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

New features!

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

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,549 answers , 19,357 comments
1,470 users with positive rep
410 active unimported users
More ...

Explanation for the minus sign in $\Omega_3$ in the Kappa symmetry of the Green - Schwarz formalism for F1 strings

+ 0 like - 1 dislike
461 views

In the Green-Schwarz formalism, for F1 strings, we have the action

$$S=S_1+S_2$$

Where

$${S_1} = - T\int_{}^{} {\sqrt { - \det \left( {{\Pi _{\alpha \mu }}\Pi _\beta ^\mu } \right)} {{\text{d}}^2}\sigma } $$

and $S_2$ is the additional action term that arises through Kappa Symmetry.

The transformations of kappa symmetry for F1 strings are intuitively related to that for D0 branes, etc. They are given by:

$$\begin{gathered} \delta {X^\mu } = {{\bar \Theta }^A}{\gamma ^\mu } \\ \delta {\Theta ^A} = - \delta {{\bar \Theta }^A}{\gamma ^\mu }{\Theta ^A} \\ \end{gathered} $$

So that:

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

From this transformation, the variation in $S_1$ by kappa symmetry, is almost trivially (not that trivial, thiough...):

$$\delta {S_1} = \frac{2}{\pi }\int_{}^{} {\sqrt { - \lambda } {\lambda ^{\alpha \beta }}\Pi _\alpha ^\mu \delta {{\bar \Theta }^A}{\gamma _\mu }{\partial _\beta }{\Theta ^A}{{\text{d}}^2}\sigma } $$

Here,

$$\begin{gathered} \lambda = \det {\lambda ^{\alpha \beta }} \\ {\lambda ^{\alpha \beta }} = {\Pi _{\alpha \mu }}\Pi _\beta ^\mu \\ \end{gathered} $$

To determine $S_2$, howefver, is not all that trivial since the simple procedure I've mentioned here, for example, is just not practical by any means.

We there fore use a 2-form, form $\Omega_2$, such that:

$${S_2} = \int_{}^{} {{\Omega _2}} = \int {{\epsilon ^{\alpha \beta }}{\Omega _{\alpha \beta }}{{\text{d}}^2}\sigma } $$

We may also (formally) define a 3-form $\Omega_3=\mbox{d}\Omega_2,$ so that by Stokes' theorem, we have it that:

$$ \int_M {{\Omega _2}} = \int_D^{} {{\Omega _3}} $$

$M$ would be the worldsheet whereas $D$ would be it's interior, ; $M=\partial D$.

There are 3 supersymmetric 1-forms prominent to us namely, ${\text{d}}{\Theta ^1},{\text{d}}{\Theta ^2},{\Pi ^\mu }$. If ${\Omega _3}$ is to be supersymmetryic, which it better be, then, we shoul'd have it that ${\Omega _3}$ involves just these 3 1-forms.

A very sensible choice of 3-form for $\Omega_3$ is:

$${\Omega _3} = A\left( {{\text{d}}{{\bar \Theta }^1}{\gamma _\mu }{\text{d}}{\Theta ^1} -{\kern 1pt} {\kern 1pt} {\kern 1pt} {\text{d}}{{\bar \Theta }^2}{\gamma _\mu }{\text{d}}{\Theta ^2}} \right){\Pi ^\mu }$$

Now, when I first learnt this, I was pretty confused at the "$-$" si1gn, (probably because of the fact that I was too confused to bother reading the next line of bbs : ) ...).

As I guess many others may also be confused at this, let me ask and answer the question:

Is this minus sign, by any chance, related to the fact that Type IIA String Theory is chiral? Does this mean that the above expression only holds for the Type IIA, and not for the Type IIB ? .

asked Sep 14, 2013 in Theoretical Physics by dimension10 (1,950 points) [ revision history ]
edited Apr 25, 2014 by dimension10
The definitions of $\Pi _{\alpha \mu },{\bar \Theta }^A$, the meaning of Kappa symmetry, are lacking. So, without reference, it is not easy to follow the context.

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Trimok
. @Trimok: $\Theta^A$ is the standard fermionic field on spacetime in the GS formalism, where as $\Pi^\mu $ is the embedding function on to superspace. I did' not define them because it's a standard convention. As for the meaning of kappa symmetry, I linked to What is Kappa Symmetry? . . .

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Dimensio1n0
You did not define $d \Theta^1, d \Theta^2$ in function of the $\Theta^A$. Here I suppose that $A=1..16$

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Trimok

1 Answer

+ 1 like - 0 dislike
No. It is not related to the Type IIA String Theory; and it holds for the Type IIB too.

(Note that by standard convention, $\Pi^\mu$ is the embedding function onto superspace given by $\Pi _\alpha ^\mu = {\partial _\alpha }{X^\mu } - {\bar \Theta ^A}{\gamma ^\mu }{\partial _\alpha }{\Theta ^A}$)

The reasjon why you have a minus sign, is that when one makes use of the definition of $\Pi^\mu$ and substitute into:

$${\Omega _3} = A\left( {{\text{d}}{{\bar \Theta }^1}{\gamma _\mu }{\text{d}}{\Theta ^1} + k{\kern 1pt} {\kern 1pt} {\kern 1pt} {\text{d}}{{\bar \Theta }^2}{\gamma _\mu }{\text{d}}{\Theta ^2}} \right){\Pi ^\mu }$$

(where $k$ has an absolute value of 1) One sees that $\Pi^\mu$ looks pretty similiar to $\left( {{\text{d}}{{\bar \Theta }^1}{\gamma _\mu }{\text{d}}{\Theta ^1} + k{\kern 1pt} {\kern 1pt} {\kern 1pt} {\text{d}}{{\bar \Theta }^2}{\gamma _\mu }{\text{d}}{\Theta ^2}} \right)$, but with $k $ being equal to $1$.

If one wants the ugly terms to disappear (which is necessary, because $\Omega_3$ is a closed surface and $\mbox{d}\Omega_3=0$), it comes from elementary elemeantary algebra, that $k=-1$.
answered Sep 14, 2013 by dimension10 (1,950 points) [ revision history ]
The problem is that you don't give the definition of $\Pi^\mu$, so I don't understand.

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Trimok
@Trimok: Done. It's a standard convention, like $X^\mu$ or $\Theta^\mu$ which is why I did not define it.

This post imported from StackExchange Physics at 2014-03-07 16:37 (UCT), posted by SE-user Dimensio1n0

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$\hbar\varnothing$sicsOverflow
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).
To avoid this verification in future, please log in or register.




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

Your rights
...