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The unbroken group is the stabilizer subgroup of the chosen vacuum expectation value:Its Lie algebra consists of generators $T$ with $T{\varphi }_0=0$. The vacuum manifold is the orbit $G{\varphi }_0\simeq G/H$ under the stated transitivity assumption. Its tangent space is the image of $T\mapsto T{\varphi }_0$, so its dimension is $r=\dim G-\dim H$.

Let $K_{ij}={\partial }_i{\partial }_{j}V({\varphi }_0)$ be the scalar mass matrix, with canonical scalar kinetic terms. Invariance of the scalar potential gives $({\partial }_{i}V)(T_a\varphi {)}_i=0$ for every $\varphi$. Differentiate with respect to ${\varphi }_j$ and evaluate at the minimum, where ${\partial }_{i}V=0$: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 ${\varphi }_0^T{T}_a{\varphi }_0=0$ by antisymmetry, the proposed condition is equivalent toIt 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 $v_a=T_a{\varphi }_0$, the variation of these conditions under the broken gauge parameters is $v_a^T{v}_b$. 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 $\varphi ={\varphi }_0+h$. To quadratic order, the covariant kinetic term containsThe middle term vanishes in this unitary gauge. The gauge-boson mass rank from a real scalar vacuum is therefore determined byThese are squared masses: the physical gauge boson masses are square roots of the matrix's nonzero eigenvalues after diagonalization. For real $c_a$,Its kernel consists exactly of the unbroken generators. With $g\ne 0$, the rank is $r$, so 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, $G=SO(2)$ andIts 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 $g^2{v}^2$ 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 $U(1{)}_{\mathrm{em}}$. 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 $⟨H⟩=(0,v_{\mathrm{EW}}/\sqrt{2}{)}^T$,The colour gauge group is unbroken. This is the Higgs mechanism, with the generic curvature condition realized by the standard Higgs potential.