Lecture notes for Introduction to Representation Theory

⇒
the expression evaluates to
since here
If
the expression evaluates to
since here
∂(g) · 1,
aga −1 B ≥ a B.
x 0
g = ,
0 y
⎩
∂1 (x)∂ 2 (y) + ∂ 1 (y)∂ 2 (x) · 1,
If
aga −1 B ≥ a B or a is an element of B multiplied by the transposition matrix.
x πy
g = , x = y
y x
the expression on the right evaluates to 0 because matrices of this type don’t have eigenvalues over
F q (and thus cannot be conjugated into B). From the definition, ∂ i (x)(i = 1, 2) is a root of unity,
so
|G|◦ν V1 , 2
, ν V1 , 2
= (q + 1) 2 (q − 1) + (q 2 − 1)(q − 1)
The last two summands come from the expansion
2 2
+ 2(q + q) (q − 1)(q − 2) + (q + q) ∂ 1 (x)∂ 2 (y)∂ 1 (y)∂ 2 (x).
2
|a + b| 2 = |a| 2 + |b| 2 + ab + ab.
x=y ∞
If
the last term is equal to
and the total in this case is
∂ 1 = ∂ 2 = µ,
(q 2 + q)(q − 2)(q − 1),
so
Clearly,
since
(q + 1)(q − 1)[(q + 1) + (q − 1) + 2q(q − 2)] = (q + 1)(q − 1)2q(q − 1) = 2|G|,
◦ν V1 , 2
, ν V1 , 2
= 2.
C µ ∧ Ind G
B C µ,µ ,
Hom G (C µ , Ind G BC µ,µ ) = Hom B (C µ , C µ ) = C (Theorem 4.33).
Therefore, Ind G
B C µ,µ = C µ W µ ; W µ is irreducible; and the character of W µ is different for distinct
values of µ, proving that W µ are distinct.
72