1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$
  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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.