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The special unitary group $SU(2)$ acts transitively on every sphere in its defining ${\mathbb{C}}^2$: for $|u{|}^2+|v{|}^2=1$, the matrix with columns $(u,v{)}^T$ and $(-\bar{v},\bar{u}{)}^T$ is in $SU(2)$. Thus any invariant real polynomial in a single commuting doublet and its complex conjugate depends only on ${\varphi }^{†}\varphi$. Through total degree four it is $V_0+m^2{\varphi }^{†}\varphi +\lambda ({\varphi }^{†}\varphi {)}^2$. A holomorphic polynomial invariant of positive degree does not exist for one doublet: ${\mathrm{Sym}}^{d}2$ is a nontrivial irreducible representation for every $d>0$.

The variables commute, so $x_i{x}_j=x_j{x}_i$. Thus $V$ has dimension six, with basis $x_1^2,x_2^2,x_3^2,x_1{x}_2,x_1{x}_3,x_2{x}_3$, even though nine coefficients occur in the redundant expression in the source. Its representation is the symmetric square of the three-dimensional permutation representation.

The identity fixes all six monomials. A three-cycle fixes none. For a transposition, say $(12)$, exactly $x_3^2$ and $x_1{x}_2$ are fixed; the remaining monomials occur in exchanged pairs. Therefore ${\chi }_{\tilde{R}}=(6,0,2)$ in the conjugacy class order of solution. The character inner products give multiplicitiesConsequentlyThere is also a concrete decomposition: the span of the three squares is one copy of the permutation representation. The span of the three mixed monomials is another, since $x_2{x}_3,x_1{x}_3,x_1{x}_2$ are permuted exactly like their omitted indices $1,2,3$. Within each, the coefficient-sum-zero vector subspace gives a copy of $D$, and the sum of its three basis monomials gives a trivial representation.

A fixed quadratic polynomial must have equal coefficients on each of the two orbits of monomials. Hence its fixed vector subspace isThe two displayed polynomial invariants are linearly independent and account for both singlets. Equivalently, one may use $x_1^2+x_2^2+x_3^2$ and $(x_1+x_2+x_3{)}^2$ as a basis. No constant term belongs to this homogeneous degree-two vector space.

Write a homogeneous quadratic polynomial as $p(x,y)=a{x}^2+bxy+c{y}^2$. Invariance under $B:(x,y)\mapsto (x,-y)$ forces $b=0$. Invariance under $A:(x,y)\mapsto (y,x)$ then forces $a=c$. Since $A,B$ generate the group, this proves that all homogeneous quadratic polynomial invariants areThis agrees with the single trivial representation in the symmetric square in solution; more precisely, coordinate polynomials lie in the symmetric square of the dual representation, which here is equivalent to $\rho$. If “quadratic” permits terms of smaller degree, arbitrary constants can be added. There are no invariant linear terms because $-I\in G$ negates every linear polynomial.

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.

The scalar fields commute, so a degree-$d$ expression in one doublet belongs to a symmetric power, rather than an arbitrary tensor product. Put $x={\varphi }_1$, $y={\varphi }_2$. The generators of a group $r={\tau }_1{\tau }_2$ and $s={\tau }_3$ act byA monomial $x^a{y}^b$ is fixed by $r$ precisely when $a-b$ is divisible by three. Imposing $s$ then pairs the exponents. Through degree four the resulting holomorphic polynomial invariants areIn particular, two explicit invariant terms are ${\varphi }_1{\varphi }_2$ and ${\varphi }_1^3+{\varphi }_2^3$. Their invariance follows directly from ${\alpha }^3=1$ and the interchange of the two components. A quartic example is $({\varphi }_1{\varphi }_2{)}^2$.

For a physical real scalar potential, complex conjugate fields are allowed as well. The Hermitian inner product ${\varphi }^{†}\varphi$ and its square are also invariant, while a holomorphic term can enter with its complex conjugate, for example $b({\varphi }_1^3+{\varphi }_2^3)+\bar{b}({\bar{\varphi }}_1^3+{\bar{\varphi }}_2^3)$. There is still no invariant linear term, because neither the doublet nor its conjugate representation contains a trivial representation. Thus the possible nonconstant total degrees areThe constant term is unrestricted.

For the fundamental representation of the special unitary group $SU(2)$, its central element $-I$ sends $\varphi \mapsto -\varphi$ and $\bar{\varphi }\mapsto -\bar{\varphi }$. Every odd-degree polynomial changes sign, excluding degrees one and three. Further, $SU(2)$ acts transitively on the unit sphere in ${\mathbb{C}}^2$: a unit vector $(u,v{)}^T$ is the first column ofAn invariant polynomial must therefore depend only on ${\varphi }^{†}\varphi$. Through quartic order the general real scalar potential isHence the possible nonconstant degrees are $2,4$; the cubic term permitted by $S_3$ is excluded by $SU(2)$.

If the notation $V(\varphi )$ is instead read as restricting to holomorphic polynomials alone, the distinction matters: a single commuting $SU(2)$ doublet has no nonconstant holomorphic polynomial invariant. The candidate contraction ${ϵ}_{ab}{\varphi }_a{\varphi }_b={\varphi }_1{\varphi }_2-{\varphi }_2{\varphi }_1$ vanishes identically. The quadratic and quartic $SU(2)$ terms above require complex conjugate fields. This also explains why counting a trivial representation in $2\otimes 2$ alone would give an incorrect holomorphic answer.

The symmetric cube of a group representation $V$ is the symmetric part of $V^{\otimes 3}$. Over the complex numbers, its character isThe projection onto symmetric tensors averages the six permutations of the three factors. The identity permutation contributes $\chi (g{)}^3$, the three transpositions each contribute $\chi (g)\chi (g^2)$, and the two three-cycles each contribute $\chi (g^3)$. Cubic polynomial invariants are fixed vectors in ${\mathrm{Sym}}^3(V^{\ast })$, whose dimension follows by averaging this character for a finite group.