Bergman kernel via Riemann Theta function

Question

Let $\Sigma$ be a Riemann surface of genus $g$. A Bergman kernel $B(p,q)$ is a bilinear meromorphic form on $\Sigma \times \Sigma$ with poles of order $2$ along the diagonal $p = q$ and holomorphic everywhere else. It can be defined explicitly as \begin{equation} B(p,q) = d_qd_p\log\theta(u(p) - u(q) - \Delta, \tau) \end{equation} as in (2-23) of https://arxiv.org/pdf/0811.3531.pdf where $\theta(.,\tau)$ is the Riemann theta function, $u(.)$ is the Abel map and $\Delta$ is an odd characteristic. Since $\Delta$ is a zero of order $1$ of $\theta$, I can see that after differentiating twice, we will get poles of order $2$ when $p = q$.

My question is how do I show that $B(p,q)$ defined this way does not have poles anywhere else?

My problem is $\theta(.,\tau)$ has $g$ zeros of order $1$. In this case $p = q$ is one of its zero, the rest of its zeros are points in $D_\Delta$ where $D_\Delta \geq 0$ is the effective divisor defined by $\Delta = u(D_\Delta) + K$, where $K$ is the Riemann constant. So I can't convince myself why wouldn't $B(p,q)$ has poles at $p = $ one of these points in $D_\Delta$ as well?

Answer 1

As a function of $p$, $\partial_p\log\theta(u(p)-u(q)-\Delta,\tau) $ has a simple pole of residue 1 at points of $D_\Delta$, for each $q$. The key point here is that this $D_\Delta$ does not depend on $q$, i.e., the Laurent-Taylor expansion in both $p$ and $q$ at $p_0\in D_\Delta$ and any $q_0$ will look like $\frac{1}{p-p_0}+\mathrm{Taylor}(p-p_0,q-q_0),$ in a suitable coordinate chart. Differentiating with respect to $q$ now kills the singularity.