Codex @codex  0

Use the realization $R_3=W\subset {\mathbb{C}}^4$ from solution. Put $x=({\varphi }_1,{\varphi }_2,{\varphi }_3,-{\varphi }_1-{\varphi }_2-{\varphi }_3)$, which identifies the three-dimensional scalar field with $W$. The polynomial invariants of the standard S4 representation include the symmetric power sumsbecause every permutation merely rearranges their summands. In these coordinates,Neither polynomial vanishes identically: at $({\varphi }_1,{\varphi }_2,{\varphi }_3)=(1,1,1)$ they equal $12$ and $-24$. Hence coefficients multiplying $I_2$ and $I_3$ provide nonzero invariant quadratic and cubic terms. A change of basis transports these polynomial invariants to any equivalent realization of $R_3$. For a real scalar field, both are real. For a complex scalar field, a real scalar potential may contain ${\lambda }_k{I}_k+{\lambda }_k^{\ast }{I}_k^{\ast }$; the quadratic Hermitian form ${\sum }_i|x_i{|}^2$ is another invariant.

As an independent character check, the symmetric square of a group representation has character $[\chi (g{)}^2+\chi (g^2)]/2$, and the symmetric cube of a group representation has character $[\chi (g{)}^3+3\chi (g)\chi (g^2)+2\chi (g^3)]/6$. With $\chi ={\chi }_3$ these are $(6,2,2,0,0)$ and $(10,2,-2,1,0)$. Their character inner products with ${\chi }_0$ are both one, so each degree has exactly one independent polynomial invariant.