Past exam of the groups and representations course of the University of Oxford / 2021 / 4 / d / Solution

The fundamental case. Set $⟨\varphi ⟩=v{e}_4$, where $v\ne 0$. Fixing this vector forces the fourth column of a unitary matrix to be $e_4$, and hence also fixes its fourth row. The remaining three-by-three block has determinant one. ThereforeEvery nonzero complex vector can be brought to this form by an SU(4) transformation: complete its normalized direction to a unitary basis and adjust the phase of a complementary basis vector to obtain determinant one. Thus a generic nonzero fundamental vacuum expectation value has a conjugate of $H_3$ as its exact unbroken subgroup. The particular embedded $H_3$ of the question is fixed only when the vacuum expectation value lies along $e_4$.

The antisymmetric case. With $ϵ=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, takeSince $VϵV^T=ϵ$ for $V\in SU(2)$, this antisymmetric tensor is fixed by $H_2$ for every $a,b$. Its transformation is $A\mapsto UA{U}^T$, so any stabilizing $U$ also preservesIf $|a|\ne |b|$, its two eigenspaces force $U=\mathrm{diag}(V,\tilde{V})$ with unitary two-by-two blocks. For each nonzero block, $VϵV^T=(\det V)ϵ$ forces the block determinant to be one. If one block is zero, the SU(4) determinant condition forces the other block determinant to be one as well. HenceThe unitary skew-diagonalization of an antisymmetric matrix shows that a generic complex $6$ has two distinct paired singular values; choosing an SU(4) rather than a $U(4)$ basis merely leaves an allowed overall phase in $a,b$. Thus its generic stabilizer is conjugate to $H_2$.

If instead $|a|=|b|\ne 0$, the form is proportional, up to unitary coordinates and phases, to a nondegenerate standard symplectic form. Its stabilizer is the compact symplectic group $USp(4)$, of dimension ten, containing $H_2$ properly. The explicit choice $a=b=1$ is a concrete counterexample to inferring an exact $H_2$ stabilizer from the existence of two singlets alone. For a real $6$, the real structure of the SU4 exterior square imposes $b=\bar{a}$ in this form, so these equal magnitudes are compulsory. Every nonzero real $6$ therefore has the larger compact symplectic group stabilizer. The zero vacuum expectation value preserves all of $SU(4)$.

The adjoint cases. The $H_2$-fixed and $H_3$-fixed traceless matrices have respectively the formsFor a Hermitian adjoint scalar field, choose $a$ real. More generally a complex $a$ gives the same stabilizers. The repeated eigenvalues implyHere $S$ means that the product of the block determinants is one. These groups contain respectively $H_2$ and $H_3$, together with an additional continuous $U(1)$. Explicit extra transformations are $\mathrm{diag}(e^{it}{I}_2,e^{-it}{I}_2)$ and $\mathrm{diag}(e^{it}{I}_3,e^{-3it})$. They commute with $X_2$ and $X_3$ for every $t$, proving that neither exact endpoint $H_2$ or $H_3$ occurs for a single adjoint vacuum expectation value.

Nor do these larger subgroups survive for a generic Hermitian traceless matrix: four distinct eigenvalues leave only the maximal torus $U(1{)}^3$. For a generic complex adjoint matrix, there need not even be a continuous stabilizer; generically its unitary centralizer is just the centre of a group ${\mathbb{Z}}_4$. Thus the displayed degenerate adjoint directions are special. All these statements concern the stabilizer of a possible vacuum expectation value; no unspecified scalar potential can be assumed to select a particular direction or a particular reality condition.