2
$\begingroup$

Let $X$ be a compact Riemann surface, $u$ a meromorphic function on $X$ with divisor supported on a set of points $\{P_1, ..., P_n\}$, and $f$ a meromorphic function on $X$ such that $f$ has no pole or zeros on $\{P_1, ..., P_n\}$. Let $d\log(u)$ be the logarithmic differential form of $u$. Do we have the following equality in $\mathbf{C}^{\times}$: $$\exp(\int_{div(f)} d\log(u)) = \prod_{i=1}^n f(P_i)^{\text{ord}_{P_i}(u)}\text{ ?}$$ Here, $\text{div}(f)$ is the divisor of $f$, and the integral is well-defined (independant of the choice of the path) up to $2\pi i \mathbf{Z}$.

This identity is true if for all $i$ we have $f(P_i)=1$, as it relates to the analytic Abel-Jacobi description of the generalized Jacobian of $X$ with respect to the divisor $(P_1)+...+(P_n)$.

$\endgroup$
3
$\begingroup$

This is true. Given two points $a,b\in X$ with $u(a)$ and $u(b)\neq 0$, one can choose a path from $a$ to $b$ and a determination of $\log u$ along that path, so that $\exp\int^b_a d\log u=u(b)/u(a)$. Thus, if $ \operatorname{div}(f)=\sum n_jQ_j$, the left-hand side is $\prod u(Q_j)^{n_j}$.

Recall that if $f,g$ are two meromorphic functions on $X$, of order $n$ and $m$ at a point $P$, the tame symbol $(f,g)_{P}$ is $(-1)^{nm}[g^{n}/f^m](P)$. Thus, if $\operatorname{div}(u)=\sum m_iP_i$, we have $(f,u)_{P_i}= f(P_i)^{-m_i}$, $(f,u)_{Q_j}= u(Q_j)^{n_j}$, and $(f,u)_{P}=1$ at the other points. Now the product formula $\prod (f,u)_P=1$ gives your equality.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.