 Solution

For an infinitesimal element of the SO(4) group, write $U=I+iϵT+O({ϵ}^2)$. Reality of $U$ makes $T$ purely imaginary; $U^{T}U=I$ gives $T^T=-T$. Together these imply $T^{†}=T$, while antisymmetry makes the trace vanish. An antisymmetric four-by-four matrix has $4(4-1)/2=6$ independent entries. Thus six imaginary, Hermitian, antisymmetric, traceless matrices span the Special orthogonal Lie algebra in this convention. The physicists' Hermitian generators differ by a factor of $i$ from the real antisymmetric matrices used for its real Lie algebra.

A nonzero real four-vector has a length and a direction. Proper rotations act transitively on the unit three-sphere, so a gauge choice aligns the scalar vacuum expectation value along the fourth coordinate: ${\Phi }_0=v{e}_4$. This chooses an orientation of the vacuum without changing its length. Its stabilizer rotates the first three coordinates and is an SO(3) group.

The supplied generators obeyHence the three sums are unbroken generators. The orthogonal field combinationsare respectively the unbroken and broken gauge directions. With $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }^A{T}^A$, direct action on the vacuum givesCompare with the vector mass term $m^2{A}_{\mu }{A}^{\mu }/2$. ThereforeThe SO(4) vector-Higgs gauge-boson spectrum uses the scalar kinetic normalization printed in the paper, $|D\Phi {|}^2$. Inserting an additional factor $1/2$ while keeping the same numerical $v$ would instead give $|gv|/\sqrt{2}$. This is a field-normalization convention, not a disagreement about the broken subgroup. The massless fields generate the residual SO(3) gauge symmetry; they are sums, not three of the six original gauge fields individually.

Multiplying the first three supplied matrices givesThey generate a copy of the SU(2) Lie algebra. The other triple generates a second copy, and the two commute. Since they span all six dimensions,This is the two commuting rotation algebras in so4 decomposition. A Lie-algebra isomorphism fixes the local group structure but not its global central identifications. The Spin(4) double cover givesThus SO(4) is not globally the direct product of the two simply connected groups. The unbroken diagonal Lie algebra integrates to SO(3), with the same simultaneous central sign already identified. This distinguishes the gauge algebra, its covering group and the actual group acting faithfully on the four-vector.