Past exam of the algebraic curves course of the University of Oxford / 2023 / 3

A divisor on an algebraic curve is a finite formal sumIts degree of a divisor is $\mathrm{deg}D={\sum }_p{n}_p$. It is an effective divisor, written $D\ge 0$, when every $n_p\ge 0$.

A meromorphic differential on a Riemann surface is a one-form represented in every local holomorphic coordinate $t$ by $a(t)dt$, where $a$ is a meromorphic function. Under a coordinate change $t=t(u)$, its coefficient becomes $a(t(u))t^{\prime }(u)$, so these expressions agree on overlapping charts. Since $t^{\prime }(u)$ is a nonvanishing holomorphic function for a valid coordinate change, the order of a zero or pole of the differential is independent of coordinate. Equivalently, for a fixed nonconstant meromorphic function $h$, any such differential can be written $fdh$ with $f$ meromorphic.

A canonical divisor is the divisor on an algebraic curveof a nonzero meromorphic differential on a Riemann surface $\eta$. If ${\eta }^{\prime }=f\eta$, then $({\eta }^{\prime })=(\eta )+(f)$, so all canonical divisors are related by linear equivalence of divisors. Their degree is

The Riemann-Roch theorem states that, for every divisor on an algebraic curve $D$ and any canonical divisor $K$ on the nonsingular projective curve $C$ of genus $g$,The term $\ell (K-D)$ measures the failure of the naive count of conditions on meromorphic functions.

Choose a nonzero meromorphic differential on a Riemann surface $\eta$ and write $K=(\eta )$. Multiplication defines a linear mapIt takes values in the holomorphic differential forms because $(f\eta )=(f)+K\ge 0$. It is injective since $\eta \ne 0$. It is surjective because, for any holomorphic differential form $\omega$, the ratio $\omega /\eta$ is a meromorphic function and has $(\omega /\eta )+K=(\omega )\ge 0$. Thus this map identifies $\mathrm{HD}(C)$ with the canonical Riemann-Roch space.

Furthermore, $\mathcal{L}(0)=\mathbb{C}$: a meromorphic function with no poles is holomorphic on the compact Riemann surface $C$, and hence constant by the maximum modulus principle. Applying the Riemann-Roch theorem with $D=0$ givesTherefore

The generator $\eta =fdh$ is a nonzero holomorphic differential form. Its canonical divisor $(\eta )$ is an effective divisor, since there are no poles. In genus one, the degree of a divisor formula in part (a) givesAn effective divisor of degree zero has every coefficient zero. ThusThis proves nonvanishing of holomorphic differentials in genus one. It concerns the differential itself: a zero or pole of the particular coefficient $f$ may be cancelled by the local behavior of $dh$.

An explicit answer, initially written where $yz\ne 0$, is the meromorphic differential on a Riemann surfaceThe ratios $x/z$ and $y/z$ are functions on $C$, so the expression does not depend on the representative of its homogeneous coordinates. It is a scalar multiple of an invariant differential on an elliptic curve.

On the complex torus $\mathbb{C}/\Lambda$, the differential $dw$ descends from $\mathbb{C}$ because translation by any element of the period lattice leaves $dw$ unchanged. It is holomorphic and nowhere zero. The given biholomorphism satisfies, wherever the displayed quotient is initially defined,Consequently, $\eta =({\Phi }^{-1}{)}^{\ast }(dw)$ extends across every apparent denominator zero and across $[0:1:0]$ as a nonzero holomorphic differential form on all of $C$.

One can also check the apparent exceptional points directly. On $z=1$, put $P(X)=4{\prod }_i(X-e_i)$, so $Y^2=P(X)$. At $Y=0$ the roots are distinct, and differentiating giveswhich is regular and nonvanishing there. At $[0:1:0]$, use $u=x/y$, $v=z/y$. The equation is $v=4{\prod }_i(u-e_{i}v)$, so $u$ is a local parameter on a smooth algebraic curve and $v=4u^3+O(u^5)$. Hencealso regular and nonvanishing.

Only distinctness of the $e_i$ is needed. The paper additionally writes $e_i\ne 0$, which is not true for every period lattice: for $\Lambda =\mathbb{Z}+\mathrm{i}\mathbb{Z}$, the symmetry $\wp (\mathrm{i}w)=-\wp (w)$ makes the half-period value at $w=(1+\mathrm{i})/2$ zero, since $\mathrm{i}w\equiv w(\mathrm{mod}\Lambda )$. This unnecessary restriction has no effect on the answer.

First fix distinct $q_1,\dots ,q_k$ and let $Q=q_1+\cdots +q_k$. A negative-degree divisor has no sections: if $0\ne f\in \mathcal{L}(D)$, then $(f)+D$ is an effective divisor of degree $\mathrm{deg}D$, since a principal divisor has degree zero, so $\mathrm{deg}D$ cannot be negative. As $\mathrm{deg}K=0$ and $\mathrm{deg}(K-Q)=-k<0$, the Riemann-Roch theorem therefore gives

Now choose distinct $p_1,\dots ,p_k$ outside the support of $Q$, and put $P=p_1+\cdots +p_k$. A function with the required zeros and poles must be a nonzero element of $\mathcal{L}(Q-P)$. Conversely, if $0\ne f\in \mathcal{L}(Q-P)$, then $(f)+Q-P$ is an effective divisor of degree zero, hence is zero. It follows that $(f)=P-Q$, giving precisely the required simple zeros and simple poles, with no others. Thus the existence question is exactly whether $\mathcal{L}(Q-P)\ne 0$.

Take a basis $f_1,\dots ,f_k$ of the Riemann-Roch space $\mathcal{L}(Q)$. Evaluation at the $p_i$ defines the linear mapwhose kernel is $\mathcal{L}(Q-P)$. The evaluation criterion for a prescribed principal divisor becomesThe crucial point is that this determinant does not vanish identically; a count of unknowns and equations alone would not prove this.

To verify it, begin with $V_0=\mathcal{L}(Q)$, of dimension $k$. Whenever $V_{j-1}\ne 0$, choose a nonzero $f\in V_{j-1}$. Its zeros are finite. Choose $p_j$ outside those zeros, the support of $Q$ and the previously chosen points. Evaluation at $p_j$ is then a nonzero linear functional on $V_{j-1}$, soAfter $k$ steps, $V_k=0$. Thus there exists a tuple of distinct points with $\Delta \ne 0$.

The entries of this determinant vary holomorphically away from the $q_i$, and are rational functions on the nonsingular projective curve. Its vanishing is therefore a proper closed exceptional locus on this configuration space; its complement is open and dense in the Zariski topology. For a generic choice of the $p_i$, $E_P$ is invertible andThis holds for every fixed tuple of distinct $q_i$, which proves the assertion for generic choices of both sets of points. At $k=1$, it is especially transparent: $\mathcal{L}(q_1)=\mathbb{C}$, so no nonzero member can vanish at a distinct $p_1$.