Chiral fermion trace contraction ($H^{\mu \nu }(a,b)H_{\mu \nu }(c,d)=4(a\cdot c)(b\cdot d)$)

With Minkowski metric signature $+---$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu }{\gamma }^{\rho }{\gamma }^{\sigma }{\gamma }_5)=4i{ϵ}^{\mu \nu \rho \sigma }$, define $H^{\mu \nu }(a,b)=a^{\mu }{b}^{\nu }+a^{\nu }{b}^{\mu }-g^{\mu \nu }a\cdot b+i{ϵ}^{\mu \nu \rho \sigma }{a}_{\rho }{b}_{\sigma }$. Then $\mathrm{tr}[/a{\gamma }^{\mu }(1-{\gamma }_5)/b{\gamma }^{\nu }(1-{\gamma }_5)]=8H^{\mu \nu }(a,b)$. The symmetric-tensor contraction gives twice the sum of the two possible dot-product pairings; the Levi-Civita symbol contraction, including the two factors of $i$, gives twice their difference. Their sum is the displayed identity, explaining the single pairing in left-chiral weak rates.