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For a gauged real scalar triplet with vacuum $v{e}_3$ and generators $(t^a{)}_{bc}=-i{ϵ}^{abc}$, the covariant kinetic term yields gauge-boson squared masses $g^2{v}^2\mathrm{diag}(1,1,0)$. Two Goldstone bosons become longitudinal vector modes through the Higgs mechanism. One radial scalar, two massive three-polarization vector fields and one massless two-polarization field remain. The physical count $3+3\cdot 2=1+2\cdot 3+2=9$ is unchanged. In unitary gauge, the scalar kinetic term includes $\frac{1}{2}{g}^2(v+h{)}^2[(B^1{)}^2+(B^2{)}^2]$.

For a real scalar triplet with $V=\frac{1}{2}{m}^2{\Phi }^2+\frac{\lambda }{4}({\Phi }^2{)}^2$, $\lambda >0$ and $m^2<0$, the vacuum sphere has radius $v^2=-m^2/\lambda$. At ${\Phi }_0=v{e}_3$, the Hessian is $\mathrm{diag}(0,0,2\lambda v^2)$. Global $O(3)\to O(2)$ breaking therefore gives two physical massless Goldstone bosons and one radial scalar with squared mass $2\lambda v^2$. The angular modes cannot be gauged away in a theory with only global rotations.

This cannot be the sole symmetry-breaking sector of the Standard Model. Its gauge-boson mass matrix has two zero eigenvalues, so besides the electromagnetic photon it predicts an extra massless neutral gauge boson. Every linear combination of $W^3$ and $B$, including the combination normally called $Z$, is massless. The observed massive Z boson is a concrete contradiction. Choosing $g,g^{\prime },m,\lambda$ differently cannot change these two zero eigenvalues.

A real scalar triplet can occur as an additional sector, but a further vacuum expectation value, for example that of a Higgs doublet, is then necessary to obtain the observed neutral gauge boson mass pattern. This does not rescue the triplet-only model posed here.

Put $r^2={\Phi }^T\Phi$. The potential has its minimum atChoose the vacuum expectation value ${\Phi }_0=(0,0,v{)}^T$. Since the weak hypercharge is zero, $Q=T^3$ on this real scalar triplet, and $T^3{\Phi }_0=0$. Neither $T^1$ nor $T^2$ annihilates ${\Phi }_0$. The stabilizer subgroup is consequentlyIn particular, spontaneous symmetry breaking has not reduced the two neutral gauge generators to a single one: $Y$ annihilates every component and remains unbroken independently of $T^3$.

The scalar mass matrix isWith canonical kinetic normalization $\frac{1}{2}(\partial h{)}^2$, the radial Higgs boson therefore hasThe two tangential zero eigenvalues are Goldstone modes, absorbed by the two massive gauge bosons through the Higgs mechanism.

For a real scalar triplet, with metric signature $(+---)$, differentiating the Lagrangian density gives $\partial \mathcal{L}/\partial {\varphi }_a=-\lambda {\varphi }_a$ and $\partial \mathcal{L}/\partial ({\partial }_{\mu }{\varphi }_a)={\partial }^{\mu }{\varphi }_a$. HenceFor a constant internal rotation parameter, the variation isEach bracketed tensor is symmetric in $a,c$, so its contraction with the antisymmetric symbol vanishes. This rotates the three field components, not the space coordinates; both the kinetic term and common mass term are invariant.

For a first-principles Noether current derivation, temporarily let $\theta$ depend on spacetime. The terms proportional to $\theta$ still cancel, leavingOn a solution of the field equations, the action variation for a compactly supported $\theta$ vanishes after integration by parts. Thus ${\partial }_{\mu }{j}^{\mu }=0$. To match the charges printed in the question, use the equivalent overall sign convention for the SO(3) Noether current of a real scalar tripletArbitrariness of $\mathbf{n}$ gives three continuity equations. Their charges are exactlyAn overall reversal of all currents is immaterial; the minus sign above keeps the generator convention distinct from the supplied charge convention.

Direct verification is also short:The mass term and the products of equal-type derivatives cancel by antisymmetry; integration by parts produces the boundary term. The charges are conserved when these integrals exist and the net boundary flux vanishes: sufficiently decaying fields on all space, periodic boundary conditions, or suitable symmetry-preserving no-flux boundaries suffice. Arbitrary open boundaries do not. A common constant $\lambda$ is necessary for this internal symmetry; its positivity is a stability issue, not needed for the algebraic conservation proof.

A real scalar triplet groups three real scalar components into a vector under SO(3). Equivalently it is the three-dimensional Adjoint representation of SU(2), whose center acts trivially. Its rotationally invariant quadratic and quartic potentials depend on $\Phi \cdot \Phi$. A nonzero vacuum direction leaves rotations around that direction unbroken.

A real scalar triplet of weak hypercharge zero with vacuum expectation value $(0,0,v)$ breaks $T^1,T^2$ but preserves both $T^3$ and the independent weak hypercharge generator. With canonical kinetic terms its gauge-boson mass matrix in $(W^1,W^2,W^3,B)$ is $g^2{v}^2\mathrm{diag}(1,1,0,0)$. The two massless neutral gauge bosons prevent this field alone from reproducing Standard Model electroweak symmetry breaking.