Advanced Algebra I
representation of finite groups,II Characters
Let ρ be a 1-dimensional representation of a group G. Then in this case ρ = χ : G → C∗. One sees that χ(st) = χ(s)χ(t) for all s, t ∈ G.
Such character is called an abelian character.
Let ˆG be the set of all 1-dimensional characters, it forms a group under the multiplication χχ0(g) := χ(g)χ0(g).
Exercise 0.1. Let G be an abelian group. Prove that G ∼= ˆG
Recall that a representation ρ : G → GL(V ) is the same as a linear action G × V → V . Suppose now that there are two representation ρ, ρ0 on V, V0 respectively. A linear transformation T : V → V0 is said to be G-invariant if it’s compatible with representations. That is,
T ρs(v) = ρ0s(T v), for all v ∈ V .
Thus an isomorphism of representation is nothing but a G-invariant bijective linear transformation.
Exercise 0.2. It’s easy to check that if T : V → V0 is G-invariant, then the ker(T ) ⊂ V and im(T ) ⊂ V0 are G-invariant subspaces.
Theorem 0.3 (Schur’s Lemma). Let ρ, ρ0 be two irreducible represen- tation of G on V, V0 respectively. And let T : V → V0 be a G-invariant linear transformation. Then
(1) Either T is an isomorphism or T = 0.
(2) If V = V0,ρ = ρ0, then T is multiplication by a scalar.
Proof. (1) Since ker(T ) is a G-invariant subspace and V is irre- ducible. One has that either ker(T ) = 0 or ker(T ) = V . Hence T is injective or T = 0. If T is injective, by looking at im(T ), One must have im(T ) = V0. Therefore T is an isomorphism.
(2) Let λ be an eigenvalue of T . One sees that T1 := T − λI is also an G-invariant linear transformation. Since ker(T1) is non-zero, one has that ker(T1) = V . Thus T1 = 0 and hence T = λI.
¤ Suppose one has T : V → V0 not necessarily G-invariant. One can produce an G-invariant linear transformation by the ”averaging process”. For T (v) = s−1T (sv), we set
T (v) :=˜ 1 g
X
s∈G
s−1T (sv).
And it’s easy to check that this is G-invariant.
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2
proof of the main theorem. (1) Let ρ, ρ0 be two irreducible repre- sentation of G on V, V0 with character χ, χ0 respectively.
Let T : V → V0 be any linear transformation. One can produce a G-invariant transformation ˜T .
Suppose first that ρ and ρ0 are not isomorphic. Then by Schur’s Lemma, ˜T = 0 for all T .
We fix bases of V, V0 and write everything in terms of matri- ces.
0 = ( ˜T )ij =X
t,k,l
(R0t−1)ik(T )kl(Rt)lj.
Take T = Eij, then one has 0 =X
t,k,l
(R0t−1)ik(Eij)kl(Rt)lj =X
t
(R0t−1)ii(Rt)jj.
Hence
< χ0, χ >=X
t,i,j
(R0t−1)ii(Rt)jj = 0.
Suppose now that ρ = ρ0, χ = χ0. The averaging process and Schur’s Lemma gives
λI = ˜T = 1 g
X
t
Rt−1T Rt.
One notice that λd = tr( ˜T ) = tr(T ).
Now we set T = Eii, then 1
d = (λI)ii = 1 g
X
t
(Rt−1)ik(Eii)kl(Rt)li =X
t
(Rt−1)ii(Rt)ii.
It follows that
< χ, χ >=X
t
X
i
(Rt−1)ii(Rt)ii =X
i
1 d = 1.
(2) A class function f on a group G is a complex value function such that f (s) = f (t) if s and t are conjugate. The space C of class function is clearly a vector space of dimension r, where r denotes the number of conjugacy classes of G. We claim that the set of character of irreducible representation form a orthonormal basis of C.
We remark that inner product can be defined on any class function.
Suppose now that φ is a class function which is orthogonal to every χi. For any character χ of an irreducible representation ρ, we can produce a linear transformation by averaging process T := 1gP
tφ(t)ρt. It’s clear that tr(T ) =< φ, χ >= 0. One sees that T : V → V is G-invariant. By Schur’s Lemma, T = λI.
But T r(T ) = 0. Thus T = 0 for any character χ.
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We apply to the regular representation ρ : G → C[G], 0 = T (e1) = 1
g X
t
φ(t)ρt(e1) = 1 g
X
t
φ(t)et.
Since et forms a basis for C[G], it follows that φ(t) = 0 for all t ∈ G and hence φ = 0.
(3) We may assume that there are r irreducible representation. And suppose that the regular representation ρ is decomposed into n1ρ1⊕ ... ⊕ nrρr. One notice that ρ(1) = g and ρ(t) = 0 for all t 6= 1. By direct computation,
di =< χρ, χi >= ni, g =< χρ, χρ>=X
i
d2i.
To prove that di|g need some extra work on the group algebra C[G] which we will do later.
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