Fierz identity for Weyl spinors in tensor currents

Question

Using Fierz identity I found that certain four-fermion operator with left $l_i$ and right-chiral $r_i$ Weyl spinors vanish

$$\bar{l}_1\sigma_{\mu\nu} r_2 \bar{r}_3 \sigma^{\mu\nu} l_4 = -\frac{3}{2} \bar{l}_1 l_4 \bar{r}_3 r_2 - \frac{1}{2}\bar{l}_1\sigma_{\mu\nu} l_4 \bar{r}_3 \sigma^{\mu\nu} r_2 - \frac{3}{2} \bar{l}_1 (-1) l_4 \bar{r}_3 (+1) r_2 $$$$= -\frac{1}{2}\bar{l}_1\sigma_{\mu\nu} l_4 \bar{r}_3 \sigma^{\mu\nu} r_2 = 0$$

Why does this operator vanish? Is that true?

Why is the product of tensor currents expressed in tensor and scalar currents only, while for such combinations of Weyl spinors $l_1$, $r_2$, $r_3$ and $l_4$ in the Fierz identity all of them vanish?

Answer 1

The simplest way to see that all products of your kind vanish is to notice that one of the objects (bilinear invariant) $T_{\mu\nu}$ is a self-dual 2-form while the other $T^{\mu\nu}$ is anti-self-dual, and their contraction without a complex conjugation has to vanish.

A self-dual (anti-self-dual) antisymmetric tensor obeys $$T_{\mu\nu} = \pm \frac i2 \epsilon_{\mu\nu\kappa\lambda} T^{\kappa\lambda}$$ and in $3+1$ dimensions, it has to have complex components.

The objects $\bar \ell \sigma \ell$ are self-dual or anti-self-dual due to the chirality of the spinors. And the inner products are zero because the 6-dimensional complex space of 2-forms may be really decomposed to mutually orthogonal self-dual and anti-self-dual (3-complex-dimensional) parts.

The claims above, when translated to maths, contain lots of signs – sometimes convention-dependent signs (and extra signs and flips of chiralities from complex conjugation and so on) – that one has to be careful about. But there are also several identities of your kind one may derive.