$\mathcal{N}=2$ susy hypermultiplet self-CPT? Is the multiplet given by
$$\left( -\frac12,0,0,\frac12 \right)$$
self-CPT conjugate?
There seems to be no common agreement upon that:


*

*Weinberg (QFT 3, page 47) and many others claim it is not, basing on $SU(2)$ $R$-symmetry reasoning 

*Terning (modern susy, page 12) and some others claim it is, provided the fermions transform in a real representation of the gauge group, regardless of $R$-symmetry considerations.
 A: It looks like Terning and Weinberg are talking about answering slightly different question. Terning is first answering the question, is there an N=2 hypermultiplet that is CPT self conjugate? The answer is yes, if the fermions transform in a real representation of the gauge group. If the gauge group representation is complex we have to add the CPT conjugate of the hypermultiplet to get a self CPT conjugate multiplet (the group representation gets complex conjugated and for a complex representation by definition this representation is not equivalent to the original), so the hypermultiplet was not self-CPT conjugate.
Weinberg does not consider a real gauge group representation. For physical reasons he considers a complex representation of the gauge group. Once again, the hypermultiplet is then not self conjugate, so he adds its CPT conjugate, but then the fermions transform in a real representation of the gauge group. For physical reasons we do not want fermions to transform  in a real representation of a gauge group, so the N=2 hypermultiplet, either by itself or by with its complex conjugate, is unappealing.
Just as an addendum, the fermions should transform under a complex representation of the gauge group because nature is chiral. Particles of opposite helicity are treated differently. A spin 1/2 particle should transform differently under a gauge transformation compared to a spin -1/2 particle. They transform under representations which are complex conjugate to one another, so having fermions transform in a real representation would ruin chirality.
Edit: I apologize, I did not see where Weinberg had taken into account the R-symmetry originally. There does some appear to be some disagreement within the literature, but the way around Weinberg's argument is if the scalars transform in a psuedoreal representation of the gauge group. An argument is illustrated in http://arxiv.org/abs/1107.0973
