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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?

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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. $USp(\mathcal{N})$.

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  • $\begingroup$ Thank you! That was really helpful $\endgroup$ Jan 14 at 20:46

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