# Is there a Riemann-Roch like result for meromorphic differentials with all periods vanishing?

The classical Riemann-Roch theorem for Riemann surfaces makes a connection between the dimension of the space of meromorphic functions and the dimension of the space of meromorphic forms. Thus it is useful for the existence of meromorphic functions on a Riemann surface with prescribed singularities (divisor).

By taking differential, a meromorphic function is equivalent to a meromorphic form with vanishing periods, up to a constant, like that $1/z$ is mapped to $z^{-2} dz$. In this sense, Riemann-Roch theorem is a relation between the dimension of meromorphic forms with all periods vanishing and that of arbitrary meromorphic forms.

However, the mapping $f \to df$ is not surjective to all meromorphic forms with vanishing periods, but to only the Abelian differentials of the second kind.

My question is: Is there a generalized Riemann-Roch theorem, about the relation between the dimension of the space of Abelian differentials of the third kind with all periods vanishing and prescribed singularities, and the dimension of a related space of meromorphic forms?

I have a concrete problem in mind: On an elliptic curve, construct a meromorphic form with three singular points and both the two periods vanishing. Around the three singular points, the local behaviours are $(1/z + O(1))dz$, $(-1/z + O(1))dz$ and $(1/z^2 + O(1))dz$. I want to know the existence and uniqueness of the meromorphic form. This question seems not to be a direct consequence of the generalized Riemann-Roch theorem conjured above, but I think they must be related. If you can answer or give a hint to the concrete problem, I also appreciate.

For existence, suppose the points are $a$, $b$ and $c$ respectively. You can use Weierstrass $P$-function centered at $c$, namely $P(z-c)$ to get a function with double pole at $c$ starting with $1/z^2$ and no residue. Then the issue is to find a function (or form, doesn't matter for elliptic curve) with poles of order $1$ at $a$ and $b$. Then the residues will have to be opposite, and you will be done. By a shift, we may assume that $a+b=0$, and then
$1/(P(z)-P(a))$
• Lev, thank you for the solution to the concrete problem. Actually this is the technical problem I encountered. But unfortunately I did not state the problem clearly (I just edited it), so The answer I would like to get is not exactly your construction. The $1/z$, $-1/z$ and $1/z^2$ singularities are for the meromorphic form, but not the corresponding integral. The corresponding integral has two logarithmic singularities and one $-1/z$ singularity, respectively. Thus I doubt if the answer is possible to be expressed by Elliptic functions in an elementary way. Thank you for your answer anyway. – Dong Wang Sep 15 '13 at 5:25