Let me summariise the discussion from Becker, Becker, Schwarz.
For the D0 brane
Taking the canonically conjugate momentum to ${X}^{\mu }$, we see that:
$${{P}_{\mu }}=\frac{\delta {{S}_{1}}}{\delta \frac{\text{d}{{X}^{\mu }}}{\text{d}\tau }}=\frac{m{{c}_{0}}}{\sqrt{-{{\Pi }_{0}}\cdot \text{ }{{\Pi }_{0}}}}\left( {{\partial }_{0}}{{X}_{\mu }}-\bar{\Theta }{{\gamma }_{\mu }}{{\partial }_{0}}\Theta \right)=\frac{m{{c}_{0}}}{\sqrt{-{{\Pi }_{0}}\cdot \text{ }{{\Pi }_{0}}}}{{\Pi }_{0\mu }}$$
The equation of motion for ${{X}^{\mu }} $ then 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,
$$P\text{ }\cdot \text{ }\gamma {{\partial }_{0}}\Theta =0$$
$${{m}^{2}}{{\partial }_{0}}\Theta =0$$
Implying that in the massive case, the fermionic field is time-unchanging across the worldline. Else, in the massless case n , the BPS bound is saturated , and implying enhanced supersymemtry. Suppose this is shown by a change to the equation of motion for the fermionic field:
$$\left(P\cdot\gamma+ m\gamma_{11} \right) \partial_0\Theta=0 $$
Then, this would only constrain half of the components of the fermionic field, as can be seen from scaring squareing the above equation ^ .:
The missing term in the action would then be:
$$ S_\kappa= -m\int \bar\Theta \gamma_{11} \partial_0\Theta \mbox{d}\tau $$
Now, to see how this is kappa symmetric,...
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, you'll see that it is equal to the following expression:
$$\delta S_1 = m\int\frac{\Pi_0\cdot\delta \Pi_0 }{\sqrt{-\Pi_0^2 } } \mbox d\tau = -2m\int \frac{\Pi_0^\mu \delta\bar\Theta \gamma_\mu \frac{\mathrm{d}\Theta}{\mathrm{d}\tau}}{\sqrt{-\Pi_0^2 } {}} = - 2 m \int \delta\bar\Theta \lambda \gamma_{11} \frac{\mbox{d} \Theta }{\mbox{d}\tau }\mbox{d}\tau $$
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
$$\delta\bar\Theta=\bar\kappa P_- $$
$$\delta X^\mu =\bar\kappa P_- \gamma^\mu \Theta $$
Which are the kappa symmetry transformations required to get the right number of fermionic degreest of freedgom and so on...
For the F1 String
Much more complicated. See here. A small article on deriving the kappa symmetric transformations for the F1 strings.
Here's how it looks like: >
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:
$$\begin{align} & \delta {{{\bar{\Theta }}}^{1}}={{{\bar{\kappa }}}^{1}}{{P}_{-}} \\ & \delta {{{\bar{\Theta }}}^{2}}={{{\bar{\kappa }}}^{2}}{{P}_{+}} \\ \end{align}$$
Q&A (4871)
