global and possibly not rational.

An Example
==========

In this section we shall show how the parametrization technique works in a simple example. The notations will be the same defined in the previous sections.

We shall consider the orthogonal linear group $G$ defined by the action of the real group O(3) in the real eight–dimensional space spanned by the independent components $x_1,\dots ,x_5$ of a symmetric and traceless tensor $Q$ and the three components $x_6,x_7,x_8$ of a polar vector $\bi{P}$. To be specific, if $O$ is a generic $3\times 3$ real orthogonal matrix, the transformation rules of the $x_i$ are obtained from the following relations:

$$Q = \frac{1}{\sqrt{2}} \left(
\begin{array}{ccc}
-{\displaystyle\frac{2}{\sqrt{3}}} x_1&  x_3 & x_4 \\
x_3                       & {\displaystyle\frac{x_1}{\sqrt{3}}} - x_2 & x_5 \\
x_4                       &  x_5                & {\displaystyle\frac{x_1}{\sqrt{3}}} + x_2 \\
\end{array}
\right)$$

$$\begin{array}{rcll}
Q'_{\alpha\beta}&=& \sum_{\gamma\,\delta=1}^3 O_{\alpha\gamma}\,
                                    Q_{\gamma\delta}\,  O_{\beta\delta}& \\
                                    &&&\alpha,\beta=1,2,3\,.    \\
P'_{\alpha}&=& \sum_{\beta=1}^3 O_{\alpha\beta}\, P_{\beta}\;,
\end{array}\label{mu2}$$

It is worthwhile to remark that the reflection $I=\mbox{\rm diag}
(-1,-1,-1)\in$ O(3) reverses only the signs of the last three coordinates $(x_6,x_7,x_8)$ and, due to the symmetry of the tensor $Q$, each $G$-orbit contains points where $Q$ takes on a diagonal form ($x_3=x_4=x_5=0)$. This remark makes much easier the determination of “typical 