Nontrivial irreducible component of permutation representation of projective general linear group of degree two on projective line

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Let K be a field. Consider the projective general linear group of degree two PGL(2,K). This has a natural action on the projective line over K, i.e., the collection of one-dimensional subspaces of the two-dimensional vector space K^2. We thus get a permutation representation of PGL(2,K) on the projective line \mathbb{P}^1(k).

The action could be described in either of these ways:

  • For any element of PGL(2,K), lift it to an element of GL(2,K), and consider the image of any one-dimensional subspace under the element of GL(2,K). Note that the image subspace does not depend on the choice of the lift, because any two lifts differ multiplicatively by a scalar matrix, which sends every subspace to itself.
  • Think of \mathbb{P}^1(K) as K \cup \{ \infty\}. For an element of PGL(2,K), consider a matrix \begin{pmatrix} a & b \\ c & d \\\end{pmatrix} that is a lift of this element. The permutation induced by this is the map z \mapsto (az + b)/(cz + d), where the value is taken to be \infty if the denominator becomes 0, and the image of \infty is taken to be a/c if c \ne 0 and to be \infty if c = 0.

When K is a finite field of size q, then this gives a permutation action of a finite group PGL(2,q) on a finite set of size q + 1. View this as a linear representation in any characteristic not dividing the order of PGL(2,q). This linear representation splits as a direct sum of a trivial representation and a nontrivial irreducible representation of degree q + 1 - 1 = q. Our goal here is to discuss this irreducible component.

Summary

Item Value
Degree of representation q
Schur index 1 in all characteristics (because the representation, being a direct summand of a permutation representation, can be realized with integer entries and hence interpreted in any characteristic).
Kernel of representation trivial subgroup. In other words, it is a faithful linear representation.
Quotient on which it descends to a faithful representation projective general linear group of degree two
Set of character values \{ q,1,0,-1\}
Characteristic zero: Ring generated: \mathbb{Z} -- ring of integers, Ideal within ring generated: whole ring, Field generated: \mathbb{Q} -- [[field of rational
Ring of realization Realized over any unital ring, by composing the representation over \mathbb{Z} with the map induced by the natural homomorphism from \mathbb{Z} to that ring.
Minimal ring of realization (characteristic zero) \mathbb{Z} -- ring of integers
Minimal ring of realization in characteristic p^k The ring of integers mod p^k, \mathbb{Z}/p^k\mathbb{Z}
Minimal field of realization Prime field in all cases.
In characteristic zero, \mathbb{Q}; in characteristic p, the field \mathbb{F}_p
Size of equivalence class under automorphisms 1
Size of equivalence class under Galois automorphisms 1
Size of equivalence class under action of one-dimensional representations by multiplication 2 if the characteristic of K is not 2, 1 if the characteristic of K is 2.

Particular cases

Field size q Underlying prime p Group PGL(2,q) Order Information on linear representation theory Description of the representation
2 2 symmetric group:S3 6 linear representation theory of symmetric group:S3 standard representation of symmetric group:S3
3 3 symmetric group:S4 24 linear representation theory of symmetric group:S4 standard representation of symmetric group:S4
4 2 alternating group:A5 60 linear representation theory of alternating group:A5 standard representation of alternating group:A5
5 5 symmetric group:S5 120 linear representation theory of symmetric group:S5 one of the five-dimensional irreducible representations

Character

FACTS TO CHECK AGAINST (for characters of irreducible linear representations over a splitting field):
Orthogonality relations: Character orthogonality theorem | Column orthogonality theorem
Separation results (basically says rows independent, columns independent): Splitting implies characters form a basis for space of class functions|Character determines representation in characteristic zero
Numerical facts: Characters are cyclotomic integers | Size-degree-weighted characters are algebraic integers
Character value facts: Irreducible character of degree greater than one takes value zero on some conjugacy class| Conjugacy class of more than average size has character value zero for some irreducible character | Zero-or-scalar lemma

Character values and interpretations

Nature of conjugacy class upstairs in GL(2,q) Eigenvalues Characteristic polynomial Minimal polynomial Size of conjugacy class Number of such conjugacy classes Total number of elements Character value Explanation (character value = number of fixed subspaces - 1)
Diagonalizable over \mathbb{F}_q with equal diagonal entries, hence a scalar \{ a, a \} where a \in \mathbb{F}_q^\ast (x - a)^2 where a \in \mathbb{F}_q^\ast x - a where a \in \mathbb{F}_q^\ast 1 1 1 q Fixes all q + 1 subspaces, so character is q + 1 - 1 = q
Diagonalizable over \mathbb{F}_{q^2}, not over \mathbb{F}_q, eigenvalues are negatives of each other. Pair of mutually negative conjugate elements of \mathbb{F}_{q^2}. All such pairs identified. x^2 - \mu, \mu a nonzero non-square Same as characteristic polynomial q(q - 1)/2 1 q(q - 1)/2 -1 No eigenvalues over \mathbb{F}_q, so no fixed subspaces, so character value is 0 - 1 = -1
Diagonalizable over \mathbb{F}_q with mutually negative diagonal entries. \{ \lambda, - \lambda \}, all such pairs identified. x^2 - \lambda^2, all identified Same as characteristic polynomial q(q + 1)/2 = (q^2 + q)/2 1 q(q + 1)/2 = (q^2 + q)/2 1 Two distinct one-dimensional eigenspaces, so we get 2 - 1 = 1.
Diagonalizable over \mathbb{F}_{q^2}, not over \mathbb{F}_q, eigenvalues are not negatives of each other. Pair of conjugate elements of \mathbb{F}_{q^2}. Each pair identified with anything obtained by multiplying both elements of it by an element of \mathbb{F}_q. x^2 - ax + b, a \ne 0, irreducible; with identification. Same as characteristic polynomial q(q - 1) (q - 1)/2 q(q -1)^2/2 = (q^3 - 2q^2 + q)/2 -1 No eigenspaces, so we get 0 - 1 = -1.
Not diagonal, has Jordan block of size two a \in\mathbb{F}_q^\ast (multiplicity 2). Each conjugacy class has one representative of each type. (x - a)^2 Same as characteristic polynomial q^2 - 1 1 q^2 - 1 0 Unique one-dimensional eigenspace, so we get 1 - 1 = 0
Diagonalizable over \mathbb{F}_q with distinct diagonal entries whose sum is not zero. \lambda, \mu where \lambda,\mu \in \mathbb{F}_q^\ast and \lambda + \mu \ne 0. The pairs \{ \lambda, \mu \} and \{ a\lambda, a\mu \} are identified. x^2 - (\lambda + \mu)x + \lambda\mu, again with identification. Same as characteristic polynomial. q(q + 1) (q - 3)/2 q(q+1)(q - 3)/2 = (q^3 - 2q^2 - 3q)/2 1 Two one-dimensional eigenspaces, so we get 2 - 1 = 1.
Total NA NA NA NA q + 2 q^3 - q