D3-Brane Actions And Gaillard-Zumino Conditions With An SO(2) Duality For SuperGravity

By George Shiber Friday, May 22, 2015Thursday, August 15, 2019 In Grand Unification Physics, M Theory, Mathematical Physics, Philosophy Of Mathematics, Philosophy Of Physics, Philosophy Of Science, Super String Theory, Supersymmetry bosons, effective D3 brane action

Nature is a mutable cloud, which is always and never the same. ~ Ralph Waldo Emerson!

Where there is life there is a pattern, and where there is a pattern there is mathematics. Once that germ of rationality and order exists to turn a chaos into a cosmos, then so does mathematics. There could not be a non-mathematical Universe containing living observers.
John D. Barrow, The Artful Universe (1995)

I last derived the D3-brane effective action in the NS5-brane geometry, via the D3-brane self-duality and Poincaré invariance properties as such:

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

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. 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

${K^{ * \dagger \mu \nu }} = \frac{{\partial {L_G}}}{{\partial {F_{\mu \nu }}}}$ , $\frac{{\partial {F_{\alpha \beta }}}}{{\partial {F_{\mu \nu }}}} = \left( {\delta _\alpha ^\mu \,\delta _{{\beta ^{\, - }}}^\nu - \delta _\beta ^\mu \,\delta _\alpha ^\nu } \right)$ 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 now is fundamentally an infinitesimal duality transformation of $F$ and $K$ and matter 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 a $SO(2)$ 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 information, 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: So,

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

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

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

and $\Xi$ 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 the 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

${S_{D3}} = \frac{1}{{4{k^2}{C_{\left[ 4 \right]}}}}{\int {\sqrt {\widetilde K} } _{\mu \nu }}^{ - 2\Phi }\left( {2{K^{\mu \nu }}{C_{\left[ 2 \right]}} + \frac{\lambda }{8} + {{\not \partial }_\mu }\Phi \,{\partial _\Phi }^\mu \Phi - {C_{\left[ 4 \right]}}^\Phi - 1{K^{ * \dagger }}_{\mu \nu }} \right){L_{G(D3)}}$