$S$-Duality with Two Vector Fields

We would like to find the same kind of Electric/Magnetic Duality in our four dimensional action (4.9). This action has two vector fields, so we can mix two equations of motion and two Bianchi-identities. In terms of the complex scalar $\lambda$, the equations of motion are

$$R_{\mu\nu} = \frac{1}{4 \lambda _2^2} \left( \partial _\mu \lambda \partial _\... ...a ^* \right) + \frac{1}{2 \sigma^2} \partial _\mu \sigma \partial _\nu \sigma +$$

$$- {\textstyle{1\over 2}}\lambda _2 \left[ \sigma \left( F_{\mu\rho}(A) F_\nu{}^\rho (A) - {\textstyle{1\over 4}}g_{\mu\nu} F(A)^2 \right) \right.$$

$$\left. + \sigma^{-1} \left( F_{\mu\rho}(B) F_\nu{}^\rho (B) - {\textstyle{1\over 4}}g_{\mu\nu} F(B)^2 \right) \right] \, ,$$ (4.24)

$$\frac{1}{\lambda _2^2} \Box \lambda + \frac{i}{\lambda _2^3}(\par... ...left( \sigma F(A)^2 + \sigma^{-1} F(B)^2 \right) + \frac{1}{2} F(A) \tilde F(B) = 0 \, ,\quad\quad$$ (4.25)

$$\frac{1}{\phi^2} \Box \sigma - \frac{1}{\sigma^3} ( \partial \sig... ...extstyle{1\over 4}}\lambda _2 \left( F(A)^2 - \frac{1}{\sigma^2} F(B)^2 \right) = 0 \, ,$$ (4.26)

$$\nabla_\mu \left( \lambda _2 \sigma F^{\mu\nu}(A) + \lambda _1 \tilde F^{\mu\nu}(B) \right) = 0 \, ,$$ (4.27)

$$\nabla_\mu \left( \lambda _2 \sigma^{-1} F^{\mu\nu}(B) + \lambda _1 \tilde F^{\mu\nu}(A) \right) = 0 \, ,$$ (4.28)

and the Bianchi-identities:

$$\nabla_\mu \tilde F^{\mu\nu}(A) = 0 \, ,$$ (4.29)

$$\nabla_\mu \tilde F^{\mu\nu}(B) = 0\, .$$ (4.30)

The procedure is the same as in the previous section, so first we define

$$G_{\mu\nu}(X) = -2 \frac{\delta {\cal{L}}}{\delta F^{\mu\nu}(X)} \quad\mbox{and}\quad \nabla_\mu G^{\mu\nu} =0$$

This implies:

$$G^{\mu\nu}(A) = \lambda _2 \sigma F^{\mu\nu}(A) + \lambda _1 \tilde F^{\mu\nu}(B) \, ,$$ (4.31)

$$G^{\mu\nu}(B) = \lambda _2 \sigma^{-1} F^{\mu\nu}(B) + \lambda _1 \tilde F^{\mu\nu}(A)\,$$ (4.32)

We can put the four 'equations' in a quadruplet an apply to it a linear transformation:

$$\left( \begin{array}{c} \tilde F(A) \\ G(B) \\ \tilde F(B) \\ G(A) \end{array} \right)' = \Omega \left( \begin{array}{c} \tilde F(A) \\ G(B) \\ \tilde F(B) \\ G(A) \end{array} \right)$$

where $\Omega$ is a $4 \times 4$ matrix. Invariance of the equations of motion gives us the transformation of the scalar fields: it turns out that the matrix $\Omega$ is divided into 4 separate parts:

$$\Omega = \left( \begin{array}{cc} A&B\\ C&D \end{array} \right)$$ (4.33)

where the diagonal parts ( A and D ) generate the following transformation of the scalars:

$$\lambda _1' + \frac{i}{2} \left( \frac{\sigma'}{\sigma} + \f... ... \right) \lambda _2' = \frac{ e + f \lambda }{ a + b \lambda }$$ (4.34)

The parameters $a,b,e,f$ are not direct components of $A$ or $D$, but rather:

$$a = A_{11} + D_{11}\, ,$$

$$b = A_{12}+D_{12}\, ,$$

$$e = A_{21} + D_{21}\, ,$$

$$f = A_{22} + D_{22} \, .$$

The off-diagonal terms give the following scalar transformation:

$$\lambda _1' + \frac{i}{2} \left( \frac{1}{\sigma' \sigma} + ... ...\right) \lambda _2' = \frac{ g + h \lambda }{ c + d \lambda }.$$ (4.35)

The parameters $c,d,g,h$ are the same linear combinations of the componenents of $C$ and $D$ as in the previous case. It is obvious that the two transformations cannot hold simultaneously. We have to make a choice for the transformation of the $\sigma$-field. There are two possibilities. The first is the trivial identity: $\sigma' = \sigma$ ; the other possibility is the discrete transformation: $\sigma' = \sigma^{-1}$.

If we choose the first possibility ( $\sigma' = \sigma$), the invariance of both the scalar and metric equations implies that the off-diagonal terms of $\Omega$ have to vanish and that the diagonal terms have to be equal ($A=D$). Furthermore the determinant of the transformation matrices has to be equal to unity. So the transformation can be reduced to two separate $SL(2,\hbox{\mybb R})$ transformations and the quadruplet can be split into two independent doublets which transform in the same manner:

$$\left( \begin{array}{c} \tilde F(A) \\ G(B) \end{array}\right)' = \omega \left( \begin{array}{c} \tilde F(A) \\ G(B) \end{array}\right) \mbox{and} \left( \begin{array}{c} \tilde F(B) \\ G(A) \end{array}\right)' = \omega \left( \begin{array}{c} \tilde F(B) \\ G(A) \end{array}\right)$$ (4.36)

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

The scalar field $\lambda$ now transforms under this $SL(2,R)$:

$$\lambda ' = \frac{c + d\lambda }{a+b\lambda }$$ (4.37)

This is the same symmetry we found in the previous section, with the difference that we have to rotate two doublets.

We could also have chosen the other possible transformation of the $\sigma$-field. In that case the diagonal terms of $\Omega$ have to vanish and the off-diagonal terms can be separated into two transformations. This time the transformation is $SL(2,\hbox{\mybb R}) \times \hbox{\mybb Z}_2$. The $SL(2,\hbox{\mybb R})$-transformation is exactly the same as is the previous case. The difference is the extra discrete transformation $\hbox{\mybb Z}_2$:

$$\sigma \enspace \rightarrow \enspace \frac{1}{\sigma} \mbox{and} A \leftrightarrow B$$ (4.38)

This is precisely the discrete subgroup of the $O(1,1)$ symmetry we found for our action, due to the reduction from a five-dimensions string effective action.