(n\) gives dimension \(n\). The missing one dimension is exactly the genus term in Riemann-Roch.