From characters to physical couplings

A character table can answer a practical model-building question: how many independent internal-symmetry contractions can a specified collection of fields have? The answer depends on the representations, the number of species, and the symmetry imposed when factors are exchanged.

This guide assumes familiarity with representations and tensor products. Its objective is to choose the correct representation space before counting its trivial constituents. All calculations below use ordinary complex representations of finite groups in characteristic zero and can be reproduced in the finite-character workbench.

1. A decomposition counts channels

Suppose $V$ and $W$ are two field multiplets. Their tensor product decomposes as

$$V\otimes W\cong\bigoplus_R m_R R.$$

The number $m_{\mathbf1}$ counts independent invariant bilinear contractions between two distinguishable multiplets, once other transformation laws have been accounted for. More generally,

$$\dim\operatorname{Hom}_G(\mathbf1,V_1\otimes\cdots\otimes V_n)$$

counts the invariant tensor channels. A character calculation obtains their number without choosing bases or writing component tensors. To write an interaction explicitly, one still chooses a basis and an invariant tensor in each channel. Character multiplicities and basis-dependent Clebsch–Gordan coefficients are different pieces of information. The mathematical starting point is character orthogonality and tensor products, Etingof et al., §§4.4–4.9.

2. Two tetrahedral triplets

Consider the real-type triplet $\mathbf3$ of $A_4$, the tetrahedral rotation group. In the class order $1$, $(12)(34)$, $(123)$, $(132)$, its character is

$$\chi_{\mathbf3}=(3,-1,0,0), \qquad |C|=(1,3,4,4).$$

The remaining irreps are $\mathbf1$, $\mathbf1'$, and $\mathbf1''$, with the latter two complex conjugate. The complete table is displayed in GroupNames: $A_4$ and the workbench.

Squaring the character gives $(9,1,0,0)$. Its inner product with the trivial character is

$$m_{\mathbf1}=\frac{9+3}{12}=1.$$

The same calculation gives one copy of each nontrivial one-dimensional irrep, while the triplet multiplicity is

$$m_{\mathbf3}=\frac{3\cdot9+3(-1)\cdot1}{12}=2.$$

Therefore

$$\boxed{\mathbf3\otimes\mathbf3 =\mathbf1\oplus\mathbf1'\oplus\mathbf1''\oplus2\,\mathbf3.}$$

The dimension check is $9=1+1+1+2\cdot3$. Only the copy of $\mathbf1$ is invariant under the whole group. The representations $\mathbf1'$ and $\mathbf1''$ carry nontrivial characters even though their dimensions are one.

Compute the two-triplet tensor product.

3. One commuting field changes the question

For one identical commuting scalar multiplet, exchange of the two copies selects the symmetric square. Using the power-map formula,

$$\begin{aligned} \operatorname{Sym}^2\mathbf3 &=\mathbf1\oplus\mathbf1'\oplus\mathbf1''\oplus\mathbf3,\\ \Lambda^2\mathbf3&=\mathbf3. \end{aligned}$$

The two triplets in the unrestricted tensor product have different exchange symmetries. At cubic order the distinction affects even the number of invariant channels:

$$\begin{aligned} \mathbf3^{\otimes3} &=2\,\mathbf1\oplus2\,\mathbf1'\oplus2\,\mathbf1''\oplus7\,\mathbf3,\\ \operatorname{Sym}^3\mathbf3 &=\mathbf1\oplus3\,\mathbf3. \end{aligned}$$

Thus three distinguishable triplets admit two internal invariant multilinear contractions, while the degree-three polynomials of one commuting triplet have one invariant. These decompositions follow directly from the displayed $A_4$ table and the symmetric-power identities in GAP, §72.11.

Degree	Distinguishable triplet factors: invariants in $\mathbf3^{\otimes n}$	One commuting triplet: invariants in $\operatorname{Sym}^n\mathbf3$
2	1	1
3	2	1
4	7	2

Strictly, homogeneous polynomial functions on a field space $V$ form $\operatorname{Sym}^n(V^*)$. The trivial multiplicity agrees with that in $\operatorname{Sym}^nV$, since taking a dual preserves the number of invariant constituents. For the real-type $A_4$ triplet, $V$ and $V^*$ are also equivalent.

Compare the cubic symmetric power with the cubic tensor power. The dedicated invariant-coupling tool provides another view of the same $A_4$ comparison.

4. Equal character matrices can hide different physics

The quaternion group $Q_8$ and the dihedral group $D_8$ of order eight both have four one-dimensional irreps and one doublet. With suitable column labels, their character matrices agree. In either group,

$$\mathbf2\otimes\mathbf2=\mathbf1\oplus a\oplus b\oplus c.$$

However, the doublets have different reality types. In $Q_8$, the six noncentral elements square to $-1$. For its doublet, $\chi(1)=2$ and $\chi(-1)=-2$, so

$$\nu_{Q_8}(\mathbf2)=\frac{2+2+6(-2)}{8}=-1.$$

In $D_8$, the two quarter turns square to the half turn, while the four reflections square to the identity. Hence

$$\nu_{D_8}(\mathbf2)=\frac{2+2+2(-2)+4(2)}{8}=+1.$$

The first doublet is quaternionic type and the second is real type. These presentations and representation types are independently tabulated in GroupNames: $Q_8$ and GroupNames: the order-eight dihedral group. The indicator criterion is given in GAP, §71.12-5.

The position of the invariant in the square exposes the difference:

Group	Symmetric square of the doublet	Exterior square of the doublet
$Q_8$	$a\oplus b\oplus c$	$\mathbf1$
$D_8$	$\mathbf1\oplus b\oplus c$	$a$

For $D_8$, the label $a$ is fixed by $a(r)=+1$, $a(s)=-1$; it is the determinant character of the geometric doublet. For $Q_8$, $a,b,c$ are the three nontrivial characters of its abelianization. These labels identify different group actions in the two rows of the table.

An identical commuting $Q_8$ doublet therefore has no invariant holomorphic quadratic polynomial formed from two copies of that doublet. Its antisymmetric invariant tensor contracts to zero against commuting components. A Hermitian bilinear $\phi^\dagger\phi$ is a different construction, involving $V^*\otimes V$, and does contain an invariant. This distinction is essential when reading a scalar-potential calculation.

Compute the $Q_8$ symmetric square, then compare $D_8$.

5. Include Lorentz indices before applying fermion statistics

An exterior power of the internal representation is appropriate when the complete set of components being exchanged is an anticommuting space with no additional indices suppressed. In a relativistic field theory, fermion exchange acts on internal, Lorentz, and species indices together.

For example, in four-dimensional Lorentzian spacetime, two left-handed Weyl fields form a scalar through

$$B_{ij}=\epsilon_{\alpha\beta}\psi_i^\alpha\psi_j^\beta.$$

The Grassmann sign and the antisymmetry of $\epsilon_{\alpha\beta}$ cancel under $i\leftrightarrow j$, giving $B_{ij}=B_{ji}$. Thus this bilinear selects a symmetric tensor in the remaining internal/species labels. This is the reason the general Weyl-fermion mass matrix is symmetric; see Dreiner, Haber, and Martin, §3.2.

Applied to a single $Q_8$ doublet, the available antisymmetric internal invariant cannot provide this symmetric bilinear. Extra species can change the answer: an antisymmetric species contraction can combine with the antisymmetric internal tensor to give the required overall symmetry. The character workbench computes the internal representation algebra; the full operator analysis must also impose Lorentz structure, charges, and field statistics.

6. A dimension does not identify a representation

The two $A_5$ triplets provide a useful diagnostic. Both have real type, but they are inequivalent. Their five-cycle values are exchanged:

$$\chi_{\mathbf3}(5A)=\varphi,\qquad \chi_{\mathbf3'}(5A)=1-\varphi, \qquad \varphi(1-\varphi)=-1.$$

The character table therefore gives

$$\mathbf3\otimes\mathbf3=\mathbf1\oplus\mathbf3\oplus\mathbf5, \qquad \mathbf3\otimes\mathbf3'=\mathbf4\oplus\mathbf5.$$

Both products have dimension nine, but only the first contains an invariant. For the mixed product, its character is $(9,1,0,-1,-1)$, which equals the sum of the $\mathbf4$ and $\mathbf5$ rows in GroupNames: $A_5$. This also checks the result without evaluating an approximate square root.

Compare the inequivalent $A_5$ triplets. For $S_4$, the workbench likewise distinguishes the standard triplet from its sign twist, which is the cubic-rotation representation.

Use the result in a calculation

Before interpreting a multiplicity, retain the exact group, irrep labels, field content, and chosen tensor symmetry. An internal invariant must still satisfy every other gauge or global charge constraint. For derivative operators, integration by parts and equations of motion introduce additional equivalences.

The workbench supplies the full decomposition, invariant multiplicity, dimension and character-reconstruction checks, mathematical conventions, and exportable provenance. The methods page explains its exact arithmetic and power maps. Reproducibility and citation explains what to retain with a research result.

The source checks and executable examples support this guide’s stated scope. Independent human review remains tracked through the site’s review method.