- Orthogonality Relations in Group Representation Theory
Introduction
This document explores the orthogonality relations in group representation theory, which are fundamental in understanding how irreducible representations of finite groups behave.
Diadic Linear Operator and Transformation Properties
We begin by defining a diadic linear operator:
$$A^{ab}_{jl} = \sum_{g \in G} D_a(g^{-1})|a,j\rangle\langle b,l|D_b(g)$$When this operator is transformed by a representation matrix $D_a(g_1)$:
$$D_a(g_1)A^{ab}_{jl} = \sum_{g \in G} D_a(g_1g^{-1})|a,j\rangle \langle b,l|D_b(g)$$Let’s substitute $g’=g_1g^{-1}$, which implies $g=g’^{-1}g_1$. This gives us:
$$\begin{align} D_a(g_1)A^{ab}_{jl} &= \sum_{g' \in G} D_a(g')|a,j\rangle\langle b,l|D_b(g'^{-1}g_1) \\ &= \sum_{g' \in G} D_a(g')|a,j\rangle\langle b,l|D_b(g')D_b(g_1) \\ &= A^{ab}_{jl}D_b(g_1) \end{align}$$Application of Schur’s Lemma
By Schur’s lemma, we can deduce that:
$$A^{ab}_{jl} = \delta_{ab} \lambda^a_{jl}I_{n_a}$$where $I_{n_a}$ is the identity matrix of dimension $n_a$.
Computing the Coefficient $\lambda^a_{jl}$
To compute $\lambda^a_{jl}$, we use the cyclic property of the trace: $\text{tr}(AB)=\text{tr}(BA)$.
From the original definition:
$$\text{tr}(A^{ab}_{jl})=\sum_{g \in G}\text{tr}(\langle b,l|D_b(g)D_a(g^{-1})|a,j\rangle)$$Hence:
$$\text{tr}(A^{ab}_{jl}) = \delta_{ab}\sum_{g \in G}\text{tr}(\langle a,l|a,j\rangle) = N \delta_{ab} \delta_{jl}$$where $N$ is the order of the group $G$.
From the Schur’s lemma form:
$$\text{tr}(A^{ab}_{jl}) = n_a \delta_{ab} \lambda^a_{jl}$$Comparing these two expressions: $$\lambda^a_{jl} = \frac{N}{n_a} \delta_{jl}$$
Orthogonality Relations
Therefore: $$\sum_{g \in G} D_a(g^{-1})|a,j\rangle\langle b,l|D_b(g) = \frac{N}{n_a}\delta_{ab}\delta_{jl}I_{n_a}$$
Taking matrix elements and Considering irreducible representation unitary: $$\langle a,k|\left(\sum_{g \in G} D_a^{\dagger}(g)|a,j\rangle\langle b,l|D_b(g)\right)|b,m\rangle$$ $$ = \frac{N}{n_a}\delta_{ab}\delta_{jl}\delta_{km}$$
This leads to the orthogonality relation:
$$\frac{n_a}{N}\sum_{g \in G} [D_a(g)]^*_{jk}[D_b(g)]_{lm} = \delta_{ab}\delta_{jl}\delta_{km}$$From the perspective of state vectors or functions of $g$, $\sqrt{\frac{n_a}{N}}[D_a(g)]_{jk}$ forms an orthonormal basis.
Corollary
By counting the number of elements in the complete orthonormal basis (state vectors or functions of $g$), we obtain:
$$N = \sum_i n_i^2$$
where $N$ is the order of $G$, and $n_i$ is the dimension of the irreducible representation $i$.