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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.

Start with the classification of finite-dimensional representations of SU2. Its irreducible complex group representations are the spin-$j$ spacesThe central element $-I$ acts on $V_j$ as $(-1{)}^{2j}$. These facts follow also by realizing $V_j$ as homogeneous polynomials of degree $2j$ in two variables: the raising and lowering operators connect all of their one-dimensional weight spaces, and the highest-weight classification supplies every irreducible.

Identify Euclidean four-space with the quaternions. The unit quaternions, each a copy of $SU(2)$, act byThe norm is multiplicative, so this is an orthogonal action. It preserves orientation because the acting group is connected. If it fixes every $q$, setting $q=1$ first gives $a=b$, and then this quaternion must commute with every quaternion; a real unit quaternion is $\pm 1$. The kernel is therefore $\{(1,1),(-1,-1)\}$.

For completeness, the differential is injective: if imaginary quaternions $u,v$ satisfy $uq-qv=0$ for every $q$, then $u=v$ is central and imaginary, hence zero. Both Lie algebras have dimension six. Thus the image contains a neighbourhood of the identity and is an open subgroup of the connected SO(4) group, so it is the whole group. This proves the Spin(4) double coverThe covering group is simply connected since each $SU(2)$ is a three-sphere.

The irreducible representations of a product of compact groups are tensor products of irreducibles of its two factors. One way to see this is to decompose an irreducible space into isotypic components for the first factor; the second commutes with the first, so only one isotypic component can occur. The multiplicity space must then be irreducible for the second factor. Hence the covering-group irreducibles are $V_{j_L}\otimes V_{j_R}$. By central parity on SU2 tensor products, the kernel element acts as $(-1{)}^{2j_L+2j_R}$. The group representation descends to $SO(4)$ precisely when that sign is positive. The representations of SO(4) from two SU2 spins are thereforeEvery finite-dimensional irreducible complex group representation of $SO(4)$ is obtained this way. Since the group is compact, these are also all its continuous irreducible unitary group representations, up to equivalence.

The Lie-algebra version makes the two spin labels visible locally. Choose rotation generators $J_i$ and generators $K_i$ mixing the fourth direction with the first three, normalized so thatThen $A_i=(J_i+K_i)/2$ and $B_i=(J_i-K_i)/2$ obey two commuting copies of $\mathfrak{su}(2)$. The quaternion quotient determines which Lie algebra representations integrate to the actual group, rather than only to its cover.

For example, $(0,0)$ is the scalar, $(1/2,1/2)$ is the four-vector, and $(1,0)$ and $(0,1)$ are the three-dimensional self-dual and anti-self-dual two-form group representations. The half-spin spaces $(1/2,0)$ and $(0,1/2)$ belong to the cover and do not descend to $SO(4)$. Restricting to rotations fixing the real quaternion axis gives the diagonal $SU(2)$, and the Clebsch-Gordan decomposition for SU2 yieldswith steps of one. For a descended group representation these diagonal spins are integers, as required for the spatial $SO(3)$ subgroup.

Tensor products $V_{j_L}\otimes V_{j_R}$ give the irreducible representations of the Spin(4) double cover. The simultaneous central sign acts by $(-1{)}^{2j_L+2j_R}$, so exactly the integer-sum pairs descend to SO(4). Their dimensions are $(2j_L+1)(2j_R+1)$. Under the diagonal spatial rotation subgroup, the Clebsch-Gordan decomposition for SU2 gives spins from $|j_L-j_R|$ through $j_L+j_R$ in steps of one.

The SO(4) group consists of real orthogonal four-by-four matrices of determinant one. It preserves orientation and the Euclidean metric. Its Spin(4) double cover identifies it with the quotient of two copies of SU(2) by their simultaneous central sign.

A real four-vector vacuum ${\Phi }_0=v{e}_4$ leaves the rotations of its perpendicular three-space unbroken. In a Hermitian generator basis split into two commuting SU(2) triples with $(T^a+T^{a+3})e_4=0$, define $A^{a,\pm }=(A^a\pm A^{a+3})/\sqrt{2}$. For scalar kinetic term $|D\Phi {|}^2$ and canonical vector kinetic terms, its vacuum value is $g^2{v}^2{\sum }_a(A^{a,-}{)}^2/2$. Comparison with $m^2{A}^2/2$ gives three massless and three mass-$|gv|$ gauge bosons.

The factor multiplying the scalar kinetic term matters: replacing it by $|D\Phi {|}^2/2$ at fixed numerical $v$ changes the massive value to $|gv|/\sqrt{2}$. The unbroken global subgroup is SO(3), rather than its simply connected SU(2) cover. The Spin(4) double cover accounts for that central identification.