Three dimensional representations of Alternating group

Question

The alternating group $A_5$ has $2$ irreducible representation of degree $3$. The characters for these representations have irrational values. I guess the ring of invariants of these representations should be known in literature but I am not able to find them. Again the matrix entries are also irrational numbers, so I can't compute the generators for the invariant ring in any of the computer algebra systems. Any reference in this direction is highly appreciated.

Answer 1

Let $G_0$ be the image of $A_5$ under one of the $3$-dimensional representations, and $G = \pm G_0$. Then $G$ is the group of symmetries of the icosahedron, which is a Euclidean reflection group (type $H_3$, Shephard-Todd #23). Thus $G$ has a polynomial invariant group, and in this case the generator degrees are $2, 6, 10$. For invariants $\phi_2, \phi_6, \phi_{10}$ we can take the Euclidean norm $x \mapsto (x,x)$, the product of six linear forms $x \mapsto (v,x)$ where $\pm v$ ranges over $6$ pairs of vertices of the icosahedron, and the product of ten linear forms $x \mapsto (v^*,x)$ where $\pm v^*$ ranges over $10$ pairs of vertices of the dual dodecahedron. The invariant ring of $G_0 \cong A_5$ can then be recovered as ${\bf C}[\phi_2, \phi_6, \phi_{10}, \phi_{15}]$ where $\phi_{15}$ is the Jacobian determinant of $\phi_2, \phi_6, \phi_{10}$ (and the product of linear forms $(e,x)$ with $\pm e$ ranging over pairs of edge centers of either the icosahedron or the dodecahedron); these generators satisfy one relation of the form $\phi_{15}^2 = P(\phi_2, \phi_6, \phi_{10})$.