Character orthogonality theorem

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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

Similar facts

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]
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]

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]
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 (ij)^{th} entry of F_{ij} is zero if i \ne j and equals 1/n if i = j. Step (6) [SHOW MORE]
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]

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]