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  Computation of Central Charges Using Seiberg-Witten Differential

+ 4 like - 0 dislike
1353 views

Hi, 

I have a question regarding computation of central charges from SW differential. Let's say that the spectral curve has the following form:

                                                                   $\lambda^2(z,u)=f(z,u) dz^2$

In which $f(z,u)$ is a polynomial of order for example three. The periods (which define the central charge of the theory) is given by the following integrals:

                          $a(u)\equiv\oint_{\gamma_1} dz\, \sqrt{f(z,u)} \qquad;\qquad a_D(u)\equiv\oint_{\gamma_2} dz\, \sqrt{f(z,u)}$

In which $\gamma_1$ and $\gamma_2$ are two cycles of torus. These cycles are defined by the branch cuts between branch points of the $\sqrt{f(z,u)}$. Here we have three branch points (and one at infinity) so we can consider three different combination of the (finite) cuts. If I denote the roots of $f(z,u)$ by $z_i(u)$, then the choice of the cuts are as follows:

- A cut from $z_1(u)$ and $z_2(u)$ and a cut from $z_3(u)$ to infinity;

- A cut from $z_2(u)$ and $z_3(u)$ and a cut from $z_1(u)$ to infinity;

- A cut from $z_1(u)$ and $z_3(u)$ and a cut from $z_2(u)$ to infinity;

My question is that which one of these choices corresponds to computation of $a(u)$ and which one corresponds to computation of $a_D(u)$ and why? In the original SW paper, it has been argued that the special choices were made to reproduce the results that are obtained previously in the paper, but what would be the choices if we want to compute central charges for a theory which may or may not have a Lagrangian description?

asked Nov 26, 2015 in Theoretical Physics by SI1989 (85 points) [ no revision ]

There are more choices of cuts, aren't there? They are obtained by fixing one of these choices and then performing a pure braid transformation.

@Ryan, I am not sure that I understand what do you mean by pure braid transformation but you are absolutely right that there are more choices.

1 Answer

+ 3 like - 0 dislike

At a generic point $u$ of the Coulomb branch, there is no natural choice of splitting of the electromagnetic lattice in electric and magnetic parts and so no natural choice of what should be called $a(u)$ and $a_D(u)$. Geometrically, there is no natural choice of a basis of 1-cycles on the torus.

Near some special points of the Coulomb branch, for example in the pure $SU(2)$ case: near infinity or near one of the two singular points at finite $u$, one of the electromagnetic coupling becomes weak and so plays a special role and it makes sense to call it $a(u)$. Geometrically, near a special point, one of the 1-cycle of the torus shrinks and so plays a special role. But going from one special point to another, the natural choice of $a(u)$ changes.

answered Dec 6, 2015 by 40227 (5,140 points) [ no revision ]

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