Exercise: reconstruct a finite tensor product

This problem turns a character table into an invariant-coupling calculation. Work over $\mathbb C$ in characteristic zero, with ordinary linear representations. A singlet below means the trivial representation, not an arbitrary one-dimensional irrep. No matrix basis is assumed until an explicit polynomial is requested.

Read characters and reality and finite tensor products first. The finite-symmetry path places this exercise between the mathematical method and its physical interpretation.

Problem A: one S₃ doublet

Let $W=\{(x_1,x_2,x_3)\in\mathbb C^3:x_1+x_2+x_3=0\}$, with $S_3$ permuting the coordinates. Use the following table, obtained by subtracting the invariant line from the permutation action. Its order and row labels match the S3 atlas record.

Class	$1$	$(12)$	$(123)$
Size	$1$	$3$	$2$
Square class	$1$	$1$	$(123)$
$\chi_{\mathbf1}$	$1$	$1$	$1$
$\chi_{\mathrm{sgn}}$	$1$	$-1$	$1$
$\chi_W$	$2$	$0$	$-1$

The last square-class entry means the class of all three-cycles: $(123)^2=(132)$ is a different element in the same class. The table agrees with Etingof et al., §4.8, printed p. 74.

1. Verify irreducibility of $W$ and completeness of the three rows.
2. Decompose $W\otimes W$, $\operatorname{Sym}^2W$, and $\Lambda^2W$. Check every character vector and total dimension.
3. Compute the Frobenius–Schur indicator of $W$.
4. How many quadratic invariant polynomials does one commuting scalar multiplet in $W$ have? Write one on the plane $x_1+x_2+x_3=0$.
5. Optional: find the invariant multiplicity in $\operatorname{Sym}^3W$ and write a cubic invariant.

Hints

Use the normalized class inner product

$$\langle\chi,\psi\rangle =\frac16\sum_C |C|\chi(C)\overline{\psi(C)}.$$

For the square, exchange of tensor factors separates equal and unequal basis indices. If the eigenvalues of $g$ on $V$ are $z_i$, then $\sum_{i,j}z_iz_j=(\sum_i z_i)^2$ and the diagonal contribution is $\sum_i z_i^2=\chi_V(g^2)$. Hence

$$\chi_{\operatorname{Sym}^2V}(g) =\frac{\chi_V(g)^2+\chi_V(g^2)}2, \qquad \chi_{\Lambda^2V}(g) =\frac{\chi_V(g)^2-\chi_V(g^2)}2.$$

For the optional cubic, average the $S_3$ permutations of three tensor factors: there are one identity, three transpositions, and two three-cycles. Their traces are respectively $\chi(g)^3$, $\chi(g)\chi(g^2)$, and $\chi(g^3)$. Thus

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

The character inner-product method is in Etingof et al., §§4.5 and 4.9; GAP’s symmetrization documentation, §72.11 explains why character values and power maps are separate inputs.

Solution A

Orthogonality and completeness

The norm of the standard character is $(4+0+2)/6=1$. Its inner products with the trivial and sign rows are both $(2-2)/6=0$. The two linear rows have norm one and inner product $(1-3+2)/6=0$. Thus the character Gram matrix is the identity. The three constructed irreps satisfy $1^2+1^2+2^2=6=|S_3|$, so no further irrep can appear in the regular representation. This uses complete reducibility in the declared category.

Tensor square

Pointwise multiplication gives $\chi_{W\otimes W}=(4,0,1)$. Its inner products with the rows are

$$\begin{aligned} m_{\mathbf1}&=(4+0+2)/6=1,\\ m_{\mathrm{sgn}}&=(4+0+2)/6=1,\\ m_W&=(8+0-2)/6=1. \end{aligned}$$

Therefore $W\otimes W=\mathbf1\oplus\mathrm{sgn}\oplus W$. Adding the rows gives $(1,1,1)+(1,-1,1)+(2,0,-1)=(4,0,1)$, which reconstructs every character value. The dimension is $4=1+1+2$.

Symmetric and exterior squares

The square map supplies $\chi_W(g^2)=(2,2,-1)$, so

$$\chi_{\operatorname{Sym}^2W}=(3,1,0),\qquad \chi_{\Lambda^2W}=(1,-1,1).$$

For the symmetric square, the three multiplicities are respectively $(3+3)/6=1$, $(3-3)/6=0$, and $6/6=1$. The exterior character is exactly the sign row. Consequently

$$\boxed{\operatorname{Sym}^2W=\mathbf1\oplus W,\qquad \Lambda^2W=\mathrm{sgn}.}$$

The dimensions are $3=2(2+1)/2$ and $1=2(2-1)/2$, adding to the tensor square’s dimension $4$. Independently, $\Lambda^2W$ is the determinant of the plane action. The full permutation representation has determinant $\mathrm{sgn}$ and its fixed line has determinant one, confirming the result.

Reality and a scalar polynomial

The Frobenius–Schur indicator is

$$\nu(W)=\frac1{|S_3|}\sum_{g\in S_3}\chi_W(g^2) =\frac{2+3\cdot2+2(-1)}6=1.$$

Thus $W$ has real type, consistently with its realization as the complexification of a real plane. The indicator convention is that of Etingof et al., §5.1, especially the calculation on printed p. 95.

A homogeneous quadratic polynomial lies in $\operatorname{Sym}^2(W^*)$. The invariant dimension equals that in $\operatorname{Sym}^2W$; here $W^*\simeq W$ as well. It is one, represented by

$$q=x_1^2+x_2^2+x_3^2=2(x_1^2+x_1x_2+x_2^2).$$

On the real plane this is a positive quadratic form. On the complexified field space it is a holomorphic polynomial, distinct from the Hermitian form $\sum_i|x_i|^2$. Symmetry does not fix its coefficient in a potential. The exterior-square sign channel is neither a singlet nor an extra quadratic scalar coupling of one commuting multiplet.

Optional cubic

The cube map gives $\chi_W(g^3)=(2,0,2)$. Substitution into the cubic formula gives $(4,0,1)$, so

$$\operatorname{Sym}^3W=\mathbf1\oplus\mathrm{sgn}\oplus W.$$

Its dimension is $4=\binom{2+3-1}{3}$. The unique trivial channel has the explicit representative

$$c=x_1^3+x_2^3+x_3^3=3x_1x_2x_3, \qquad x_1+x_2+x_3=0.$$

The equality follows by expanding $(x_1+x_2+x_3)^3$. Both expressions are symmetric under coordinate permutations, verifying the invariant directly.

Problem B: why the Q₈ doublet behaves differently

Write $Q_8=\{1,z,\pm u,\pm v,\pm w\}$, with $z=-1$, $u^2=v^2=w^2=z$, and $uv=w=-vu$. Let $D$ be the faithful complex doublet. The class order and linear-character labels below are those of the Q8 record, with $u,v,w$ its quaternionic generators.

Class	$1$	$z$	$\{\pm u\}$	$\{\pm v\}$	$\{\pm w\}$
Size	$1$	$1$	$2$	$2$	$2$
Square class	$1$	$1$	$z$	$z$	$z$
$\mathbf1$	$1$	$1$	$1$	$1$	$1$
$a$	$1$	$1$	$1$	$-1$	$-1$
$b$	$1$	$1$	$-1$	$1$	$-1$
$c$	$1$	$1$	$-1$	$-1$	$1$
$D$	$2$	$-2$	$0$	$0$	$0$

The linear characters factor through $Q_8/\{1,z\}\simeq C_2\times C_2$. The doublet can be realized by determinant-one matrices

$$\rho(u)=\begin{pmatrix}i&0\\0&-i\end{pmatrix},\qquad \rho(v)=\begin{pmatrix}0&1\\-1&0\end{pmatrix},\qquad \rho(w)=\rho(u)\rho(v).$$

Squaring the first two matrices and multiplying them in both orders verifies the presentation; taking traces produces the last row. This construction and table are treated in Etingof et al., §§4.3(3) and 4.8.

Compute $D\otimes D$, $\operatorname{Sym}^2D$, $\Lambda^2D$, and $\nu(D)$. Does one commuting complex scalar in $D$ have a holomorphic quadratic invariant? Does it have a Hermitian quadratic invariant?

Solution B

The doublet norm is $(4+4)/8=1$, and the squared dimensions sum to $1+1+1+1+4=8$. The tensor-square character is $(4,4,0,0,0)$. Its multiplicity against each linear row is $(4+4)/8=1$, while its multiplicity against $D$ is $(8-8)/8=0$. Therefore

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

The square map gives $\chi_D(g^2)=(2,2,-2,-2,-2)$. Hence

$$\begin{aligned} \chi_{\operatorname{Sym}^2D}&=(3,3,-1,-1,-1)=\chi_a+\chi_b+\chi_c,\\ \chi_{\Lambda^2D}&=(1,1,1,1,1)=\chi_{\mathbf1}. \end{aligned}$$

This checks all columns, not only the dimensions. The symmetric and exterior dimensions are respectively $3$ and $1$, and

$$\nu(D)=\frac{2+2+2(-2)+2(-2)+2(-2)}8=-1.$$

So the doublet is quaternionic type despite its real-valued character. Its singlet lies in the antisymmetric square. This also follows from the matrices: their determinants are one, so they preserve the alternating form $\epsilon=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$, which spans $\Lambda^2D^*$.

For commuting components, $\phi^T\epsilon\phi=\phi_1\phi_2-\phi_2\phi_1=0$. There is no holomorphic quadratic invariant of a single doublet. By contrast, $\phi^\dagger\phi$ uses $D^*\otimes D$ and is invariant under these unitary matrices. Distinct scalar species can also have the nonvanishing holomorphic contraction $\phi^T\epsilon\eta$. No statement about a relativistic fermion bilinear follows from the internal exterior square alone; its Lorentz and species indices must also be included.

Reproduce and compare

The solutions above were derived from the tables and group relations. Use the workbench to check the implementation and retain its exact exports:

Calculation	Expected result
S3 tensor square	$\mathbf1\oplus\mathrm{sgn}\oplus W$
S3 symmetric square	$\mathbf1\oplus W$
S3 exterior square	$\mathrm{sgn}$
S3 symmetric cube	$\mathbf1\oplus\mathrm{sgn}\oplus W$
Q8 symmetric square	$a\oplus b\oplus c$
Q8 exterior square	$\mathbf1$

For the method in reusable notation, continue to finite-character formulas. For an application with a complete field specification, continue to finite internal symmetries.