The Dirac-Born-Infeld Action, D-Branes and Spacetime Symmetries

Ex nihilo nihil fit ~ Parmenides

 

In this post, in order to solve the Wheeler-deWitt problem-of-time I discussed in my last post, and since string/M-theoretic braneworld cosmology offers the only solution in a unified C*-Heisenberg-algebraic sense, I must first analyse a field theory on a D-brane sigma world-volume and study the Dirac-Born-Infeld (DBI) action. Let me start by giving a proof that the DBI action has a symmetry under the transformation

\left\{ {\begin{array}{*{20}{c}}{\delta {A_a} = {\Lambda _a} - {\varepsilon ^c}{{\not F}_{ca}} + {\Lambda _k}{{\not \partial }_a}{\Phi ^k}}\\{\delta \,{\Phi ^i} = {\varepsilon ^i} - {\varepsilon ^c}{{\not \partial }_c}{\Phi ^i}}\end{array}} \right.

of full

Diff\left( M \right){\dot \propto ^\dagger }\Omega _{closed}^2\left( M \right)

on a generalized Riemannian structure {C_ + } \subset \Im M

First, I must prove the following relation

{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}g\,{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}\not S\widetilde {\not F} = \sqrt {{\rm{det}}{{\left( {{\varphi ^ * }\left( {g + B} \right) - F} \right)}_{ab}}}

with g the Riemannian metric on \Im M

g \in {\Im ^ * }M \otimes {\Im ^ * }M

and \not S\widehat {\not F} the metric on {\not L_{\widehat {\not F}}}

\not S\widehat {\not F} \in {\not L^ * }_{\widehat {\not F}} \otimes {\not L^ * }_{\widehat {\not F}}

Now, since \varphi _\Phi ^ * is the pullback of the embedding map defined by the field \Phi , we have

{\varphi _\Phi }:\underbrace \to _{pullback}M

Thus, the determinants on the l. h. s. are those of the D \times D matrices, while the determinant on the r.h.s. is that of the \left( {p + 1} \right) \times \left( {p + 1} \right) matrix distinguished by the index ab. Now, the r. h. s. of

{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}g\,{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}\not S\widetilde {\not F} = \sqrt {{\rm{det}}{{\left( {{\varphi ^ * }\left( {g + B} \right) - F} \right)}_{ab}}}

is the Lagrangian {L_{DBI}} of the Dirac-Born-Infeld  action and can be proven by combining the following relations of various determinants

1) Let s be the symmetric part of t given by

\left\{ {\begin{array}{*{20}{c}}{{t^{ij}} = {E^{ij}}}\\{t_a^j = - {E_{ak}}{E^{kj}}}\end{array}} \right.

and

\left\{ {\begin{array}{*{20}{c}}{t_b^i = {E^{ik}}{E_{kb}}}\\{{t_{ab}} = {E_{ab}} - {E_{ak}}{E^{kl}}{E_{lb}}}\end{array}} \right.

and {t^{ij}} = {E^{ij}} the inverse matrix of {E_{ij}}, then we have

\det s = \det g{\left( {\det {t^{ij}}} \right)^2}

2) From

\begin{array}{l}\not S\widehat {\not F}{ = _{{\rm{def}}}}s - \left( {a - \widehat {\not F}} \right){s^{ - 1}}\left( {a - \widehat {\not F}} \right)\\ \in \Gamma \left( {{L^ * } \otimes {L^ * }} \right)\end{array}

and the definition of \not S\widehat {\not F}, we have

\det s\,{\rm{det}}\,\not S\widehat {\not F} = {\left( {\det {t_{\widehat {\not F}}}} \right)^2}

3) Now, by using the explicit expression for {t_{\widehat {\not F}}}, it follows that

{\rm{det}}\,{t_{\widehat {\not F}}} = \det {t^{ij}}{\rm{det}}{\left( {\varphi _\Phi ^ * \left( {g + B} \right) - F} \right)_{ab}}

Using these relations in 1), 2), 3), one can prove the representation of the DBI action given in

{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}g\,{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}\not S\widetilde {\not F} = \sqrt {{\rm{det}}{{\left( {{\varphi ^ * }\left( {g + B} \right) - F} \right)}_{ab}}}

which is

\begin{array}{l}{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}g\,{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}\not S\widehat {\not F} = \frac{1}{{\sqrt {\det {t^{ij}}} }}{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}s\,{\rm{de}}{{\rm{t}}^{\frac{1}{4}}}\not S\widehat {\not F}\\ = \frac{1}{{\sqrt {\det {t^{ij}}} }}\sqrt {\det {t_{\widehat {\not F}}}} = \sqrt {{\rm{det}}{{\left( {\varphi _\Phi ^ * (E) - F} \right)}_{ab}}} \end{array}

The integral of the scalar density agrees with the DBI action

{S_{DBI}} = \int_{\varphi \Phi \left( \Sigma \right)} {\sqrt {{\rm{det}}\left( {\varphi _\Phi ^ * {{\left( {g + B - F} \right)}_{ab}}} \right)} } d{x^0} \wedge ... \wedge d{x^p}

when evaluated on the leaf \varphi \,\Phi \left( \Sigma \right) of {L_{\widehat {\not F}}} at {x^i} = {\Phi ^i}(x),  hence, the DBI action is invariant under the world-volume diffeomorphism on the D-brane.

Now I am in a position to prove that the DBI action is not only invariant under the world-volume diffeomorphism but also under the full target space diffeomorphism and the B-field gauge transformation. Towards this end,

{S_{DBI}} = \int_{\varphi \Phi \left( \Sigma \right)} {\sqrt {{\rm{det}}\left( {\varphi _\Phi ^ * {{\left( {g + B - F} \right)}_{ab}}} \right)} } d{x^0} \wedge ... \wedge d{x^p}

can be re-written as an integral over the target space M as

{S_{DBI}} = {\int_M {\widehat {\not L}} _{DBI}}{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}({x^a})} \right)d{x^0} \wedge ... \wedge d{x^{D - 1}}

where

{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}({x^a})} \right)

is a Dirac delta function interpreted as a distribution along {x^i} directions. The infinitesimal transformation of the full diffeomorphism and the B-field gauge transformation are parametrized by \varepsilon + \Lambda .  So the transformation of the integrand {\widehat {\not L}_{DBI}} is gotten from that of \det g

\begin{array}{l}\delta \det g = - \,{\varepsilon ^M}{{\not \partial }_M}\det g + \det g\\ \cdot \left\{ { - 2{{\not \partial }_M}{\varepsilon ^M}} \right\}\end{array}

and that of \not S\widehat {\not F}

\begin{array}{l}\delta \det \not S\widehat {\not F} = - \,{\varepsilon ^M}{{\not \partial }_M}{\rm{det}}\,\not S\widehat {\not F} + {\rm{det}}\,\not S\widehat {\not F}\\ \cdot \left\{ { - 2{{\not \partial }_c}{\varepsilon ^c} + 2{{\not \partial }_k}{\varepsilon ^k} - 4{{\widehat {\not F}}^k}_c{{\not \partial }_k}{\varepsilon ^c}} \right\}\end{array}

And the glorious result is

\delta {\widehat {\not L}_{DBI}} = - \,{\varepsilon ^M}{\not \partial _M}{\widehat {\not L}_{DBI}} - \left( {{{\not \partial }_c}{\varepsilon ^c} + {{\not \partial }_c}{\Phi ^k}{{\not \partial }_k}{\varepsilon ^c}} \right){\widehat {\not L}_{DBI}}

with

{{{\not \partial }_c}{\varepsilon ^c} + {{\not \partial }_c}{\Phi ^k}{{\not \partial }_k}{\varepsilon ^c}}

being central to full diffeomorphism. This is the expected result since {\widehat {\not L}_{DBI}}d{x^0} \wedge ... \wedge d{x^p} is a section of {\rm{det}}\left( {{\Delta ^ * }} \right)

On the other hand, the delta function transforms as

\begin{array}{l}\delta \left[ {{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}} \right)} \right] = - \,{\varepsilon ^M}{{\not \partial }_M}\\ \cdot \left[ {{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}} \right)} \right] - \\\left( {\not \partial _k^M{\varepsilon ^k} - {{\not \partial }_k}{\varepsilon ^c}{{\not \partial }_c}{\Phi ^k}} \right){\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}} \right)\end{array}

Now, combining, we get

\begin{array}{l}\delta \left[ {{{\widehat {\not L}}_{DBI}}{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}} \right)} \right] = \\ - {{\not \partial }_M}\left[ {{\varepsilon ^M}{{\widehat {\not L}}_{DBI}}{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}} \right)} \right]\end{array}

as desired, hence, the transformation of the integrand in the DBI action

{S_{DBI}} = {\int_M {\widehat {\not L}} _{DBI}}{\delta ^{\left( {D - p - 1} \right)}}\left( {{x^i} - {\Phi ^i}({x^a})} \right)d{x^0} \wedge ... \wedge d{x^{D - 1}}

is a total derivative and the DBI action is invariant under full target space diffeomorphisms and B-field gauge transformations and the invariance within the static gauge is what will allow us to further analyse the Wheeler-deWitt problem-of-time.

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