SO(2) Duality, Type IIB SuperGravity and the Super-D3-Brane Action

By George Shiber Friday, September 1, 2017Thursday, August 15, 2019 In Dp-Branes, M Theory, Mathematical Physics, Mathematics, Supersymmetry D3-branes, supergravity

In this post, I shall analyze certain relations between holomorphic properties of Dp-brane actions and SO(2)-duality of type IIB supergravity. Specifically, I will show that the super D3-brane action:

$S = {S_{DBI}} + {S_{WZ}}$

in Type IIB SUGRA background satisfies the Gaillard-Zumino duality condition and exhibits exact self-duality. Dp-branes are p + 1 dimensional Ramond-Ramond charged dynamical hypersurfaces ${\Sigma _{p + 1}}$ that open strings end on and admit perturbative worldsheet description in terms of open strings satisfying Dirichlet boundary conditions in p + 1 dimensions. Naturally, for 4-D spacetime physics, D3 branes are especially important for string-phenomenology due to mirror symmetry on Calabi-Yau 3-folds where they holomorphically wrap Fukaya super-Lagrangians. The D3-brane effective action in the NS5-brane geometry, given that it satisfies D3-brane self-duality and Poincaré invariance, is given by:

${S_{D3}} = {g_s}{{\rm T}_3}\left[ { - \int {{d^4}} \varsigma \,{e^{ - \Phi }}\sqrt { - \,{\rm{det}}\left( {{G_{\mu \nu }} + {F_{\mu \nu }}} \right)} + i\int {\left( {{C_{\left[ 4 \right]}} + F \wedge {C_{\left[ 2 \right]}}} \right)} } \right]$

with

${{\rm T}_3} = \frac{1}{{{g_s}\,l_s^{\left[ 3 \right]}{{\left( {2\,\pi \,{l_s}} \right)}^2}}}$

${{\rm T}_3}$ being the D3-brane tension, and ${C_{\left[ 4 \right]}}$ , ${C_{\left[ 2 \right]}}$ are the RR-4 and RR-2 exterior forms, and generally, the DBI action is:

${S_{DBI}} = - {T_D}\int {{d^{p + 1}}} \sigma {e^{ - \phi }}\sqrt { - \det \left( {\mathcal{G} + F} \right)}$

and the D-brane WZ action is given by:

$\int_{{\Sigma _{p + 1}}} {{C^{RR}}} \wedge {e^F}$

In order for the effective action to be integrable with fields in 2nd-quantized form, one must work under the Gaillard-Zumino Condition: that is – using 8-loop counterterms with superspace torsion:

${S^8} \sim {k^{14}}\int_{T8} {{d^4}} x\,{d^{32}}\theta {B_{er}}\,{\rm E}\,{T_{ijk\alpha }}\left( {\chi ,\theta } \right){T^{ * \dagger ijk\alpha }}\left( {\chi ,\theta } \right){{\rm T}_{mn{l^\alpha }}}{T_{{\vartheta _i}(\chi ,\theta )}}^{ * \dagger mnl}$

where ${T_{ijk\alpha }}$ is the superfield torsion. One starts with a Lagrangian:

${L_G}\left( {{F_{\mu \nu }},{g_{\mu \nu }},{\Phi ^A}} \right) = \sqrt { - {g_s}} L\left( {{F_{\mu \nu }},{g_{_{\mu \nu }}},{\Phi ^A}} \right)$

in D = 4, with a dependence on a $U(1)$ gauge field strength ${F_{\mu \nu }}$ , metric ${g_{\mu \nu }}$ , and matter field ${\Phi ^A}$ . So, we now have:

${K^{*\dag \mu \nu }} = \frac{{\partial {L_G}}}{{\partial {F_{\mu \nu }}}}$

$\frac{{\partial {F_{\alpha \beta }}}}{{\partial {F_{\mu \nu }}}} = \left( {\delta _\alpha ^\mu {\mkern 1mu} \delta _{{\beta ^{{\mkern 1mu} {\mkern 1mu} - }}}^\nu - \delta _\beta ^\mu {\mkern 1mu} \delta _\alpha ^\nu } \right)$

and the Hodge dual components for the tensor ${K_{\mu \nu }}$ are given by:

$K_{\mu \nu }^{ * \dagger } = \frac{1}{2}{\eta _{\mu \nu }}^{\rho \sigma }{k_{\rho \sigma }}$

$K_{\mu \nu }^{ * \dagger } = - {K_{\mu \nu }}$

The Gaillard-Zumino condition is an infinitesimal duality transformation of $F$ and $K$ and fermionic transformation given by:

$\delta \left( {{\Gamma _k}} \right) = \left( {\begin{array}{*{20}{c}}\alpha &\beta \\\gamma &\delta \end{array}} \right)\left( {\begin{array}{*{20}{c}}F\\K\end{array}} \right)$

$\delta \,{\Phi ^A} = {\xi ^A}(\Phi )$

$\delta {g_{\mu \nu }} = 0$

Now, the Lagrangian must transform as:

$\delta {L_G} = \frac{1}{4}\left( {\gamma F\,{F^{ * \dagger }} + \beta K\,{K^{ * \dagger }}} \right)$

and one has an $SO(2)$ transformation given by $\delta F = \lambda k$ , $\delta K = - \lambda F$ , and so the Lagrangian is given by:

$\delta {L_G} = \frac{1}{2}\,\frac{{\partial L}}{{\partial F}}\delta {F_{\mu \nu }} + \frac{{\partial L}}{{\partial {\phi ^A}}} = \frac{\lambda }{2}{\widetilde K^{\mu \nu }}{K_{\mu \nu }} + {\delta _\Phi }L$

and by D3-brane self-duality, it follows that:

$\frac{\lambda }{4}\left( {F \cdot {F^{ * \dagger }} + K \cdot {K^{ * \dagger }}} \right) + {\delta _\Phi }{L_G} = 0$

Now we are in a position to analyse the D3-brane action, and for simplicity but with no loss of generality, without scalar supergravity backgrounds. Let ${X^M}$ be a bosonic brane coordinatization in D = 10 flat target bulk space $\left( {M,g,R,...} \right)$ and ${\theta _{A\alpha }}$ its fermionic partner given by the Majorana-Weyl spinor index $\alpha$ with N = 2 SUSY index $A$ . The D3 action for the brane coordinates $\left( {X,\theta } \right)$ and worldvolume gauge field ${A_\mu }$ must have Kappa symmetry and we require N = 2 SUSY. Hence:

$S = \int {{d^4}} \sigma {L_G}^{DBI} + \int {{d^4}} \sigma {L_G}^{WZ}$

where

${L_G}^{DBI} = - \sqrt { - {\rm{det}}\left( {{G_{\mu \nu }} + {F_{\mu \nu }}} \right)}$

${G_{\mu \nu }} = \prod _\mu ^M{\prod _{\nu M}}$

${F_{\mu \nu }} = {\partial _{\left[ {\mu A\nu } \right]}} + \Omega _{\mu \nu }^3$

$\Omega _{\mu \nu }^j = \overline \theta {\not \prod _{\left[ {\mu \,{\tau _{i\,{\partial _\nu }}}} \right]}}^{ * \dagger }\theta {\rm{ ,}}\;{\rm{ }}j = 1\,,2$

with ${\tau _i}$ being the Pauli matrices and act on the N = 2 SUSY indices. The 1-form defined by:

$\prod _\mu ^M \equiv d{x^M} + \overline {\theta \,} {\Gamma ^M}d\theta \equiv d{\sigma ^\mu }\prod _\mu ^M$

and

$\prod _\mu ^M = {\partial _\mu }{\chi ^{{M^\mu }}} - \overline \theta \,{\Gamma ^M}{\partial _\mu }d\theta = d{\chi ^M} + \frac{1}{2}\overline \theta \,{\Gamma ^M}d\theta$

By use of exterior differential forms on the bulk, with an RR pull-back 2-form ${C_{\left[ 2 \right]}}$ and 4-form ${C_{\left[ 4 \right]}}$ , we get:

${L^{WZ}} = {C_{\left[ 2 \right]}}F + {C_{\left[ 4 \right]}}$

${C_{\left[ 2 \right]}} = \overline \theta {\not \prod _{{\phi _i}}}{\tau _i}d\theta = {\Omega _1}$

and

${C_{\left[ 4 \right]}} = \Xi \,\; - \frac{1}{2}{\Omega _{\,1}}\Omega {\,_3}$

where $\Xi$ is given by:

$\begin{array}{c}\Xi = \frac{1}{6}\overline \theta {{\not \prod }^3}{\tau _3}\,{\tau _1}\,d\theta - \frac{1}{{12}}\overline \theta \left( {{{\not \prod }^2}{{\not \beta }_0} + \not \prod {\beta _0}\not \prod + {{\not \beta }_0}{{\not \prod }^2}} \right){\tau _3}\,{\tau _1}\,d\theta \\ + \frac{1}{{18}}\overline \theta \left( {\not \prod \not \beta _0^2 + {{\not \beta }_0}\not \prod } \right)\tau {\,_3}{\tau _1}\,d\theta - \frac{1}{{12}}\overline \theta \not \prod {\tau _{\left[ {1,d\theta \overline {\theta \,{{\not \beta }_{0{\tau _3}}}} } \right]}}\,d\theta \\ - \frac{1}{{24}}\overline \theta \not \beta _0^3\,{\tau _3}\,{\tau _1}\,d\theta \end{array}$

with ${\beta _0} \equiv \overline \theta \,\Gamma d\theta$ . To check whether the Gaillard-Zumino condition is met, we must calculate the first 2 terms of the condition:

so, $\widetilde K$ is given by:

${\widetilde K^{\mu \nu }} = \frac{{\partial {L_G}}}{{\partial {F_{\mu \nu }}}} = \frac{{\sqrt { - G} }}{{\sqrt { - {G_\Gamma }} }}\left( {{F^{\mu \nu }} + \Upsilon \widetilde {{F^{\mu \nu }}}} \right) + {}^ * {C_{\left[ 2 \right]}}^{\mu \nu }$

where I have made an explicit use of the determinant formula for the four-by-four matrix:

${G_F} \equiv {\rm{det}}\left( {G + F} \right) = G\left( {1 + \frac{1}{2}{F^{\mu \nu }}{F_{\mu \nu }} + {\Upsilon ^2}} \right)$

$\Upsilon \equiv \frac{1}{4}{F_{\mu \nu }}{\widetilde F^{\mu \nu }}$

and by Hodge duality, $K$ can be derived as:

${K_{\mu \nu }} = - \frac{1}{2}{\eta _{\mu \nu \rho \sigma }}{\widetilde K^{\rho \sigma }} = \frac{{\sqrt { - G} }}{{\sqrt { - {G_F}} }}\left( {F_{\mu \nu }^{ * \dagger } + \Upsilon {{\not F}_{\mu \nu }}} \right) + C_{\left[ 2 \right]}^{\mu \nu }$

and by the GKP-Witten relation for the D3-brane action:

${Z_{CFT}} = {e^{ - {S_{D3}}}}({\phi _i})$

one gets

${C_{\left[ 2 \right]}}F = \frac{1}{4}{d^4}\sigma \,{\varepsilon ^{\mu \nu \rho \sigma }}C_{\left[ 2 \right]}^{\mu \nu }{F_{\rho \sigma }}$

and by conjugation, one derives the essential identity:

${K_{\mu \nu }}{\widetilde K^{\mu \nu }} + {F_{\mu \nu }}{\widetilde F^{\mu \nu }} = - 2{\widetilde F^{\mu \nu }}\Omega _{\mu \nu }^3\,\widetilde \Omega _3^{\mu \nu } + {\widetilde K^{\mu \nu }}C_{\left[ 2 \right]}^{\mu \nu } - {\widetilde C_{\left[ 2 \right]}}_{\mu \nu }{C_{\left[ 2 \right]}}_{\mu \nu }$

A few remarks are in order now on the bosonic truncation of the D3-brane action. Note in the above equation, the right-hand-side vanishes completely, and so the Lagrangian transforms accordingly as:

$\delta {L_G} = \frac{\lambda }{2}F\widetilde F$

The supersymmetry situation under the Gaillard-Zumino condition can now be considered for the matter field contributions. For a D3-brane, the matter fields transform as:

$\delta \theta = \lambda \frac{{i{\tau _2}}}{2}\theta$

$\delta \chi = 0$

and hence we get:

$\delta {\prod _\mu }^M = \delta {G_{\mu \nu }} = \delta \,\Xi = 0$

$\delta \,{\Omega _{\mu \nu }}^3 = - \lambda \,{\Omega _{\mu \nu }}^1$

$\delta \,{\Omega _{\mu \nu }}^1 = \lambda \,{\Omega _{\mu \nu }}^3$

while noting that the Majorana-Weyl fermions $\left( {{\theta _1},{\theta _2}} \right)$ and $\left( {{\Omega ^1},{\Omega ^3}} \right)$ transform as $SO(2)$ group doublet and the gauge variation of the total Lagrangian with respect to matter fields transforms as:

${\delta _\Phi }{L_G} = {\delta _\theta }L = \frac{1}{2}\frac{{\partial L}}{{\partial {F_{\mu \nu }}}} + \frac{1}{2}{\widetilde F^{\mu \nu }}\delta {C_{\left[ 2 \right]}}^{\mu \nu } + \delta {\widetilde C_{\left[ 4 \right]}}$

where ${\widetilde C_{\left[ 4 \right]}}$ is the Hodge dual of ${C_{\left[ 4 \right]}}$ , with:

${C_{\left[ 4 \right]}} = {d^4}\sigma \frac{1}{{4!}}{\varepsilon ^{\mu \nu \rho \sigma }}{C_{\left[ 4 \right]}}^{\mu \nu \rho \sigma } \equiv {d^4}\sigma \sqrt { - {G_{\mu \nu }}} {\widetilde C_{\left[ 4 \right]}} \cdot {C_{\left[ 2 \right]}} = \Omega$

and the Poincaré invariance of D3-branes transfers to $\Xi$ and induces a relation on the differential forms:

$\delta \,\Omega = \delta \left[ {\frac{1}{2}{C_{\left[ 2 \right]}}\,{\Omega _3} + {C_{\left[ 4 \right]}}} \right] = \frac{\lambda }{2}\left( { - {{({C_{\left[ 2 \right]}})}^2} + {{({\Omega _3})}^2}} \right) + {\delta _\theta }{C_{\left[ 4 \right]}} = 0$

Combining the last 3 equations, we get:

$\frac{\lambda }{4}\left( {F\widetilde F} \right. + k{g_s}\left. {\widetilde K} \right) + {\delta _\Phi }{L_G} = \frac{\lambda }{4}\left( { - {C_{\left[ 4 \right]}}^{\mu \nu } + \,\,{\Omega _{\mu \nu }}^3\,\widetilde {{\Omega _3}^{\mu \nu }}} \right) + \delta {\widetilde C_{\left[ 4 \right]}} = 0$

And so the D3-brane self-duality and Poincaré invariance are satisfied by the Gaillard-Zumino condition. To second-quantize the D3-brane action, one must lift the $SO(2)$ duality by an introduction of a dilaton $\phi$ and axion ${\chi ^a}$ living on constant background fields on the bulk space. The redefined Lagrangian, via D3-brane 4-dimensional worldvolume topological tori-throating becomes:

${L_{G(D3)}} \equiv {\widetilde L_G}\left( {F,\,{\chi ^a}\theta \,;\,\phi ,\chi } \right) = {L_G}\left( {{e^{ - \phi /2}}F,\,{\chi ^a},\theta } \right) - \frac{1}{4}{\chi ^a}F\widetilde F$

and by such tori redefinition, we obtain:

${S_{D3}} = \frac{1}{{4{k^2}{C_{\left[ 4 \right]}}}}{\int {{{\sqrt K }_{\mu \nu }}} ^{ - 2\Phi }}\left( {2{K^{\mu \nu }}{C_{\left[ 2 \right]}} + \frac{\lambda }{8} + {\partial _\mu }\Phi {\mkern 1mu} {\partial _\Phi }^\mu \Phi - {C_{\left[ 4 \right]}}^\Phi - 1{K^{*\dagger }}_{\mu \nu }} \right){\rm I}$

with:

${\rm I} \equiv {L_{G(D3)}}$

Now, the super D3-brane action in type IIB SUGRA background with varying dilaton and axion has an SL(2, Z) self-duality, which, given the general Dp-action:

$S = {S_{DBI}} + {S_{WZ}}$

where the Dirac-Born-Infeld action is given by:

${S_{DBI}} = - \int_{{M_{p + 1}}} {{d^{p + 1}}} \sigma {e^{\frac{{p - 3}}{4}}}\sqrt { - \det \left( {{G_{ij}} + {e^{ - \frac{1}{2}\phi }}{F_{ij}}} \right)}$

and the Wess-Zumino action is given by:

${S_{WZ}} = \int_{{M_{p + 1}}} {{d^{p + 1}}} {e^F} \wedge C = \int_{{M_{p + 1}}} {{\Omega _{p + 1}}} = \int_{{M_{p + 1}}} {{I_{p + 2}}}$

with the DBI and WZ terms:

${S_{DBI}} = - \int_{{M_4}} {{d^4}} \sigma \sqrt { - \det \left( {{G_{ij}} + {e^{ - \frac{\phi }{2}}}{F_{ij}}} \right)}$

${S_{WZ}} = \int_{{M_4}} {\left( {{C_4} + {C_2} \wedge F + \frac{1}{2}{C_0}F \wedge F} \right)} = \int_{{M_4}} {{\Omega _4}}$

necessitates that we introduce the Lagrangian multiplier term:

${S_{\tilde H}} = \int_{{M_4}} {{d^4}} \sigma \frac{1}{2}{\tilde H^{ij}}\left( {{F_{ij}} - 2{\partial _i}{A_j}} \right)$

In order to derive the self-duality of the super D3-brane action in a type IIB supergravity background, where we have:

${\tilde H^{ij}} = {\hat \varepsilon ^{ijkl}}{\partial _k}{B_l}$

we substitute the self-dual vector potential ${B_i}$ in:

$S = {S_{DBI}} + {S_{WZ}}$

thus allowing us to derive the action:

$S' = {S_1} + {S_{2D}}$

where we have:

$\begin{array}{l}{S_1} = \int_{{M_4}} {\left[ { - {d^4}} \right.} \sigma \sqrt { - \det \left( {{G_{ij}} + {e^{ - \frac{\phi }{2}}}{F_{ij}}} \right)} \\ + \left( {{C_2} + \tilde F} \right) \wedge F + \frac{1}{2}{C_0}\left. {F \wedge F} \right]\end{array}$

and

${S_{2D}} = \int_{{M_4}} {\left( {{C_4} + \tilde F \wedge {b_2}} \right)}$

with

$\tilde F = dB$

Now we solve the Euler-Lagrange equations for ${F_{ij}}$ and plug the solution in the action $S'$ .

In the Lorentz frame with:

${G_{ij}} = {\eta _{ij}} = {\rm{diag}}\left( { - 1,1,1,1} \right)$

and $F$ block-diagonal:

${F_{ij}} = \left( {\begin{array}{*{20}{c}}0&{{F_{01}}}&0&0\\{ - {F_{01}}}&0&0&0\\0&0&0&{{F_{23}}}\\0&0&{ - {F_{23}}}&0\end{array}} \right)$

thus allowing us to derive the action:

$S' = {S_1} + {S_{2D}}$

in the local Lorentz frame as:

$\begin{array}{l}S' = \int_{{M_4}} {{d^4}} \sigma \left[ { - \sqrt {\left( {1 - {e^{ - \phi }}F_{01}^2} \right)\left( {1 + {e^{ - \phi }}F_{23}^2} \right)} } \right.\\ + {\left( {{C_2} + \tilde F} \right)_{23}}{F_{01}} + {\left( {{C_2} + \tilde F} \right)_{01}}{F_{23}}\\ + {C_0}{F_{01}}{F_{23}} + {C_{0123}} + \left. {{{\tilde F}_{01}}{b_{23}} + {{\tilde F}_{23}}{b_{01}}} \right]\end{array}$

Solutions to the EM for ${F_{01}}$ and ${F_{23}}$ give us:

${F_{01}} = - \frac{1}{A}\left( {{e^\phi }{C_0}b + a\sqrt {\frac{{A - {b^2}}}{{A + {a^2}}}} } \right)$

and

${F_{23}} = - \frac{1}{A}\left( {{e^\phi }a - b\sqrt {\frac{{A + {b^2}}}{{A - {a^2}}}} } \right)$

with:

$\left\{ {\begin{array}{*{20}{c}}{A \equiv {e^{ - \phi }} + {e^\phi }C_0^2}\\{a \equiv {{\left( {{C_2} + \tilde F} \right)}_{23}}}\\{b \equiv {{\left( {{C_2} + \tilde F} \right)}_{01}}}\end{array}} \right.$

Plugging them in the local Lorentz frame action above, we get the dual action:

${S_D} = - \int_{{M_4}} {\sqrt { - \det \left[ {{G_{ij}} + \frac{1}{{\sqrt {{e^{ - \phi }} + {e^\phi }C_0^2} }}\left( {{{\tilde F}_{ij}} + {C_{2ij}}} \right)} \right]} } + \int_{{M_4}} {{\Omega _D}}$

with

${\Omega _D} = {C_4} + {b_2} \wedge \tilde F - \frac{1}{2}\frac{{{e^\phi }{C_0}}}{{{e^{ - \phi }} + {e^\phi }C_0^2}}\left( {\tilde F \wedge {C_2}} \right) \wedge \left( {\tilde F + {C_2}} \right)$

Now in light of the following dilaton, axion, and form-potentials:

${e^{ - \phi '}} = \frac{1}{{{e^{ - \phi }} + {e^\phi }C_0^2}}$

${C'_0} = \frac{{{e^\phi }{C_0}}}{{{e^{ - \phi }} + {e^\phi }C_0^2}}$

and

$\left\{ {\begin{array}{*{20}{c}}{b' = - {C_2}}\\{{{C'}_2} = {b_2}}\\{{{C'}_4} - {b_2} \wedge {C_2}}\end{array}} \right.$

our dual action is given by:

${S_D} = - \int_{{M_4}} {{d^4}\sigma \,\Theta + \int_{{M_4}} \Xi }$

where $\Theta$ and $\Xi$ are stand-ins, respectively, for:

$\Theta \equiv \sqrt { - \det \left( {{G_{ij}} + {e^{ - \frac{{\phi '}}{2}}}{{\tilde F'}_{ij}}} \right)}$

and

$\Xi \equiv \left( {{{C'}_4} + {{C'}_2} \wedge \tilde F' + \frac{1}{2}{{C'}_0}\tilde F' \wedge \tilde F'} \right)$

By dualization, then under the transformations above of the dilaton, axion, and form-potentials, as well as:

$\left\{ {\begin{array}{*{20}{c}}{\tilde F \to F}\\{F \to \tilde F}\end{array}} \right.$

we get an equivalence between:

$S = {S_{DBI}} + {S_{WZ}}$

and

${S_D} = - \int_{{M_4}} {{d^4}\sigma \,\Theta + \int_{{M_4}} \Xi }$

To show that the super D3-brane action satisfies an $SL\left( {2,R} \right)$ self-duality in a Type IIB SUGRA background, one introduces the axio-dilatonic variable:

$\tau = {C_0} + i{e^{ - \phi }}$

Then the following transformations:

${e^{ - \phi '}} = \frac{1}{{{e^{ - \phi }} + {e^\phi }C_0^2}}$

${C'_0} = \frac{{{e^\phi }{C_0}}}{{{e^{ - \phi }} + {e^\phi }C_0^2}}$

are expressible as:

$\left\{ {\begin{array}{*{20}{c}}{\tau \to \tau ' = - \frac{1}{\tau } \in SL\left( {2,R} \right)}\\{S = \left( {\begin{array}{*{20}{c}}0&1\\{ - 1}&0\end{array}} \right)}\end{array}} \right.$

By holomorphicity and the modularity of the axio-dilaton, we get an $SL\left( {2,R} \right)$ doublet:

$\left\{ {\begin{array}{*{20}{c}}{\left( {{b_2}, - {C_2}} \right)}\\{\left( {\tilde F, - F} \right)}\end{array}} \right.$

that transforms according to a D7/M2 monodromy. To derive the axion-shift:

$C \to {C_0} + 1$

corresponding to the $SL\left( {2,R} \right)$ member:

$\tau \to \tau ' = \tau + 1$

one shows that in the super D-string action the super D3-brane action:

$S = {S_{DBI}} + {S_{WZ}}$

where the Dirac-Born-Infeld part is given by:

${S_{DBI}} = - \int_{{M_{p + 1}}} {{d^{p + 1}}} \sigma {e^{\frac{{p - 3}}{4}}}\sqrt { - \det \left( {{G_{ij}} + {e^{ - \frac{1}{2}\phi }}{F_{ij}}} \right)}$

and the Wess-Zumino one by:

${S_{WZ}} = \int_{{M_{p + 1}}} {{d^{p + 1}}} {e^F} \wedge C = \int_{{M_{p + 1}}} {{\Omega _{p + 1}}} = \int_{{M_{p + 1}}} {{I_{p + 2}}}$

is invariant under the following transformations:

$\left\{ {\begin{array}{*{20}{c}}{{C_0} \to {C_0} + 1}\\{{b_2} \to {b_2}}\\{{C_2} \to {C_2} \to - F = {C_2} + {b_2} - F}\\{{C_4} \to {C_4} + \frac{1}{2}F \wedge F}\\{F \to F}\end{array}} \right.$

and

$\left\{ {\begin{array}{*{20}{c}}{{C_0} \to {C_0} + 1}\\{{b_2} \to {b_2}}\\{{C_2} \to {C_2} + {b_2}}\\{{C_4} \to {C_4} + \frac{1}{2}{b_2} \wedge {b_2}}\\{F \to F}\end{array}} \right.$

It is clear now that the super D3-brane action:

$S = {S_{DBI}} + {S_{WZ}}$

and the type-IIB Lagrangian multiplier term:

${S_{\tilde H}} = \int_{{M_4}} {{d^4}} \sigma \frac{1}{2}{\tilde H^{ij}}\left( {{F_{ij}} - 2{\partial _i}{A_j}} \right)$

and the ${H_{\left( . \right)}}/{R_{\left( {1/3/5} \right)}}$ constraints they impose, are invariant under the first set of transformations and up to a topological term, under the second set.

We define now the dual field strength ${K_{ij}}$ by the Hodge antisymmetric tensor construction:

$\left\{ {\begin{array}{*{20}{c}}{ * {K^{ij}} = \frac{{\partial L}}{{\partial {F_{ij}}}}}\\{\frac{{\partial {F_{kl}}}}{{\partial {F_{ij}}}} = \delta _k^i\delta _l^j - \delta _l^i\delta _k^j}\end{array}} \right.$

where we have:

$* {K_{ij}} \equiv \frac{1}{2}\varepsilon _{ij}^{kl}{K_{kl}}\,,\, * * {K_{ij}} = - {K_{ij}}$

Since we are working with the Levi-Civita symbol in 4 dimensions, it follows that:

$\left( {{K_{ij}},{F_{ij}}} \right)$

transforms in accordance with:

$\left\{ {\begin{array}{*{20}{c}}{\tilde F \to \tilde F' = a\tilde F - bF}\\{F \to - c\tilde F + dF}\end{array}} \right.$

for a monodromy action:

$S = \left( {\begin{array}{*{20}{c}}{1 + \alpha }&\beta \\\gamma &{1 - \alpha }\end{array}} \right)$

given by:

$\left\{ {\begin{array}{*{20}{c}}{\delta {K_{ij}} = + \,\alpha {K_{ij}} + \beta {F_{ij}}}\\{\delta {F_{ij}} = - \,\alpha {F_{ij}} + \beta {K_{ij}}}\end{array}} \right.$

A sufficient condition for the $SL\left( {2,R} \right)$ invariance of the field equation and the invariance of the world-volume energy-momentum tensor is for the Gaillard-Zumino duality equation:

$\frac{\gamma }{4} * {K^{ij}}{K_{ij}} - \frac{\beta }{4} * {F^{ij}}{F_{ij}} - \frac{\alpha }{2} * {K^{ij}}{F_{ij}} + {\delta _\Phi }L = 0$

to hold for any $\alpha$ , $\beta$ , $\gamma$ in $SL\left( {2,R} \right)$ , where ${\delta _\Phi }$ is the world-volume Abelian gauge field variation. We can now show that the D3-brane action on type IIB supergravity background described by the action:

$S = {S_{DBI}} + {S_{WZ}}$

under the above ${H_{\left( . \right)}}/{R_{\left( {1/3/5} \right)}}$ constraints satisfies the Gaillard-Zumino duality equation under $SL\left( {2,R} \right)$ duality and $SO\left( 2 \right)$ spinor-action.

We can now write the super D3-brane Lagrangian in terms of component fields as:

$L = {L_{DBI}} + {\varepsilon ^{ijkl}}\left( {\frac{1}{{24}}{C_{ijkl}} + \frac{1}{4}{C_{ij}}{{\tilde F}_{kl}} + {C_0}{{\tilde F}_{ij}}{{\tilde F}_{kl}}} \right)$

where:

${L_{DBI}} = \Pi _G^{\tilde F} = - \sqrt G \Omega _{\tilde F}^\phi$

with:

$\Pi _G^{\tilde F} \equiv - \sqrt { - \det \left( {{G_{ij}} + {e^{ - \frac{\phi }{2}}}{{\tilde F}_{ij}}} \right)}$

$\Omega _{\tilde F}^\phi \equiv \sqrt {1 + \frac{{{e^{ - \phi }}}}{2}{{\tilde F}_{ij}}{F^{ij}} - \frac{{{e^{ - 2\phi }}}}{{16}}{{\left( {{{\tilde F}_{ij}} * {{\tilde F}^{ij}}} \right)}^2}}$

and $G = \det {G_{ij}}$ . It follows that the dual field strength defined by:

is given by:

$* {K^{ij}} = \frac{{\partial {L_{DBI}}}}{{\partial {{\tilde F}_{ij}}}} + * {C^{ij}} + {C_0} * {{\tilde F}^{ij}}$

${K_{ij}} = - {\left( { * \frac{{\partial {L_{DBI}}}}{{\partial F}}} \right)_{ij}} + {C_{ij}} + {C_0}\tilde F$

The $SL\left( {2,R} \right)$ transformations for:

$S = \left( {\begin{array}{*{20}{c}}{1 + \alpha }&\beta \\\gamma &{1 - \alpha }\end{array}} \right)$

acting on the fields and k-forms are thus given by:

$\delta {C_0} = 2\alpha {C_0} + \beta - \gamma \left( {C_0^2 - {e^{ - 2\phi }}} \right)$

$\delta \phi = 2\gamma {C_0} - 2\alpha$

$\delta {K_{ij}} = \alpha {K_{ij}} + \beta {F_{ij}}$

$\delta {F_{ij}} = - \alpha {F_{ij}} + \gamma {K_{ij}}$

$\delta {b_{ij}} = - \alpha {b_{ij}} + \gamma {C_{ij}}$

$\delta {C_{ij}} = \alpha {C_{ij}} + \beta {b_{ij}}$

as well as

$\delta {C_4}\frac{\beta }{2}{b_2} \wedge {b_2} + \frac{\gamma }{2}{C_2} \wedge {C_2}$

and the $SO\left( 2 \right)$ spinor-rotation is given by:

$\delta \theta = - \frac{{\gamma {e^{ - \phi }}}}{2}\Im \theta$

$\delta {\partial _\alpha } = \frac{{\gamma {e^{ - \phi }}}}{2}{\left( {\Im \partial } \right)_\alpha }$

$\delta E = \frac{{\gamma {e^{ - \phi }}}}{2}\Im E$

$\delta \bar E = - \frac{{\gamma {e^{ - \phi }}}}{2}\bar E\Im$

which necessitate the invariance of the Type IIB SUGRA constraints. We then find that ${\delta _\Phi }$ is given by:

$\begin{array}{l}{\delta _\Phi }L = \frac{1}{2}\frac{{\partial L}}{{\partial {b_{ij}}}}\delta {b_{ij}} + \frac{1}{2}\frac{{\partial L}}{{\partial {C_{ij}}}}\delta {C_{ij}} + \frac{{\partial L}}{{\partial \phi }}\delta \phi + \\\frac{{\partial L}}{{\partial {C_0}}}\delta {C_0} + \frac{1}{{24}}\frac{{\partial L}}{{\partial {C_{ijkl}}}}\delta {C_{ijkl}}\end{array}$

The Gaillard-Zumino duality constraint now dictates that equation:

$\frac{\gamma }{4} * {K^{ij}}{K_{ij}} - \frac{\beta }{4} * {F^{ij}}{F_{ij}} - \frac{\alpha }{2} * {K^{ij}}{F_{ij}} + {\delta _\Phi }L = 0$

is satisfied for arbitrary variations of $\alpha$ , $\beta$ , and $\gamma$ . Therefore their coefficients identically vanish. By appropriate substitutions, we get the following three variant equations:

$- \frac{1}{4} * {K^{ij}}{K_{ij}} - \frac{1}{2}\frac{{\partial L}}{{\partial {b_{ij}}}}{b_{ij}} + \frac{1}{2}\frac{{\partial L}}{{\partial {C_{ij}}}}{C_{ij}} - 2\frac{{\partial L}}{{\partial \phi }} + 2{C_0}\frac{{\partial L}}{{\partial {C_0}}} = 0$

$- \frac{1}{4} * {F^{ij}}{F_{ij}} + \frac{1}{2}\frac{{\partial L}}{{\partial {C_{ij}}}}{b_{ij}} + \frac{{\partial L}}{{\partial {C_0}}} + \frac{1}{8}{\varepsilon ^{ijkl}}{b_{ij}}{b_{kl}} = 0$

and

$\frac{1}{4} * {K^{ij}}{K_{ij}} + \frac{1}{2}\frac{{\partial L}}{{\partial {b_{ij}}}}{C_{ij}} + 2{C_0}\frac{{\partial L}}{{\partial \phi }} - \left( {C_0^2 - {e^{ - 2\phi }}} \right)\frac{{\partial L}}{{\partial {C_0}}} + \frac{1}{8}{\varepsilon ^{ijkl}}{C_{ij}}{C_{kl}} = 0$

The first two equations are trivially satisfied for $\beta$ and $\alpha$ in light of the DBI-identity:

$\frac{{\partial {L_{DBI}}}}{{\partial \phi }} = - \frac{1}{4}\frac{{\partial {L_{DBI}}}}{{\partial {F_{ij}}}}{\tilde F_{ij}}$

which holds by virtue of the fact that the dilaton $\phi$ is contained in ${L_{DBI}}$ in the form of

${e^{ - \frac{\phi }{2}\tilde F}}$

which in turn implies that we have a $\gamma$ -reduction equation:

$\frac{1}{2}{\varepsilon ^{ijkl}}\left( {\frac{{\partial {L_{DBI}}}}{{\partial {F_{ij}}}}\frac{{\partial {L_{DBI}}}}{{\partial {F_{kl}}}} + {e^{ - 2\phi }}{{\tilde F}_{ij}}{{\tilde F}_{kl}}} \right) = 0$

Thus we have established that $SL\left( {2,R} \right)$ self-duality and $SO\left( 2 \right)$ spinor-action imply that the super D3-brane action:

$S = {S_{DBI}} + {S_{WZ}}$

in the most general Type IIB SUGRA background satisfies the Gaillard-Zumino duality condition and exhibits exact self-duality. As we shall see in upcoming posts, this plays an essential role in how one geometrically derives the Standard Model of physics from F-theory.