Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The following is in the context of the ${\cal N}=2$ supersymmetry in $1+1$ dimensions - which is probably generically constructed as a reduction from the ${\cal N}=1$ case in $3+1$ dimensions.

  • In the $\pm$ notation what is the definition of ${\cal D}_+$ and ${\cal D}_{-}$, which I understand from context to be the gauge covariant superderivatives. (..It would be great if someone can relate them to the usual definition in the notation of say Wess and Bagger..)

  • So what is the meaning/motivation of defining a twisted chiral superfield as, $\Sigma = \{\bar{{\cal D}}_{+}, {\cal D}_{-}\} $ (..naively this looks like an operator and not a field - I guess there is some way of arguing that the derivative terms which are not evaluated on something actually go to zero..)

I am guessing that in the above context it will be helpful if someone can explain as to what is meant by the following decomposition/reduction of the gauge field from $3+1$ dimensions,

$\sum _ {\mu = 0}^3 A_\mu dx^\mu = \sum _{\mu =0} ^1 A_\mu dy^\mu + \sigma (dy^2-idy^3) + \bar{\sigma}(dy^2+idy^3)$ ?

  • From the above (does it/how does it) follow that one can write $\Sigma$ as,

$\Sigma = \sigma + \theta\lambda + \theta \bar{\theta}(F+iD)$

(..where I am not sure if $F,D,\sigma$ are real or complex scalar fields...and $\lambda$ is a Weyl fermion..)

  • What is the R-charge of this twisted chiral super field? (..from some consistency conditions I would think that its 2..but I am not sure..)

I guess that the R-symmetry transformations act as,

  • The "right" R symmetry keeps $\theta^-$s invariant and maps, $\theta^+ \mapsto e^{i\alpha}\theta^+$, $\bar{\theta}^+ \mapsto e^{-i\alpha}\bar{\theta}^+$

  • The "left" R-symmetry keeps $\theta^+$ invariant and maps, $\theta^- \mapsto e^{-i\alpha}\theta^-$, $\bar{\theta}^- \mapsto e^{i\alpha}\bar{\theta}^+$.

Though I am not sure and like to understand as to why one wants to think of these two different R-symmetry groups as having two different origins - one coming from the rotation symmetry of the two spatial dimensions of the original $\cal{N}=1$, $1+3$ theory and another coming from R-symmetry of the $\cal{N}=1$, $U(1)$ gauge theory.

share|improve this question
add comment

1 Answer 1

After dimensional reduction from 4 to 2 dimensions, it is convenient to simply label the last two remaining dimensions as $+$ and $-$ instead of 1 and 2. So, basically you have ${\cal D}_- = {\cal D}_1$ and ${\cal D}_+ = {\cal D}_2$.

As for a motivation for twisted chiral superfields, I'm going to quote Witten [http://arxiv.org/abs/hep-th/9301042]:

Sigma models containing both chiral and twisted chiral superfields are quite lovely. Since mirror symmetry turns chiral multiplets into twisted chiral multiplets, it is likely that consideration of appropriate models containing multiplets of both types is helpful for understanding mirror symmetry.

The introduction given by Witten on twisted chiral superfields in the above paper should cover most of your questions.

I am curious though, where did you find your equations? I'm a bit confused by the F-Term in your twisted chiral superfield, as I thought it was common practice to use the WZ-gauge for these types of fields?

share|improve this answer
    
Thanks for the reply. My notation is that from this lecture of Witten, math.ias.edu/QFT/spring/witten13.ps. Look at the bottom of page 11 and top of page 12. –  user6818 Apr 5 '12 at 22:56
    
Can you kindly help about the last part of my question about the existence of two "different" R-symmetries. For example isn't there a mismatch between what these two R-transformations do as written in say Page 14 of the paper you have linked to and say, what is there on Page 59 of these notes, arxiv.org/pdf/hep-th/0504147v1.pdf. May be if you can make explicit the R-symmetry transformation that is being alluded to in the first paragraph of the "Symmetries" section of the paper that you linked to. –  user6818 Apr 5 '12 at 23:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.