# GS Formalism

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String Theory
Prior to the First Superstring Revolution
Early History S-Matrix Theory
Regge Trajectory
Bosonic String Theory Worldsheet
String
Bosonic String Theory
String Perturbation Theory
Tachyon Condensation
Supersymmetric Revolution Supersymmetry
RNS Formalism
GS Formalism
BPS
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First Superstring Revolution GSO Projection
Type II String Theory
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Second Superstring Revolution T-Duality
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(2,0) Theory
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String Phenomenology

GS String Theory, or the GS Superstring Theory was an early attempt to include fermions in String Theory. "GS" stands for Green-Schwarz. In contrast to the RNS String Theory, the GS String Theory automatically has spacetime supersymmetry, but not worldsheet supersymmetry.

## Introduction to Spacetime SupersymmetryEdit this section

Instead of the “fake” fermionic fields in the RNS String Theory, which are really spacetime vectors, the GS String Theory has “actual” bosonic and fermionic fields ${X^\mu },\Theta$.

The Spacetime Supersymmetric transformations are given by:

\begin{align} & \delta {{\Theta }^{Aa}}\leftrightarrow {{\varepsilon }^{Aa}} \\ & \delta {{X}^{\mu }}\leftrightarrow {{{\bar{\varepsilon }}}^{A}}{{\gamma }^{\mu }}{{\Theta }^{A}} \\ \end{align}

This is actually intuitively related to the worldsheet supersymmetry of the RNS String Theory. Thsee are also also the transformations of superspace. It is to be noted that here, $\gamma^\mu$ is the Dirac Gamma Matrix on the background spacetime. It is relatively trivial to show that:

\begin{align} & [{{\delta }_{1}},{{\delta }_{2}}]{{X}^{\mu }}=-2\bar{\varepsilon }_{1}^{A}{{\gamma }^{\mu }}\varepsilon _{2}^{A}={{a}^{\mu }} \\ & [{{\delta }_{1}},{{\delta }_{2}}]{{\Theta }^{A}}=0 \\ \end{align}

This means that the commutator bracket of infinitesimal Supersymmetric transformations, translates the bosonic field by $a^\mu$, and leaves the fermionic field untouched. These transformations, combined with the stanadard Poincaire transformations, give rise to the Super-Poincaire transformations, forming the Super-Poincaire Group.

Clearly, to be consistent with the transformations of superspace, the standard D0 brane action would have to be changed in such a way that ${{X}^{\mu }}$ is replaced with the field:

$\Pi _{0}^{\mu }={{\partial }_{0}}{{X}^{\mu }}-{{\bar{\Theta }}^{A}}{{\gamma }^{\mu }}{{\partial }_{0}}{{\Theta }^{A}}$

The subscript of 0 on the Left-Hand-Side indicates that only time-derivatives are taken. This will be especially clear when we discuss spacetime Supersymmetry for strings and other D-branes. The D0-brane action then becomes:

${{S}_{1}}=-m\int_{{}}^{{}}{\sqrt{-{{\Pi }_{0}}\cdot {{\Pi }_{0}}}\text{d}\tau }$

This is invariant under super-poincaire transformations (and diffeomorphisms, of course). These D0-branes are the same D0-branes that appear in the Type IIA String Theory. The $\mathsf{\mathcal{N}}=2$ supersymmetry of the Type IIA String Theory is interpreted as having 2 spinor fermionic coordinates ${{\Theta }^{1a}},{{\Theta }^{2a}}$. Since Type IIA String Theory is a chiral theory, these have opposite chirality. In other words, due to the GSO Projection, we have:

\begin{align} & {{\Theta }^{1}}=\frac{1}{2}\left( 1+{{\gamma }^{11}} \right)\Theta \\ & {{\Theta }^{2}}=\frac{1}{2}\left( 1-{{\gamma }^{11}} \right)\Theta \\ \end{align}

## Kappa Symmetry and D0 BranesEdit this section

The total (Kappa-Symmetric) action is:

 Kappa Symmetry for D0 Branes $S=-m\left( \int_{{}}^{{}}{\sqrt{-{{\Pi }_{0}}\cdot {{\Pi }_{0}}}\text{d}\tau }+\int{{\bar{\Theta }}}{{\gamma }_{11}}{{\partial }_{0}}\Theta \text{d}\tau \right)$
Proof of Kappa Symmetry for D0 branes

Taking the canonically conjugate momentum to ${{X}^{\mu }}$, ' ${{P}_{\mu }}=\frac{\delta {{S}_{1}}}{\delta \frac{\text{d}{{X}^{\mu }}}{\text{d}\tau }}=\frac{m{{c}_{0}}}{\sqrt{-{{\Pi }_{0}}\cdot \text{ }{{\Pi }_{0}}}}\left( {{\partial }_{0}}{{X}_{\mu }}-\bar{\Theta }{{\gamma }_{\mu }}{{\partial }_{0}}\Theta \right)=\frac{m{{c}_{0}}}{\sqrt{-{{\Pi }_{0}}\cdot \text{ }{{\Pi }_{0}}}}{{\Pi }_{0\mu }}$

The equation of motion for ${{X}^{\mu }}$ implies that:

${{\partial }_{0}}{{P}_{\mu }}=0$

From squaring the equation for the canonically conjugate momentum, we can say that:

${{P}^{2}}=-{{m}^{2}}c_{0}^{2}$

Meanwhile, the equation of motion for $\Theta$ is:

$P\text{ }\cdot \text{ }\gamma {{\partial }_{0}}\Theta =0$

${{m}^{2}}{{\partial }_{0}}\Theta =0$

So, in the massive case, the fermionic field $\Theta$ is unchanging throughout time. Else, a saturation of the BPS, implying enhanced supersymemtry. Suppose this is shown by a change to the equation of motion for $\Theta$, as:

$\left(P\cdot\gamma+ m\gamma_{11} \right) \partial_0\Theta=0$

This would only constrain half the components of $\Theta$/, which can be seen from squaring it/.

The missing contribution would then be, :

$S_\kappa= -m\int \bar\Theta \gamma_{11} \partial_0\Theta \mbox{d}\tau$

The sign of this action is arbitrary. If we take it to be negative, as we have done here, then the sign of the corresponding action for the antimatter would be positive. The total action would then be:

$S=-m\left( \int_{{}}^{{}}{\sqrt{-{{\Pi }_{0}}\cdot {{\Pi }_{0}}}\text{d}\tau }+\int{{\bar{\Theta }}}{{\gamma }_{11}}{{\partial }_{0}}\Theta \text{d}\tau \right)$

This action actually has Kappa Symmetry. To show this,

The variation $\delta\Theta$, and $\delta X^\mu$ are related by the transformations:

$\delta X^\mu = \bar\Theta \gamma^\mu \delta\Theta = -\delta \bar\Theta \gamma^\mu\Theta$

If you work out the transformations for $\Pi^\mu_0$ , you see that w

$\delta\Pi_0^\mu = - 2 \delta \bar{\Theta} \gamma^\mu \frac{\mbox{d} \Theta}{\mbox{d} \tau }$

So, under these transformations, what happens to $S_1$? . If you work it out (quite trivially), you'll see that it is equal to the following expression:

$\delta S_1 = m\int\frac{\Pi_0\cdot\delta \Pi_0 }{\sqrt{-\Pi_0^2 } } \mbox{d} \tau = -2m\int \frac{\Pi_0^\mu \delta\bar\Theta \gamma_\mu \frac{\mathrm{d}\Theta}{\mathrm{d}\tau}}{\sqrt{-\Pi_0^2 } } = - 2 m \int \delta\bar{\Theta} \lambda \gamma_{11} \frac{\mbox{d} \Theta }{\mbox{d}\tau }\mbox{d}\tau$

Ok, so now,  !

$\delta S_2 = - 2 m \int \delta\bar\Theta \lambda \frac{\mbox{d} \Theta }{\mbox{d}\tau }\mbox{d}\tau$

We also obwvservwe that $\lambda ^2=1$, so that this can be used to derive the familiar projection23:

$P_\pm=\frac12 (1\pm\gamma)$

So, adding up these actions results in a

$\delta\bar\Theta=\bar\kappa P_-$

$\delta X^\mu =\bar\kappa P_- \gamma^\mu \Theta$

## Spacetime Supersymmetry for StringsEdit this section

The Nambu-Goto Action is given by:

$S=-T\int_{{}}^{{}}{\sqrt{-\det \left( {{\partial }_{\alpha }}{{X}^{\mu }}{{\partial }_{\beta }}{{X}_{\mu }} \right)}{{\text{d}}^{2}}\sigma }$

In analogy with the action for the Supersymmetric D0 brane, the action for the Supersymmetric string would become:

${{S}_{1}}=-T\int_{{}}^{{}}{\sqrt{-\det \left( {{\Pi }_{\alpha \mu }}\Pi _{\beta }^{\mu } \right)}{{\text{d}}^{2}}\sigma }$

Where we define:

$\Pi _{\alpha }^{\mu }={{\partial }_{\alpha }}{{X}^{\mu }}-{{\bar{\Theta }}^{A}}{{\gamma }^{\mu }}{{\partial }_{\alpha }}{{\Theta }^{A}}$

In a way analogous to the Pi mu nought for the D0 - branes.

## Kappa Symmetry and D1 branesEdit this section

The total action (with $\mathcal N=2$ supersymmetry) is therefore:

 Kappa Symmetry for D1 Branes \begin{align} & S=-T\int_{{}}^{{}}{\sqrt{-\det \left( {{\Pi }_{\alpha \mu }}\Pi _{\beta }^{\mu } \right)}{{\text{d}}^{2}}\sigma } \\ & \text{ }+\frac{1}{\pi }\int{{{\text{d}}^{2}}}\sigma {{\varepsilon }^{\alpha \beta }}\left( {{{\bar{\Theta }}}^{1}}{{\gamma }^{\mu }}{{\partial }_{\alpha }}{{{\bar{\Theta }}}^{1}}\text{ }{{{\bar{\Theta }}}^{2}}{{\gamma }_{\mu }}{{\partial }_{\beta }}{{\Theta }^{2}}-i{{\partial }_{\alpha }}{{X}^{\mu }}\left( {{{\bar{\Theta }}}^{1}}{{\gamma }_{\mu }}{{\partial }_{\beta }}{{\Theta }^{1}}-{{\Theta }^{2}}{{\gamma }_{\mu }}{{\partial }_{\beta }}{{\Theta }^{2}} \right) \right)\text{ } \\ \end{align}

Proof of Kappa Symmetry for D1 branes

The Kappa Symmetric transformations would be:

$\delta {{X}^{\mu }}={{\bar{\Theta }}^{A}}{{\gamma }^{\mu }};\delta {{\Theta }^{A}}=-\delta {{\bar{\Theta }}^{A}}{{\gamma }^{\mu }}{{\Theta }^{A}}$

So that:

$\delta \Pi _{\alpha }^{\mu }=-2\delta {{\bar{\Theta }}^{A}}{{\gamma }^{\mu }}{{\partial }_{\alpha }}{{\Theta }^{A}}$

It is also clear that:

$\delta {{S}_{1}}=\frac{2}{\pi }\int_{{}}^{{}}{\sqrt{-\lambda }{{\lambda }^{\alpha \beta }}\Pi _{\alpha }^{\mu }\delta {{{\bar{\Theta }}}^{A}}{{\gamma }_{\mu }}{{\partial }_{\beta }}{{\Theta }^{A}}{{\text{d}}^{2}}\sigma }$

If we let

\begin{align} & \lambda =\det {{\lambda }^{\alpha \beta }} \\ & {{\lambda }^{\alpha \beta }}={{\Pi }_{\alpha \mu }}\Pi _{\beta }^{\mu } \\ \end{align}

Now, to determine ${{S}_{2}}$, we see that:

${{S}_{2}}=\int_{{}}^{{}}{{{\Omega }_{2}}}=\int{{{\mathsf{\epsilon }}^{\alpha \beta }}{{\Omega }_{\alpha \beta }}{{\text{d}}^{2}}\sigma }$

Switching to the exterior derivative ("d") notation,

$\int_{M}{{{\Omega }_{2}}}=\int_{D}^{{}}{{{\Omega }_{3}}}$

${{\Omega }_{2}}$ here is a 2-form, independent of the worldsheet metric. Introducing a three-form ${{\Omega }_{3}}=\text{d}{{\Omega }_{2}}$, we obtain the desired equation through Stokes' Theorem\\. $M=\partial D$ is the boundary of $D$/ . In 10 dimensions, a Majorana spinor satisfies:

${{\Omega }_{3}}=A\left( \text{d}{{{\bar{\Theta }}}^{1}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{1}}\text{+}k\text{d}{{{\bar{\Theta }}}^{2}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{2}} \right){{\Pi }^{\mu }}$

Where $k$ is a real number with an absolute value of 1. In order to ensure that $\Omega_3$ is closed, i.e., that $\mbox{d}\Omega_3=0$, $k=-1$ which can be trivjially seen from explicitly writing the superspace embedding function $\Pi^\mu$ in terms of $X^\mu$ and $\Theta^A$.

\begin{align} & \delta {{\Omega }_{3}}=2A\left( \text{d}\delta {{{\bar{\Theta }}}^{1}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{1}}-\text{d}\delta {{{\bar{\Theta }}}^{2}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{2}} \right){{\Pi }^{\mu }}-2A\left( \text{d}{{{\bar{\Theta }}}^{1}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{1}}-\text{d}{{{\bar{\Theta }}}^{2}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{2}} \right)\delta {{{\bar{\Theta }}}^{A}}{{\gamma }^{\mu }}\text{d}{{\Theta }^{A}} \\ & \text{ }=2A\left( \text{d}\delta {{{\bar{\Theta }}}^{1}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{1}}-\text{d}\delta {{{\bar{\Theta }}}^{2}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{2}} \right){{\Pi }^{\mu }}-2A\left( \delta {{{\bar{\Theta }}}^{1}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{1}}-\delta {{{\bar{\Theta }}}^{2}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{2}} \right){{\Pi }^{\mu }} \\ & \text{ }=\text{d}\left( 2A\left( \delta {{{\bar{\Theta }}}^{1}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{1}}-\delta {{{\bar{\Theta }}}^{2}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{2}} \right){{\Pi }^{\mu }} \right) \\ & \delta {{\Omega }_{2}}=2A\left( \delta {{{\bar{\Theta }}}^{1}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{1}}-\delta {{{\bar{\Theta }}}^{2}}{{\gamma }_{\mu }}\text{d}{{\Theta }^{1}} \right) \\ \end{align}

$\delta {{S}_{2}}=2A\int{{{\varepsilon }^{\alpha \beta }}\left( \delta {{{\bar{\Theta }}}^{1}}{{\gamma }_{\mu }}{{\partial }_{\alpha }}{{\Theta }^{1}}-\delta {{{\bar{\Theta }}}^{2}}{{\gamma }_{\mu }}{{\partial }_{\alpha }}{{\Theta }^{2}} \right)\Pi _{\beta }^{\mu }{{\text{d}}^{2}}\sigma }$

${{S}_{2}}$ itself would be given by:

${{S}_{2}}=\frac{1}{\pi }\int{{{\text{d}}^{2}}}\sigma {{\varepsilon }^{\alpha \beta }}\left( {{{\bar{\Theta }}}^{1}}{{\gamma }^{\mu }}{{\partial }_{\alpha }}{{{\bar{\Theta }}}^{1}}\text{ }{{{\bar{\Theta }}}^{2}}{{\gamma }_{\mu }}{{\partial }_{\beta }}{{\Theta }^{2}}-i{{\partial }_{\alpha }}{{X}^{\mu }}\left( {{{\bar{\Theta }}}^{1}}{{\gamma }_{\mu }}{{\partial }_{\beta }}{{\Theta }^{1}}-{{\Theta }^{2}}{{\gamma }_{\mu }}{{\partial }_{\beta }}{{\Theta }^{2}} \right) \right)\text{ }$

\begin{align} & S=-T\int_{{}}^{{}}{\sqrt{-\det \left( {{\Pi }_{\alpha \mu }}\Pi _{\beta }^{\mu } \right)}{{\text{d}}^{2}}\sigma } \\ & \text{ }+\frac{1}{\pi }\int{{{\text{d}}^{2}}}\sigma {{\varepsilon }^{\alpha \beta }}\left( {{{\bar{\Theta }}}^{1}}{{\gamma }^{\mu }}{{\partial }_{\alpha }}{{{\bar{\Theta }}}^{1}}\text{ }{{{\bar{\Theta }}}^{2}}{{\gamma }_{\mu }}{{\partial }_{\beta }}{{\Theta }^{2}}-i{{\partial }_{\alpha }}{{X}^{\mu }}\left( {{{\bar{\Theta }}}^{1}}{{\gamma }_{\mu }}{{\partial }_{\beta }}{{\Theta }^{1}}-{{\Theta }^{2}}{{\gamma }_{\mu }}{{\partial }_{\beta }}{{\Theta }^{2}} \right) \right)\text{ } \\ \end{align}

$\delta S=\frac{4}{\pi }\int_{{}}^{{}}{{{\varepsilon }^{\alpha \beta }}\left( \delta {{{\bar{\Theta }}}^{1}}{{P}_{+}}{{\gamma }_{\mu }}{{\partial }_{\alpha }}{{\Theta }^{1}}-\delta {{{\bar{\Theta }}}^{2}}{{P}_{-}}{{\gamma }_{\mu }}{{\partial }_{\alpha }}{{\Theta }^{2}} \right)\Pi _{\beta }^{\mu }{{\text{d}}^{2}}\sigma }\text{ }$

Here,

\begin{align} & {{P}_{\pm }}=\frac{1\pm \gamma }{2} \\ & \gamma =-\frac{{{\varepsilon }^{\alpha \beta }}\Pi _{\alpha }^{\mu }\Pi _{\beta }^{\nu }{{\gamma }_{\mu }}{{\gamma }_{\nu }}}{\sqrt{-\lambda }} \\ \end{align}

We finally conclude that this modified action is invariant under the following kappa symmetric transformations:

\begin{align} & \delta {{{\bar{\Theta }}}^{1}}={{{\bar{\kappa }}}^{1}}{{P}_{-}} \\ & \delta {{{\bar{\Theta }}}^{2}}={{{\bar{\kappa }}}^{2}}{{P}_{+}} \\ \end{align}

## Relation with RNS String TheoryEdit this section

GS String Theory is equivalent to a GSO Projected RNS String Theory. The specific consistent String Theory that this is equivalent to depends on the chirality and number of components of the fermionic fields.