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

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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.

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 $w$).

EDIT: 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 that $\frac{\partial\lambda}{\partial v}$ is a holomorphic one form on the curve, where $\lambda = xdz$ as asked in exercise 3ii. The way to see this, I think, is that the residue of $\lambda$ at the points $z_i$ seems not to be independent of the parameters $v,w$.

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

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user sam
A humble question : what is the precise meaning of $dz^k$ ?

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user Trimok
I am not a mathemathetician, so my notation may be wrong but I believe $dz^k\equiv dz\otimes dz\otimes ... \otimes dz$ k times. Therefore it lives in the k'th power of the cotangent bundle of the Riemann sphere. Now I am speculating, but I think one way of looking at this is to say that for every branch of the Riemann sphere associated with a k'th order pole, we have a separate cotangent bundle, and on each we consider a $dz$.

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user sam
@Trimok: It looks like what is meant by a degree $k$ differential form is a section of the $k$-th tensor power. If $z$ is a local holomorphic coordinate, then $dz$ is a local basis (over the holomorphic or meromorphic functions) of the space of differentials, usually denoted $\Omega$. From this vector bundle, new vector bundles can be constructed by locally applying operations in multilinear algebra, taking tensor powers, exterior powers, duals, etc. Where $dz$ is a local basis for $\Omega$, $dz\otimes\cdots\otimes dz = dz^k$ is a local basis for the $k$-th tensor power $\Omega^{\otimes k}$.

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user doetoe
@Trimok, cont'd: writing $\zeta$ (rather than $z'$) for a local coordinate around $\infty$, and $z$ for the coordinate centered at 0, so that $\zeta = z^{-1}$, an ordinary meromorphic function $f(z)$ becomes $f(\zeta^{-1})$ in the new coordinates. The chosen basis differential $dz$ turns into $d\zeta^{-1} = -\zeta^{-2}d\zeta$ in the new coordinates and the $k$-th tensor power of $dz$ transforms as described above: $dz^k = (-1)^k\zeta^{2k}d\zeta^k$.

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user doetoe
So doetoe, would it sound reasonable to you that the reason a k-th tensor power of the cotangent bundle is used to describe the different sheets of the Riemann sphere? Can I view it as something like the cotangent bundle of the universal cover of the (k-th order) punctured Riemann sphere?

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user sam
I don't know if there is such an interpretation but that would be interesting. I usually view it as something abstract, like a tensor product of vector spaces. I'll think about it.

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user doetoe
@user30803: sorry, I didn't see your first comment, which I essentially repeated (had had the page open in a tab since this morning)

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user doetoe
@user30803: there could be a nice geometrical interpretation along the lines you describe. Could you elaborate on your intuition?

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user doetoe
@doetoe: I also posted a related question on mathoverflow: mathoverflow.net/questions/144478/…. The answer I got there was that a residue of a k-differential is $f(0)$ if for instance one has $f(z)\frac{dz^k}{z^k}$. My interpretation of this is that you split up your degree k-pole in k degree 1 poles on different sheets and simultaneuosly perform a simple pole contour integration with a simple $dz$ on each sheet. (then maybe the residue on each sheet should be $|Res_{0}|=f(0)^{1/k}$.

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user sam
@sam : I don't understand how exactly you have applied the Riemann-Roch theorem : $l(D) - l(K-D) = deg(D) + 1 - g$, in your examples.

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user Trimok
Take as a divisor $D=k(z_1)+...+k(z_n)$.I want to calculate $dim(L^{(1)}(D))\equiv i(D)$ I took a slightly different version: $$l(D')=deg(D')+1-g$$ where $D'=K+D$ with $K=-2k(z_i)$ for some $i$. The fact that $i(D')=0$ comes from $deg(D')=0>2g-1$. Now $l(D')=i(D)$.

This post imported from StackExchange Physics at 2014-03-21 17:02 (UCT), posted by SE-user sam

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