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

Question

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 ? .

Comments

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

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. @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

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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

Answer 1

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 $k=-1$.

Comments

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

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@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