$R$ charge of the chiral multiplet in $2+1$ dimensions These are two examples that I am puzzled by,

*

*One can see in this paper on page 16 that for ${\cal N} =2$ theory on $2+1$ the $R$-charge of the $\phi$ and the $\psi$ is determined to be $\frac{1}{2}$ and $-\frac{1}{2}$. How?

Further in this case since the supercharges are in a vectorial (self-conjugate!) representation of the $R$-symmetry group why is the $R$-charge of $\psi$ and $\psi^*$ different? Since post radial quantization $\bar{Q}=S$ it follows that $Q$ and $\bar{Q}$ transform in exactly the same way under $SO(2)$ and hence shouldn't the $R$-charges $\psi$ and $\bar{\psi}$ been the same?

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*One can see in this paper at the bottom of page 7 a statement about how the scalar component of the ${\cal N}=3$ chiral multiplet in $2+1$ dimensions is in a $SU(2)_R$ triplet. So does this (what?) determine the $R$-charge of the scalar field and other components in the multiplet?

What is puzzling is why did the authors in the second paper think of the $R$-symmetry group as $SU(2)$ for ${\cal N}=3$ rather than the $SO(3)$ that would be gotten by definition?
I would have thought that the $R$-charges of a field are the same as the eigenvalue under the Cartan subalgebra of the $R$-symmetry group of the supercharge(s) which acted on the lowest helicity state state of the multiplet. But isn't the $R$-charge of the lowest helicity state of a multiplet merely a convention? Can't the scalar field be always chosen to be uncharged ($R=0$) under the $SO({\cal N})$ $R$-symmetry of $2+1$?
 A: Note that we have $spin(1,2)\cong SL(2,\mathbb{R})$ so we can have real spinors $(\chi_{\alpha})^{*}=\chi_{\alpha}$, with $\alpha=1,2$. The supersymmetry algebra is then
$$
\{q^{i}_{\alpha},q^{j}_{\beta}\}=2\delta^{ij}\sigma^{m}_{\alpha\beta}p_{m}
$$
in this case you see that the $R$-symmetry must be $SO(n)_{R}$. However, for $SO(2)_{R}$ we can define
$$
q_{\alpha}\equiv \frac{1}{2}\left(q_{\alpha}^{1}+iq_{\alpha}^{2}\right),\qquad \bar q_{\alpha}\equiv \frac{1}{2}\left(q_{\alpha}^{1}-iq_{\alpha}^{2}\right)
$$
such that $(q_{\alpha})^{*}=\bar q_{\alpha}$. Now, $q_{\alpha}$ have charge $+1$, and $\bar q_{\alpha}$ charge $-1$, under $U(1)_{R}\cong SO(2)_{R}$.
If you construct your multiplet using a scalar field $\phi$ with $U_{R}$-charge $+\frac{1}{2}$ as
$$
q_{\alpha}\phi=0,\quad \psi_{\alpha}=\bar q_{\alpha} \phi
$$
the superpartner $\psi_{\alpha}$ will have $U(1)_{R}$ charge $\left(+\frac{1}{2}-1\right)=-\frac{1}{2}$. The same apply to the complex conjugations:
$$
\bar q_{\alpha}\phi^{*}=0,\quad \psi^{*}_{\alpha}=q_{\alpha}\phi^{*}
$$
So all your confusion is about the definition of $U(1)_{R}\cong SO(2)_{R}$ charges. Objects with well defined values for this charge are necessarily complex. Usually people just say $SO(2)_{R}$ charge meaning the $U(1)_{R}$ and this does not generate any confusion since for $SO(2)$ group of matrices does not make sense to talk about charges without doing $SO(2)\cong U(1)$.
