Lecture notes for Introduction to Representation Theory

If ∂ 1 = ⇒ ∂ 2 , let z = xy −1 , then the last term of the summation is
2 2 2
(q + q) ∂ 1 (z)∂ 2 (z) = (q + q) ∂ 1
(z) = (q + q)(q − 1) ∂ 1
(z).
∂ 2 ∂ 2
x=y ∞ x;z=1 ∞ z=1 ∞
Since
∂1
(z) = 0,
×
zF q
because the sum of all roots of unity of a given order m > 1 is zero, the last term becomes
∂ 2
2 2
−(q + q)(q − 1) ∂ 1
(1) = −(q + q)(q − 1).
The difference between this case and the case of ∂ 1 = ∂ 2 is equal to
z=1 ∞
∂ 2
−(q 2 + q)[(q − 2)(q − 1) + (q − 1)] = |G|,
so this is an irreducible representation by Lemma 4.27.
To prove the third assertion of the theorem, we look at the characters on hyperbolic elements
and note that the function
∂ 1 (x)∂ 2 (y) + ∂ 1 (y)∂ 2 (x)
determines ∂ 1 , ∂ 2 up to permutation.
4.24.4 Complementary series representations
Let F q 2 ∩ F q be a quadratic extension F q ( ∀ π), π F q \ F 2
q . We regard this as a 2-dimensional vector
space over F q ; then G is the group of linear transformations of F q 2 over F q . Let K → G be the cyclic
group of multiplications by elements of F × ,
q 2
For λ : K ⊃ C × a homomorphism, let
x πy
K = { }, |K| = q 2 − 1.
y x
Y ξ = Ind G
K C ξ .
This representation, of course, is very reducible. Let us compute its character, using the Mackey
formula. We get
x 0
ν = q(q − 1)λ(x);
0 x
ν(A) = 0 for A parabolic or hyperbolic;
x πy
x πy
x πy
q
ν = λ + λ .
y x y x y x
The last assertion holds because if we regard the matrix as an element of F q 2 , conjugation is an
automorphism of F q 2 over F q , but the only nontrivial automorphism of F q 2 over F q is the q th power
map.
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