Finite internal symmetries in physical models

A finite symmetry constrains interactions only after its action on each field has been specified. The relevant calculation is the dimension of an invariant tensor space, with identical-field statistics imposed before interpreting the answer. A group name or a list of irrep dimensions alone cannot determine a Lagrangian.

We will construct the potential of one real scalar triplet of $A_4$ in four-dimensional Lorentzian spacetime. The symmetry acts internally, leaves the spacetime point fixed, and is treated as a global symmetry. There are no other charges, fermions or derivatives in the potential. Scalar components commute; each has canonical mass dimension one. The character calculation uses ordinary complex representations in characteristic zero, obtained here by complexifying the real triplet.

Fix the physical action

Write $\phi=(x,y,z)^{\mathsf T}$ and choose the generators

$$S=\begin{pmatrix}1&0&0\\0&-1&0\\0&0&-1\end{pmatrix}, \qquad T=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}.$$

Thus $S:(x,y,z)\mapsto(x,-y,-z)$ and $T:(x,y,z)\mapsto(y,z,x)$. They obey $S^2=T^3=(ST)^3=I$ and generate twelve matrices: the three cyclic permutations combined with the four diagonal sign choices having product $+1$. This is the faithful real triplet of the tetrahedral group $A_4$. The same basis is given in Altarelli–Feruglio, §3, equation (29), printed p. 11; their primed generators are our $S,T$.

Both matrices are orthogonal, so the canonical kinetic term $\tfrac12\partial_\mu\phi^{\mathsf T}\partial^\mu\phi$ is invariant. Although these matrices also describe spatial rotations of a tetrahedron, their role in this model is internal: all three components are Lorentz scalars. Using the same abstract group as a point group acting on spatial coordinates would be a different physical setup.

The triplet is of real type, so three real component fields are possible. This does not hold for an isolated complex-type irrep: an $A_4$ field in $\mathbf1'$ cannot be one real scalar, since the nontrivial generator acts by a complex cube root of unity. Its complex conjugate transforms in $\mathbf1''$. The reality guide explains the distinction; the types and character values are in the GroupNames $A_4$ table.

Count polynomials, then construct them

Degree-$d$ polynomial functions on a field space $V$ form $\operatorname{Sym}^d(V^*)$. Here the invariant real inner product identifies $V^*\simeq V$. Complexifying commutes with taking invariants of this real finite-group action, so the complex character calculation gives the real dimension of the invariant polynomial space too.

For any representation $W$, averaging its action projects onto the invariant subspace:

$$P_G=\frac1{|G|}\sum_{g\in G}\rho_W(g),\qquad P_G^2=P_G,\qquad \dim W^G=\operatorname{tr}P_G.$$

The equality $P_G^2=P_G$ follows because each group element appears $|G|$ times in the double sum. Hence a character calculation determines whether the explicit invariant polynomials below span the required space. Etingof et al., Theorem 4.5.1, printed p. 68 gives this projection argument and character orthogonality.

In class order $1,2,3A,3B$, the triplet character is $(3,-1,0,0)$ and the class sizes are $(1,3,4,4)$. For three distinguishable triplets, the tensor character is the cube of this character, so

$$\dim(V^{\otimes3})^{A_4} =\frac{3^3+3(-1)^3}{12}=2.$$

For one commuting triplet we instead need the symmetric cube. Its character is

$$\chi_{\operatorname{Sym}^3V}(g) =\frac{\chi(g)^3+3\chi(g)\chi(g^2)+2\chi(g^3)}6.$$

Using the class power maps gives $(10,-2,1,1)$, and therefore

$$\dim(\operatorname{Sym}^3V)^{A_4} =\frac{10-6+4+4}{12}=1.$$

The formula is the cycle-type sum for the symmetrizing projector on three slots; see GAP, §§72.11-1–2. The same method gives the following low-degree result.

Degree $d$	Invariants in $V^{\otimes d}$	Invariants in $\operatorname{Sym}^dV$	Polynomial basis for one real triplet
$1$	$0$	$0$	None
$2$	$1$	$1$	$I_2=x^2+y^2+z^2$
$3$	$2$	$1$	$I_3=xyz$
$4$	$7$	$2$	$I_2^2,\quad I_4=x^4+y^4+z^4$

Each displayed polynomial is invariant under both $S$ and $T$, which establishes invariance under their generated group. The two quartics are independent: their ratio is different on $(1,0,0)$ and $(1,1,1)$. Their number agrees with the character count, so the list is complete through degree four. It is not a claimed set of generators for the entire invariant ring.

Compare the cubic tensor power and cubic symmetric power, then calculate the quartic symmetric power. The calculator counts channels. The component polynomials above are the additional basis choice needed to write the model.

See the missing cubic explicitly

Let $a,b,c$ be three distinct triplet scalar species. Two invariant multilinear contractions are

$$\begin{aligned} D(a,b,c)&=\sum_{\sigma\in S_3} a_{\sigma(1)}b_{\sigma(2)}c_{\sigma(3)},\\ E(a,b,c)&=\sum_{\sigma\in S_3}\operatorname{sgn}(\sigma) a_{\sigma(1)}b_{\sigma(2)}c_{\sigma(3)}. \end{aligned}$$

Here the permutation sum runs over component indices $1,2,3$, not group elements of $A_4$. The first expression is symmetric under exchanging species slots; the second is a determinant. Even sign changes and cyclic coordinate permutations preserve $D$, while determinant-one matrices preserve $E$. Swapping $a$ and $b$ leaves $D$ unchanged and negates $E$, so they are independent. The character multiplicity two shows they exhaust the invariant cubic contractions for these three species.

Now identify the fields:

$$D(\phi,\phi,\phi)=6xyz,\qquad E(\phi,\phi,\phi)=0.$$

The second equality follows from the antisymmetric tensor contracting commuting components. It explains precisely why the tensor-power answer two overcounts the cubic coupling of one scalar multiplet. For fermions, exchange acts on the complete Lorentz, internal and species space. Applying an exterior power only to internal indices is generally insufficient; see fields, Lorentz symmetry and statistics.

Write the potential and identify its limits

Up to an irrelevant additive constant, the most general real polynomial potential of degree at most four for this field content is

$$U(\phi)=\frac{m^2}{2}I_2+\kappa I_3 +\frac{\lambda}{4}I_2^2+\frac{\eta}{4}I_4.$$

All coefficients are real; their mass dimensions are $2,1,0,0$ respectively. The factors $1/2$ and $1/4$ are choices of coupling normalization. This enumerates power-counting renormalizable potential terms for the specified four-dimensional real scalar theory. Complex scalar triplets, extra charges or additional species require a new polynomial space, including conjugate fields where appropriate.

Symmetry allows the coefficients but does not fix them. For example, with $r^2=I_2>0$, the inequalities

$$\frac13\le\frac{I_4}{I_2^2}\le1$$

follow by applying Cauchy–Schwarz to $x^2,y^2,z^2$ and by expanding $I_2^2$. Thus the quartic is strictly positive in every nonzero direction exactly when both $\lambda+\eta/3>0$ and $\lambda+\eta>0$. These strict conditions make the potential grow to $+\infty$ at large field. Boundary cases with flat quartic directions need the cubic and quadratic terms examined separately; symmetry does not establish stability.

There is also an accidental extra symmetry at this truncation: exchanging $x$ and $y$ preserves every term above, although that reflection is not in our determinant-one $A_4$ action. Therefore this potential alone cannot be used to assert that the exact internal symmetry is only $A_4$. The degree-six polynomial

$$J_6=(x^2-y^2)(y^2-z^2)(z^2-x^2)$$

is invariant under $S,T$ but changes sign under that reflection. A term proportional to $J_6$ can distinguish these actions in an effective theory; it is outside the renormalizable potential just counted.

From a field value to a residual subgroup

For a proposed nonzero expectation value $v$, its unbroken subgroup within $A_4$ is $H_v=\{g:\rho(g)v=v\}$. In the stated basis, $v\propto(1,1,1)$ has stabilizer $\langle T\rangle\simeq C_3$, whereas $v\propto(1,0,0)$ has stabilizer $\langle S\rangle\simeq C_2$. The result follows by applying the twelve signed permutation matrices. It describes the residual subgroup if the dynamics selects that field value. Establishing a minimum requires solving the stationary equations, checking fluctuations and comparing competing extrema. Counting invariants does not do those steps.

Use the appropriate representation problem

The finite-symmetry atlas separates the abstract group from its particular action. Before interpreting a count, retain the field content, ordinary/projective category, coefficient field, statistics, and every other symmetry charge.

An antiunitary action on quantum states is antilinear: $A(c\psi)=\overline c\,A\psi$. It is not an ordinary complex-linear representation of the kind used by this character workbench. Time reversal, spatial symmetries and projective state actions need their own physical and mathematical specification.

Finally, invariance of a classical coupling is not a quantum anomaly test, and does not authorize gauging a global symmetry. The fermion measure and possible background-field obstructions concern additional data. For examples of mixed discrete anomaly conditions, see Ishimori et al., §14.1, printed pp. 108–110; that discussion is separate from the invariant-polynomial calculation here.

Continue with the finite learning path, the character method reference, or further finite tensor examples. Export the group, irrep, power operation and character evidence along with a numerical count; retain any component basis and coupling normalization separately.