Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 13 (Curves and Their Jacobians)

Let $X$ be a connected compact Riemann surface of genus $g$. Choose an oriented symplectic basis $a_1,\dots ,a_g,b_1,\dots ,b_g$ of $H_1(X,\mathbb{Z})$, with $a_i\cdot b_j={\delta }_{ij}$. For a closed complex differential form $\alpha$, write $A_i(\alpha )={\int }_{a_i}\alpha$ and $B_i(\alpha )={\int }_{b_i}\alpha$. The Riemann bilinear relations begin with the identityTo prove it, cut $X$ along the chosen cycles to obtain a fundamental polygon $P$. On this simply connected polygon, a closed differential form has a primitive $F$, with $dF=\alpha$. Stokes theorem gives ${\int }_P\alpha \wedge \beta ={\int }_{\partial P}F\beta$. The two copies of each edge have opposite orientations, and their values of $F$ differ by the period acquired in passing between them. Pairing the copies of the $a_i$ and $b_i$ edges gives, respectively, $-B_i(\alpha )A_i(\beta )$ and $A_i(\alpha )B_i(\beta )$. Summing proves the identity; the convention $a_i\cdot b_i=1$ fixes its sign.

For holomorphic one-forms $\omega ,\eta$, their wedge product is zero because locally both are multiples of $dz$. Thus the first holomorphic Riemann bilinear relations isFor a nonzero holomorphic one-form $\omega =f(z)dz$, the second relation follows by applying the bilinear identity to $\omega ,\overline{\omega }$:In particular, a holomorphic one-form with all $a$-periods zero must vanish. Since $\dim H^0(X,K_X)=g$, the map taking a holomorphic one-form to its $a$-periods is an isomorphism. We can therefore choose normalized holomorphic one-forms ${\omega }_1,\dots ,{\omega }_g$ with ${\int }_{a_i}{\omega }_j={\delta }_{ij}$. Put $B_{ij}={\int }_{b_i}{\omega }_j$. The first relation applied to ${\omega }_j,{\omega }_k$ gives $B_{jk}=B_{kj}$, and the second applied to ${\sum }_j{c}_j{\omega }_j$ gives $2{\overline{c}}^{,t}(\mathrm{Im}B)c>0$ for $c\ne 0$. ConsequentlyThese conclusions make the construction of the Jacobian variety work. Integration identifies the image of $H_1(X,\mathbb{Z})$ in $H^0(X,K_X{)}^{\ast }$ with the period lattice $\Lambda ={\mathbb{Z}}^g+B{\mathbb{Z}}^g$. The real-linear map $(m,n)\mapsto m+Bn$ is invertible: its imaginary part is $Yn$, and $Y$ is invertible. Hence $\Lambda$ is a discrete lattice of full real rank $2g$, andis a compact complex torus, rather than a non-Hausdorff quotient. Changing the symplectic basis changes the displayed coordinates but not this intrinsic quotient.

The Riemann bilinear relations also supply its polarization of a complex torus. Use the Hermitian form $H(z,w)=z^t{Y}^{-1}\overline{w}$, linear in the first argument, and set $E=-\mathrm{Im}H$. Symmetry of $B$ givesThus $E$ is an integral positive Riemann form on a complex torus. Its matrix on the displayed lattice basis is the unimodular symplectic matrix. The corresponding positive holomorphic line bundle, equivalently the one constructed from theta automorphy in Solution 4, gives a principal polarization of a complex torus. By the Kodaira embedding theorem this makes the complex torus projective, hence an abelian variety. The normalization $E=-\mathrm{Im}H$ here is important because $H$ was chosen linear in its first argument. If $g=0$, the spaces of holomorphic one-forms and periods are zero and the Jacobian variety is a point.

Fix $p_0\in X$ and use the normalized holomorphic one-forms and period lattice of Solution 1. The Abel-Jacobi map of a compact Riemann surface isChanging paths changes the vector by a period, so this is well defined and holomorphic. Adding these vectors gives a symmetric map on $X^d$ and therefore the Abelian sum map $u_d:X^{(d)}\to \mathrm{Jac}(X)$. Locally, even where points coincide, symmetric holomorphic functions descend to the elementary symmetric coordinates of the symmetric product of a curve. For any divisor $D={\sum }_p{n}_{p}p$ of degree zero, define $u(D)={\sum }_p{n}_{p}u(p)$. Abel's theorem says that $D$ is a principal divisor if and only if $u(D)=0$. Equivalently, two effective divisors of the same degree have the same Abel sum exactly when they are linearly equivalent.

We first establish the residue version of the bilinear identity. Let $\eta$ be a meromorphic differential on a Riemann surface with only simple poles and total residue zero, and let $F={\int }_{p_0}^p\omega$ on the cut polygon, for a holomorphic one-form $\omega$. Excise small disks about the poles. Since $d(F\eta )=\omega \wedge \eta =0$ off the poles, the outer boundary integral is the sum of the positive small-circle integrals. The latter are $2\pi i{\mathrm{res}}_p(\eta )F(p)$. Pairing the outer edges as in Solution 1 therefore givesThe paths and the cut polygon are fixed together in this equality. Changing them changes the Abel vector by its period lattice, which is precisely the ambiguity we need.

For the sufficiency direction, we need a differential of the third kind with prescribed residues $n_p$. Let $S$ be the reduced support of $D$, consisting of $s$ points. When $s>0$, the Riemann-Roch theorem gives $h^0(K_X(S))=g-1+s$. The residue map from this space has kernel $H^0(K_X)$, of dimension $g$: a differential with at most a simple pole and residue zero has no pole. Its image therefore has dimension $s-1$. The residue theorem puts the image inside the $s-1$ dimensional space of tuples with sum zero, so it is all that space. In particular there is a meromorphic differential on a Riemann surface $\eta$ with residues $n_p$. Subtracting a combination of the normalized holomorphic one-forms makes all its $a$-periods zero. The bilinear identity then sayswhere on the right we temporarily choose the representative determined by the paths. If $D=0$, we simply take $\eta =0$.

Suppose now that $u(D)=0$ in the Jacobian variety, so the chosen vector is $m+Bn$, with $m,n\in {\mathbb{Z}}^g$. Replace $\eta$ byIts $a$-periods are $-2\pi i{n}_j$ and its $b$-periods are $2\pi i{m}_i$. The periods around the punctures are $2\pi i{n}_p$. These loops together generate the homology of the punctured Riemann surface, so every period of ${\eta }^{\prime }$ lies in $2\pi i\mathbb{Z}$. Consequently $f(p)=\exp ({\int }^p{\eta }^{\prime })$ is a single-valued nonzero holomorphic function off the support of $D$. At a support point, ${\eta }^{\prime }=n_{p}dt/t+$ a holomorphic differential, so $f=t^{n_p}$ times a nonvanishing holomorphic function. It extends meromorphically and has divisor exactly $D$. This proves sufficiency, including negative coefficients.

Conversely, if $D=\mathrm{div}(f)$, take $\eta =df/f$. Its residues are the coefficients of $D$, while its $a$- and $b$-periods are integer multiples of $2\pi i$, by the winding numbers of $f$ along these loops. Writing them as $2\pi ia$ and $2\pi ib$, the residue bilinear identity gives $u(D)=b-Ba\in \Lambda$. Thus $u(D)=0$ in the Jacobian variety, proving necessity and henceHere the displayed necessity of degree zero is also the residue theorem applied to $df/f$.

There is a useful geometric version of the conclusion. For an effective divisor $D$, the fiber of $u_d$ through $D$ consists precisely of the effective divisors linearly equivalent to $D$. A nonzero section of ${\mathcal{O}}_X(D)$ is a meromorphic function $f$ with $\mathrm{div}(f)+D\ge 0$, and its zero divisor as a section is $\mathrm{div}(f)+D$. Two such sections give the same effective divisor exactly when their ratio is a nonzero constant. Thus the fiber is the complete linear system of a divisor $|D|=\mathbb{P}{H}^0(X,{\mathcal{O}}_X(D))$.

Let $D={\sum }_p{r}_{p}p\in X^{(d)}$. The derivative of the Abelian sum map is most naturally stated using principal parts. The tangent space of the symmetric product of a curve at $D$ is $H^0({\mathcal{O}}_D(D))$, and the tangent space of the Jacobian variety is $H^0(K_X{)}^{\ast }$. With these identifications,Its dual is restriction of holomorphic one-forms to the length-$d$ divisor scheme. In particular,The last equality is the Riemann-Roch theorem. These formulas apply to repeated points as well as distinct ones.

Here is a direct proof of the repeated-point Abel differential formula. At a multiplicity-$r$ point, choose a local coordinate $t$ with $t(p)=0$. A nearby local divisor is described by the roots $t_1,\dots ,t_r$, or, without making an ordering choice, by their elementary symmetric functions $e_1,\dots ,e_r$:These $e_k$ are smooth local coordinates on the symmetric product of a curve, including at the repeated divisor. Under a first-order deformation, minus the logarithmic variation of this defining polynomial is the principal partThe sign can be checked at $r=1$: moving the point from $0$ to $ϵa$ gives $v=a/t$, and the Abel integral changes by $a\omega (p)$.

Write $\omega =({\sum }_{j\ge 0}{c}_j{t}^j)dt$ and its local primitive as $F(t)={\sum }_{j\ge 0}{c}_j{t}^{j+1}/(j+1)$. The local contribution to the Abelian sum map is ${\sum }_{i}F(t_i)$. Newton's identities, linearized at $e_1=\cdots =e_r=0$, giveSubstituting in the primitive therefore givesSumming over the support proves the theorem. In particular, a multiplicity-$r$ point tests the first $r$ coefficients of a holomorphic one-form, not just its value. The residue pairing between principal parts of order at most $r$ and differential jets through order $r-1$ is perfect. Thus the annihilator of the derivative is precisely the holomorphic one-forms vanishing to order at least $r_p$ at every $p$, namely $H^0(K_X(-D))$.

There is also a sheaf cohomology description that explains its kernel geometrically. The sequencegives a connecting homomorphism $\delta :H^0({\mathcal{O}}_D(D))\to H^1({\mathcal{O}}_X)$. Represent a principal part by meromorphic lifts on coordinate disks; their differences on overlaps give its Čech cohomology class. Under Serre duality, pairing this class with $\omega$ is the sum of the residues of $v_p\omega$, so $\delta$ is the displayed derivative. Exactness shows that its kernel is $H^0({\mathcal{O}}_X(D))/\mathbb{C}$, of dimension $h^0({\mathcal{O}}_X(D))-1$, in agreement with the tangent directions to the complete linear system of a divisor in Solution 2.

Two consequences will be useful below. If $d\ge 2g-1$, then $K_X(-D)$ has negative degree and no nonzero sections, so $u_d$ is a submersion everywhere. Its image is open, and it is also closed because $X^{(d)}$ is compact. Since the Jacobian variety is connected, $u_d$ is surjective. Thus the points $u(p)$ generate the Jacobian variety as a group. Together with Abel's theorem, this identifies the degree-zero Picard group with the Jacobian variety: every line bundle has a meromorphic section and thus a divisor representation, injectivity is Abel's theorem, and surjectivity follows from $u_d(D)-du(p_0)$.

For $g\ge 1$, one can choose $g-1$ distinct points imposing independent conditions on $H^0(K_X)$: at each step a surviving nonzero holomorphic one-form has only finitely many zeros, so choose a point where it does not vanish. Their sum $D$ has $h^0(K_X(-D))=1$ and hence $h^0({\mathcal{O}}_X(D))=1$. The derivative has rank $g-1$ there. Consequently $W_{g-1}=u_{g-1}(X^{(g-1)})$ is an irreducible subvariety of dimension $g-1$. This also covers $g=1$, when $X^{(0)}$ and $W_0$ are points.

For a symmetric matrix $B$ with $Y=\mathrm{Im}B$ positive definite, the Riemann theta function isOn any compact set of $z$, the absolute value of the summand is bounded by $\exp (-c|n{|}^2+C|n|)$ for some $c>0$, so the series and all its derivatives converge uniformly there. It is therefore entire. Its constant Fourier coefficient in $\mathrm{Re}z$ is $1$, so it is not identically zero. Reindexing $n$ gives its parity and automorphy laws:Although the theta function is not itself a function on $J={\mathbb{C}}^g/\Lambda$, these factors define a holomorphic line bundle on $J$, and its zeros define a divisor $\Theta$ there.

For the period matrix of $X$, the Riemann vanishing theorem asserts that a fixed class $\kappa \in J$, the vector of Riemann constants in the convention used here, satisfiesWe prove both its support statement and its reduced divisor structure. This will make the multiplicity calculation in Solution 5 independent of any assumed multiplicity theorem.

The essential contour calculation is the theta pullback zero-divisor identity. Put $f_z(p)=\theta (u(p)-z,B)$, regarding this as a local expression for a section on $X$. If it is not identically zero, let $D_z$ be its zero divisor. We claimUse the cut polygon from Solution 1 and choose cuts avoiding the zeros. There ${\alpha }_z=d_p\log f_z$ is a meromorphic differential whose residues are the zero multiplicities. Under continuation by an $a_j$ period, $u$ changes by the integer vector $e_j$, so ${\alpha }_z$ is unchanged. Under a $b_j$ period, $u$ changes by $B{e}_j$, so automorphy gives ${\alpha }_z\mapsto {\alpha }_z-2\pi i{\omega }_j$. Pairing opposite boundary sides therefore givesFor example, the two $a_j$ sides, traversed with opposite orientations and separated by a $b_j$ translation, contribute ${\int }_{a_j}({\alpha }_z-({\alpha }_z-2\pi i{\omega }_j))$. This fixes the sign of the count.

For the first moment of the zeros, the residue theorem on the polygon gives, component by component,This is interpreted modulo the period lattice, with the lifts to the polygon fixed. Vary $z$ without letting zeros cross the cuts, and put $h=\delta \log f_z$. The coordinate $u_k$ does not vary, so integration by parts givesThere is no endpoint term: the closed polygon boundary returns both the Abel integral and $h$ to their starting values. Across an $a_j$ translation $h$ is unchanged, while across a $b_j$ translation it changes by $2\pi i\delta z_j$. Pairing the boundary edges consequently gives ${\int }_{\partial P}{\omega }_{k}h=-2\pi i{\sum }_j\delta z_j{\int }_{a_j}{\omega }_k=-2\pi i\delta z_k$. Thus $\delta u_g(D_z)=\delta z$, and $u_g(D_z)-z$ is constant. The calculation uses weighted zeros, so their collisions cause no problem. The parameters where $f_z$ is identically zero form a proper analytic subset: they are given by the vanishing of all its local Taylor coefficients, and parameters outside $\Theta$ are not in it. Its complement in the connected complex manifold $J$ is connected, so the constant is one class $\kappa$. This proves the pullback identity globally, with continuation of the zero divisors where necessary.

We must account for the parameters with identically zero pullback, rather than discard them. No irreducible component $C$ of $\Theta$ can consist entirely of such parameters. Otherwise parity would imply $C-u(p)\subset \Theta$ for every $p\in X$. The image of $C\times X$ under subtraction is irreducible, is contained in $\Theta$, and contains $C$ when $p=p_0$. Since $C$ is an irreducible component of the hypersurface, that image must be $C$. Hence $C-u(p)=C$ for all $p$. The surjectivity of the large-degree Abelian sum map, proved in Solution 3, says that these translations generate all of $J$. This would make the nonempty proper subset $C$ invariant under every translation, which is impossible.

A generic point $z$ of each component of $\Theta$ therefore has nonzero pullback. Since $\theta (-z)=\theta (z)=0$, its zero divisor contains $p_0$. Write $D_z=p_0+D^{\prime }$, with $D^{\prime }$ effective of degree $g-1$. The pullback identity gives $u_{g-1}(D^{\prime })=z+\kappa$. Each component of $\Theta$ is consequently contained in $W_{g-1}-\kappa$. By Solution 3 the latter is irreducible of dimension $g-1$, so any such component equals it. The zero locus is nonempty for $g>0$: any parameter with nonzero pullback has $g$ pullback zeros. We have proved equality of supports, in both directions.

Finally the automorphy factors make the First Chern class of the theta line bundle primitive. On the two-torus spanned by $a_i,b_j$, the logarithm of the $b_j$ factor changes by $-2\pi i{\delta }_{ij}$ after the $a_i$ translation; equivalently the transition for local frames has opposite winding. Thus $c_1(a_i,b_j)={\delta }_{ij}$, while $c_1(a_i,a_j)=c_1(b_i,b_j)=0$. In particular some integral pairing equals $1$. As its support is the single irreducible hypersurface $W_{g-1}-\kappa$, the zero divisor is $m(W_{g-1}-\kappa )$ for a positive integer $m$. Its integral First Chern class would then be divisible by $m$. Primitivity forces $m=1$, proving the displayed equality as reduced divisors.

The sign convention is $\kappa =u_g(D_z)-z$, so the translate is $W_{g-1}-\kappa$; a convention naming $-\kappa$ the Riemann constant reverses that displayed sign. For $g=1$, $\kappa =(1+B)/2$ modulo periods and the zero locus is the corresponding single point. For $g=0$, the empty-lattice theta series is $1$, its zero locus is empty, and the degree-$-1$ effective locus is empty; no negative symmetric product is required.

Identify ${\mathrm{Pic}}^{g-1}(X)$ with the Jacobian variety using the base point and the translation of Solution 4. A line bundle $L$ of degree $g-1$ corresponds to the point $z=u(D)-\kappa$ for any divisor representation $L\simeq {\mathcal{O}}_X(D)$. The Riemann singularity theorem isIn particular, the theta divisor is singular exactly at the degree-$g-1$ line bundles with at least two independent sections. We prove the equality, including the upper bound on multiplicity.

Choose a fixed effective divisor $A$ of degree $N\ge g$, and a local holomorphic family $L_{\eta }$ representing nearby points of ${\mathrm{Pic}}^{g-1}(X)$. Such a family can be constructed locally from the holomorphic exponential sequence, by varying line-bundle transition functions through exponentials of Čech cohomology representatives. By Serre duality, $H^1(L_{\eta }(A))$ is dual to $H^0(K_X\otimes L_{\eta }^{-1}(-A))$, whose degree is $g-1-N<0$, so it vanishes. The Riemann-Roch theorem gives $h^0(L_{\eta }(A))=N$. By cohomology and base change for line bundles on a curve, these spaces form a holomorphic vector bundle $E$ of rank $N$. The spaces $H^0(L_{\eta }(A){|}_A)$ form another holomorphic vector bundle $F$ of rank $N$. Evaluation gives a holomorphic square matrix $T(\eta )$ locally, with exact sequenceConsequently $\det T(\eta )=0$ exactly when $L_{\eta }$ has a nonzero section, namely on the support of the theta divisor.

At the point $L$, put $r=h^0(L)=h^1(L)$, the equality being Riemann-Roch theorem. An invertible minor of size $N-r$ persists near this point. Holomorphic row and column operations, or the Schur complement, givewhere $M$ is an $r\times r$ holomorphic matrix. This is the Schur complement presentation of a theta singularity. Its linear term in a tangent direction $\xi \in H^1({\mathcal{O}}_X)$ is the cup product mapTo see this concretely, vary transition functions $g_{ij}$ of $L$ to $g_{ij}(1+ϵ{\xi }_{ij})$. A section $s_i$ lifts to $s_i+ϵs_i^{\prime }$ precisely when its first-order compatibility equation can be solved; the obstruction is the cocycle ${\xi }_{ij}{s}_j$. This is $\xi ⌣s$. Passing to the kernel and cokernel of $T(0)$ gives exactly the derivative of the reduced matrix $M$, proving the assertion without assuming a multiplicity formula.

Before extracting the order of $\det M$, we verify that $\det T$ is a reduced equation of the theta divisor. Generic points of $W_{g-1}$ have $h^0(L)=1$ by Solution 3. At such a point, both $H^0(L)$ and $H^0(K_X\otimes L^{-1})$ are one-dimensional. Their nonzero sections $s,t$ have nonzero product $st\in H^0(K_X)$. Choose $\xi \in H^1({\mathcal{O}}_X)$ with $⟨\xi ,st⟩\ne 0$ under Serre duality. The scalar $d{M}_0(\xi )$ is nonzero, so $\det T$ has a simple zero generically on the irreducible theta divisor. Since its support is exactly that irreducible hypersurface, it defines the reduced divisor. Solution 4 independently proved the same reduced structure for the Riemann theta function. On the smooth ambient Jacobian variety, local equations of the same reduced divisor differ by a nonvanishing holomorphic unit. Hence the orders of the local theta equation and $\det M$ agree everywhere.

Every entry of $M$ vanishes at the point, so each term in its determinant has order at least $r$. This proves ${\mathrm{mult}}_z\Theta \ge r$. To obtain equality, we exhibit an invertible cup-product direction for a line bundle on a curve. Set $V=H^0(L)$ and $W=H^0(K_X\otimes L^{-1})$, both of dimension $r$. There are $r$ distinct points $p_1,\dots ,p_r$ at which evaluation is an isomorphism for $V$: choose them inductively, since a surviving nonzero section has only finitely many zeros. The same holds for $W$. The nonvanishing of each evaluation determinant defines a nonempty open subset of the irreducible variety $X^r$. Their intersection, also avoiding the diagonals, is therefore nonempty. Choose points in it and local coordinates and compatible trivializations of $L$ and $K_X\otimes L^{-1}$ at each point.

Let $\xi$ be the connecting homomorphism class of the meromorphic principal parts ${\lambda }_i/t_i$ at these points, with all ${\lambda }_i\ne 0$. The residue form of Serre duality givesIf $S$ and $U$ are the two invertible evaluation matrices, the matrix of this pairing is $S^t\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_r)U$, and is invertible. Thus the cup product map $\xi ⌣-:H^0(L)\to H^1(L)$ is an isomorphism. Along the corresponding analytic parameter line,with nonzero leading coefficient. The order is at most $r$, completing the proof of the Riemann singularity theorem.

This argument also identifies the tangent cone to a theta divisor: its degree-$r$ equation is $\det (\xi ⌣-)$, which is not the zero polynomial. For $r=1$, the tangent hyperplane is $⟨\xi ,st⟩=0$, in agreement with the annihilator description of the derivative of the Abelian sum map. For $r\ge 2$ all first derivatives vanish, proving singularity. In genus $1$ the theta divisor is a single smooth point of multiplicity $1$. In genus $0$ the effective degree-$-1$ locus is empty, so there is no singularity statement at a point of it.