Tensor products and dual representations

For representations on V and W, define the group action on a pure tensor by applying the group element to both factors, then extend linearly. The group multiplication law holds because it holds separately in each factor. A basis of tensor products has one vector per ordered pair of basis vectors, so dimensions multiply. The trace of a tensor-product matrix is the product of the traces, giving a pointwise product formula for characters. The resulting representation need not be irreducible even when both factors are.

To decompose a tensor product, multiply character values class by class and take weighted inner products with the irreducible table. This method avoids constructing a matrix of size equal to the product of dimensions for every group element. It identifies how many copies occur, though not a preferred basis in each copy. Such decompositions organize possible symmetry types of products, bilinear operations, and coupled degrees of freedom. They are exact finite-dimensional statements here, without approximation or a probabilistic independence assumption.

A dual vector is a linear functional. Define the action on a functional by precomposing with the inverse group action, so evaluating a transformed functional on a transformed vector returns the original scalar. In dual bases the matrix is the transpose of the inverse, not generally the conjugate transpose. Its trace is the original character at the inverse group element. Since finite-order complex eigenvalues lie on the unit circle, this equals the complex conjugate of the original character.

After choosing an invariant Hermitian form and orthonormal coordinates, the original matrices are unitary. Then the inverse transpose equals the entrywise complex conjugate matrix. It still need not equal the adjoint matrix, which is instead the inverse. This difference is visible in dimension greater than one whenever a unitary matrix is not symmetric. The conjugate representation and the dual representation are equivalent in this unitary setting, but that equivalence should not be confused with a basis-independent identity between their underlying spaces.

The tensor square has an operator that swaps its two factors. This swap commutes with the group action, so its positive and negative eigenspaces are invariant. Over the complex numbers, dividing identity plus or minus swap by two gives the corresponding projections. Their images are the symmetric and alternating squares, with dimensions respectively $d(d+1)/2$ and $d(d-1)/2$. A swap eigenvalue concerns the two factors, not the sign of a group permutation acting on the original space.

If the eigenvalues of a group element on V are listed, the symmetric square contains their pairwise products with repeated indices allowed, while the alternating square uses distinct unordered pairs. Expanding the square of their sum separates these two lists. The correction term is the sum of the squared eigenvalues, which is the character at the square of the group element. This derives the familiar half-sum and half-difference formulas rather than guessing them from dimensions at the identity.