Orthonormality of unitary irreducible representations

Let $Γ 𝛼 𝑗 𝑘 \in ℂ [𝐺]$ be concrete realizations of irreps $Γ 𝛼 : 𝐺 \to ℂ 𝑑 𝛼$ for each $𝛼 \in ˆ 𝐺$ . Then ${\sqrt 𝑑 𝛼 Γ 𝛼 𝑖 𝑗}$ with $𝛼 \in ˆ 𝐺$ , $1 \leq 𝑖, 𝑗 \leq 𝑑 𝛼$ form an Orthonormal basis of the Group ring under a certain inner product.¹ rep In particular

(Γ 𝛼 𝑗 𝑘 | Γ 𝛽 𝑗' 𝑘') = 1 | 𝐺 | \sum 𝑔 \in 𝐺 ―――― Γ 𝛼 𝑗 𝑘 (𝑔) Γ 𝛽 𝑗' 𝑘' (𝑔) = 1 𝑑 𝛼 𝛿 𝛼 𝛽 𝛿 𝑗 𝑗' 𝛿 𝑘 𝑘'

Should be changed

I think its more productive to view these as elements of $ℂ 𝐺$

Proof of orthonormality

Let $𝐀 : ℂ 𝑑 𝛼 \to ℂ 𝑑 𝛽$ be an arbitrary linear map. Define a linear map
$˜ 𝐀 = 1 | 𝐺 | \sum 𝑔 \in 𝐺 Γ 𝛽 (𝑔) 𝐀 Γ 𝛼 (𝑔) - 1 = 1 | 𝐺 | \sum 𝑔 \in 𝐺 Γ 𝛽 (𝑔) 𝐀 ―――― Γ 𝛼 (𝑔)$
It follows that for any $ℎ \in 𝐺$
$Γ 𝛽 (ℎ) ˜ 𝐀 = 1 | 𝐺 | \sum 𝑔 \in 𝐺 Γ 𝛽 (ℎ 𝑔) 𝐀 Γ 𝛼 (𝑔) - 1 = 1 | 𝐺 | \sum 𝑔 \in 𝐺 Γ 𝛽 (𝑔) 𝐀 Γ 𝛼 (ℎ - 1 𝑔) - 1 = 1 | 𝐺 | \sum 𝑔 \in 𝐺 Γ 𝛽 (𝑔) 𝐀 Γ 𝛼 (𝑔 - 1 ℎ) = 1 | 𝐺 | \sum 𝑔 \in 𝐺 Γ 𝛽 (𝑔) 𝐀 Γ 𝛼 (𝑔) - 1 Γ 𝛼 (ℎ) = ˜ 𝐀 Γ 𝛼 (ℎ)$
so $˜ 𝐀$ is an intertwining operator and therefore by Schur’s lemma either $˜ 𝐀 = 𝐎$ or $𝛼 = 𝛽$ , so $𝐀 = ˜ 𝐀 = 𝑐 𝐈$ with $𝑐 = t r 𝐀 / 𝑑 𝛼$ . Combining these two possibilities gives
$˜ 𝐀 = 1 𝑑 𝛼 t r (𝐀) 𝛿 𝛼 𝛽 𝐈$
If we chose $𝐀$ to $𝐴 𝑝 𝑞 = 𝛿 𝑝 𝑘' 𝛿 𝑞 𝑘$ then $t r 𝐴 = 𝛿 𝑘 𝑘'$ , thus
$˜ 𝐴 𝑗 𝑗' = 1 𝑑 𝛼 𝛿 𝛼 𝛽 𝛿 𝑗 𝑗' 𝛿 𝑘 𝑘' = 1 | 𝐺 | \sum 𝑔 \in 𝐺 Γ 𝛽 (𝑔) 𝐀 ―――― Γ 𝛼 (𝑔) = 1 | 𝐺 | \sum 𝑔 \in 𝐺 Γ 𝛽 𝑗 𝑘 (𝑔) ――――― Γ 𝛼 𝑗' 𝑘' (𝑔) = (Γ 𝛼 𝑗 𝑘 | Γ 𝛽 𝑗' 𝑘')$
thus the matrix elements fulfil the orthonormality condition.

Proof of spanning set

Let $𝐴 = s p a n {\sqrt 𝑑 𝛼 Γ 𝛼 𝑗 𝑘}$ , so we must prove $𝐴 = ℂ [𝐺]$ . The tensor product of irreps is reducible, which for concrete reälizations may be stated as
$Γ 𝛼 𝑖 𝑗 (𝑔) Γ 𝛽 𝑘 𝑙 (𝑔) = \sum 𝜇, 𝑝, 𝑞 𝑇 𝜇 𝑗 𝑝 Γ 𝛼 𝜇 𝑝 𝑞 (𝑦) ――― 𝑇 𝜇 𝑘 𝑞$
for some numbers $𝑇 𝜇 𝑗 𝑝$ . Thus the point-wise product of basic functions is in $𝐴$ , and by distributivity $𝐴$ is closed under point-wise multiplication. Since Irreps collectively distinguish group elements, it follows from Finite version that $𝐴 = ℂ [𝐺]$ . From this and above, ${\sqrt 𝑑 𝛼 Γ 𝛼 𝑖 𝑗}$ is an orthonormal basis under $(\cdot | \cdot)$ .

Since the number of basis elements equals the dimension of the vector space, it follows that Square sum of irrep dimensions is given by

\sum 𝛾 \in ˆ 𝐺 (𝑑 𝛾) 2 = | 𝐺 |

tidy | en | SemBr

1996, Representations of finite and compact groups, §III.1 ↩

Orthonormality of unitary irreducible representations

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Orthonormality of unitary irreducible representations

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