 Solution

For a differentiable curve $R(t)$ in the SO(3) group with $R(0)=I$, differentiating $R(t{)}^{T}R(t)=I$ gives $X^T+X=0$, where $X=R^{\prime }(0)$. Conversely, if $X$ is a real skew-symmetric matrix, $e^{tX}$ is an orthogonal matrix and has determinant $e^{t\mathrm{tr}X}=1$. HenceThe displayed generators of the SO(3) Lie algebra areThey are linearly independent and span the three independent entries of a real skew-symmetric matrix. Equivalently, $T_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The vector triple-product identity then givesThus the structure constants of a Lie algebra in this basis are

Use a real, antisymmetric convention throughout the SO(3) vector Higgs model. With gauge coupling $g$, the gauge covariant derivative and gauge field strength areThere is no factor of $i$ in these formulas because $T_a$ are real antisymmetric generators. For a local non-Abelian gauge transformation ${\Phi }^{\prime }=R\Phi$, with $R(x)$ in the SO(3) group, the compatible transformation of the gauge potential isIt gives $D_{\mu }^{\prime }{\Phi }^{\prime }=R{D}_{\mu }\Phi$ and $F_{\mu \nu }^{\prime }=R{F}_{\mu \nu }{R}^{-1}$. These covariance laws fix the relative signs in the gauge covariant derivative and gauge field strength.

An SO(3) invariant quartic scalar potential isSince $(R\Phi {)}^T(R\Phi )={\Phi }^T\Phi$, this scalar potential is gauge-invariant. Its minimum is the sphere ${\Phi }^T\Phi =v^2$, rather than a preferred direction in the internal vector space.

Writing ${\mathbf{A}}_{\mu }=(A_{\mu 1},A_{\mu 2},A_{\mu 3})$, the cross product form of the gauge covariant derivative is $D_{\mu }\Phi ={\partial }_{\mu }\Phi +g{\mathbf{A}}_{\mu }\times \Phi$. In components this isCorrespondingly, the gauge field strength has components

A covariantly constant vector Higgs field has constant norm, becauseThe last equality uses antisymmetry of the gauge potential. On a connected region write this norm as $\varphi \ge 0$. If $\varphi =0$, the vector Higgs field is identically zero. If $\varphi >0$, the SO(3) group acts transitively on unit vectors, so choose a smooth local non-Abelian gauge transformation rotating $\Phi /\varphi$ to ${\mathbf{e}}_3$. In this unitary gauge, $\Phi =(0,0,\varphi {)}^T$ with $\varphi$ constant. This is a local construction, and is global on a trivial contractible region; a nontrivial bundle can require more than one gauge patch.

For nonzero $\varphi$, substituting the constant aligned vector Higgs field into $D_{\mu }\Phi =0$ gives $g\varphi (A_{\mu 2},-A_{\mu 1},0{)}^T=0$. For nonzero gauge coupling, the most general compatible gauge potential and gauge field strength arewhere $a_{\mu }$ is an arbitrary real one-form. Thus the surviving gauge group is the $SO(2)$ stabilizer subgroup of ${\mathbf{e}}_3$. A covariantly constant vector Higgs field does not force the gauge field strength to vanish: it only requires the curvature to annihilate that field.

For the Higgs mechanism, choose a vacuum of the SO(3) invariant quartic scalar potential, so the constant norm is $v$. The canonically normalized kinetic term then containsThe two broken-direction gauge bosons have mass $gv$ (or $|g|v$ if the sign of $g$ is unrestricted), while the $T_3$ gauge boson remains massless. The two angular Goldstone bosons supply the massive vectors' longitudinal gauge-boson polarizations. The radial fluctuation remains a physical scalar field, with $m_h^2=2\lambda v^2$ for this normalization. The condition $D_{\mu }\Phi =0$ selects backgrounds without broken-direction gauge potentials; fluctuations about such a vacuum exhibit the two massive modes. Constancy of the norm alone does not require it to minimize the scalar potential.