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 $\varphi_1$ and $\varphi_2$ 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:

$\sum_{g \in G} \chi_1(g)\overline{\chi_1(g)} = |G|$

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:

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

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:

$\langle \chi_1, \chi_1 \rangle = 1$

and

$\langle \chi_1, \chi_2 \rangle = 0$

Statement over general fields

Let $G$ be a finite group and $k$ a field whose characteristic does not divide the order of $G$. Let $\varphi_1$ and $\varphi_2$ 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:

$\sum_{g \in G} \chi_1(g)\chi_1(g^{-1}) = d|G|$

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:

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

Then, if $\chi_1, \chi_2$ are the characters of inequivalent irreducible representations, we get:

$\langle \chi_1, \chi_1 \rangle = d$

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:

$\langle \chi_1, \chi_2 \rangle = 0$

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.

Related facts

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

Similar facts

• Grand orthogonality theorem is a stronger version that deals with matrix entries rather than traces.

Facts used

1. Schur's lemma
2. Trace of inverse is complex conjugate of trace

Proof

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

• Hard step: We first prove the orthogonality of distinct irreducible characters with the bilinear version.
• Hard step: Next, we handle the normality in case of an algebraically closed field and the bilinear version.
• 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 of orthogonality of distinct irreducible characters

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:

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

$\chi_1,\chi_2$ are characters of inequivalent irreducible representations $\varphi_1, \varphi_2$ of $G$ over $k$. The degree of $\varphi_1$ is $m$ and the degree of $\varphi_2$ is $n$.

To prove: $\langle \chi_1, \chi_2 \rangle_G = 0$.

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. We allow $1 \le i \le m, 1 \le j \le n$	--	--	--	--
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 $(ij)^{th}$ entry of $F_{ij}$ is zero.			Step (4)
6	The $(ij)^{th}$ entry of $F_{ij}$ equals $\frac{1}{|G|} \sum_{g \in G} \varphi_1(g)_{ii}\varphi_2(g^{-1})_{jj}$.			Step (2)	Follows by simplifying the matrix multiplication.
7	$\frac{1}{|G|} \sum_{g \in G} \varphi_1(g)_{ii}\varphi_2(g^{-1})_{jj} = 0$ for $1 \le i \le m, 1 \le j \le n$.			Steps (5), (6)
8	$\frac{1}{|G|} \sum_{g \in G} \chi_1(g)\chi_2(g^{-1}) = 0$		$\chi_1, \chi_2$ are the characters of $\varphi_1,\varphi_2$ respectively.	Step (7)	[SHOW MORE] Double sum Step (7) for $1 \le i \le m$ and $1 \le j \le n$. Now use that $\chi_1(g) = \sum_{i=1}^m \varphi_1(g)_{ii}$ and $\chi_2(g^{-1}) = \sum_{j=1}^n \varphi_2(g^{-1})_{jj}$.

Proof of normality of characters when the field is algebraically closed

For simplicity, we drop the subscript $1$ when referring to the character and simply call it $\chi$.

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:

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

$\chi$ is the character of an irreducible linear representation $\varphi$ of $G$ over $k$. Suppose the degree of $\varphi$ is $n$.

To prove: $\langle \chi, \chi\rangle_G = 1$.

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. We allow $1 \le i \le n, 1 \le j \le n$	--	--	--	--
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 $(ij)^{th}$ entry of $F_{ij}$ is zero if $i \ne j$ and equals $1/n$ if $i = j$.			Step (6)	[SHOW MORE] 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 $1/n$. In particular, the $(ij)^{th}$ entry, being a diagonal entry, is $1/n$.
8	The $(ij)^{th}$ entry of $F_{ij}$ equals $\frac{1}{|G|} \sum_{g \in G} \varphi(g)_{ii}\varphi(g^{-1})_{jj}$.			Step (2)	Follows by simplifying the matrix multiplication.
9	We have $\frac{1}{|G|} \varphi(g)_{ii}\varphi(g^{-1})_{jj} = 1/n$ if $i = j$ and $\frac{1}{|G|} \varphi(g)_{ii}\varphi(g^{-1})_{jj} = 0$ if $i \ne j$			Steps (7), (8)	Step-combination direct
10	$\frac{1}{|G|} \sum_{g \in G} \chi(g)\chi(g^{-1}) = 1$		$\chi$ is the character of $\varphi$.	Step (9)	[SHOW MORE] Double sum Step (9) for $1 \le i \le n$ and $1 \le j \le n$. Now use that $\chi(g) = \sum_{i=1}^m \varphi(g)_{ii}$ and $\chi(g^{-1}) = \sum_{j=1}^n \varphi(g^{-1})_{jj}$.

For the Hermitian inner product over the complex numbers

The Hermitian inner product coincides with the bilinear form when both inputs are characters. See trace of inverse is complex conjugate of trace.

Case of a splitting field

Suppose $k$ is a splitting field for $G$ but is not necessarily algebraically closed. Consider the algebraic closure $\overline{k}$. Because $k$ is a splitting field for $G$, all the irreducible representations over $\overline{k}$ are in fact realized inside $k$, and in particular any irreducible representation of $k$ remains irreducible over $\overline{k}$.

We can thus prove the statement about normality over $\overline{k}$, and then note that the inner product remains the same when we restrict to a subfield.

Case of a non-splitting field

The orthogonality result continues to hold (because the proof does not use any assumptions about the field).

We need to show the analogue of normality. This basically just follows by writing the decomposition over a splitting field and then using the bilinearity of the bilinear form and simplifying. 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.