# Grand orthogonality theorem

This article gives the statement, and possibly proof, of a basic fact in linear representation theory.
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## Statement

### Statement over complex numbers

Suppose $G$ is a finite group. Let $\mathbb{C}$ denote the field of complex numbers. For each equivalence class of irreducible linear representation of $G$ over $\mathbb{C}$, choose a basis such that the representation is unitary, i.e., the image lies inside $U(n,\mathbb{C})$. Note that this can be done, because linear representation of finite group over complex numbers has invariant Hermitian inner product.

Now, consider the functions from $G$ to $\mathbb{C}$ obtained as the matrix entries for these representations. (There are $n^2$ functions for each representation of degree $n$).

Consider the usual Hermitian inner product on the space of complex-valued functions on $G$:

$\langle f_1, f_2 \rangle = \frac{1}{|G|} \sum_{g \in G} f_1(g) \overline{f_2(g)}$

Then:

1. Any two matrix entry functions are orthogonal with respect to the inner product described above.
2. The inner product of any matrix entry function with itself equals $\frac{1}{n}$ where $n$ is the degree of the representation from which it is picked.

### Statement over general fields

Suppose $G$ is a finite group. Let $k$ be a splitting field for $G$ such that the characteristic of $k$ does not divide the order of $G$. For every equivalence class of irreducible linear representation of $G$ over $k$, choose a basis. Now consider the functions from $G$ to $k$ obtained as the matrix entries of these representations.

Consider the bilinear form for functions on the group:

$\langle f_1, f_2 \rangle_G = \frac{1}{|G|} \sum_{g \in G} f_1(g)f_2(g^{-1})$

Then:

1. Matrix entries for distinct irreducible representations are orthogonal to each other, i.e., the inner product is zero.
2. Matrix entries for the same irreducible representation have inner product $1/n$ iff the entries are transposes of each other, and have inner product zero otherwise.

## Facts used

1. Schur's lemma

## Proof

We first provide the proof using the bilinear form for general splitting fields, then discuss how this can be used to deduce the proof for the Hermitian inner product if we use unitary matrices over the complex numbers.

### Proof of orthogonality of matrix entries from inequivalent irreducible representations

This part of the proof does not require the field to be a splitting field.

Given: A finite group $G$, a field $k$ whose characteristic does not divide the order of $G$. For functions $f_1,f_2: G \to k$ define:

$\langle f_1, f_2 \rangle_G = \frac{1}{|G|} f_1(g)f_2(g^{-1})$

$\varphi_1, \varphi_2$ are inequivalent irreducible representations of $G$ over $k$ of degrees $m,n$ respectively.

Pick $a,i \in \{ 1,2,\dots, m \}$ and $b,j \in \{ 1,2,\dots, n \}$ ($a,i$ are allowed to be equal to each other, $b,j$ are allowed to be equal to each other).

Define:

$f_1(g) := \varphi_1(g)_{ai}, \qquad f_2(g) := \varphi_2(g)_{jb}$

To prove: $\langle f_1, f_2 \rangle_G = 0$

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Denote by $E_{ij}$ the $m \times n$ matrix with a 1 in the $ij^{th}$ entry and 0s elsewhere. -- -- -- --
2 Define $F_{ij} = \frac{1}{|G|} \sum_{g \in G} \varphi_1(g)E_{ij}\varphi_2(g^{-1})$. $F_{ij}$ is a $m \times n$ matrix. $\varphi_1$ has degree $m$, $\varphi_2$ has degree $n$, so the matrix multiplication makes sense. Step (1) --
3 $F_{ij}$ is a homomorphism of representations from $\varphi_2$ to $\varphi_1$ Step (2) PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
4 $F_{ij}$ is the zero matrix. Fact (1) (note that the roles of $\varphi_1, \varphi_2$ are interchanged from the statement of Fact (1)) $\varphi_1, \varphi_2$ are inequivalent irreducible representations. Step (3)
5 The $(ab)^{th}$ entry of $F_{ij}$ is zero. Step (4)
6 The $(ab)^{th}$ entry of $F_{ij}$ equals $\frac{1}{|G|} \sum_{g \in G} \varphi_1(g)_{ai}\varphi_2(g^{-1})_{jb}$. Step (2) Follows by simplifying the matrix multiplication.
7 $\frac{1}{|G|} \sum_{g \in G} \varphi_1(g)_{ai}\varphi_2(g^{-1})_{jb} = 0$, so $\langle f_1, f_2 \rangle_G = 0$. Definitions of $f_1, f_2$, inner product Steps (5), (6) Step-combination direct for first version, use definitions of $f_1, f_2$ for second version.

### Proof of orthogonality and norm of distinct matrix entries from the same representation when the field is algebraically closed

We assume the field to be algebraically closed. After that, we will discuss why the result holds for other splitting fields.

Given: A finite group $G$, an algebraically closed field $k$ whose characteristic does not divide the order of $G$. For functions $f_1,f_2: G \to k$ define:

$\langle f_1, f_2 \rangle_G = \frac{1}{|G|} f_1(g)f_2(g^{-1})$

$\varphi$ is an irreducible representation of degree $n$ and $a,i,j,b$ are all elements of $\{ 1,2,\dots, n\}$ (with the restriction that ).

Define:

$f_1(g) := \varphi(g)_{ai}, \qquad f_2(g) := \varphi(g)_{jb}$

To prove: If $a = b$ and $i = j$ (so $f_1$ and $f_2$ correspond to mutually transpose matrix entries), we have:

$\langle f_1, f_2 \rangle_G = \frac{1}{n}$

Otherwise, we have:

$\langle f_1, f_2 \rangle_G = 0$

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Denote by $E_{ij}$ the $n \times n$ matrix with a 1 in the $ij^{th}$ entry and 0s elsewhere. -- -- -- --
2 Define $F_{ij} = \frac{1}{|G|} \sum_{g \in G} \varphi(g)E_{ij}\varphi(g^{-1})$. $F_{ij}$ is a $n \times n$ matrix. Step (1) --
3 $F_{ij}$ defines a homomorphism of linear representations from $\varphi$ to itself. Step (2) PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]
4 $F_{ij}$ is a scalar matrix. Fact (1) $k$ is algebraically closed, $\varphi$ is irreducible. Step (3) Step-fact combination direct
5 The trace of $F_{ij}$ is the same as the trace of $E_{ij}$. Step (2) [SHOW MORE]
6 If $i = j$, then $F_{ij}$ is the scalar matrix with diagonal entries $1/n$. Otherwise, it is the zero matrix. Steps (4), (5)
7 The $(ab)^{th}$ entry of $F_{ij}$ is $1/n$ if $a = b$ and $i = j$, and is 0 otherwise. Step (6) If $i \ne j$, then $F_{ij}$ is the zero matrix, so all its entries are zero. If $i = j$, then it is a diagonal matrix with diagonal entries $1/n$ and zero elsewhere. The $(ab)^{th}$ entry is thus $1/n$ if $a = b$ and zero otherwise.</toggledisplay>
8 The $(ab)^{th}$ entry of $F_{ij}$ equals $\frac{1}{|G|} \sum_{g \in G} \varphi(g)_{ai}\varphi(g^{-1})_{jb}$. Step (2) Follows by simplifying the matrix multiplication.
9 We have $\frac{1}{|G|} \varphi(g)_{ai}\varphi(g^{-1})_{jb} = 1/n$ if $i = j$ and $a = b$. Also, $\frac{1}{|G|} \varphi(g)_{ii}\varphi(g^{-1})_{jj} = 0$ otherwise. Steps (7), (8)
10 We get the desired conclusion Step (9) Re-interpret Step (9) in terms of $f_1,f_2$.

### Proof for a splitting field

To prove the result for a splitting field, we embed the splitting field in an algebraically closed field, show the result there, and then note that all the inner product values remain the same upon restriction to a subfield containing the matrix entries.

### Proof for unitary matrices

A unitary matrix is a matrix whose inverse is its conjugate-transpose. In particular, if $\varphi(g)$ is unitary, then:

$\varphi(g)^{-1} = \overline{\varphi(g)^T}$

In particular:

$\varphi(g)^{-1}_{jb} = \overline{\varphi(g)}_{bj}$

Thus:

$\varphi(g)_{ai}\varphi(g^{-1})_{jb} = \varphi(g)_{ai}\overline{\varphi(g)}_{bj}$

This equality allows us to go back and forth between the two formulations.