# How do central charges affect $R$-symmetry group in extended SUSY?

When examining a $${\cal N}=1$$ SUSY one finds that the corresponding $$R$$-symmetry group is simply $$U(1)$$. On the other hand, when considering extended SUSY (i.e. $${\cal N}>1$$) the largest possible group is $$U({\cal N})$$, but that depend on the central charges, which are defined by the anticommutator

$$\{Q_{\alpha}^I,Q_{\beta}^{J}\} = \varepsilon_{\alpha\beta} Z^{IJ}.$$

in fact, in presence of non-vanishing central charges one can prove that the $$R$$-symmetry group reduces to $$USp({\cal N})$$, the compact version of the symplectic group $$Sp({\cal N})$$, $$USp({\cal N}) \simeq U({\cal N}) ∩ Sp({\cal N})$$. Why is that?

Well, now your $$R$$-transformations have to preserve $$Z^{IJ}$$. Because $$\epsilon_{\alpha\beta}$$ is antisymmetric, $$Z^{IJ}$$ should also be antisymmetric so that their product was symmetric under the exchange $$(I,\alpha)\leftrightarrow (J,\beta)$$ just like the anticommutator.

Of course if you have sufficiently large number of supercharges $$\mathcal{N}$$ you may introduce $$Z^{IJ}$$ that will only have, for example $$Z^{12}=-Z^{21}$$ with vanishing other components. In that case the $$R$$-symmetry group will be larger. But (for even $$\mathcal{N}$$) you may consider the situation when there are no such combinations of the generators $$c_I Q^I_\alpha$$ that $$c_I Z^{IJ}=0$$. Then $$Z^{IJ} dq_I\, dq_J$$ is a skew-symmetric nondegenerate bilinear form, i.e. the symplectic form. Therefore the $$R$$-symmetries should belong to $$Sp(\mathcal{N})$$.

But they also should belong to $$U(\mathcal{N})$$. Therefore the $$R$$-symmetry group should be an intersection of $$U(\mathcal{N})$$ and $$Sp(\mathcal{N})$$, i.e. $$USp(\mathcal{N})$$.

• Thank you! That was really helpful Jan 14 at 20:46