D3-Brane Self-Duality And Super-Hamiltonian Action Under Poincaré Invariance

Well, a big question is how did the universe begin. And we, cannot answer that question. Some people think that the big bang is an explanation of how the universe began, IT IS NOT. The big bang is a theory of how the universe evolved from a split second after whatever brought it into existence. And the reason why we’ve been unable to look right back at time zero, to figure out how it really began; is that conflict between Einstein’s ideas of gravity and the laws of quantum physics. So, STRING THEORY may and will be able to – it hasn’t yet; we’re working on it today – feverishly. It may be able to answer the question, how did the universe begin. And I don’t know how it’ll affect your everyday life, but to me, if we really had a sense of how the universe really began, I think that would, really, alert us to our place in the cosmos in a DEEP way. ~ Brian Greene!

I listed the 4 essential properties of D3-branes, namely: i) the propagation of a D3-brane through spacetime generates a 4-dimensional worldvolume that has 4-dimensional Poincaré invariance, ii) the string worldsheet generating the graviton via quantum fluctuation can be topologically compactified on the boundary of its corresponding Ad{S_4} \times {S^4} space, iii) D3-branes have constant axion and dilaton fields, and for the purposes of this post, iv) D3-branes are self-dual. Thus, the gravitonic D3-brane action with a super-Lagrangian coupling can be derived as

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


    \[{L_{G(D3)}} \equiv \widetilde {{L_G}}\left( {{F_{\mu \nu }},{\chi ^a},\theta ;\,\phi ,\chi } \right) = {L_G}\left( {{e^{ - \Phi /2}}{F_{\mu \nu }},{\chi ^a},\theta } \right) - \frac{1}{4}{\chi ^a}{F_{\mu \nu }}{\widetilde F_{\mu \nu }}\]

However, as I showed, one must exhibit the self-duality of the D3-brane in the Hamiltonian setting. It is my aim in this post to provide the proof. One can always lift an SO(2) duality to an S(2,\Re ) duality by introducing the D3-brane dilaton \phi and axion \chi which are constant background fields. Then, one can re-define an SO(2) duality Lagrangian as such:

    \[\widetilde L\left( {F,X,\theta ;\phi ,\chi } \right) = {L_G}\left( {{e^{ - \Phi /2}}F,X,\theta } \right) - \frac{1}{4}\chi F\,\widetilde F\]

with \widetilde F = {e^{ - \Phi /2}}F. From the above SO(2) dual Lagrangian, the D3-brane Hamiltonian action can be derived as

    \[S = \int {{d^4}} \sigma {L^{DBI}} + \int {{L^{WZ}}} - \int {\frac{1}{2}} \,\chi {F^2} = \int {{d^4}} \sigma L_G^{{\rm{Total}}}\]


    \[\left\{ {\begin{array}{*{20}{c}}{{L^{DBI}} = - \sqrt { - {\rm{det}}\left( {{G_{\mu \nu }} + {F_{\mu \nu }}} \right)} }\\{{G_{\mu \nu }} = \prod _\mu ^M{\prod _\mu }M}\\{{F_{\mu \nu }} = {{\not \partial }_{\left[ {\mu A\nu } \right]}} + \Omega _{\mu \nu }^3}\\{\Omega _{\mu \nu }^i = {{\widetilde {\theta {{\not \prod }_{{{\left[ {\mu {\tau _j}{{\not \partial }_\nu }} \right]}^\theta }}}}}^{ * \dagger }}}\end{array}} \right.\]

(j = 1,3) and {\tau _i} are the Pauli matrices cohomologically acting on the supersymmetric group indices, and {L^{WZ}} is the Wess-Zumino Lagrangian satisfying the Matsubara condition, and is given by

    \[\left\{ {\begin{array}{*{20}{c}}{{L^{WZ}} = {C^{(2)}}{F_{\mu \nu }} + C(4)}\\{{C^{(2)}} = \widetilde \theta {{\not \prod }^{ * \dagger }}{\tau _i}\,d\theta = {\Omega _1}}\\{{C^{(4)}} = \Xi \,\; - \frac{1}{2}{\Omega _1}\,{\Omega _3}}\end{array}} \right.\]

where {C^{(2)}} and {C^{(4)}} are RR-2 and RR-4 differential forms, and \Xi represents the Kappa symmetry of the gauge bundle of the D3-brane’s Ad{S_4} \times {S^4} topology. Now, let

    \[\left\{ {\begin{array}{*{20}{c}}{\left( {{X^M},{P^M}} \right)}\\{\left( {\theta ,\,{\pi _\theta }} \right)}\\{\left( {{A_\mu },{E^\mu }} \right)}\end{array}} \right.\]

be a canonical conjugate set for the super-Kahler phase space variables, and define the critical 3-dimensional anti-symmetric tensor

    \[{\varepsilon ^{ijk}} = {\varepsilon ^{0ijk}}\]

and introduce the de Rham variables

    \[\left\{ {\begin{array}{*{20}{c}}{{{\not B}^i} = \frac{1}{2}{\varepsilon ^{ijk}}{F_{ik}} = {e^{ - \Phi /2}}{B^i} + \frac{1}{2}{\varepsilon ^{ijk}}\Omega _{jk}^3}\\{{B^i} = \frac{1}{2}{\varepsilon ^{ijk}}{F_{ij}}\;;\,\;\left( {i,j,k = 1,2,3} \right)}\\{{{\not P}_M} \equiv \frac{{\partial {L^{DBI}}}}{{\partial \prod _0^M}} = P_M^{ * \dagger } - {e^{\phi /2}}{{\left( {E + \chi B} \right)}^i} \cdot \frac{{\partial {F_{0i}}(\prod ,\widetilde \theta )}}{{\partial \prod _0^M}} - \frac{{\partial {L^{WZ}}\left( {\prod ,F,\widetilde \theta } \right)}}{{\partial \prod _0^M}}}\\{{{\not \Sigma }^i} \equiv \frac{{\partial {L^{DBI}}}}{{\partial {F_{0i}}}} = {e^{\phi /2}}{{\left( {E + \chi B} \right)}^i} - \frac{{\partial {L^{WZ}}\left( {\prod ,F,\widetilde \theta } \right)}}{{\partial {F_{0i}}}}}\end{array}} \right.\]

where {\not \Sigma ^i} transforms as

    \[\delta {\not \Sigma ^i} = \not \partial {L^{{\rm{Total}}}}/\not \partial {F_{0i}}\]

One then finds that the constraints of the system to be given by

U(1) symmetry constraint

{\not \Sigma ^0} = 0

{\partial _i}{\not \Sigma ^i} = 0


– the p + 1 diffeomorphism constraints

    \[\left\{ {\begin{array}{*{20}{c}}{{\varphi _i} \equiv \not P \cdot {\prod _i} + {{\not \Sigma }^{ij}}{F_{ij}} = \not P \cdot {\prod _i} + \,{\varepsilon _{ijk}}\,{{\not \Sigma }^j}{{\not B}^k} = 0}\\{{\varphi _0} \equiv \frac{1}{2}\left[ {{{\not P}^2} + \gamma + {\gamma _{ij}}\left( {{{\not \Sigma }^i}{{\not \Sigma }^j} + {{\not B}^i}{{\not B}^j}} \right)} \right] = 0}\end{array}} \right.\]


the fermionic constraints

    \[\begin{array}{c}\psi \equiv {\pi _0} - {{\not P}_M}\frac{{\not \partial \prod _o^M}}{{\not \partial \widetilde \theta }} - {e^{\phi /2}}{\left( {E + \chi \not B} \right)^i}\frac{{\not \partial {F_{0i}}\left( {\prod ,\widetilde \theta } \right)}}{{\not \partial \widetilde \theta }}\\ - \frac{{\not \partial {L^{WZ}}\left( {\prod ,F,\widetilde \theta } \right)}}{{\not \partial \widetilde \theta }}\end{array}\]

with {\gamma _{ij}} being the spatial part of the Ad{S_4} \times {S^4} metric, and its determinant being {\gamma ^{WZ}}. One must now show the Poincaré invariance of the bosonic constraints

    \[\left( {{\varphi _0},{\varphi _i}} \right)\]

and the supersymmetric covariance of the fermionic constraint \psi under the SL(2,\Re ) transformation of \left( {\not B,\not \Sigma } \right) and \left( {\phi ,\chi } \right) corresponding to the SO(2) fermionic field rotation holds. Then one gets

    \[\left( {\begin{array}{*{20}{c}}{\not B}\\{\not \Sigma }\end{array}} \right) = {V^{ - 1}}\left( {\begin{array}{*{20}{c}}B\\{{{\not \Sigma }^{ - 1}}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{{\Omega ^3}}\\{ - \,{\Omega ^1}}\end{array}} \right)\]


    \[{\left( {{\Omega ^l}} \right)^i} \equiv {\varepsilon ^{ijk}}\widetilde \theta \not \prod _i^{ * \dagger }{\tau _l}\,\not \partial \theta \]

with V being an SL(2,\Re )/SO(2) matrix satisfying

    \[V = {e^{\Phi /2}}\left( { - {\chi ^l} \cdot {e^{ - \phi }}} \right)\]

and transforms as

    \[\delta V \to \Lambda V{\rm O}{(\Lambda )^{ - 1}}\]


    \[\Lambda = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right) \in SL(2,\Re )\]


    \[ad - bc = 0\]

and {\rm O}(\Lambda ) being an SO(2) transformation satisfying

    \[{\rm O}{(\Lambda )^{ - 1}} = \left( {\begin{array}{*{20}{c}}{\cos \lambda }&{\sin \lambda }\\{ - \sin \lambda }&{ - \cos \lambda }\end{array}} \right) \in SO(2)\]


    \[{V^{ - 1}} = {e^{\phi /2}}\left( {\begin{array}{*{20}{c}}{{e^{ - \phi }}}&0\\\chi &1\end{array}} \right) \to {\rm O}(\Lambda )\left( {\begin{array}{*{20}{c}}{{\Omega ^3}}\\{ - {\Omega ^1}}\end{array}} \right)\]

with {\tau _2} rotation \theta \to {\rm O}(\Lambda )\theta.

To prove D3-brane self-duality, we must consider

    \[\begin{array}{c}{{\not P}_M} = \not P - \left( {{{\not \Sigma }^i}\frac{{\not \partial {F_{01}}\left( {\prod ,\widetilde \theta } \right)}}{{\not \partial \prod _0^M}} + {{\not B}^i}\frac{{\not \partial C_{ij}^{(2)}\left( {\prod ,\widetilde \theta } \right)}}{{\not \partial \prod _o^M}}} \right) - \\\left( {\frac{1}{2}{\varepsilon ^{ijk}}C_{ij}^{(2)}\frac{{\not \partial {F_{0i}}\left( {\prod ,\widetilde \theta } \right)}}{{\not \partial \prod _0^M}} + \frac{{\not \partial }}{{\not \partial \prod _0^m}}\sqrt { - {G_{\mu \nu }}} {C^{(4)}}\left( {\prod ,\widetilde \theta } \right)} \right)\end{array}\]

while noting that all terms in the above expression and \not P itself are invariant under Poincaré duality transformations. So, {\not P_M}, the conjugate of \not P, is also likewise invariant, hence,

    \[\frac{{\not \partial }}{{\not \partial \prod _0^M}}\sqrt { - {G_{\mu \nu }}} {C^{(4)}}\left( {\prod ,\widetilde \theta } \right)\]

reduces to

    \[\begin{array}{c}\left( {{{\not \Sigma }^i}\frac{{\not \partial {F_{0i}}}}{{\not \partial \prod _0^M}} + \frac{1}{2}{\varepsilon _{ijk}}{{\not B}^i}\frac{{\not \partial {C^{{{(2)}^{jk}}}}}}{{\not \partial \prod _0^M}}} \right) \cdot {{\not \Sigma }^i}\widetilde {\theta \,}{\Gamma _M}{\tau _3}\,{{\not \partial }_i}\theta = \left( {{{\not B}^i},{{\not \Sigma }^i}} \right) \cdot \\\left( {\begin{array}{*{20}{c}}{\widetilde {\theta \,}{\Gamma _M}{\tau _1}{{\not \partial }_i}\theta }\\{\widetilde \theta \,{\Gamma _M}{\tau _3}\,\not \partial \theta }\end{array}} \right) = \widetilde {\theta \,}\Gamma {\tau _{{0^i}}}{{\not \partial }_i}\widetilde \theta \end{array}\]


    \[\tau _0^i = {\not \Sigma ^i}{\tau _3} + {\not B^i}{\tau _1}\]

holds. In terms …