Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 18 / 4 / Solution

Assume $X$ is a connected compact Riemann surface of geometric genus $g\ge 1$. Work first in ${\mathrm{Pic}}^{g-1}(X)$, translating back to the Jacobian variety by tensoring with ${\mathcal{O}}_X(-(g-1)p_0)$ when desired. PutThis is the image of the Abelian sum map from $X^{(g-1)}$. The Riemann singularity theorem asserts that this image is a reduced Cartier divisor and thatHere multiplicity means the order of a local defining equation in the maximal ideal of the smooth ambient Jacobian variety. Consequently the singular points are precisely the classes with at least two independent sections. We will prove the multiplicity equality, including the nonvanishing of its proposed leading term.

Choose an effective divisor $E$ of degree $n\ge g$, and a local holomorphic family $\mathcal{L}$ of degree-$g-1$ line bundles about $L$ on $X\times U$, where $U$ is a small neighborhood in ${\mathrm{Pic}}^{g-1}(X)$. Such a family is obtained by varying the Čech cocycles of a line bundle. Since $\mathrm{deg}{L}^{\prime }(E)>2g-2$ for all $L^{\prime }\in U$, Serre duality gives $H^1(L^{\prime }(E))=0$; the Riemann-Roch theorem gives $h^0(L^{\prime }(E))=n$. The cohomology and base change for line bundles on a curve then identifiesas vector bundles of rank $n$. Evaluation defines a matrix $M:A\to B$. For each $L^{\prime }$, its exact sequence isThus $f=\det M$ vanishes exactly on $\Theta$.

The function $f$ is not identically zero. To exhibit a degree-$g-1$ line bundle with no section, choose $g$ distinct points $p_1,\dots ,p_g$ giving independent evaluation functionals on $H^0(K_X)$. Such points can be chosen successively because a nonzero holomorphic differential form cannot vanish everywhere. For $F=\sum p_i$, the Riemann-Roch theorem gives $h^0({\mathcal{O}}_X(F))=1$. Choose $q$ outside $F$. The unique section of ${\mathcal{O}}_X(F)$ does not vanish at $q$, so ${\mathcal{O}}_X(F-q)$ has no section. Therefore the evaluation determinant genuinely cuts out a hypersurface. Its support is irreducible, since it is the image of the irreducible symmetric product of a curve $X^{(g-1)}$.

Fix now $L\in \Theta$, and put $r=h^0(L)=h^1(L)$, the equality following from the Riemann-Roch theorem. The matrix $M(L)$ has an invertible $(n-r)$ by $(n-r)$ block. After invertible holomorphic row and column operations, its local form iswhere $N$ is $r$ by $r$. More explicitly, eliminate the invertible block by its Schur complement; the discarded determinant is a holomorphic unit. Hence $f$ is a unit times $\det N$. Every entry of $N$ lies in the maximal ideal, so its order is at least $r$.

The linear term of $N$ is the cup-product mapTo verify this, deform the transition functions as $g_{ij}(1+\epsilon {\xi }_{ij})$ and try to lift a section $s$ to $s+\epsilon s^{\prime }$. Its first-order gluing equation has right side ${\xi }_{ij}{s}_j$, whose Čech cohomology class is exactly the obstruction to the lift. The kernel-to-cokernel derivative of the evaluation matrix gives that same obstruction. Under Serre duality, if $s_1,\dots ,s_r$ and $t_1,\dots ,t_r$ are bases of $H^0(L)$ and $H^0(K_X\otimes L^{-1})$, thenWe must show that this matrix is invertible for some $\xi$; otherwise the lower bound on multiplicity would not be an equality.

For each of these two $r$-dimensional section spaces, there are $r$ distinct points with independent evaluations. Choose them successively, using a section in the remaining kernel to find the next point. The set of such $r$-tuples is a nonempty open subset of $X^r$. The two open sets intersect because $X^r$ is irreducible, and we may avoid their diagonals. Choose $q_1,\dots ,q_r$ in that intersection and trivialize $L$ and $K_X$ there. Both matricesare invertible. Let $\xi \in H^0(K_X{)}^{\ast }=H^1({\mathcal{O}}_X)$ be the sum of the evaluation functionals at these points, with any nonzero weights $c_k$. Thenis invertible. It follows that the degree-$r$ polynomial $\det N_1$ is nonzero, so ${\mathrm{ord}}_{L}f=r$. This proves the needed invertible cup-product direction for a line bundle on a curve by an explicit construction.

We still need to identify this determinant hypersurface with the reduced image, rather than a multiple of it. Choose $g-1$ distinct points imposing independent conditions on $H^0(K_X)$. For their sum $D$, the Riemann-Roch theorem gives $h^0(D)=h^0(K_X-D)=1$. At its class, the calculation above gives order one. The determinant hypersurface has irreducible support, so its sole divisor on an algebraic curve coefficient is one everywhere. A local hypersurface in a smooth space with these divisor on an algebraic curve coefficients is reduced. Thus it is exactly the reduced image $W_{g-1}$, and the multiplicity formula holds for that image itself.

The argument gives the useful additional descriptionFor $r=1$, its nonzero linear equation is $⟨\xi ,st⟩=0$. This agrees with the derivative of the Abelian sum map: if $\mathrm{div}(s)=D$, then $st$ is the unique holomorphic differential form vanishing along $D$. For $r\ge 2$, the local equation has no linear term and the point is singular. When $g=1$, $W_0$ is the single trivial line bundle in ${\mathrm{Pic}}^0(X)$, with multiplicity one; the argument includes this case using the empty effective divisor. In genus zero there is no $W_{g-1}$ of effective divisors on an algebraic curve, so that hypothesis is necessary.