Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

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

+ 1 like - 0 dislike
2104 views

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
asked Oct 10, 2013 in Theoretical Physics by sam (45 points) [ no revision ]
Most voted comments show all comments
@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
Most recent comments show all comments
@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

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$y$\varnothing$icsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...