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  Intuition for S-duality

+ 6 like - 0 dislike
1064 views

first of all, I need to confess my ignorance with respect to any physics since I'm a mathematician. I'm interested in the physical intuition of the Langlands program, therefore I need to understand what physicists think about homological mirror symmetry which comes from S-duality. This question is related to my previous one Intuition for Homological Mirror Symmetry.

As I have heard everything starts with an $S$-duality between two $N= 4$ super-symmetric Yang-Mills gauge theories of dimension $4$, $(G, \tau)$ and $(^{L}G, \frac{-1}{n_{\mathfrak{g}}\tau})$, where $\tau = \frac{\theta}{2\pi} + \frac{4\pi i}{g^2}$, $G$ is a compact connected simple Lie group and $n_{\mathfrak{g}}$ is the lacing number (the maximal number of edges connecting two vertices in the Dynkin diagram) . And, then the theory would be non-perturbative, since it would be defined "for all" $\tau$, because amplitudes are computed with an expansion in power series in $\tau$

So I need to understand what this would mean to a physicist.

1) First of all, what's the motivation form the Yang-Mills action and how should I understand the coupling constants $\theta$ and $g$?

2) How can I get this so called expansion in power series with variable $\tau$ of the probability amplitude?

3) What was the motivation to start looking at this duality? A creation of an everywhere defined (in $\tau$) gauge theory, maybe?

Thanks in advance.


This post imported from StackExchange Physics at 2015-03-17 04:40 (UTC), posted by SE-user user40276

asked Mar 6, 2015 in Theoretical Physics by user40276 (140 points) [ revision history ]
edited Mar 17, 2015 by dimension10

Re 'And, then the theory would be non-perturbative, since it would be defined "for all" τ, because amplitudes are computed with an expansion in power series in τ':

Actually, to a physicist, such a power-series expansion is the hallmark of (the outcome of) a perturbative theory: Such power series typically correspond to some perturbation calculated to (arbitrarily) high order. A base in the coefficient corresponds to a physical coupling constant and causes such approaches to become invalid for large (e.g. unity) coupling constants as the power series no longer converges.

Please take this comment with a grain of salt: I am myself from a foreign field (to theoretical physics) as I am a mere quantumoptics experimental physicist curious about expanding my mental horizon. This is just my first "that's usually like this" association

1 Answer

+ 5 like - 0 dislike

  First of all, what's the motivation form the Yang-Mills action and how should I understand the coupling constants θ and g?

The Yang-Mills action functional is supposed to be a local functional on the space of \(G\)-principal connections \(\nabla\) on a 4-dimensional pseudo-Riemannian manifold \((X,g)\), where "local" means that it is given by the integral of a differential 4-form depending on \(\nabla \) (the local Lagrangian 4-form). Now the connection gives rise to its curvature 2-form \(F_\nabla\) which locally takes values in the Lie algebra of \(G\). But by standard assumption the Lie group \(G\) here carries a Killing form invariant polynomial \(\langle -,-\rangle\), and so evaluating any two Lie algebra valued 2-forms inside that gives a plain differential 4-form.

Accordingly, there are thus two kinds of differential 4-forms naturally assigned to \(\nabla\) here, namely

  1. \(\langle F_\nabla \wedge F_\nabla\rangle\)
  2. \(\langle F_\nabla \wedge \star_g F_\nabla \rangle\)

where \(\star_g\) denotes the Hodge star operator of the given metric.

The Yang-Mills action functional is simply 

\( \nabla \mapsto \frac{1}{g^2 }\int_X F_\nabla \wedge \star F_\nabla \;+\; i \theta \int_X F_\nabla \wedge F_\nabla\)

the linear combination of the integral of these two possible local Lagrangian 4-forms. The coefficients in the linear combination are traditionally parameterized by the "coupling constant" \(g \) and the theta angle \(\theta\).

That this should satisfy something like S-duality is easily motivated from (and was historically motivated from) looking at the special case that \(G = U(1)\) (Maxwell theory, electromagnetism). Maybe best look at the original references.

answered Mar 17, 2015 by Urs Schreiber (6,095 points) [ revision history ]
edited Mar 18, 2015 by Urs Schreiber

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