# R Symmetry

1. Mar 8, 2008

### BenTheMan

Is there a nice, cute way to see what R symmetry is? I mean, N=1 SUSY has a global $$U(1)_R$$ symmetry, which is a charge carried by the supercharges, right? And a spontnaeously broken $$U(1)_R$$ is a sufficient (but not necessary) condition for broken global SUSY. (Counter-example is O'Raifeartaigh type models with broken SUSY but an intact $$U(1)_R$$.)

Is there anything else that I'm missing, or have I screwed something up?

2. Mar 14, 2008

### ophase

I'm not sure. I memorized that R-parity odd particles(Pr=-1) are sparticles and R-parity even particles (Pr=+1) are ordinary particles.
Do you mean a model while you'r saying "a cute way" ??

3. Mar 14, 2008

### BenTheMan

Well, R-symmetry and R-parity are two different things, that are only sometimes related.

I think I just need to read some more of Seiberg's papers. I think he has a set of lecture notes where he outlines this in some more detail.

4. Jul 19, 2008

### JayFsd

Could you post references to the Seiberg papers you refere to?

5. Jul 19, 2008

### Haelfix

R Symmetries is basically an invariance of a supersymmetry algebra free of central charges under a group U(N) of internal symmetries. So for N=1 SuSy you have a U(1) invariance. This implies that in the lagrangian the left and right handed parts of the gaugino fields will have invariant transformations by phases (the left will transform with a negative phase and the right with a positive phase)

This symmetry can be spontaneously broken or violated by anomalies, and so forth depending on the specifics of the theory. There are interesting relationships with things like the Witten index, and this symmetry is important in probing interesting nonperturbative effects. Also the existence of R symmetry is a necessary condition for SuSy breaking.

But what you wrote also sounds right. A spontaneously broken R symmetry is a sufficient condition for Susy breaking at least for most generic classes.