I have a bit of confusion with the following consideration. Generally, to impose some BPS condition (i.e. to check if some multiplet preserves some SUSY charges) one imposes to zero the SUSY variations of fermions belonging to multiplet

$$\langle [Q,\psi] \rangle = 0$$

This is because variations of bosons is trivially zero because such variations are proportional to a Grassmann variables and, if we want to preserve Lorentz group, this must be zero. This is more general

If the theory is Lorentz invariant only scalar can have a non zero vev.

Then

1. SUSY variation of gravitino (spin 3/2) must be something proportional to $\frac{3}{2}\otimes\frac{1}{2}=1\oplus 2$ so, by the consideration above it should be always zero if the Lorentz group is not broken;
2. If I have some defect (branes ecc) then the Lorentz group is broken. In this case, in order to check the SUSY invariance condition, also the vev of bosonic variation is apparently allowed to be different form zero.

That conclusions seem to be strange. I think I'm wrong.

1. Your conclusion is true only for $N=1$ supersymmetry. If I have extended supersymmetry, then I can start from the graviton state and end up in a scalar state.
• You will need enough supersymmetry to create a scalar particle by acting on the gravitino, so basically $N=8$ in $d=4$. May 12 '16 at 0:14