 Solution

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.