The D3-Brane, Self-Duality, the Super-Hamiltonian Action and the M5-Brane

The importance of D3-branes derives partly from their role in F-theory due to D3/D7-brane systems in the presence of orientifolds, yielding an effective 4D N = 1 F-action, and D3-branes naturally correspond to M5-branes in M-theory compactified on a Type-II-A/B torus, and instantons that arise from D3-branes wrapping internal divisors on a Calabi-Yau 3-fold correspond to M5-brane instantons in M-theory compactified on an elliptically fibered Calabi-Yau 4-fold. Moreover, a unique property of D3-branes is self-duality. Hence, for the M5- $\mathfrak{g}_{{\text{Lie}}}^{3}$ algebra to close, the D3 worldvolume super-Hamiltonian must exhibit manifest Poincaré symmetry. Note that Dp-brane solutions, for p = 1, 2, 3, 4, preserving 1/2 SUSY, have general form:

$\displaystyle ds_{{Dp}}^{2}={{\Omega }^{{-1}}}\left[ {d{{t}^{2}}-ds_{p}^{2}} \right]-\Omega dx_{{5-p}}^{2}$

$\displaystyle {{e}^{{2\phi }}}={{\Omega }^{{1-p}}}$

$\displaystyle F_{{01...pm}}^{A}={{\partial }_{m}}{{H}^{A}}$

$\displaystyle {{M}_{{Kah}}}^{{AB}}={{\text{I}}_{{\text{Re}{{\text{p}}_{G}}}}}^{{AB}}+2{{\Omega }^{{-1}}}{{H}^{A}}{{H}^{B}}$

where the M5-brane action in a D = 11 SUGRA background is given by:

$\displaystyle \begin{array}{l}S=2\int_{{{{M}_{6}}}}{{d{{x}^{6}}}}\left[ {\sqrt{{-\det \left( {{{g}_{{\mu \nu }}}+i{{{\tilde{H}}}_{{\mu \nu }}}} \right)}}} \right.\\+\frac{{\sqrt{{-g}}}}{{4{{{\left( {\partial a} \right)}}^{2}}}}{{\partial }_{\lambda }}a{{{\tilde{H}}}^{{\lambda \mu \nu }}}\left. {{{H}_{{\mu \nu \rho }}}{{\partial }^{\rho }}_{a}} \right]-\int_{{{{M}_{6}}}}{{{{C}_{6}}}}+{{H}_{3}}\wedge {{C}_{3}}\end{array}$

with:

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {{{{\tilde{H}}}^{{\rho \mu \nu }}}\equiv \frac{1}{{6\sqrt{{-g}}}}{{\epsilon }^{{\rho \mu \nu \lambda \sigma \tau }}}{{H}_{{\lambda \sigma \tau }}}} \\ {{{{\tilde{H}}}_{{\lambda \sigma \tau }}}\equiv \frac{{{{\partial }^{\rho }}a}}{{\sqrt{{{{{\left( {\partial a} \right)}}^{2}}}}}}{{{\tilde{H}}}_{{\rho \mu \nu }}}} \\ {g=\det {{g}_{{\mu \nu }}}} \\ {{{\epsilon }^{{0...5}}}=-{{\epsilon }_{{0...5}}}=1} \end{array}} \right.$

and the $M5$ -brane action takes the following form:

$\displaystyle S=\int_{{{{{\hat{M}}}_{6}}}}{{{{d}^{6}}x}}\left( {-\frac{{\sqrt{{-g}}}}{6}{{{\tilde{\mathcal{L}}}}_{{M5}}}^{{F,G}}\left( {\tilde{F}} \right)} \right)-\int_{{{{{\hat{M}}}_{6}}}}{{\left( {{{C}_{6}}+H\wedge {{C}_{3}}} \right)}}$

where:

$\displaystyle {{\widetilde{\mathcal{L}}}_{{M5}}}^{{F,G}}\left( {\tilde{F}} \right)\doteq \left( {{{{\tilde{G}}}^{{\mu \nu \rho }}}{{G}_{{\mu \nu \rho }}}+3{{{\tilde{F}}}^{{\mu \nu \rho }}}{{F}_{{\mu \nu \rho }}}} \right)+2{{\mathcal{L}}_{{M5}}}\left( {F,G} \right)$

with:

$\displaystyle \begin{array}{l}{{\mathcal{L}}_{{M5}}}=-\frac{1}{{36\left( {1+{{G}^{2}}} \right)}}{{\epsilon }^{{{{\mu }_{1}}}}}{{^{{{{\mu }_{2}}}}}^{{{{\mu }_{3}}}}}{{^{{{{\mu }_{4}}}}}^{{{{\mu }_{5}}}}}^{{{{\mu }_{6}}}}{{G}_{{{{{^{{{{\mu }_{1}}}}}}^{{{{\mu }_{2}}}}}^{{{{\mu }_{3}}}}}}}{{F}_{{{{\mu }_{4}}\nu \lambda }}}{{F}_{{{{\mu }_{5}}}}}^{{\lambda \kappa }}{{F}_{{{{\mu }_{6}}\kappa }}}^{\nu }\\+\frac{1}{{1+{{G}^{2}}}}\sqrt{{-\det \left( {{{g}_{{\mu \nu }}}+\frac{1}{2}{{{\left( {F+G} \right)}}_{{\mu \rho \sigma }}}{{{\left( {F+G} \right)}}_{{{{\nu }^{{\rho \nu }}}}}}} \right)}}\end{array}$

Now, a D3-brane has a super-Yang-Mills worldvolume theory and at strong coupling, it constitutes a black brane solution of type II supergravity. Hence, by D3 self-duality, the bosonic D3-brane action with a super-Lagrangian coupling is given by:

${S_{D3}} = \frac{1}{{4{k^2}{C_{\left[ 4 \right]}}}}\int {\sqrt {\widetilde {{k_{\mu \nu }}^{ - 2\Phi }}} } \left( {2{k^{\mu \nu }}{C_{\left[ 2 \right]}} + \frac{\lambda }{8} + {{\partial }_\mu }\Phi \,\partial _\phi ^\mu \Phi - C_{\left[ 4 \right]}^\Phi - 1{K_{\mu \nu }}^{ * \dagger }} \right){L_{G(D3)}}$

with:

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

and where the Ramond-Ramond gauge-coupling sector is given by the action:

$\displaystyle \mathcal{L}_{G}^{{Loc}}=\sum\limits_{{b=1}}^{{N-1}}{{\frac{1}{{2g_{b}^{2}}}}}{{\int{\text{d}}}^{2}}\theta {{W}^{\alpha }}{{W}_{\alpha }}{{\delta }^{2}}\left( {\left( {1-{{e}^{{ib\phi }}}} \right)z} \right)$

and the Ramond-Ramond term being:

$\displaystyle {{S}_{{CS}}}=\frac{{{{T}_{p}}}}{2}\int\limits_{{{{\Sigma }_{{p+1}}}}}{{C\wedge \text{Tr}}}\left( {{{e}^{{F/2\pi }}}} \right)$

which yields the Type-IIB Calabi-Yau three-fold superpotential:

$\displaystyle {{V}_{W}}=\int\limits_{X}{{{{G}_{3}}}}\wedge {{\Omega }_{3}}+\sum\limits_{{i=1}}^{{{{h}^{{1,1}}}}}{{{{A}_{i}}}}\left( {\left( {{{e}^{{-\phi }}}+i{{C}_{0}}} \right),U} \right){{e}^{{-a\left( {{{e}^{{-\phi }}}{{\tau }_{i}}+i{{\rho }_{i}}} \right)}}}$

and where the topologically mixed Yang-Mills action is given by:

$\displaystyle {{\mathcal{L}}_{{TYM}}}\equiv -\frac{1}{4}e{{\tilde{F}}_{{\mu \nu }}}^{M}{{\tilde{F}}^{{\mu \nu N}}}{{\hat{M}}_{{MN}}}+\kappa {{\mathcal{L}}_{{CS}}}$

with the corresponding Chern-Simons action:

$\displaystyle {{S}_{{CS}}}=\frac{{{{T}_{p}}}}{2}\int\limits_{{{{\Sigma }_{{p+1}}}}}{{C\wedge \text{ch}}}\left( {\tilde{F}} \right)\wedge \sqrt{{\frac{{\hat{A}\left( {{{R}_{T}}} \right)}}{{\hat{A}\left( {{{R}_{N}}} \right)}}}}$

and where the Ramond-Ramond coupling-term:

$\displaystyle {{S}_{{CS}}}=\frac{{{{T}_{p}}}}{2}\int\limits_{{{{\Sigma }_{{p+1}}}}}{{C\wedge \text{Tr}}}\left( {{{e}^{{\tilde{F}/\pi }}}} \right)$

has variational action:

$\displaystyle \begin{array}{c}\delta {{\mathcal{L}}_{{TYM}}}=\left( {\Theta _{F}^{\kappa }-\Xi _{D}^{M}} \right)\delta {{A}_{\mu }}^{M}+\\5{{d}^{{MKN}}}{{\partial }_{K}}\left( {\tilde{\Theta }_{F}^{\kappa }+\mathcal{H}} \right)\Delta {{B}_{{\mu \nu N}}}+\vartheta \left( {\delta {{g}_{{\mu \nu }}}} \right)+\vartheta \left( {\delta {{{\hat{M}}}_{{MN}}}} \right)\end{array}$

However, one must exhibit the self-duality of the D3-brane in the Hamiltonian metaplectic setting in order for the M5- $\mathfrak{g}_{{\text{Lie}}}^{3}$ algebra to close. To that end, I will derive that as well as the induced D-brane and M5-brane Hamiltonians. First we note that 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, we can re-define an $SO(2)$ -dual 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:

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

where:

$\left\{ {\begin{array}{*{20}{c}}{{L^{DBI}} = - \sqrt { - {\rm{det}}\left( {{G_{\mu \nu }} + {F_{\mu \nu }}} \right)} }\\{{G_{\mu \nu }} = \prod _\mu ^M{\prod _\mu }M}\\{{F_{\mu \nu }} = {{\partial }_{\left[ {\mu A\nu } \right]}} + \Omega _{\mu \nu }^3}\\{\Omega _{\mu \nu }^i = {{\widetilde {\theta {{\prod }_{{{\left[ {\mu {\tau _j}{{\partial }_\nu }} \right]}^\theta }}}}}^{ * \dagger }}}\end{array}} \right.$

$(j = 1,3)$ 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:

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {{{L}^{{WZ}}}={{C}^{{(2)}}}{{F}_{{\mu \nu }}}+{{C}^{{(4)}}}} \\ {{{C}^{{(2)}}}=\tilde{\theta }{{\prod }^{{*\dagger }}}{{\tau }_{i}}d\theta ={{\Omega }_{1}}} \\ {{{C}^{{(4)}}}=\Xi -\frac{1}{2}{{\Omega }_{1}}{{\Omega }_{3}}} \end{array}} \right.$

where ${C^{(2)}}$ and ${C^{(4)}}$ are the RR-2 and RR-4 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:

$\left\{ {\begin{array}{*{20}{c}}{{{B}^i} = \frac{1}{2}{\varepsilon ^{ijk}}{F_{ik}} = {e^{ - \Phi /2}}{B^i} + \frac{1}{2}{\varepsilon ^{ijk}}\Omega _{jk}^3}\\{{B^i} = \frac{1}{2}{\varepsilon ^{ijk}}{F_{ij}}\;;\,\;\left( {i,j,k = 1,2,3} \right)}\\{{{P}_M} \equiv \frac{{\partial {L^{DBI}}}}{{\partial \prod _0^M}} = P_M^{ * \dagger } - {e^{\phi /2}}{{\left( {E + \chi B} \right)}^i} \cdot \frac{{\partial {F_{0i}}(\prod ,\widetilde \theta )}}{{\partial \prod _0^M}} - \frac{{\partial {L^{WZ}}\left( {\prod ,F,\widetilde \theta } \right)}}{{\partial \prod _0^M}}}\\{{{\Sigma }^i} \equiv \frac{{\partial {L^{DBI}}}}{{\partial {F_{0i}}}} = {e^{\phi /2}}{{\left( {E + \chi B} \right)}^i} - \frac{{\partial {L^{WZ}}\left( {\prod ,F,\widetilde \theta } \right)}}{{\partial {F_{0i}}}}}\end{array}} \right.$

where ${\Sigma ^i}$ transforms as:

$\delta {\Sigma ^i} = \partial {L^{{\rm{Total}}}}/\partial {F_{0i}}$

One then finds that the constraints of the system are given by:

– $U(1)$ symmetry constraint

– ${\Sigma ^0} = 0$

– ${\partial _i}{\Sigma ^i} = 0$

– the p + 1 diffeomorphism constraints:

$\displaystyle {{\varphi }_{i}}\equiv {{{\tilde{P}}}_{M}}\cdot \prod\nolimits_{i}{{+{{\Sigma }^{{ij}}}}}{{F}_{{ij}}}={P}'\cdot \prod\nolimits_{i}{{+{{\varepsilon }_{{ijk}}}}}{{\Sigma }^{j}}{{B}^{k}}=0$

$\displaystyle {{\varphi }_{0}}\equiv \frac{1}{2}\left[ {{{P}^{2}}+\tilde{\gamma }+{{\gamma }_{{ij}}}\left( {{{\Sigma }^{i}}{{\Sigma }^{j}}+{{B}^{i}}{{B}^{j}}} \right)} \right]=0$

as well as:

– the fermionic constraints:

$\begin{array}{c}\psi \equiv {\pi _0} - {{P}_M}\frac{{\partial \prod _o^M}}{{\partial \widetilde \theta }} - {e^{\phi /2}}{\left( {E + \chi B} \right)^i}\frac{{\partial {F_{0i}}\left( {\prod ,\widetilde \theta } \right)}}{{\partial \widetilde \theta }}\\ - \frac{{\partial {L^{WZ}}\left( {\prod ,F,\widetilde \theta } \right)}}{{\partial \widetilde \theta }}\end{array}$

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 that the supersymmetric covariance of the fermionic constraint $\psi$ under the $SL(2,\Re )$ transformation of $\left( {B, \Sigma } \right)$ and $\left( {\phi ,\chi } \right)$ corresponding to the $SO(2)$ fermionic field rotation holds. Note that we get:

$\left( {\begin{array}{*{20}{c}}{B}\\{\Sigma }\end{array}} \right) = {V^{ - 1}}\left( {\begin{array}{*{20}{c}}B\\{{{\Sigma }^{ - 1}}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{{\Omega ^3}}\\{ - \,{\Omega ^1}}\end{array}} \right)$

and:

${\left( {{\Omega ^l}} \right)^i} \equiv {\varepsilon ^{ijk}}\widetilde \theta \prod _i^{ * \dagger }{\tau _l}\,\partial \theta$

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:

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

with:

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

and

$ad - bc = 0$

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

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

and:

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

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

To establish D3-brane self-duality, we must consider:

$\begin{array}{c}{{P}_M} = P - \left( {{{\Sigma }^i}\frac{{\partial {F_{01}}\left( {\prod ,\widetilde \theta } \right)}}{{\partial \prod _0^M}} + {{B}^i}\frac{{\partial C_{ij}^{(2)}\left( {\prod ,\widetilde \theta } \right)}}{{\partial \prod _o^M}}} \right) - \\\left( {\frac{1}{2}{\varepsilon ^{ijk}}C_{ij}^{(2)}\frac{{\partial {F_{0i}}\left( {\prod ,\widetilde \theta } \right)}}{{\partial \prod _0^M}} + \frac{{\partial }}{{\partial \prod _0^m}}\sqrt { - {G_{\mu \nu }}} {C^{(4)}}\left( {\prod ,\widetilde \theta } \right)} \right)\end{array}$

while noting that all terms in the above expression as well as $P$ are invariant under Poincaré duality transformations. So, ${P_M}$ , the conjugate of $P$ , is also likewise invariant, hence:

$\frac{{\partial }}{{\partial \prod _0^M}}\sqrt { - {G_{\mu \nu }}} {C^{(4)}}\left( {\prod ,\widetilde \theta } \right)$

reduces to:

$\begin{array}{c}\left( {{{\Sigma }^i}\frac{{\partial {F_{0i}}}}{{\partial \prod _0^M}} + \frac{1}{2}{\varepsilon _{ijk}}{{B}^i}\frac{{\partial {C^{{{(2)}^{jk}}}}}}{{\partial \prod _0^M}}} \right) \cdot {{\Sigma }^i}\widetilde {\theta \,}{\Gamma _M}{\tau _3}\,{{\partial }_i}\theta = \left( {{{B}^i},{{\Sigma }^i}} \right) \cdot \\\left( {\begin{array}{*{20}{c}}{\widetilde {\theta \,}{\Gamma _M}{\tau _1}{{\partial }_i}\theta }\\{\widetilde \theta \,{\Gamma _M}{\tau _3}\,\partial \theta }\end{array}} \right) = \widetilde {\theta \,}\Gamma {\tau _{{0^i}}}{{\partial }_i}\widetilde \theta \end{array}$

where:

$\tau _0^i = {\Sigma ^i}{\tau _3} + {B^i}{\tau _1}$

holds. In terms of differential forms:

$P \equiv \frac{{\partial {L^{DBI}}}}{{\partial \prod _0^M}}$

reduces to:

$\begin{array}{c}{C^{{{(2)}^{0,i}}}}\frac{{\partial {F_{0i}}}}{{\partial \prod _0^M}} + \frac{{\partial }}{{\partial \prod _0^M}}\sqrt { - {G_{\mu \nu }}} {C^{(4)}} \to \frac{1}{2}{\left[ { - \left( {\widetilde \theta {\Gamma _M}{\tau _{\left[ {3d\theta } \right]}}\,d\theta } \right)\left( {\widetilde \theta \prod {\tau _1}d\theta } \right)} \right]_3} + \\{\left[ {\frac{{\partial {C^{(4)}}}}{{\partial \prod _0^M}}} \right]_3} = \frac{1}{2}{\left[ { - \widetilde {\theta \,}{\Gamma _M}{\tau _{\left[ {3d\theta } \right]}}d\theta \left( {\widetilde \theta \prod {\tau _1}d\theta } \right)} \right]_3} + {\left[ {\frac{{\partial }}{{\partial \prod _0^M}}\left( {{C^{(4)}} + \frac{1}{2}{\Omega _1}\,{\Omega _3}} \right)} \right]_3}\end{array}$

where ${\left[ {} \right]_3}$ represents a spatial 3-form coefficient of $\left[ {} \right]$ , and given the Poincaré invariance of ${\tau _1}$ and ${\tau _3}$ under $\theta$ rotation, one can finish the proof by utilizing the gauge invariance of $P$ and by using the identity:

$\Xi = \left[ {{C^{(4)}} + \frac{1}{2}{\Omega _1}\,{\Omega _3}} \right]$

Diffeomophically, $B$ and $\Sigma$ essentially appear in the irreducible representation of $SO(2)$ , and thus, ${\varphi _0}$ and ${\varphi _i}$ also have Poincaré invariance. Now, the supersymmetric covariance of the fermionic constraints with the above $P$ is:

$\begin{array}{c}\psi = {\pi _\theta } + {{P}_M}\left( {\widetilde \theta \,{\Gamma ^M}} \right) - \left( {{{\Sigma }^i}\frac{{\partial {F_{0i}}}}{{\partial \widetilde \theta }} + \frac{1}{2}{\varepsilon _{ijk}}{{B}^i}\frac{{\partial {C^{{{(2)}^i}}}}}{{\partial \widetilde \theta }}} \right) - \\\left( {{C^{{{(2)}^{0i}}}}\frac{{\partial {F_{0i}}}}{{\partial \widetilde \theta }} + \frac{{\partial \sqrt { - {G_{\mu \nu }}} {C^{(4)}}}}{{\partial \widetilde \theta }}} \right)\end{array}$

where one finds by Gaussian functional reduction:

$\left( {{{\Sigma }^i}\frac{{\partial {F_{0i}}}}{{\partial \widetilde \theta }} + \frac{1}{2}{\varepsilon _{ijk}}{{B}^i}\frac{{\partial {C^{{{(2)}^{jk}}}}}}{{\partial \widetilde \theta }}} \right) = \frac{1}{2}\widetilde {\theta \,}{\Gamma _{\tau _0^i}}{\partial _i}\theta \cdot \widetilde \theta \,\Gamma - \widetilde {\theta \,}{\Gamma _{\tau _0^i}}{\prod _i}$

and:

${C^{(2)}}\frac{{\partial {F_{0i}}}}{{\partial \widetilde \theta }} + \frac{{\partial }}{{\partial \widetilde \theta }}\sqrt { - {G_{\mu \nu }}} {C^{(4)}} = {\left[ { - \frac{1}{2}\left( {\widetilde \theta \prod {\tau _1}d\theta } \right)\left\{ {\left( {\widetilde \theta \,{\Gamma _M}{\tau _3}d\widetilde \theta } \right)\frac{1}{2}\widetilde \theta \,{\Gamma ^M} - \widetilde \theta \prod {\tau _3}} \right\} + \frac{{\partial \Xi }}{{\partial \widetilde \theta }}} \right]_3}$

Now, given that ${\pi _\theta }$ transforms as:

$\displaystyle ({{\pi }_{\theta }}\to {{\pi }_{\theta }}\text{O}{{(\Lambda )}^{{{{T}_{3}}}}})$

where ${T_3}$ is the D3-brane tension, the fermionic expression $\psi$ covariantly transforms as:

$\displaystyle (\psi \to \psi \text{O}{{(\Lambda )}^{{{{T}_{3}}}}})$

Now we must exhibit the self-duality of the D3-brane Hamiltonian, given the gauge field, as usual, ${A_\mu }$ , acting on the automorphic group of $Ad{S_4} \times {S^4}$ , and the D = 3 metric ${g_{ij}}$ replaced by the D3-brane worldvolume 4-D metric $\gamma _{ij}^{D3}$ , expressed in terms of brane coordinates $\left( {\chi ,\theta } \right)$ . Note that for any function $R(p,q)$ on the super-Kähler phase-space, we have:

$\delta R = - \left[ {R,W} \right]$

with:

$W = \lambda \int {{d^3}} \sigma \sqrt {\gamma _{ij}^{D3}} \left[ {\frac{1}{2}\frac{{\Sigma }}{{\sqrt {\gamma _{ij}^{D3}} }}D_{il}^{ - 1}\frac{{\Sigma }}{{\sqrt {\gamma _{ij}^{D3}} }} + \frac{1}{2}{A_i}{D^{ij}} + {\pi _\theta }\frac{{i{\tau _2}}}{2}\theta } \right]$

where $D_{il}^{ - 1}$ is a tensor acting on a vector as such:

$D_{il}^{ - 1} \equiv {({\widetilde \Delta ^{ - 1}})_i}^k{\nabla ^j}{\eta _{jkl}} = {\eta _{ijk}}{\nabla ^k}({\widetilde \Delta ^{ - 1}})_l^j$

with:

${\eta _{ikl}} = {\varepsilon _{jkl}}\sqrt {\gamma _{ij}^{D3}}$

being the covariant anti-symmetric constant tensor, and is the inverse of:

${D^{jk}} = {\eta ^{jlk}}{\nabla _l} = {\nabla _l}{\eta ^{jlk}}$

In the projective geometry of $Ad{S_4} \times {S^4}$ , we then have:

$\left\{ {\begin{array}{*{20}{c}}{D_{im}^{ - 1}{D^{mk}} = {\rm O}_i^k(\nabla )}\\{{\rm O}_i^k(\nabla ) = \delta _i^k - {\nabla _i}({\Delta ^{ - 1}}){\nabla ^k}}\\{{D^{im}}D_{mk}^{ - 1} = {\rm O}_k^i(\nabla )}\end{array}} \right.$

noting that ${\rm O}_i^k(\nabla )$ longitudinally projects out the components of the covariant derivative ${\nabla _i}$ .

Now, in the D3-brane 4-dimensional curved worldspace, the Laplacian operator $(\widetilde \Delta )_j^i$ is given by:

$(\widetilde \Delta )_j^i = \Delta \delta _j^i - R_j^i{T_3}$

with $R_j^i$ being the Ricci tensor. One then derives, by use of the Atiyah–Singer index theorem, the fact that $W$ generates the desired ${A_i}$ duality transformation for the gauge field, and generates the $SO(2)$ rotation of $\theta$ :

$\delta {A_l} = \lambda \left( {{{\widetilde \Delta }^{ - 1}}} \right)_l^k{\nabla ^j}{\varepsilon _{jkm}}\Sigma = D_{lm}^{ - 1}\left( {\lambda \frac{{\Sigma }}{{\sqrt {\gamma _{ij}^{D3}} }}} \right)$

with:

$\left\{ {\begin{array}{*{20}{c}}{\delta \Sigma = - \lambda {\varepsilon ^{ijk}}{{\partial }_j}{A_k}}\\{\delta \theta = \lambda \frac{{i{\tau _2}}}{2}\theta }\\{\delta {\pi _\theta } = - {\pi _\theta }\lambda \frac{{i{\tau _2}}}{2}}\\{\delta \chi = 0}\\{\delta P = 0}\end{array}} \right.$

and so we obtain:

$\left\{ {\begin{array}{*{20}{c}}{\delta B = \lambda {{\Sigma }^{ * \dagger }}}\\{\delta {{\Sigma }^{ * \dagger }} = - \lambda B}\end{array}} \right.$

where:

${\Sigma ^{ * \dagger }} = \Sigma - \sqrt {\gamma _{ij}^{D3}} \left( {{\Delta ^{ - 1}}} \right)\left( {{\nabla _{lm}}\frac{{\Sigma }}{{\sqrt {\gamma _{ij}^{D3}} }}} \right)$

We are now in a position to derive the Dp/M5-brane Hamiltonian. The bosonic D-brane Lagrangian has the general form:

$\displaystyle \mathcal{L}={{\dot{X}}^{m}}{{P}_{m}}+{{\dot{V}}_{i}}{{E}^{i}}+\dot{\psi }{{T}_{{Dp}}}-H$

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {{{P}_{m}}\quad \quad c.m.\quad \quad {{X}^{m}}} \\ {{{E}^{i}}\quad \quad c.m.\quad \quad V} \end{array}} \right.$

where $\psi$ is the Hodge dual of a p-form potential and $H$ is the Hamiltonian density given as such:

$\displaystyle H={{\psi }^{i}}{{\mathcal{T}}_{i}}+{{V}_{t}}\tilde{K}+{{s}^{i}}{{\mathcal{H}}_{i}}+\lambda \mathcal{H}$

with:

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {{{\mathcal{T}}_{i}}=-{{\partial }_{i}}{{T}_{{Dp}}}} \\ {\tilde{K}=-{{\partial }_{i}}{{{\bar{E}}}^{i}}+{{{(-)}}^{{p+1}}}{{T}_{{Dp}}}S} \\ {S=*{{{\left( {\tilde{R}{{e}^{{\tilde{F}}}}} \right)}}_{p}}} \\ {{{\mathcal{H}}_{i}}={{{\bar{P}}}_{a}}E_{i}^{a}+{{{\bar{E}}}^{j}}{{{\tilde{F}}}_{{ij}}}} \\ {E_{i}^{a}=E_{m}^{a}{{\partial }_{i}}{{X}^{m}}} \\ {\mathcal{H}=\frac{1}{2}\left[ {{{{\bar{P}}}^{2}}+{{{\bar{E}}}^{j}}{{{\bar{E}}}^{j}}{{{\tilde{G}}}_{{ij}}}+{{T}_{{Dp}}}{{e}^{{-2\phi }}}\det \left( {{{{\tilde{G}}}_{{ij}}}+{{{\tilde{F}}}_{{ij}}}} \right)} \right]} \end{array}} \right.$

Hence we have constant brane-tension, abelian p-form gauge transformations, gauge field transformations satisfying Gauss’s law, and that ${{\mathcal{H}}_{a}}$ and $\mathcal{H}$ generate worldvolume diffeomorphisms and time translations. Since the RR field strength $R=dC-C\wedge {{H}_{3}}$ with WZ coupling acts as a Gaussian-law source, we can derive the following:

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {{{{\bar{P}}}_{a}}={{E}_{a}}^{m}\left( {{{P}_{m}}+{{E}^{i}}{{Z}^{*}}{{{\left( {{{i}_{m}}B} \right)}}_{i}}+{{T}_{{Dp}}}{{{\tilde{C}}}_{m}}} \right)} \\ {{{{\tilde{C}}}_{m}}=*\left( {{{Z}^{*}}\left( {{{i}_{m}}C} \right)\wedge {{e}^{{\tilde{F}}}}} \right)} \\ {{{{\bar{E}}}^{i}}={{E}^{i}}+T{{{\tilde{C}}}^{i}}} \\ {{{{\tilde{C}}}^{i}}={{{\left[ {*{{{\left( {\tilde{C}{{e}^{{\tilde{F}}}}} \right)}}_{{p-1}}}} \right]}}^{i}}} \end{array}} \right.$

and hence the Hamiltonian phase-space constraint is quartic, as required. The Hamiltonian for the M5-brane breaks $SO\left( {1,5} \right)$ into $SO\left( 5 \right)$ . Working with the worldvolume metric ${{\tilde{G}}_{{ij}}}$ and its inverse $\tilde{G}_{5}^{{ij}}$ , we get the following:

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {{{{\bar{H}}}^{{ij}}}=\frac{1}{{6\sqrt{{\det {{{\tilde{G}}}_{5}}}}}}{{\varepsilon }^{{ij{{k}_{1}}{{k}_{2}}{{k}_{3}}}}}{{\mathcal{H}}_{{{{k}_{1}}{{k}_{2}}{{k}_{3}}}}}} \\ {\det \left( {{{{\tilde{G}}}_{{\mu \nu }}}+{{{\bar{H}}}_{{\mu \nu }}}} \right)=\left( {{{{\tilde{G}}}_{{00}}}-{{{\tilde{G}}}_{{0i}}}\tilde{G}_{5}^{{ij}}{{{\tilde{G}}}_{{0j}}}} \right){{{\det }}^{5}}\left( {\tilde{G}-\bar{H}} \right)} \end{array}} \right.$

with ${{{\tilde{G}}}_{5}}$ the determinant of the worldvolume form ${{{\tilde{G}}}_{{ij}}}$ and:

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {{{{\det }}^{5}}\left( {\tilde{G}-\bar{H}} \right)=\det \left( {{{{\tilde{G}}}_{{ij}}}-{{{\bar{H}}}_{{ij}}}} \right)} \\ {{{{\bar{H}}}_{{ij}}}={{{\tilde{G}}}_{{\mu \nu }}}{{{\tilde{G}}}_{{jl}}}{{{\bar{H}}}^{{kl}}}} \end{array}} \right.$

Thus, the phase-space M5-brane bosonic Lagrangian is given by:

$\displaystyle \mathcal{L}={{\dot{X}}^{m}}{{P}_{m}}+\frac{1}{2}{{\Pi }^{{ij}}}{{\dot{V}}_{{ij}}}-\lambda \mathcal{H}-{{s}^{i}}{{\mathcal{H}}_{i}}+{{\sigma }_{{ij}}}{{\tilde{K}}^{{ij}}}$

where:

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {{{P}_{m}}} \\ {{{{\tilde{C}}}_{m}}} \\ {{{\Pi }^{{ij}}}} \end{array}} \right.$

are given as such:

$\displaystyle {{P}_{m}}={{E}_{m}}^{a}{{{\tilde{P}}}_{a}}+{{T}_{{M5}}}{{{\tilde{C}}}_{m}}$

$\displaystyle {{{\tilde{C}}}_{m}}=*\left[ {{{Z}^{{\bar{\circ }}}}\left( {{{i}_{m}}{{C}_{6}}} \right)-\frac{1}{2}{{Z}^{{\bar{\circ }}}}\left( {{{i}_{m}}{{C}_{3}}} \right)\wedge \left( {{{{\tilde{C}}}_{3}}+2{{\mathcal{H}}_{3}}} \right)} \right]$

$\displaystyle {{\Pi }^{{ij}}}=\frac{1}{4}{{T}_{{M5}}}{{\varepsilon }^{{ij{{k}_{1}}{{k}_{2}}{{k}_{3}}}}}{{\partial }_{{{{k}_{1}}}}}{{V}_{{{{k}_{2}}{{k}_{3}}}}}$

By the Bianchi identify for $d{{V}_{{\left( {...} \right)}}}$ , we have:

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {\Pi ={{T}_{{M5}}}{{*}_{H}}\left( {dV} \right)} \\ {d{{*}_{H}}\Pi =0} \end{array}} \right.$

and the last three functionals in:

$\displaystyle \mathcal{L}={{\dot{X}}^{m}}{{P}_{m}}+\frac{1}{2}{{\Pi }^{{ij}}}{{\dot{V}}_{{ij}}}-\lambda \mathcal{H}-{{s}^{i}}{{\mathcal{H}}_{i}}+{{\sigma }_{{ij}}}{{\tilde{K}}^{{ij}}}$

are given by:

$\displaystyle \mathcal{H}=\frac{1}{2}\left[ {{{\mathcal{P}}^{2}}+T_{{M5}}^{2}{{{\det }}^{5}}\left( {\tilde{G}+\tilde{H}} \right)} \right]$

$\displaystyle {{\mathcal{H}}_{i}}={{\partial }_{i}}{{X}^{m}}{{P}_{m}}+{{T}_{{M5}}}\left( {{{V}_{i}}-{{{{\tilde{C}}'}}_{i}}} \right)$

$\displaystyle \tilde{K}={{\Pi }^{{ij}}}-\frac{1}{4}{{T}_{{M5}}}{{\varepsilon }^{{ij{{k}_{1}}{{k}_{2}}{{k}_{3}}}}}{{\partial }_{{{{k}_{1}}}}}{{V}_{{{{k}_{2}}{{k}_{3}}}}}$

and they correspond, respectively, to generating-functionals for time-translation, worldvolume diffeomorphism and self-duality, with the following relations holding:

$\displaystyle \left\{ {\begin{array}{*{20}{c}} {{{V}_{i}}=\frac{1}{{24}}{{\varepsilon }^{{{{i}_{1}}{{i}_{2}}{{i}_{3}}{{i}_{4}}{{i}_{5}}}}}{{H}_{{{{i}_{3}}{{i}_{4}}{{i}_{5}}}}}{{H}_{{{{i}_{1}}{{i}_{2}}}}}} \\ {{{\mathcal{P}}_{a}}={{E}_{a}}^{m}{{P}_{m}}+{{T}_{{M5}}}\left( {{{V}^{i}}{{\partial }_{i}}{{X}^{m}}{{E}_{m}}^{b}{{\eta }_{{ba}}}-{{{{\tilde{C}}'}}_{a}}} \right)} \\ {{{{{\tilde{C}}'}}_{a}}=E_{a}^{m}{{{{\tilde{C}}'}}_{m}}} \end{array}} \right.$

In summary, the M5- $\mathfrak{g}_{{\text{Lie}}}^{3}$ algebra closes and the D3-brane self-dual Hamiltonian action is invariant under linear $SL(2,\Re )$ transformation:

$\left( {\begin{array}{*{20}{c}}{B}\\{\Sigma }\end{array}} \right) \to \Lambda \left( {\begin{array}{*{20}{c}}{B}\\{\Sigma }\end{array}} \right)$