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
Contents
Name
This result is known as the first orthogonality theorem, character orthogonality theorem or row orthogonality theorem.
Statement
Statement over complex numbers
Let be a finite group and denote the field of complex numbers. Let denote the complex conjugate of . Then, if and are two inequivalent irreducible linear representations, and and are their characters, we have:
and:
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 . This is a -vector space in a natural way, with basis being the indicator functions of elements of . Consider the Hermitian inner product on this vector space given by:
Then, the characters form an orthonormal set of functions with respect to this basis. In other words, if are the characters of inequivalent irreducible representations, we get:
and
Statement over general fields
Let be a finite group and a field whose characteristic does not divide the order of . Let and be two inequivalent irreducible linear representations of over and let and denote their characters. Then, the following are true:
And:
where if the field is a splitting field for (for instance, if is sufficiently large for , viz., contains all the roots of where is the exponent of ).
When is not a splitting field, is the number of irreducible constituents (with multiplicities) of when taken over a splitting field containing .
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 , define the following inner product:
Then, if are the characters of inequivalent irreducible representations, we get:
where if is a splitting field for . In general, where are the multiplicites of pairwise distinct irreducible constituents of . Also:
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 ) 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
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 , a field whose characteristic does not divide the order of . For functions define:
are characters of inequivalent irreducible representations of over . The degree of is and the degree of is .
To prove: .
Proof:
Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | Denote by the matrix with a 1 in the entry and 0s elsewhere. We allow | -- | -- | -- | -- |
2 | Define . is a matrix. | has degree , has degree , so the matrix multiplication makes sense. | Step (1) | -- | |
3 | is a homomorphism of representations from to | Step (2) | PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] | ||
4 | is the zero matrix. | Fact (1) (note that the roles of are interchanged from the statement of Fact (1)) | are inequivalent irreducible representations. | Step (3) | |
5 | The entry of is zero. | Step (4) | |||
6 | The entry of equals . | Step (2) | Follows by simplifying the matrix multiplication. | ||
7 | for . | Steps (5), (6) | |||
8 | are the characters of respectively. | Step (7) | [SHOW MORE] |
Proof of normality of characters when the field is algebraically closed
For simplicity, we drop the subscript when referring to the character and simply call it .
Given: A finite group , an algebraically closed field whose characteristic does not divide the order of . For functions define:
is the character of an irreducible linear representation of over . Suppose the degree of is .
To prove: .
Proof:
Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explanation |
---|---|---|---|---|---|
1 | Denote by the matrix with a 1 in the entry and 0s elsewhere. We allow | -- | -- | -- | -- |
2 | Define . is a matrix. | Step (1) | -- | ||
3 | defines a homomorphism of linear representations from to itself. | Step (2) | PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE] | ||
4 | is a scalar matrix. | Fact (1) | is algebraically closed, is irreducible. | Step (3) | Step-fact combination direct |
5 | The trace of is the same as the trace of . | Step (2) | [SHOW MORE] | ||
6 | If , then is the scalar matrix with diagonal entries . Otherwise, it is the zero matrix. | Steps (4), (5) | |||
7 | The entry of is zero if and equals if . | Step (6) | [SHOW MORE] | ||
8 | The entry of equals . | Step (2) | Follows by simplifying the matrix multiplication. | ||
9 | We have if and if | Steps (7), (8) | Step-combination direct | ||
10 | is the character of . | 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 is a splitting field for but is not necessarily algebraically closed. Consider the algebraic closure . Because is a splitting field for , all the irreducible representations over are in fact realized inside , and in particular any irreducible representation of remains irreducible over .
We can thus prove the statement about normality over , 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]