# N=2 Dualities; k-differentials on the riemann sphere and spectral curves

Currently I am working on my masters thesis about dualities in QFT and their geometric realizations. As of now, I am trying to understand the article 'N=2 Dualities" by Davide Gaiotto. On the internet I found some exercises related to the article (http://www.sns.ias.edu/pitp2/2010files/Gaiotto-Problems.pdf). My questions are about some of these exercises, which come from physics however are probably better answerable by mathematicians.

I will shortly summarize the exercise and then put my question forward. The full exercise is reachable via the link above.

Exercise 1: We first look at degree k meromorphic differentials with poles of order $k$ at $n$ points $z_i$ on the Riemann sphere: $\phi_k(z)=F(z)dz^k$. Here $F(z)$ is a rational function on the complex plane. If we want to know the behaviour of $\phi_k(z)$ at $\infty$ we change coordinates to $z'=1/z$ under which the k-differential transforms as $\phi_k(z')=F(1/z')(-dz'/(z')^2)^k$. Furthermore, these differentials are required to have fixed residues $\alpha_i$ on each $z_i$. The question then is how big the dimension is of the space of these k-differentials.

First of all it is unclear to me what precisely is meant with $$\phi_k(z) \approx \frac{\alpha_i}{(z-z_i)^k} dz^k +...$$. My interpretation is that $\phi_k(z)$ may be written as a fraction of two polynomials $f(z)/g(z)$ where $g(z)$ has $k^{th}$ order zeroes at $n$ points $z_i$ and $f(z)$ has $k(n-2)$ zeroes (to get the correct degree of the divisor of a k-differential on the Riemann Sphere, namely $-2k$). The zeroes of $f(z)$ we may choose freely (as long as we satisfy the fixed residues $\alpha_i$).

First I attempted to solve this with Riemann Roch. This led me to a counting of $k(n-2)+1$ free parameters, however this doesn't account for the fixed residues I think. Then, with fixed residues, I reasoned it should be $(k-1)(n-2)$ by counting the free parameters for a k-differential. For n k'th order poles one has $(n-2)k$ zeroes to freely choose (in order that the degree of the divisor of the k-differential is $-2k$) and one constant $c$ multiplying $f(z)$ .

However, for fixed residues, one has to subtract $n-1$ parameters (not $n$ since the residues sum to zero), which leads to the total of $(k-1)(n-2)$ free parameters. T his would also be the dimension of the vector space of k-differentials with n k'th order poles with fixed residues, since we can look at all linearly independent $F(z)'s$, ie different degrees of the polynomial $f(z)$ which may look like $\prod^l_{i=1}c(z-u_i)$ for $l\in {1,2,..,n}$, $u_i$ a zero and $c$ a constant.

Does anybody know if this counting and way of looking at the $F_i(z)$'s is correct?

3i)I guess my problems with this question depend very much on the definitions in question 1.

I tried to solve a simple example with $k=2$ and $n=3$:

$$x^2 + F_1(z) x + F_2(z) = 0$$

with

$$F_1(z)=\frac{c(z-u)}{(z-z_1)(z-z_2)(z-z_3)}$$

and

$$F_2(z)=\frac{d(z-v)(z-w)}{(z-z_1)^2(z-z_2)^2(z-z_3)^2}$$

According to my calculation in 1ii) it follows that only $w$ is a free parameter in this equation; $c$ and $u$ are completely determined by the fixed residues of $F_1$ and $d$ and $v$ are determined by the fixed residues of $F_2$ (and are functions of $w$).

When I try to solve this equation (with a change of variables to $y = x(z-z_1)(z-z_2)(z-z_3))$ with Mathematica, the expressions become very complicated and it does not follow that $x$ is independent of $w$ (as asked to prove in exercise 3ii)).

Does someone has an idea what mistakes I am making? Thanks, Sam

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