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.

  • 1
    $\begingroup$ The 3-d complex reps of $A_5$ are here: web.mat.bham.ac.uk/atlas/v2.0/alt/A5/#reps $\endgroup$
    – Nick Gill
    Commented Jul 5, 2016 at 14:27
  • 1
    $\begingroup$ Can't you compute these with Magma (probably Sage/Singular too)? $ $ C:=CharacterTable(AlternatingGroup(5)); IR:=InvariantRing(MatrixGroup(GModule(C[2]))); PrimaryInvariants(IR); Algebra(IR); FreeResolution(IR); HilbertSeries(IR); $\endgroup$ Commented Jul 5, 2016 at 17:17

1 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 $\phi_{15}^2 = P(\phi_2, \phi_6, \phi_{10})$.

  • 2
    $\begingroup$ Can you clarify the wording of your third line: "Thus $G_0$ has a polynomial invariant group ..."? (Here $G$ has order $120=2 \cdot 6 \cdot 10$ and has a polynomial algebra of invariants with generators of those degrees.). $\endgroup$ Commented Jul 5, 2016 at 17:25
  • $\begingroup$ Sorry, typo: $G$ has a polynomial invariant group. Fixed. Thanks. $\endgroup$ Commented Jul 5, 2016 at 17:56
  • 1
    $\begingroup$ Thanks for the answer. The part which is not clear to me is that how do you recover the invariants of $G_0$ from G ? Could you please give me a reference ? $\endgroup$
    – Mathew
    Commented Jul 5, 2016 at 20:20
  • $\begingroup$ You're welcome. It's a general fact about Euclidean reflection groups $G$ that the invariant ring, call it $R$, is polynomial, and the orientation-preserving subgroup $G_0$ has invariants $R[\Delta]$ where $\Delta$ is the Jacobian of the generators of $R$, is the square root of an element of $R$, and is the product of linear forms vanishing on the planes of reflection. Reference $-$ not sure what to suggest; it must be in the relevant chapter of Bourbaki Groupes et Algèbres de Lie, but that must be overkill. $\endgroup$ Commented Jul 5, 2016 at 21:33
  • $\begingroup$ $A_5$ also has a $5$ dimensional irreducible representation. Can the matrices be viewed as a subgroup of some finite reflection group so that the ring of invariants can be computed easily ? $\endgroup$
    – Mathew
    Commented Jul 6, 2016 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.