= Solution
The unbroken group is the <stabilizer subgroup> of the chosen <vacuum expectation value>:
$$
\boxed{H=\{h\in G:h\phi_0=\phi_0\}.}
$$
Its <Lie algebra> consists of generators $T$ with $T\phi_0=0$. The <vacuum manifold> is the orbit $G\phi_0\simeq G/H$ under the stated transitivity assumption. Its tangent space is the image of $T\mapsto T\phi_0$, so its dimension is $r=\dim G-\dim H$.
Let $K_{ij}=\partial_i\partial_jV(\phi_0)$ be the <scalar mass matrix>, with canonical scalar <kinetic terms>. Invariance of the <scalar potential> gives $(\partial_iV)(T_a\phi)_i=0$ for every $\phi$. Differentiate with respect to $\phi_j$ and evaluate at the minimum, where $\partial_iV=0$:
$$
K_{ji}(T_a\phi_0)_i=0.
$$
Thus every independent tangent direction is a zero <eigenvector> of $K$. These are the <Goldstone directions in the scalar mass matrix>; for a global symmetry they are $r$ physical massless <Goldstone bosons>. For a generic nondegenerate minimum transverse to the orbit, they are precisely all the zero eigenvalues. Without that nondegeneracy assumption, additional massless scalars are possible.
For a local <gauge group>, use the <orthogonal unitary-gauge slice near a scalar vacuum>. Since $\phi_0^TT_a\phi_0=0$ by antisymmetry, the proposed condition is equivalent to
$$
(\phi-\phi_0)^TT_a\phi_0=0.
$$
It sets the gauge-orbit components of the scalar fluctuation to zero. This is an admissible local <unitary gauge>: on a basis of independent broken tangent vectors $v_a=T_a\phi_0$, the variation of these conditions under the broken gauge parameters is $v_a^Tv_b$. This <Gram matrix> is invertible on that basis, so the <implicit function theorem> selects the required gauge transformation near the vacuum. Unbroken transformations remain as a residual $H$ gauge symmetry; no global gauge slice through zeros of the scalar field is being asserted.
Write $\phi=\phi_0+h$. To quadratic order, the covariant <kinetic term> contains
$$
\frac12(\partial_\mu h)^T\partial^\mu h+gA_\mu^a(T_a\phi_0)^T\partial^\mu h+\frac12g^2A_\mu^aA^{b\mu}(T_a\phi_0)^TT_b\phi_0.
$$
The middle term vanishes in this <unitary gauge>. The <gauge-boson mass rank from a real scalar vacuum> is therefore determined by
$$
\boxed{(M_A^2)_{ab}=g^2(T_a\phi_0)^TT_b\phi_0.}
$$
These are squared masses: the physical <gauge boson> masses are square roots of the matrix's nonzero eigenvalues after diagonalization. For real $c_a$,
$$
c_a(M_A^2)_{ab}c_b=g^2\left|\sum_ac_aT_a\phi_0\right|^2\geq0.
$$
Its kernel consists exactly of the unbroken generators. With $g\ne0$, the rank is $r$, so \b[exactly $\dim G-\dim H$ gauge-field combinations become massive]. Each removed scalar <Goldstone mode> supplies the extra longitudinal polarization of a massive <gauge boson>.
The remaining physical scalar fluctuations lie in the orthogonal complement of the orbit tangent space. They have no massless modes if $K$ is positive definite on that complement. The unconditional absence of massless scalars does not follow from the printed assumptions. For a concrete counterexample, take a real doublet, $G=SO(2)$ and
$$
V(\phi)=\lambda(\phi_1^2+\phi_2^2-v^2)^4,\qquad \lambda>0,\quad v>0,\quad\phi_0=(v,0).
$$
Its entire minimum set is one circle orbit, as required, but its <Hessian matrix> vanishes at the minimum. Gauging the rotation gives one vector of squared mass $g^2v^2$ and removes the angular scalar. The radial scalar remains massless at quadratic order, since its potential starts at fourth order in the radial fluctuation. This is an <accidental massless radial mode beyond Goldstone modes>. Thus the intended no-massless-scalar conclusion needs nonzero transverse curvature.
In the <Standard Model>, the <Higgs doublet> contains four real components and breaks $SU(2)_L\times U(1)_Y$ to $U(1)_{\rm em}$. Three <Goldstone bosons> become the longitudinal components of $W^\pm$ and $Z$, while the photon remains massless. The remaining radial <Higgs boson> is massive for the ordinary quartic potential with positive curvature. With $\langle H\rangle=(0,v_{\rm EW}/\sqrt2)^T$,
$$
m_W=gv_{\rm EW}/2,\qquad m_Z=v_{\rm EW}\sqrt{g^2+g'^2}/2,\qquad m_h^2=2\lambda v_{\rm EW}^2.
$$
The colour gauge group is unbroken. This is the <Higgs mechanism>, with the generic curvature condition realized by the standard Higgs potential.
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