Grand orthogonality theorem

This article gives the statement, and possibly proof, of a basic fact in linear representation theory.
View a complete list of basic facts in linear representation theory OR View all facts related to linear representation theory

This article describes an orthogonality theorem. View a list of orthogonality theorems

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.

Related facts

• Sum of squares of degrees of irreducible representations equals order of group
• Matrix entries form a basis for the space of all functions

Related facts

• Character orthogonality theorem
• Column orthogonality theorem

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] One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.
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] One of the people editing this page intended to fill in this information at a later stage, but hasn't gotten around to doing it yet. If you see this placeholder for a long time, file an error report at the error reporting page.
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] Since the trace is invariant under conjugation, each $\varphi (g)E_{ij}\varphi (g^{-1})$ has the same trace as $E_{ij}$. Since trace is additive, the average of these also has the same trace.
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 $i\neq j$, 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:

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

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.