S-duality of Einstein-Maxwell-Dilaton theory Consider theory with action
$$S = \int d^D x \sqrt{-g} (R - \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2k!} e^{a \phi} F^2 _{[k]} ) $$
where $\phi$ is dilaton and $F_{[k]}$ is electromagnetic $k$-form.
S-duality is the symmetry of this action
$$g_{\mu \nu} \to g_{\mu \nu} \ , \ \ F \to e^{- a \phi} \star F \ , \ \ \phi \to - \phi $$
I cannot understand why we have to use this transformation in order to get , for example, magnetic solution if electric one is already known. Why cannot we use only $F \to \star F \ , \ \ \phi \to \phi $ transformation?
Moreover, the equations of motion for magnetic solution are
$$\partial _\mu (\sqrt{-g} e^{a \phi} F^ {\mu \alpha_2 ... \alpha_k } ) = 0$$
And it is claimed that magnetic solution of this equation (for diagonal radially symmetric metric) is
$$F_{[k]} = \frac{P}{R^{D-2}} d\theta_1 \wedge ... \wedge d\theta_k$$
But I cannot understand why it does not depend on dilaton via $e^{- a \phi}$ as electric solution does.
 A: At the moment I have no time to include more details, but maybe the following helps already: 
Hodge-Dualisation in the action is in fact a subtle business. Note that you cannot simply plug in $F= \star G$ (where $G$ is now my dual field strength tensor). Instead you have to impose a constraint in the action to make sure that the Bianchi identity of $F$, namely $dF=0$, holds true. So, you one should add by hand a Lagrange-multiplier term like $\chi \wedge dF$, where $\chi$ is a $(D-k-1)$-form and it will turn out that $\chi$ is the dual gauge field so that $G \sim d\chi$. In this procedure one can show that the couplings reverse, so strong couplings go to weak coupling and vice versa. This is why it's called S-duality. What you should find after dualisation would be $e^{-a\phi}G \wedge \star G$ and you can undo the minus sign by $\phi \to - \phi$.
The last equation follows from the Bianchi-identity $dF=0$ together with rotational symmetry of space. So, no dilatonic prefactor occurs.
psm  
