Generalized Electromagnetic Duality

Can we find such a symmetry for the equations of motions of our four-dimensional action? Let us first look at the action with only one vector field. We have to leave out the modulus field, $\sigma=1$, and take the vectors to be equal. The action then becomes

$${\cal L} = -R + {\textstyle{1\over 2}}(\partial \phi)^2 + {\... ...ver 4}}e^{-\phi} F^2 - {\textstyle{1\over 4}}\Psi F \tilde F.$$ (4.13)

This becomes even more symmetric if we introduce the complex scalar field $\lambda$ as,

$$\lambda = \Psi + i e^{-\phi} \equiv \lambda _1 + i \lambda _2.$$ (4.14)

With this complex field the action becomes

$${\cal L} = -R + \frac{\partial \lambda \partial \lambda ^{*}... ...}}\lambda _2 F^2 - {\textstyle{1\over 4}}\lambda _1 F \tilde F$$ (4.15)

The field equations of this action take on the following form. First the equation for the metric

$$R_{\mu\nu} = \frac{1}{4 \lambda _2^2} \left( \partial _\mu ... ...\nu{}^\rho - {\textstyle{1\over 4}}g_{\mu\nu} F^2 \right] \, ,$$ (4.16)

for the complex scalar,

$$\frac{1}{\lambda _2^2} \Box \lambda + \frac{i}{\lambda _2^3}... ...\lambda )^2 + \frac{i}{4} F^2 + \frac{1}{4} F \tilde F =0 \,,$$ (4.17)

and finaly for the vector field

$$\nabla_\mu \left( \lambda _2 F^{\mu\nu} + \lambda _1 \tilde F^{\mu\nu} \right) = 0 \, .$$ (4.18)

And the Bianchi-identity for the vector field is

$$\nabla_\mu \tilde F^{\mu\nu} = 0$$ (4.19)

Again we want to rotate the vector field eqution and its Bianchi-identity. First define the tensor

$$G_{\mu\nu} = -2 \frac{\delta {\cal{L}}}{\delta F^{\mu\nu}} ,$$

then the equation of motion of the vector field can be written as

$$\nabla_\mu G^{\mu\nu} = 0.$$

Itstead of interchanging the two equations, we can look for a continuous symmetry. Make the following $SL(2,\hbox{\mybb R})$ transformation [22] [24],

$$\left( \begin{array}{c} \tilde F \\ G \end{array} \right)' =... ...ega \left( \begin{array}{c} \tilde F \\ G \end{array} \right),$$ (4.20)

where $\omega \in SL(2,\hbox{\mybb R})$, or explicitly,

$$\omega = \left( \begin{array}{cc} a&b\\ c&d \end{array} \right) \quad\mbox{and}\quad ad-bc =1.$$ (4.21)

If we demand invariance of the equations of motion, the complex scalar has to transform as

$$\lambda ' = \frac{c + d\lambda }{a+b\lambda }.$$ (4.22)

The equation for the scalar is clearly invariant, the vector equation are invariant by conctruction, but the metrical equation transforms to itself plus an extra term, given by

$$-\frac{\lambda _1 (\lambda _2)^2 }{\vert l\vert^2} \left( 2 ... ...rho - g_{\mu\nu} F_{\rho\sigma} \tilde F^{\rho\sigma} \right).$$ (4.23)

This extra term, however, can be shown to vanish identically in four dimensions, showing that $SL(2,\hbox{\mybb R})$ is a genuine symmetry of the equations of motion. This symmetry is often called $S$-Duality. In The early days the complex scalar field was called the Superfield, that is why the symmetry is called $S$-Duality.

A special case of this symmetry is the discrete transformation

$$\lambda \to \frac{-1}{\lambda } \quad \mbox{and} \quad \tilde F \to G, \quad G \to -\tilde F,$$

which interchanges magnetic and electric charges. This means that if we can find a electically charged black hole solution, we can also find a magnetically charged one using this transformation. Another important aspect of this symmetry is the fact that the Dilaton in string theory can be interpreted as the coupling constant, so the transformation $\lambda _2=e^{-\phi} \to (\lambda _2)^{-1}$ takes a strong coupling theory to a weak coupling theory. This is why $S$-Duality is sometimes refered to as strong/weak coupling duality [22].