In the presence of massless chiral fermions, a $\theta$ term in can be rotated away by an appropriate chiral transformation of the fermion fields, because due to the chiral anomaly, this transformation induces a contribution to the fermion path integral measure proportional to the $\theta$ term Lagrangian.

$$\psi_L \rightarrow e^{i\alpha }\psi_L$$

$${\mathcal D}\psi_L {\mathcal D}\overline{\psi_L}\rightarrow {\mathcal D} \psi_L {\mathcal D}\overline{\psi_L} \exp\left(\frac{i\alpha g N_f}{64 \pi^2}\int F \wedge F\right)$$

So the transformation changes $\theta$ by $C \alpha g N_f $ ($g$ is the coupling constant, $N_f$ the number of flavors).

The gluons have the same coupling to the right and left handed quarks, and a chiral rotation does not leave the mass matrix invariant. Thus the QCD $\theta$ term cannot be rotated away.

The $SU(2)_L$ fields however, are coupled to the left handed components of the fermions only, thus both the left and right handed components can be rotated with the same angle, rotating away the $\theta$ term without altering the mass matrix.

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