Character orthogonality theorem

This fact is related to: linear representation theory
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This article describes an orthogonality theorem. View a list of orthogonality theorems

Name

This result is known as the first orthogonality theorem, character orthogonality theorem or row orthogonality theorem.

Statement

Statement over complex numbers

Let $G$ be a finite group and $\mathbb{C}$ denote the field of complex numbers. Let $\overline{z}$ denote the complex conjugate of $z$. Then, if ${\rho }_1$ and ${\rho }_2$ are two inequivalent irreducible linear representations, and ${\chi }_1$ and ${\chi }_2$ are their characters, we have:

${\sum }_{g\in G}{\chi }_1(g)\overline{{\chi }_2(g)}=0$

and:

${\sum }_{g\in G}{\chi }_1(g)\overline{{\chi }_1(g)}=\left|G\right|$

Statement over complex numbers in terms of inner product of class functions

For information on the Hermitian inner product, see Inner product of functions#Hermitian inner product over the complex numbers

Consider the space of complex-valued functions $G\to \mathbb{C}$. This is a $\mathbb{C}$-vector space in a natural way, with basis being the indicator functions of elements of $G$. Consider the Hermitian inner product on this vector space given by:

$⟨f_1,f_2⟩=\frac{1}{\left|G\right|}{\sum }_{g\in G}{f}_1(g)\overline{f_2(g)}$

Then, the characters form an orthonormal set of functions with respect to this basis. In other words, if ${\chi }_1,{\chi }_2$ are the characters of inequivalent irreducible representations, we get:

$⟨{\chi }_1,{\chi }_1⟩=1$

and

$⟨{\chi }_1,{\chi }_2⟩=0$

Statement over general fields

Let $G$ be a finite group and $k$ a field whose characteristic does not divide the order of $G$. Let ${\rho }_1$ and ${\rho }_2$ be two inequivalent irreducible linear representations of $G$ over $k$ and let ${\chi }_1$ and ${\chi }_2$ denote their characters. Then, the following are true:

${\sum }_{g\in G}{\chi }_1(g){\chi }_2(g^{-1})=0$

And:

${\sum }_{g\in G}{\chi }_1(g){\chi }_1(g^{-1})=d\left|G\right|$

where $d=1$ if the field $k$ is a splitting field for $G$ (for instance, if $k$ is sufficiently large for $G$, viz., contains all the $m^{th}$ roots of $1$ where $m$ is the exponent of $G$).

When $k$ is not a splitting field, $d$ is the number of irreducible constituents (with multiplicities) of ${\chi }_1$ when taken over a splitting field containing $k$.

Statement over general fields in terms of inner product of class functions

For more on this inner product definition, see Inner product of functions#Bilinear form

For functions $f_1,f_2:G\to k$, define the following inner product:

$⟨f_1,f_2⟩=\frac{1}{\left|G\right|}{\sum }_{g\in G}{f}_1(g)f_2(g^{-1})$

Then, if ${\chi }_1,{\chi }_2$ are the characters of inequivalent irreducible representations, we get:

$⟨{\chi }_1,{\chi }_1⟩=d$

where $d=1$ if $k$ is a splitting field for $G$. In general, $d=d_1^2+d_2^2+\dots +d_r^2$ where $d_1,d_2,\dots ,d_r$ are the multiplicites of pairwise distinct irreducible constituents of ${\chi }_1$. Also:

$⟨{\chi }_1,{\chi }_2⟩=0$

Interpretation in characteristic zero and prime characteristic

In characteristic zero, both sides are being viewed as elements in a field of characteristic zero.

In prime characteristic, however, the inner product is taking values modulo the prime characteristic, hence is not actually an integer, whereas the right side (1, 0, or $d$) is an integer, which needs to be reduced modulo the prime to be interpreted on the other side.

Relation between the Hermitian inner product and the bilinear inner product

See inner product of functions#Relation between the definitions. The upshot is that both inner products are different but it does not matter if the input functions are characters.

Consequences

• Orthogonal projection formula: This is a formula that uses the character of a finite-dimensional representation to determine the irreducible constituents of the representation and their multiplicities.
• Column orthogonality theorem
• Character determines representation in characteristic zero

Facts used

1. Schur's lemma

Proof

The proof is somewhat hard to directly do in its most general case, so we proceed in steps:

• We first handle the case of an algebraically closed field and the bilinear version. (This is where most of the hard work happens)
• We also note that for the complex numbers, the statement for the bilinear version implies the statement for the Hermitian version.
• We then note that since any splitting field embeds inside an algebraically closed field, and the representations do not split further, the result remains valid in any splitting field.
• Finally, we tackle the case of a non-splitting field.

Proof over an algebraically closed field

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:

$⟨f_1,f_2{⟩}_G=\frac{1}{\left|G\right|}{f}_1(g)f_2(g^{-1})$

${\chi }_1,{\chi }_2$ are characters of inequivalent irreducible representations of $G$ over $k$.

To prove: $⟨{\chi }_1,{\chi }_1{⟩}_G=1$ and $⟨{\chi }_1,{\chi }_2{⟩}_G=0$.

Proof: Let ${\phi }_1$ be an irreducible representation with character ${\chi }_1$ and ${\phi }_2$ be an irreducible representation with character ${\chi }_2$. Suppose the degree of ${\phi }_1$ is $m$ and the degree of ${\phi }_2$ is $n$.

We first tackle the case of $⟨{\chi }_1,{\chi }_2{⟩}_G$.

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 $i{j}^{th}$ entry and 0s elsewhere. We allow $1\le i\le m,1\le j\le n$	--	--	--	--
2	Define $F_{ij}=\frac{1}{\left|G\right|}{\sum }_{g\in G}{\phi }_1(g)E_{ij}{\phi }_2(g^{-1})$. $F_{ij}$ is a $m\times n$ matrix.		${\phi }_1$ has degree $m$, ${\phi }_2$ has degree $n$, so the matrix multiplication makes sense.	Step (1)	--
3	$F_{ij}$ is a homomorphism of representations from ${\phi }_2$ to ${\phi }_1$			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 ${\phi }_1,{\phi }_2$ are interchanged from the statement of Fact (1))	${\phi }_1,{\phi }_2$ are inequivalent irreducible representations.	Step (3)
5	The $(ij{)}^{th}$ entry of $F_{ij}$ is zero.			Step (4)
6	The $(ij{)}^{th}$ entry of $F_{ij}$ equals $\frac{1}{\left|G\right|}{\sum }_{g\in G}{\phi }_1(g{)}_{ii}{\phi }_2(g^{-1}{)}_{jj}$.			Step (2)	Follows by simplifying the matrix multiplication.
7	$\frac{1}{\left|G\right|}{\sum }_{g\in G}{\phi }_1(g{)}_{ii}{\phi }_2(g^{-1}{)}_{jj}=0$ for $1\le i\le m,1\le j\le n$.			Steps (5), (6)
8	$\frac{1}{\left|G\right|}{\sum }_{g\in G}{\chi }_1(g){\chi }_2(g^{-1})=0$		${\chi }_1,{\chi }_2$ are the characters of ${\phi }_1,{\phi }_2$ respectively.	Step (7)	Double sum Step (7) for $1\le i\le m$ and $1\le j\le n$. Now use that ${\chi }_1(g)=\sum _{i=1}^m{\phi }_1(g{)}_{ii}$ and ${\chi }_2(g^{-1})=\sum _{j=1}^n{\phi }_2(g^{-1}{)}_{jj}$.