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The BRST charge generates BRST transformations. It has odd Grassmann parity and is nilpotent, $Q^2=0$.

A closed fermion loop carries an additional minus sign from the odd Grassmann parity of the fields in the Wick theorem contraction expansion. Equivalently, integrating out Grassmann fields gives a determinant, whose logarithmic interaction expansion has the fermionic sign relative to a bosonic inverse determinant. This sign must be included in addition to propagator and vertex phases.

For adjoint fields whose coefficients have Grassmann parity, $X\times Y=-(-1{)}^{|X||Y|}Y\times X$. Thus two odd ghost fields have a symmetric adjoint bracket, and $c\times c$ need not vanish. The graded Jacobi identity and this symmetry are the algebraic mechanisms behind BRST nilpotence.

The Weyl fields have odd Grassmann parity. With the given epsilon convention, write ${\chi }_{\alpha }=(a,b)$ and ${\psi }_{\alpha }=(c,d)$, where all four components anticommute. Raising an index gives ${\chi }^{\alpha }=(b,-a)$ and ${\psi }^{\alpha }=(d,-c)$. HenceThe minus sign from the antisymmetric epsilon tensor is canceled by exchanging the two Grassmann-odd components. Therefore the Weyl spinor bilinear exchange identity isFor commuting numerical spinors the exchange sign would instead be negative. The answer here uses fermionic Weyl fields, as appropriate to the supersymmetric model.