same for $\theta_{12}^u$ we arrive immediately to $\theta_{12} =
\theta_{12}^d - \theta_{12}^u = \hat \theta_{12} ( 1 +{\mathcal{O}}(\varepsilon^2))$.

Moreover, it is easy to check that the effect of the transformation to canonical Kähler for the right handed fields on the left handed mixings is usually negligible. To see this, we consider the limit of trivial left handed Kähler and nontrivial right-handed Kähler. Then, we consider the diagonalisation of the Hermitian matrix $H^{t}$, $$H^{t}=Y^{t}(Y^{t})^{\dagger }=\hat{Y}N_{R}N_{R}^{\dagger }\hat{Y}^{\dagger }=\hat{Y}K^{-1}\hat{Y}^{\dagger }=\hat{V}_{L}^{\dagger }\hat{M}_{f}\hat{V}_{R}K^{-1}\hat{V}_{R}^{\dagger }\hat{M}_{f}V_{L}\equiv \hat{V}_{L}^{\dagger }\hat{M}_{f}\tilde{K}^{-1}\hat{M}_{f}\hat{V}_{L}$$where we have written $\hat{Y}=\hat{V}_{L}^{\dagger }\hat{M}_{f}\hat{V}_{R}$ and reabsorbed the right-handed rotation in $\tilde{K}^{-1}$, i.e. we have written the inverse of the Kähler in the basis of right handed mass eigenstates. Now it is trivial to see that the matrix diagonalising $H^{t}$ will be the product of $V_{L}$ with the matrix diagonalising $\hat{M}_{f}\tilde{K}^{-1}\hat{M}_{f}$. As we have seen $\hat{M}_{f}$ are approximately equal to the eigenvalues of the total Yukawa matrix, this implies that $\hat{M}_{f}\tilde{K}^{-1}\hat{M}_{f}$ is strongly hierarchical and then the mixing angles diagonalising this matrix will be, $$\tilde{\theta}_{i3}\simeq \frac{\displaystyle{m_{i}m_{3}(\tilde{K}^{-1})_{i3}}}{\displaystyle{m_{3}^{2}(\tilde{K}^{-1})_{33}}}\qquad \tilde{\theta}_{12}\simeq \frac{\displaystyle{m_{1}m_{2}(\tilde{K}^{-1})_{12}}}{\displaystyle{m_{2}^{2}\left( (\tilde{K}^{-1})_{22}-\frac{|(\tilde{K}^{-1})_{23}|^{2}}{(\tilde{K}^{-1})_{33}}\right) }}$$therefore these contributions are suppressed both by the smallness of off-diagonal entries in the Kähler with respect to diagonal ones and by ratios of fermion masses. This last suppression is 