Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 18 / 4

Put $n=3g-3$. To specify the Abel map of an algebraic curve into $X={\mathrm{Pic}}^{g-1}(C)$, useThe canonical-class twist is necessary: an untwisted sum has degree $3g-3$, rather than $g-1$. Let $L={\mathcal{O}}_C(2K_C)$. The Riemann-Roch theorem and Serre duality give $h^0(L)=n$ and $h^1(L)=0$.

On $C\times C^n$, write ${\Gamma }_i$ for the graph of the $i$th coordinate, $\mathcal{D}={\sum }_i{\Gamma }_i$, and $\pi$ for projection to $C^n$. The locally free sheaf $E_L={\pi }_{\ast }(L{|}_{\mathcal{D}})$ has rank $n$; at repeated points it records values and jets. Evaluation gives a morphism of rank-$n$ vector bundlesIts kernel at a tuple is $H^0(C,2K_C-\sum z_j)$. By the Riemann-Roch theorem, this space has the same dimension as $H^0(C,\sum z_j-K_C)$, since both line bundles have degree $g-1$. Thus its determinant vanishes exactly along ${\alpha }^{\ast }\Theta$. This is also an equality of Cartier divisors with multiplicities: the evaluation complex computes the sheaf cohomology of $L(-\mathcal{D})$; Serre duality identifies its determinant equation, up to an invertible factor, with the defining theta evaluation equation for $K_C\otimes L^{-1}(\mathcal{D})$.

To calculate the determinant line, set ${\mathcal{D}}_i={\Gamma }_1+\cdots +{\Gamma }_i$. Restriction to the extra graph givesProjection is exact on these finite relative divisors. On ${\Gamma }_i\cong C^n$, the first term is ${\mathrm{pr}}_i^{\ast }L\otimes \mathcal{O}(-{\sum }_{j<i}{\Delta }_{ij})$, where ${\Delta }_{ij}=\{z_i=z_j\}$. Taking determinants inductively yields the evaluation determinant on a product of curves:The determinant of the fixed space $H^0(C,L)$ contributes only a constant line. ThereforeThe diagonal subtraction is essential: ordinary evaluation determinants vanish when columns coincide, whereas restriction to a repeated-point Cartier divisor uses jets. The restriction exact sequences account for those factors in every characteristic.

Let $P$ now denote the degree-zero torsion line bundle represented by the point of $A[N]$, and put $L_P={\mathcal{O}}_C(2K_C)\otimes P$. Again the Riemann-Roch theorem gives $h^0(L_P)=n$ and $h^1(L_P)=0$. For translation, use ${\Theta }_P=\Theta +P$ as a translated subset. With ordinary $T_P(x)=x+P$, this is $T_{-P}^{\ast }\Theta$; the notation $t_P^{\ast }\Theta =\Theta +P$ therefore uses $t_P=T_{-P}$.

Let $s_P$ and $s_0$ be the evaluation determinants for the two chosen bases, interpreted as sections of $\det E_{L_P}$ and $\det E_L$. The kernel for the first determinant is $H^0(C,2K_C+P-\sum z_j)$. By Serre duality and the Riemann-Roch theorem its dimension is $h^0(C,\sum z_j-K_C-P)$. Thus the determinant vanishes along ${\alpha }^{\ast }{\Theta }_P$, including multiplicities by the same determinant-complex argument as in part (i). The quotient is a rational sectionwith zero-minus-pole Cartier divisor ${\alpha }^{\ast }({\Theta }_P-\Theta )$. The diagonal factors in the two determinants cancel.

Choose a trivialization $u:P^{\otimes N}\stackrel{\sim }{\to }{\mathcal{O}}_C$. This is possible because $P\in A[N]$. Taking the $N$th tensor power of $r$ and using ${⨂}_j{\mathrm{pr}}_j^{\ast }u$ produces a scalar rational function $R$ on $C^n$ withOn $X$, the Theorem of the square and $NP=0$ implyConsequently there is a rational function $f_P\in k(X{)}^{\times }$ with $\mathrm{div}(f_P)=N{\Theta }_P-N\Theta$. The map $\alpha$ is surjective, since every degree-$3g-3$ line bundle has a section by the Riemann-Roch theorem. The two rational functions ${\alpha }^{\ast }{f}_P$ and $R$ have the same Cartier divisor on the smooth projective connected product $C^n$; their quotient is an everywhere invertible regular function and hence a constant. Rescale $f_P$ to remove that constant. Away from the diagonals, ordinary evaluation giveswhere the $N$th power is interpreted with the chosen torsion trivialization. Equality as rational functions then extends from this dense open set to the whole product.

The torsion trivialization in a determinant quotient cannot be omitted if one uses raw meromorphic representatives. Explicitly, write $P={\mathcal{O}}_C(p)$ and choose $h$ with $\mathrm{div}(h)=Np$. If $r_{\mathrm{raw}}$ is the quotient of the evaluation matrices in these meromorphic frames, thenThus the displayed formula suppresses bundle trivializations; read as an unframed quotient of arbitrary meromorphic representatives, it needs precisely this extra product. Changes of bases or rescaling $u$ only change the allowable nonzero scalar normalization of $f_P$.