Tensor product of group representations

For representations $V$ and $W$ of a group $G$, the tensor product representation is the tensor product $V\otimes W$ with diagonal action

The subrepresentation of $V\otimes V$ fixed by exchange of the two tensor factors. In characteristic zero the projector is $(I+\tau )/2$, and its character is $[\chi (g{)}^2+\chi (g^2)]/2$. Real coordinate-quadratic perturbations of a nonmagnetic electronic multiplet transform in this symmetric square.

For a finite-dimensional irreducible complex representation $V$, invariant tensors in $V\otimes V$ correspond to equivariant maps $V^{\ast }\to V$. Schur's lemma makes their dimension one if $V$ is self-dual and zero otherwise. In particular a real-character irreducible representation has precisely one trivial copy in its tensor square. For arbitrary complex irreducibles, the unconditional statement is that $V\otimes V^{\ast }$ contains the trivial representation once.

For a $G$-representation $V$ and an $H$-representation $W$ over the same field, $V⊠W$ is the representation of $G\times H$ on $V\otimes W$ given by $(g,h)(v\otimes w)=gv\otimes hw$. Its character is $(g,h)\mapsto {\chi }_V(g){\chi }_W(h)$. Over a splitting field in characteristic zero the external products of irreducible representations form all irreducible representations of $G\times H$: character orthogonality gives irreducibility, and their squared degrees sum to $|G||H|$, proving exhaustion. Unlike the tensor product of two representations of one group, the two factors here act independently.

Identify the product with endomorphisms of the defining space. Special unitary group conjugation fixes scalar matrices and preserves the complementary traceless subspace. For $N\ge 2$, the latter is the complex Adjoint representation of a Lie algebra, which is irreducible because ${\mathfrak{sl}}_N$ is simple. With fundamental Dynkin index $1/2$, the trace-Casimir identity gives $C_2(\mathbf{N})=(N^2-1)/(2N)$; the tensor-product Casimir trace identity then gives $C_2(\mathbf{Adj})=N$.

A $(j,k)$ tensor has one fundamental representation factor for every upper index and a dual factor for every lower index. Under $A\in SU(n)$, upper factors transform by $A$ and lower factors by $A^{-1}$; unitarity identifies the latter with conjugate fundamental matrices. Thus complex conjugation exchanges $(j,k)$ and $(k,j)$. Index permutations and contraction with an invariant tensor commute with the action, producing invariant subspaces and decomposition maps.

An invariant tensor is fixed by every element of the represented group, so it spans or belongs to a trivial group representation. The special unitary group preserves its identity tensor and the upper and lower volume tensors. Such tensors make tensor contraction equivariant and permit invariant inner products and real structures.

The tensor-product weight diagram has six distinct nonzero differences of the three defining weights and a zero weight of multiplicity three. Identify the tensor product with endomorphisms of ${\mathbb{C}}^3$ under conjugation. Its scalar line is a singlet, while the traceless matrices give the irreducible adjoint octet, with two zero weights. The root commutators connect all of the traceless matrix units, proving irreducibility rather than inferring it solely from a drawing.