- For a general ${\cal N}$ the R-symmetry group is $U({\cal N})$ but for the ${\cal N}=2$ case why is it $SU(2)$ ? I guess it is again different for ${\cal N}=4$. How does one understand this?
One denotes the generators of the R-symmetry group by $B^i$ and let $\alpha$ be the spinor index and let $a,b$ be the R-charge indices in a ${\cal N}=2$ theory. Then one writes the defining equations as,
$[Q_{\alpha a}, B^i] = -\frac{1}{2} (\tau ^i)_a ^b Q_{\alpha b}$ and $[\bar{Q}_{\dot{\alpha}}^a, B^i] = \frac{1}{2} \bar{Q}_{\dot{\alpha}}^b (\tau ^i)_b ^a$
- But it is not clear to me as to why over and above the aforementioned equations there should be yet another charge $R$ (called the $U(1)_R$ charge) with the defining equations,
$[Q_{\alpha a},R] = Q_{\alpha a}$ and $[\bar{Q}_{\dot{\alpha} a}, R] = - \bar{Q}_{\dot{\alpha} a}$
I would like to know why the above charge should exist in ${\cal N}=2$ supersymetry separate from the R-symmetry.
Is there an analogue for this $R$ for other values of ${\cal N}$ ?
- Typically this R-charge as defined above is anomalous. How does one see that? Also what is the analogous statement for R-symmetry?
No comments:
Post a Comment