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?

• 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 $2+1$?

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