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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.