Finite-character formulas

For a finite group $G$ and ordinary finite-dimensional complex representations, a character table gives tensor multiplicities by an inner product. Symmetric powers, exterior powers, and reality indicators also require the class containing each power $g^k$. Keep the class sizes and power maps with the character values.

This reference fixes characteristic zero and counts complex representation multiplicities. It collects the formulas for quick use; the finite-character methods page documents exact arithmetic and the released calculation contract. The character lesson and finite-symmetry path provide the surrounding explanation.

Inner products and multiplicities

Let $C$ run over conjugacy classes, $g_C\in C$ be a representative, and $|C|$ its size. Put $\chi_V(C)=\operatorname{tr}\rho_V(g_C)$ and use an inner product linear in the first argument:

$$\boxed{\quad \langle f,h\rangle_G =\frac1{|G|}\sum_C |C|\,f(C)\overline{h(C)}. \quad}$$

For ordinary characters, $\overline{\chi(C)}=\chi(g_C^{-1})$. The elementary operations are

$$\begin{aligned} \chi_{V\oplus W}&=\chi_V+\chi_W,& \chi_{V\otimes W}&=\chi_V\chi_W,\\ \chi_{V^*}&=\overline{\chi_V},& \chi_{V^{\otimes n}}&=\chi_V^{\,n}. \end{aligned}$$

The irreducible characters $\chi_R$ are orthonormal. Therefore

$$\begin{aligned} m_R(V)&=\langle\chi_V,\chi_R\rangle_G,\\ m_R(V\otimes W)&=\frac1{|G|}\sum_C|C|\, \chi_V(C)\chi_W(C)\overline{\chi_R(C)},\\ \dim V^G&=\langle\chi_V,1\rangle_G =\frac1{|G|}\sum_C|C|\,\chi_V(C). \end{aligned}$$

See Etingof et al., §4.4 and Theorem 4.5.1, printed pp. 67–68. GAP §72.8-5, ScalarProduct, gives the equivalent inverse-element formula for ordinary characters.

The invariant-count formula is the trace of the averaging projector

$$P_G=\frac1{|G|}\sum_{g\in G}\rho_V(g).$$

Multiplying by any $\rho_V(h)$ permutes the summands, so its image consists of invariant vectors; on those vectors it is the identity. This explains both the class-size measure and why the result is an integer. Averaging equally over classes would generally give a different, incorrect answer.

Conjugation check. For $C_3=\langle r\rangle$ and $\omega=e^{2\pi i/3}$, the character $\chi(r)=\omega$ has values $(1,\omega,\omega^2)$. Its norm is $1$, since every term $\chi(g)\overline{\chi(g)}$ is $1$. Omitting conjugation instead gives $(1+\omega+\omega^2)/3=0$. The tensor square is $\chi^2$, while $\chi\otimes\overline\chi=1$.

Power maps and symmetric or exterior powers

Write $C^{[k]}$ for the class containing $g_C^k$; the brackets distinguish this class map from a set product. Define

$$p_k(C)=\chi_V(C^{[k]}),\qquad h_n(C)=\chi_{\operatorname{Sym}^nV}(C),\qquad e_n(C)=\chi_{\Lambda^nV}(C).$$

With $h_0=e_0=1$, Newton’s recurrences are

$$\boxed{\begin{aligned} n h_n(C)&=\sum_{k=1}^n p_k(C)h_{n-k}(C),\\ n e_n(C)&=\sum_{k=1}^n(-1)^{k-1}p_k(C)e_{n-k}(C). \end{aligned}}$$

After computing the class values, apply the same irreducible-character inner product. In particular,

$$\begin{aligned} \chi_{\operatorname{Sym}^2V}(g)&=\frac{\chi_V(g)^2+\chi_V(g^2)}2,\\ \chi_{\Lambda^2V}(g)&=\frac{\chi_V(g)^2-\chi_V(g^2)}2,\\ \chi_{\operatorname{Sym}^3V}(g)&= \frac{\chi_V(g)^3+3\chi_V(g)\chi_V(g^2)+2\chi_V(g^3)}6,\\ \chi_{\Lambda^3V}(g)&= \frac{\chi_V(g)^3-3\chi_V(g)\chi_V(g^2)+2\chi_V(g^3)}6. \end{aligned}$$

These identities follow directly from eigenvalues. If $\alpha_1,\ldots,\alpha_d$ are the eigenvalues of $\rho_V(g)$, then

$$H_g(t)=\sum_{n\geq0}h_n(g)t^n=\prod_{j=1}^d(1-t\alpha_j)^{-1}, \qquad E_g(t)=\sum_{n\geq0}e_n(g)t^n=\prod_{j=1}^d(1+t\alpha_j).$$

The coefficient of $t^n$ selects unordered products with repetition for $H_g$ and distinct-index products for $E_g$. Logarithmic differentiation gives $tH'_g/H_g=\sum_{k\geq1}\chi_V(g^k)t^k$ and $tE'_g/E_g=\sum_{k\geq1}(-1)^{k-1}\chi_V(g^k)t^k$, yielding the recurrences. The equivalent permutation-symmetrization formula and operations are documented in GAP §§72.11-1–72.11-5.

A matrix of irreducible character values alone need not determine the power maps. Preserve their actual group-theoretic definition; see GAP §73.1. The $D_8,Q_8$ comparison below gives a concrete failure of that inference.

At the identity these formulas must reproduce

$$\dim V^{\otimes n}=d^n,\qquad \dim\operatorname{Sym}^nV=\binom{d+n-1}{n},\qquad \dim\Lambda^nV=\binom dn.$$

All zeroth powers are the trivial representation, including for $V=0$. For $n>0$, powers of the zero representation are zero; $\Lambda^nV=0$ for $n>d$. Handle these cases directly rather than using a binomial expression with an unintended convention.

Worked decomposition: the standard S₃ doublet

Take $W$ to be the permutation representation on $\mathbb C^3$ with the fixed line removed. Counting fixed coordinates gives its character as $(2,0,-1)$. The classes and the irreducibles are

Representative	$1$	$(12)$	$(123)$
Class size	$1$	$3$	$2$
Trivial character	$1$	$1$	$1$
Sign character	$1$	$-1$	$1$
$\chi_W$	$2$	$0$	$-1$
$\chi_W(g^2)$	$2$	$2$	$-1$
$\chi_W(g^3)$	$2$	$0$	$2$

The power rows follow from the orders two and three of the representatives. For $W\otimes W$, the character is $(4,0,1)$. Its multiplicities against the trivial, sign, and doublet rows are

$$\frac{4+2}{6}=1,\qquad \frac{4+2}{6}=1,\qquad \frac{8-2}{6}=1.$$

The symmetric and alternating square characters are $(3,1,0)$ and $(1,-1,1)$, respectively. Thus

$$\begin{aligned} W\otimes W&\cong\mathbf1\oplus\mathrm{sgn}\oplus W,\\ \operatorname{Sym}^2W&\cong\mathbf1\oplus W,\\ \Lambda^2W&\cong\mathrm{sgn}. \end{aligned}$$

The cubic formulas give $\chi_{\operatorname{Sym}^3W}=(4,0,1)$ and $\chi_{\Lambda^3W}=0$. Hence $\operatorname{Sym}^3W\cong\mathbf1\oplus\mathrm{sgn}\oplus W$, with dimension four, while the third exterior power vanishes because $\dim W=2$. These are consequences of the displayed rows, independently checkable without a stored decomposition.

Reproduce the symmetric square or inspect the S₃ record. The character exercise develops a related invariant calculation.

Frobenius–Schur indicators

For an irreducible ordinary complex representation $R$, the second indicator is

$$\boxed{\quad\nu_2(R)=\frac1{|G|}\sum_C|C|\,\chi_R(C^{[2]}).\quad}$$

Indicator	Type of the complex irrep	Invariant structure
$+1$	Real	Nondegenerate symmetric complex bilinear form; equivariant antilinear $J$ with $J^2=+1$
$0$	Complex	$R\not\cong R^*$; no invariant nondegenerate complex bilinear form
$-1$	Quaternionic	Nondegenerate alternating complex bilinear form; equivariant antilinear $J$ with $J^2=-1$

The antilinear structures in the table are normalized; the complex representation dimension is retained. Both the real and quaternionic cases have real-valued characters. See Etingof et al., §5.1, Definitions 5.1.1 and 5.1.4, Problem 5.1.2 and GAP §71.12-5, Indicator.

Subtracting the square-character formulas and averaging yields

$$\nu_2(V)=\dim(\operatorname{Sym}^2V)^G-\dim(\Lambda^2V)^G.$$

This identity also holds for reducible $V$, but its value is then additive over the irreducible constituents and is not a single reality-type classification. For example, a sum of two real irreps has indicator $2$. The derivation and irreducible case appear in the proof of Etingof et al., Theorem 5.1.5, printed pp. 94–95.

Equal character matrices, different indicators: D₈ and Q₈

Here $D_8$ is the order-eight square-symmetry group, with $r^4=s^2=1$ and $srs=r^{-1}$; GroupNames calls it $D_4$. The quaternion group is $Q_8=\{\pm1,\pm i,\pm j,\pm k\}$ with $i^2=j^2=k^2=-1$.

Their faithful complex doublets both have character $(2,-2,0,0,0)$ in the following order:

Group	Class representatives	Class sizes	Character of the square
$D_8$	$1,r^2,r,s,rs$	$1,1,2,2,2$	$2,2,-2,2,2$
$Q_8$	$1,-1,i,j,k$	$1,1,2,2,2$	$2,2,-2,-2,-2$

The final row differs because every noncentral quaternion element squares to $-1$, while the dihedral reflections square to $1$. The group presentations and doublet characters agree with the database entries GroupNames: $D_4$ and GroupNames: $Q_8$; class labels have been explicitly reordered above.

Direct substitution gives

$$\nu_2(D_8,\mathbf2)=\frac{2+2-4+4+4}{8}=+1, \qquad \nu_2(Q_8,\mathbf2)=\frac{2+2-4-4-4}{8}=-1.$$

The square decompositions make the distinction visible. In the workbench’s row labels, $a,b,c$ are the three nontrivial linear characters, with $a(r)=1$, $a(s)=-1$ for $D_8$, and $a(i)=1$, $a(j)=-1$ for $Q_8$:

$$\begin{array}{c|cc} &\operatorname{Sym}^2\mathbf2&\Lambda^2\mathbf2\\ \hline D_8&\mathbf1\oplus b\oplus c&a\\ Q_8&a\oplus b\oplus c&\mathbf1 \end{array}$$

Both full tensor squares equal $\mathbf1\oplus a\oplus b\oplus c$. The difference lies in which terms are symmetric. As a separate check, the $Q_8$ doublet embeds in $SU(2)$ and has trivial determinant, so its top exterior power is trivial. The geometric $D_8$ doublet has determinant $-1$ on reflections, giving $a$ instead.

Compare the Q₈ exterior square with the D₈ exterior square.

Checks and limits

For a proposed decomposition $V\cong\bigoplus_R m_R R$, verify nonnegative integral $m_R$, $\sum_Rm_R\dim R=\chi_V(1)$, and reconstruction $\sum_Rm_R\chi_R(C)=\chi_V(C)$ on every class. Dimension alone cannot distinguish the $D_8$ and $Q_8$ results above. The complete ordinary table also satisfies $\sum_R(\dim R)^2=|G|$ and character orthogonality.

The workbench supports $C_3,S_3,D_8,Q_8,A_4,S_4,A_5$, with power degree at most $12$. The interface accepts $1$–$12$ copies of a selected irrep as a direct sum, within the input-dimension bound of $60$; powers act on that whole sum. Distinct copies have mixed terms: $\operatorname{Sym}^2(V\oplus W)=\operatorname{Sym}^2V\oplus(V\otimes W)\oplus\operatorname{Sym}^2W$. Preserve exported conventions and class orders when comparing results.

These formulas do not classify modular representations or projective representations with a specified multiplier. Nor do their singlet counts automatically give a basis of physical operators: identical-field statistics apply to the complete internal and spacetime representation, and derivatives or other equivalences require additional treatment. See fields, Lorentz symmetry, and statistics and from characters to couplings.