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.

  • $\begingroup$ What is your definition of "magnetic" and "electric" solution here? If your definition of "magnetic solution" is just solving that equation of motion there, then what is your question in "I cannot understand why it does not depend on dilaton"? Either your thing there solves the equation or not. $\endgroup$
    – ACuriousMind
    Commented Mar 20, 2016 at 11:40
  • $\begingroup$ Well, as I understand, magnetic solution is the solution of equation of motion. Here we mush take dual equation. I don't understand why "dual" here means not only Hodge star, but also multiplying with $e^{-a \phi}$. $\endgroup$
    – newt
    Commented Mar 20, 2016 at 12:47

1 Answer 1


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.



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