= 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 $x=\phi_1$, $y=\phi_2$. The <generators of a group> $r=\tau_1\tau_2$ and $s=\tau_3$ act by
$$
r:(x,y)\mapsto(\alpha x,\alpha^{-1}y),\qquad s:(x,y)\mapsto(y,x).
$$
A <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> are
$$
\begin{array}{c|l}
\text{degree}&\text{basis}\\\hline
0&1\\
1&\text{none}\\
2&xy\\
3&x^3+y^3\\
4&x^2y^2
\end{array}
$$
In particular, two explicit invariant terms are $\phi_1\phi_2$ and $\phi_1^3+\phi_2^3$. Their invariance follows directly from $\alpha^3=1$ and the interchange of the two components. A quartic example is $(\phi_1\phi_2)^2$.
For a physical real <scalar potential>, <complex conjugate> fields are allowed as well. The <Hermitian inner product> $\phi^\dagger\phi$ and its square are also invariant, while a holomorphic term can enter with its <complex conjugate>, for example $b(\phi_1^3+\phi_2^3)+\bar b(\bar\phi_1^3+\bar\phi_2^3)$. 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 are
$$
\boxed{S_3:\quad 2,3,4.}
$$
The constant term is unrestricted.
For the <fundamental representation> of the <special unitary group> $SU(2)$, its central element $-I$ sends $\phi\mapsto-\phi$ and $\bar\phi\mapsto-\bar\phi$. 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 of
$$
\begin{pmatrix}u&-\bar v\\v&\bar u\end{pmatrix}\in SU(2).
$$
An invariant <polynomial> must therefore depend only on $\phi^\dagger\phi$. Through quartic order the general real <scalar potential> is
$$
\boxed{V=V_0+m^2\phi^\dagger\phi+\lambda(\phi^\dagger\phi)^2,
\qquad V_0,m^2,\lambda\in\mathbb R.}
$$
Hence the possible nonconstant degrees are $2,4$; the cubic term permitted by $S_3$ is excluded by $SU(2)$.
If the notation $V(\phi)$ 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 $\epsilon_{ab}\phi_a\phi_b=\phi_1\phi_2-\phi_2\phi_1$ vanishes identically. The quadratic and quartic $SU(2)$ terms above require <complex conjugate> fields. This also explains why counting a <trivial representation> in $\mathbf2\otimes\mathbf2$ alone would give an incorrect holomorphic answer.
Back to article page
