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Over an algebraically closed field of characteristic zero, every three-dimensional perfect Lie algebra is isomorphic to the sl2 Lie algebra. Indeed the trace of a commutator gives $\mathrm{tr}(\mathrm{ad}x)=0$ for all $x$, because each element is a sum of Lie brackets. If every $\mathrm{ad}x$ were a nilpotent linear map, the Engel theorem would make the Lie algebra nilpotent, contradicting its constant nonzero lower central series of a Lie algebra. Some $\mathrm{ad}x$ therefore has a nonzero eigenvalue $\lambda$. Its other eigenvalues are $0,-\lambda$, since it kills $x$ and has zero trace. They are distinct, so its eigenvectors give a basis $h,e,f$ after scaling $x$ to $h=2x/\lambda$, with $[h,e]=2e$ and $[h,f]=-2f$. The Jacobi identity puts $[e,f]$ in the zero eigenspace $kh$. Its coefficient cannot vanish, since otherwise the derived algebra would lie in $ke\oplus kf$. Rescaling $f$ gives $[e,f]=h$. Thus this basis is an sl2 triple, proving the claimed Lie algebra isomorphism.

Write $H={\varphi }_1$ and $P={\varphi }_2$, both Hermitian matrices. Through quadratic order around the chosen vacuum expectation value,The trace of a commutator vanishes. The needed matrix traces are thereforeUsing $m^2=-(\alpha +N\beta )C^2$, all linear terms cancel. The quadratic scalar potential reduces towith no $P^2$ term. The kinetic term issince its mixed term is the trace of a commutator. A Hermitian matrix has $N^2$ independent real components, so $P$ describes exactly $N^2$ massless scalar modes.

To determine the other masses, take $T_0=1/\sqrt{N}$ and traceless Hermitian $T_A$ with $\mathrm{tr}(T_A{T}_B)={\delta }_{AB}$, and expand $H=h_0{T}_0+{\sum }_{A=1}^{N^2-1}{h}_A{T}_A$. These are canonically normalized real scalar fields. The scalar mass matrix is diagonal in this basis:Thus $H$ contains one massive singlet and a degenerate multiplet of $N^2-1$ massive fields in the Adjoint representation of the unbroken diagonal $SU(N)$. Both squared masses are strictly positive for the specified parameters. The $P$ fields are the $N^2$ Goldstone bosons associated with the broken generators: $N^2-1$ traceless relative left-right rotations and one overall phase rotation.

Goldstone's theorem guarantees those massless directions. It does not alone forbid additional accidental massless modes; the positive scalar mass matrix computed here supplies the further conclusion that every $H$ component is massive. The fields describe a global symmetry, so these Goldstone bosons are physical rather than being absorbed into massive gauge bosons.

For a perfect Lie algebra $\mathfrak{s}$, every element is a sum of Lie brackets. The Adjoint representation satisfies $\mathrm{ad}[x,y]=[\mathrm{ad}x,\mathrm{ad}y]$, and the trace of a commutator is zero. ThereforeSome $\mathrm{ad}x$ must have a nonzero eigenvalue. Otherwise, over the algebraically closed field $k$, every $\mathrm{ad}x$ would be a nilpotent linear map, and the Engel theorem would imply that $\mathfrak{s}$ is a nilpotent Lie algebra. This contradicts perfection: the lower central series of a Lie algebra ${\gamma }_1=\mathfrak{s}$, ${\gamma }_{r+1}=[\mathfrak{s},{\gamma }_r]$ would equal $\mathfrak{s}\ne 0$ at every step instead of terminating.

Choose such an $x$. Since $[x,x]=0$, zero is an eigenvalue of $\mathrm{ad}x$. Its trace is zero and its dimension is three, so the remaining eigenvalues are $\lambda ,-\lambda$ with $\lambda \ne 0$. These three eigenvalues are distinct in characteristic zero, hence $\mathrm{ad}x$ is diagonalizable. Put $h=2x/\lambda$, and choose nonzero eigenvectors $e,f$ withThe zero eigenspace is $kh$, so $h,e,f$ form a basis. Applying the Jacobi identity givesThus $[e,f]=ch$ for some $c\in k$. If $c=0$, all Lie brackets of basis vectors lie in $⟨e,f⟩$, contradicting $[\mathfrak{s},\mathfrak{s}]=\mathfrak{s}$. Therefore $c\ne 0$, and replacing $f$ by $f/c$ gives $[e,f]=h$.

We have constructed an sl2 triple that is a basis. The linear map to the sl2 Lie algebra sendingpreserves all Lie brackets of basis vectors and is bijective. HenceThis is the classification of three-dimensional perfect Lie algebras without appealing to a classification of all simple Lie algebras.

Taking the trace of the proposed commutator identity would givea contradiction for positive integer $n$ over the real numbers. No such finite real matrices exist. This is the trace of a commutator obstruction; it also applies to complex finite matrices.