Spontaneous symmetry breaking of a spinor / vector field Why does SSB deal only with scalar fields and not with fermion or vector fields?
My professor told me that it's closely related to the Lorentz invariance of the theory, but I don't understand at all the meaning of that.
 A: Essentially you are asking why only scalars are allowed to develop a vacuum expectation value (VEV).
A scalar (as the name suggests) does not point to any direction -it has spin 0- therefore it can have a VEV without breaking the Lorentz symmetry.
On the other hand, a boson with higher spin, e.g. a vector (spin 1)  would spontaneously break Lorentz by singling out a 'direction', and this is experimentally very constrained as no sign of Lorenz violation has ever been observed.
For fermions, we can apply the same arguments although there could be a mathematical reason forbidding them from taking VEV, see for instance this discussion (it would be nice if someone could develop this further here).
However, the scalar field need not be fundamental. As long as you construct an object which transforms trivially under Lorentz (this could be formed by a pair of more fundamental fermions for instance), it can have a VEV and, eventually, spontaneously break some symmetry depending on its other charges. 
