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What is kappa symmetry?

+ 8 like - 0 dislike
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On page 180 David McMohan explains that to obtain a (spacetime) supersymmetric action for a GS superstring one has to add to the bosonic part

$$ S_B = -\frac{1}{2\pi}\int d^2 \sigma \sqrt{h}h^{\alpha\beta}\partial_{\alpha}X^{\mu}\partial_{\beta}X_{\mu} $$

the fermionic part

$$ S_1 = -\frac{1}{2\pi}\int d^2 \sigma \sqrt{h}h^{\alpha\beta}\Pi_{\alpha}^{\mu}\Pi_{\beta}{\mu} $$

plus a long and unwieldy term $S_2$ due to the so called local kappa symmetry which has to be preserved. This $S_2$ term is not further explained or derived.

So can somebody at least roughly explain to me what this kappa symmetry is about and what its purpose is from a physics point of view?

asked Apr 15, 2013 in Theoretical Physics by Dilaton (4,175 points) [ revision history ]
Be warned, it's a technically very complex thing with limited physical implications. See e.g. intro to arxiv.org/abs/hep-th/9908045 for some background. Surprising that David McMahon chose this topic/formalism in a "demystified book". The kappa-symmetry is a local fermionic symmetry on the world sheet whose task is to remove the excessive number of spinor components of the Green-Schwarz "covariant" string down to 8 physical transverse fermions (8+8 on left/right). It may be done in some backgrounds - in others, the right known constructions don't start with a manifestly covariant start.

This post imported from StackExchange Physics at 2014-03-09 16:22 (UCT), posted by SE-user Luboš Motl
Thanks @LubošMotl for this comment and link. David McMohan just said that there is this additional $S_2$ contribution to the action due to kappa symmetry but considered it to be inappropriate to explain this further in a demystivied book ... ;-). This picked me and is why I asked here to see and give it a try if not somebody, like you for example :-P, could explain it in a way such that I can get it.

This post imported from StackExchange Physics at 2014-03-09 16:22 (UCT), posted by SE-user Dilaton
An interesting discussion of this can be found at Becker, Becker, Schwarz . (pg 156 and beyond . )

This post imported from StackExchange Physics at 2014-03-09 16:22 (UCT), posted by SE-user Dimensio1n0
Thanks for the hint @Dimension10, I have even downloaded that :-)

This post imported from StackExchange Physics at 2014-03-09 16:22 (UCT), posted by SE-user Dilaton
For people without a copy of Mcmohan or BBS or... : This is the unwieldy term: $$S_\kappa=\frac1\pi\int\mbox{d}^2\xi \varepsilon^{\alpha\beta}\left( \overline{\Psi}^- \gamma ^\mu\partial_\alpha\Psi^-\mbox{ } \overline{\Psi}^+ \gamma_\mu \partial_\beta \Psi^+ -i\partial_\alpha X^\mu\left( \overline{\Psi}^- \gamma_\mu \partial_\beta\Psi^--\Psi^+ \gamma _\mu\partial_\beta \Psi^+ \right) \right)$$ .

This post imported from StackExchange Physics at 2014-03-09 16:22 (UCT), posted by SE-user Dimensio1n0
Actually, it isn't so "unwieldy" . If you notice, there are some elegant symetries in this term .

This post imported from StackExchange Physics at 2014-03-09 16:22 (UCT), posted by SE-user Dimensio1n0
@Dimension10 yep, the term is not unwieldy to look at , but unwieldy to type without typo ... ;-)

This post imported from StackExchange Physics at 2014-03-09 16:22 (UCT), posted by SE-user Dilaton

3 Answers

+ 4 like - 0 dislike

On general super-target spaces the $\kappa$-symmetry of the Green-Schwarz action functional is indeed a bit, say, in-elegant. But a miracle happens as soon as the target space has the structure of a super-group (notably if it is just super-Minkowski spacetime with its canonical structure of the super-translation group over itself): in that case the Green-Schwarz action functional is just a supergeometric analog of the Wess-Zumino-Witten functional with a certain exceptional super-Lie algebra cocycle on spacetime playing the role of the B-field in the familiar WZW functional. It turns out that this statement implies and subsumes $\kappa$-symmetry in these cases.

Moreover, this nicely explains the brane scan of superstring theory: a Green-Schwarz action functional for super-$p$-branes on super-spacetime exists precisely for each exceptional super-Lie algebra cocycle on spacetime. Classifying these yields all the super-$p$-branes...

... or almost all of them. It turns out that some are missing in the "old brane scan". For instance the M2-brane is there (is given by a $\kappa$-symmetric Green-Schwarz action functional) but the M5-brane is missing in the "old brane scan". Physically the reason is of course that the M5-brane is not just a $\sigma$-model, but also carries a higher gauge field on its worldvolume: it has a "tensor multiplet" of fields instead of just its embedding fields.

But it turns out that mathematically this also has a neat explanation that corrects the "old branee scan" of $\kappa$-symmetric Green-Schwarz action functional in its super-Lie-theoretic/WZW interpretation: namely the M5-brane and all the D-branes etc. do appear as generalized WZW models as soon as one passes from just super Lie algebras to super Lie n-algebras. Using this one can build higher order WZW models from exceptional cocycles on super-$L_\infty$-algebra extensions of super-spacetime. The classification of these is richer than the "old brane scan", and looks like a "bouquet", it is a "brane bouquet"... and it contains precisely all the super-$p$-branes of string M-theory.

This is described in a bit more detail in these notes:

The brane bouquet diagram itself appears for instance on p. 5 here. Notice that this picture looks pretty much like the standard "star cartoon" that everyone draws of M-theory. But this brane bouquet is a mathematical theorem in super $L_\infty$-algebra extension theory. Each point of it corresponds to precisely one $\kappa$-symmetric Green-Schwarz action functional generalized to tensor multiplet fields.

This post imported from StackExchange Physics at 2014-03-09 16:22 (UCT), posted by SE-user Urs Schreiber
answered Aug 9, 2013 by Urs Schreiber (5,085 points) [ no revision ]
+ 2 like - 0 dislike

Since no other answer has turned up so far, I decided the Lubos Motl's comment is good enough to make a start and I hope he does not mind when I make what he said a CW answer:

Be warned, it's a technically very complex thing with limited physical implications. See e.g. [this](Be warned, it's a technically very complex thing with limited physical implications. See e.g. this intro for some background. Surprising that David McMahon chose this topic/formalism in a "demystified book". The kappa-symmetry is a local fermionic symmetry on the world sheet whose task is to remove the excessive number of spinor components of the Green-Schwarz "covariant" string down to 8 physical transverse fermions (8+8 on left/right). It may be done in some backgrounds - in others, the right known constructions don't start with a manifestly covariant start.

answered Jun 10, 2013 by Dilaton (4,175 points) [ revision history ]
+ 1 like - 0 dislike

 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:

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

answered Aug 28, 2013 by dimension10 (1,950 points) [ revision history ]
edited Mar 10, 2014 by dimension10

Thanks for this cool answer (will have to reread it) and for saving my questions BTW :-). At the top of post there seems to be some kind of a LaTex issue with the \renewcommand . Not sure if this should be reported as a bug ?
 

@Dilaton No, it's not a bug, MathJaX does not permit using newcommands "math". 

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