# Polynomial with icosahedral symmetry

I am interested in polynomials with icosahedral symmetry.

https://arxiv.org/pdf/1308.0955.pdf

says that

$$p(v,w)=v^{11}w+11v^6w^6-vw^{11}$$

has icosahedral symmetry. Each term is homogeneous of degree $$12$$. Moreover, within each term the power of $$v$$ and the power of $$w$$ are congruent mod $$5$$. That implies $$p$$ is invariant under the following generator of the binary icosahedral subgroup of $$SU_2$$ $$\begin{bmatrix} e^{\frac{2\pi i}{5}} & 0\\ 0 & e^{-\frac{2\pi i}{5}} \end{bmatrix}= \begin{bmatrix} \zeta_5 & 0\\ 0 & \overline{\zeta_5} \end{bmatrix}$$ In other words, the symmetry \begin{align*} v \mapsto \zeta_5 v \\ w \mapsto \overline{\zeta_5} w \end{align*} fixes $$p$$. Apparently $$p$$ is also invariant under $$\begin{bmatrix} -\zeta_5+\overline{\zeta_5} & \zeta_5^2-\overline{\zeta_5}^2\\ \zeta_5^2-\overline{\zeta_5}^2 & \zeta_5-\overline{\zeta_5} \end{bmatrix}$$ and these two matrices generate the entire 120 element binary icosahedral subgroup $$2I \cong SL_2(\mathbb{F}_5)$$ of $$SU_2$$.

Are there any lower degree polynomials in two complex variables $$v,w$$ with icosahedral symmetry? How about polynomials in three real variables $$x,y,z$$ that are invariant with respect to the icosahedral subgroup $$I \cong A_5 \cong PSL_2(\mathbb{F}_5)$$ of $$SO_3(\mathbb{R})$$?

Let $$G = PSL_2(\mathbb{F}_5)$$. Assuming that you are working over a field of characteristic $$0$$ (in fact, it would be enough to work over a field of characteristic not dividing $$\# G$$), it is a theorem of Klein, proved in his book on the Icosahedron, that the invariant ring $$K[v,w]^G$$ is generated by the polynomials $$\begin{gather*} v^{11} w - 11v^6w^6 - vw^{11}\\ v^{20} + 228 v^{15} w^5 + 494 v^{10} w^{10} - 228 v^5 w^{15} + w^{20} \\ v^{30} - 522 v^{25} w^5 - 10005 v^{20} w^{10} - 10005 v^{10} w^{20} + 522 v^5 w^{25} + w^{30} \end{gather*}$$
This can be proved by computing the Poincare series using Molien's formula (or by computer algebra e.g., Magma's FundamentalInvariants).
• Thanks so much, this is exactly what I wanted! Is there any similar result for the case of $G=PSL_2(\mathbb{F}_5)$ and the invariant ring $\mathbb{R}[x,y,z]^G$? Also two minor questions: is there a reason you wrote the icosahedral group $PSL_2(\mathbb{F}_5)$ in your question when the action seems to be associated to $SU_2$ and the binary icosahedral group $SL_2(\mathbb{F}_5)$? Also the reference I looked at had a positive sign in front of the $11v^6w^6$ while your answer has a negative sign. Is that difference significant? Dec 29, 2021 at 21:08
• The difference in the sign comes from me accidentally swapping the roles of $v$ and $w$ (then taking $-p$) - I should have followed your convention, sorry. Writing $PSL_2(\mathbb{F}_5)$ was a mistake, it came from the fact I was thinking geometrically (about the action on $\mathbb{P}^1$) when I wrote this answer, in this context $-I$ acts trivially. Dec 29, 2021 at 23:18
• Certainally there will be a similar result for the action on $\mathbb{R}[x,y,z]$, since Hilbert-Noether showed that the invariants are a finitely generated $\mathbb{R}$-algebra. One can work it out once you have an explicit representation of your group - the book of Benson "Polynomial Invariants of Finite Groups" is a good reference. Dec 29, 2021 at 23:21
• My representation is just the standard representation of $SO_3(\mathbb{R})$ acting in $\mathbb{R}^3$ (really the restriction of that action to the icosahedral subgroup). Thanks for the reference I'll look into it! Dec 30, 2021 at 15:08
• If you work with the larger group of all symmetries of the icosahedron, including the orientation reversing ones, then the ring of invariants is a polynomial ring. (The ring of invariants for any reflection group is a polynomial ring.) The degrees of the generators for this reflection group are $2$, $6$ and $10$. The additional generator in degree $15$ should be the product of the $15$ hyperplanes which contain a pair of antipodal edges of the icosahedron. Mar 11, 2022 at 14:00