Conjugacy classes in GL2Fq

5 Pages·2015·0.084 MB·English

by  Sam Evens

CONJUGACY CLASSES IN GL(2,F )
q
SAM EVENS
1. discussion
Let F = F be a finite field with q = pr elements. We describe the conjugacy classes
q
in G = GL(2,F).
Let A M(2,F) be a 2 2 matrix with entries in F. The choice of A makes V = F2
∈ ×
into a module over the polynomial ring F[t]. Indeed, for p(t) F[t] and v V, we
∈ ∈
set p(t) v = p(A) v, where for p(t) = a ti, p(A) = a Ai. This makes V into
i i
· ·
a F[t]-module. Since V is a finite dimePnsional F-moduPle, V is certainly a finitely
generated F[t]-module. By the classification of finitely generated modules over the
PID F[t], it follows that there exist irreducible monomials p (t) and positive integers e
i i
such that V = ⊕ji=1F[t]/(pi(t)ei) for some j. Since dim(F[t]/(pi(t)ei)) = eideg(pi), it
follows that the dimension of V as a F-vector space of dimension e deg(p ). Hence,
i i
2 = eideg(pi). It follows that there are three cases. P
P
Remark 1.1. Case (1) j = 2, deg(p ) = deg(p ) = 1 and e = e = 1,
1 2 1 2
Case (2) j = 1, deg(p ) = 1 and e = 2, or
1 1
Case (3) j = 1, deg(p ) = 2 and e = 1.
1 1
Incase(1),ifdeg(p ) = 1,thenp (t) = (t a )forsomea F. Thenifv = 1+(t a )
i i i i i i
− ∈ −
in F[t]/(t a ), then A v = t v = a +t a +(t a ) = a v , so A has eigenvalue
i i i i i i i i
− · · − − ·
a on V. With respect to the basis v ,v , A has matrix
i 1 2
a 0
1
(cid:18)0 a2(cid:19)
In case (2), p (t) = t a with a F as above, and if v denote the equivalence class
1 1
− ∈
t a + ((t a )2) F[t]/(t a)2, then (A a) v = (t a) v = 0, so v is an
1 1 1 1
− − ∈ − − · − ·
eigenvector with eigenvalue a. If we let v = 1+(t a)2, then (A t) v = v . Hence,
2 2 1
− − ·
the matrix of A with respect to the basis v ,v is
1 2
a 1
(cid:18)0 a(cid:19)
Note that in cases (1) and (2), the matrix A has an eigenvalue on V. In case (3),
the matrix A has no eigenvalues on V. Indeed, since p (t) is degree 2 and irreducible,
1
a nonzero vector v of V = F[t]/(p (t)) is a class a(t)+(p (t)) where a(t) is a nonzero
1 1
1
2 SAMEVENS
polynomial of degree 0 or 1. Suppose A v = cv for some c F. Then t a(t)+(p (t)) =
1
· ∈ ·
c a(t)+(p (t)) in V. It follows that p (t) divides (t c) a(t) in F[t]. Since in a PID,
1 1
· − ·
irreducible elements are prime, it follows that p (t) divides t c or a(t), which is
1
−
impossible since deg(p (t)) = 2 > 1 deg(t c) and the same for a(t). Hence, A
1
≥ −
cannot have an eigenvalue on V.
In case (3), let v V be a nonzero vector. Since A has no eigenvalue on V, the
∈
vector A v is not in F v, so v,A v are linearly independent. We let v = v and
1
· · { · }
v = A v. By the Cayley-Hamilton theorem, A2 Tr(A)A + Det(A) = 0. Hence,
2
· −
A v = A2 v = Det(A)v +Tr(A)v . It follows that the matrix of A with respect
2 1 1 2
· · −
to this basis is
0 det(A)
−
(cid:18)1 trace(A)(cid:19)
We discuss a way to use field extensions to construct matrices with no eigenvalues
on V. Let L/F be a quadratic extension, so the field L has q2 elements. Then L is a
F-vector space with dim (L) = 2, so if we choose a F-basis v ,v of L, we may identify
F 1 2
L with F2. For α L, define T : L L by T (x) = α x. Since L is commutative and
α α
∈ → ·
F L, it follows that T is a F-linear map from L to L, and hence gives an element
α
⊂
of M(2,F). If α L×, then since Tα Tα−1 = T1 = id, Tα is invertible so gives an
∈ ◦
element of G. If α F, then T = αid is a scalar matrix.
α
∈
Remark 1.2. Suppose α L F. Since F is finite, L/F is Galois, and we let
∈ −
Gal(L/F) = id,σ . We let trace(α) = α + σ(α) and norm(α) = α σ(α), and
{ } ·
note that since trace(α) and norm(α) are fixed by the Galois group, they are in F. Let
m (x) be the minimal polynomial of α over F, which is necessarily of degree 2 since
α
[F(α) : F] = 1 and [F(α) : F] divides [L : F] = 2. By standard arguments, σ(α) is
6
a root of m (x), and m (x) = (x α)(x σ(α)) = x2 trace(α)x+norm(α). Thus,
α α
− − −
α2 = trace(α)α norm(α). Since α F, the set 1,α is a F-basis of L. Using the
− 6∈ { }
above quadratic identity for α, we see that the matrix of T with respect to the above
α
basis is
0 norm(α)
dα = (cid:18)1 −trace(α) (cid:19).
We conclude that Tr(T ) = trace(α) and Det(T ) = norm(α). Further, since L =
α α
F(α) = F[t]/(m (t)). Since m (t) is irreducible over F, and T acts on F[t]/(m (t))
α α α α
it follows from the analysis of case (3) above that T has no eigenvalues on L = F2.
α
Finally, a calculation shows that the characteristic polynomial of d is m (t) = (t
α α
−
α)(t σ(α)), it follows that d has eigenvalues α and σ(α) in L.
α
−
For g G, let C := xgx−1 : x G be the conjugacy class of g.
g
∈ { ∈ }
Proposition 1.3. The conjugacy classes in G = GL(2,F) have representatives:
(1a) For x F×, let
∈
x 0
a =
x (cid:18)0 x(cid:19)
CONJUGACY CLASSES IN GL(2,Fq) 3
Each conjugacy class C = a , and C = C if and only if x = y. In particular,
ax { x} ax ay
there are q 1 distinct conjugacy classes of type C .
− ax
(2) For x F×, let
∈
x 1
b =
x (cid:18)0 x(cid:19)
The conjugacy class C has q2 1 elements, and C = C if and only if x = y. In
bx − bx by
particular, there are q 1 distinct conjugacy classes of type C .
− bx
(1b) For x,y F× with x = y, let
∈ 6
x 0
c =
x,y (cid:18)0 y(cid:19)
The conjugacy class c has q2 +q elements and C = C if and only if (u,v) =
x,y cx,y cu,v
(x,y) or (u,v) = (y,x).
(3) Let L/F be a degree 2 field extension and let α L F, and recall trace(α) and
∈ −
norm(α) from Remark 1.2. We consider the matrix
0 norm(α)
dα = (cid:18)1 −trace(α) (cid:19)
The conjugacy class c has q2 q elements, and for α,β L F, c = c if and
dα − ∈ − dα dβ
only if β = α or β = σ(α). In particular, there are (q2 q)/2 distinct conjugacy classes
−
of type c .
dα
Proof. In the cases (1), (2), (3) discussed in Remark 1.1, we found a basis of V = F2
such that g can be put into one of the forms stated in the Proposition. By linear
algebra, we know that if A is a matrix which is written as B with respect to a basis,
then A is conjugate to B via the change of basis matrix. It follows that g can be put
into one of the discussed forms, and it remains to check the assertions concerning the
number of elements in each class and equalities between different classes.
In case (1), the element g has two eigenvalues a and a . If a = a = x, then g is
1 2 1 2
multiplication by x, so g = a . Since a is in the center of G, the remaining assertions
x x
are clear. If x = a = a = y, then g is conjugate to c . Let D be the collection of all
1 2 x,y
6
matrices of the form
a 0
.
(cid:18)0 d(cid:19)
An easy calculation shows that the centralizer C (c ) = D. Since D = (q 1)2,
G x,y
| | −
it follows that the cardinality C = |G| = q2 +q. Recall that if two matrices are
| cx,y| |D|
conjugate, theyhavethesameeigenvalues, countedwithmultiplicity. Hence, c isnot
x,y
conjugate to cx′,y′ unless the sets x,y and x′,y′ coincide. Let w be the nontrivial
{ } { }
permutation matrix
0 1
(cid:18)1 0(cid:19)
4 SAMEVENS
By an easy calculaton, wc w−1 = c , and this takes care of case (1b) above.
x,y y,x
In case (2), the assertions follow from since conjugate matrices have the same eigen-
values, and the fact that the centralizer of b is the collection of matrices of the form
x
u v
(cid:18)0 u(cid:19)
so there are q2 q elements in the centralizer.
−
In case (3), suppose d and d are conjugate in G. Then they are conjugate in
α β
GL(2,L), so d and d have the same eigenvalues over L. Thus, the sets α,σ(α) and
α β
{ }
β,σ(β) coincide, so β = α or β = σ(α). It remains to compute the centralizer of d .
α
{ }
Defineφ : L× GL(2,F)bysendingα LtothematrixofT withrespecttothebasis
α
→ ∈
1,α of L. It follows from definitions that φ is an injective group homomorphism, so
{ }
thatφ(L×) C (d ). Theremainingassertionfollowsonceweknowφ(L×) = C (d ).
G α G α
⊂
This can be verified by a counting argument; if for some α L F, C (d ) >
G α
∈ − | |
q2 1 = L× , then the conjugacy class C would have fewer than q2 q elements.
− | | dα −
Then a sum over all conjugacy classes of G would have fewer than G elements, by
| |
a computation using the classification of conjugacy classes and the previously known
cases. Alternatively, consider a matrix A of the form
0 x
(cid:18)1 y(cid:19)
with x = 0. Consider a matrix B written as
6
a b
.
(cid:18)c d(cid:19)
The condition that AB = BA implies the following four equations:
(1)b = xc; (2)ax+by = xd; (3)a+yc = d; (4)b+yd = cx+dy.
Equation (4) follows from Equation (1) and may be omitted. Since x = 0, c is de-
6
termined by b, and d is determined by a and b. Thus, the number of B GL(2,F)
∈
centralizing A is bounded by the number of choices of a and b subject to the require-
ment that a and b cannot both be 0. This shows that C (d ) q2 1, and proves
G α
| | ≤ −
the claim. A more conceptual proof is discussed in the next remark. (cid:3)
Remark 1.4. The best way to prove that C (d ) = φ(L×) is to use known facts from
G α
the theory of central simple algebras. By definition, a finite dimensional F-algebra A
is called a central simple algebra if the center of A is F, and A has no proper two-
sided ideals (i.e., A is a simple ring). It is an easy exercise to show that M(2,F)
is a central simple F-algebra. According to Chapter IV (Brauer group) of Milne’s
notes on class field theory (see http://www.jmilne.org/math/CourseNotes/cft.html),
dim (A) = n2 for some n, and if E is a subring of A that is a field and dim (E) = n,
F F
then the centralizer of E in A is E (see Corollary 3.4 of Milne). It follows that if we
regard our field L as a subring of M(2,F), then the centralizer of L in GL(2,F) is
CONJUGACY CLASSES IN GL(2,Fq) 5
L GL(2,F) = L×. But for α L F, L = F +Fα, so the centralizer of L in G is
∩ ∈ −
the centralizer of α in G. Since d is the matrix of α, it follows that C (d ) = L×.
α G α
Remark 1.5. In Fulton and Harris, the authors describe the conjugacy classes in G =
GL(2,F) under the assumption that the characteristic p of F is odd. In this case, by
the quadratic formula, L = F[√ǫ] for some element ǫ of F that is not a square. Their
description is the same as ours except that they replace the conjugacy classes of d
α
with classes of d , where d is the matrix
x,y x,y
x yǫ
,
(cid:18)y x(cid:19)
with x and y not both zero. We note that d is the matrix of multiplication by
x,y
α = x+y√ǫ with respect to the basis 1,√ǫ of L over F. For α L F, y = 0,
{ } ∈ − 6
and it is not difficult to see that d has eigenvalues α and x y√ǫ, which are not in
x,y
−
F. We see that Tr(d ) = 2x and Det(d ) = x2 y2ǫ. Hence, as in the discussion of
x,y x,y
−
case (3) in Remark 1.2, the matrix of d with respect to a suitable basis is
x,y
0 (x2 y2ǫ)
dα = (cid:18)1 − 2−x (cid:19).
Hence, d is conjugate to d , and if we replace d with d , we see that d is
x,y α x,y x,−y x,−y
conjugate to d also. This shows that the discussion in Fulton and Harris when p is
α
odd is equivalent to our discussion.
Department of Mathematics, University of Notre Dame, Notre Dame, IN 46556
E-mail address: [email protected]