Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  What is kappa symmetry?

+ 8 like - 0 dislike
1901 views

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 (6,240 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 (6,095 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 (6,240 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 (1,985 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". 

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsO$\varnothing$erflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...