a^{(3)}_2 &=& \sqrt{1/2}\ \beta_3\ \sin \gamma _3 +i
\frac{\beta_2}{\sqrt{\beta_2^2+2 \beta_3^2}}\  u \sin \chi \cr
a^{(3)}_3 &=& w \sin \vartheta \left[ \cos \gamma_3 + 
(\sqrt{15}/2)\ \sin \gamma_3 \right]\cr
  &+& i\ w \cos \vartheta \left[ \cos \gamma_3 - 
(\sqrt{15}/2)\ \sin \gamma_3 \right] \cr
&\approx& w\ (\sin \vartheta +i \cos \vartheta )
\label{E:2.2}\end{aligned}$$ With this choice, valid in situations close to the axial symmetry, the tensor of inertia turns out to be diagonal [*up to the first order*]{} in the [*small*]{} quantities describing the non-axial deformations.

In the Eq.s \[E:2.2\] the variables $\gamma_2$ and $\gamma_3$ are still employed, in order to keep some transparency with respect to the standard expressions used to describe the quadrupole [@bohr52] or the octupole deformation alone [@wexler99]. However, it is more convenient to substitute them with expressions involving the variables $u, \chi$ and a new variable $u_0$: neglecting second-order and higher-order terms, $$\begin{aligned}
\gamma_2 &=& \frac{\sqrt{10}\beta_3}{\beta_2
\sqrt{\beta_2^2+5\beta_3^2}} u \cos \chi 
+\frac{f(\beta_2,\beta_3)}{\sqrt{\beta_2^2+5\beta_3^2}} u_0 \\*
\gamma_3&=&-\frac{\sqrt{2}\beta_2}{\beta_3
\sqrt{\beta_2^2+5\beta_3^2}} u \cos \chi 
+\frac{\sqrt{5} f(\beta_2,\beta_3)}{\sqrt{\beta_2^2+5\beta_3^2}} u_0 \nonumber
\label{E:2.3}\end{aligned}$$ It is possible to show that a definite value of the angular-momentum component $K$ along the intrinsic axis 3 and a definite parity can be associated to the degrees of freedom corresponding to the variables 