# Number of supercharges in ${\cal N}\ge 1$ SUSY in $d\ge4$ dimensions

I have a couple of questions concerning supercharges and superalgebras:

1. In four-dimensions, the minimal spinor representations are Weyl spinors. These have 4 real degrees of freedom (dof). This means minimal SUSY is ${\cal N}=1$ with 4 supercharges and we can also consider ${\cal N}=2$ with 8 supercharges. Obviously N=2 will require 2 such Weyl spinors and then consistency with the superalgebra forces them to form an $SU(2)$ doublet under the $R$-symmetry group. Is this correct?

2. "In five dimensions, the minimal spinor representations are symplectic Majorana. In $D$ spacetime dimensions, these spinors have $\frac{1}{2} \times 2^{\lfloor D/2 \rfloor}$ complex degrees of freedom. In five dimensions, this corresponds to four real degrees of freedom and so, to accommodate the eight supercharges of ${\cal N}=2$ in $d=5$, we require a pair of symplectic Majorana spinors, transforming as a doublet under the $SU(2)$ $R$-symmetry group of the five-dimensional superalgebra."

This kind of makes sense to me but I'm wondering how we know that there must be 8 supercharges for such a theory? Is it by relation to the ${\cal N}=2$, $d=4$ theory above - all ${\cal N}=2$ theories must have the same number since they can be related by dimension reduction? Or is it something to do with spinor reps?

3. In $d=5$, it is not possible to have ${\cal N}=1$ i.e. ${\cal N}=2$ is the minimal SUSY. However, if an individuat Majorana spinor has 4 cpts, surely we could accommodate 4 supercharges? Why do they always need to come in pairs - in 4d, we were happy to have an individual Weyl spinor in order to have 4 supercharges and ${\cal N}=1$ SUSY?

## 1 Answer

1. Yes

2. In 5 dimensions, the R-Symmetry group is $SU(2)$. Any spinor, therefore, has to transform under $SU(2)$ as $\delta_{SU(2)} (\Lambda^{ij}) \lambda^i = - \Lambda^{i}{}_{j} \lambda^j$. You cannot consider a spinor $\psi$ without the $SU(2)$ index, i.e. $\delta_{SU(2)} (\Lambda^{ij}) \psi =0$. The algebra would not close on such a field, thus $\psi$, if not an $SU(2)$ doublet, is not a representation of $F^2 (4)$, which is the superalgrabra you are looking for.

3. It is not possible in 5d, because there is no superalgebra that contains Poincare $\times U(1)$ as a subgroup in that dimension. You can only have Poincare $\times SU(2)$. In 4d, however, there is such a supergroup, thus you can have N=1 supersymmetry.