6Proof of orthogonality
II Representation Theory
6 Proof of orthogonality
We will do the proof in parts.
Theorem
(Row orthogonality relations)
.
If
ρ
:
G → GL
(
V
) and
ρ
0
:
G →
GL
(
V
0
) are two complex irreducible representations affording characters
χ, χ
0
respectively, then
hχ, χ
0
i =
(
1 if ρ and ρ
0
are isomorphic representations
0 otherwise
.
Proof.
We fix a basis of
V
and of
V
0
. Write
R
(
g
)
, R
0
(
g
) for the matrices of
ρ
(
g
)
and ρ
0
(g) with respect to these bases respectively. Then by definition, we have
hχ
0
, χi =
1
|G|
X
g∈G
χ
0
(g
−1
)χ(g)
=
1
|G|
X
g∈G
X
1≤i≤n
0
1≤j≤n
R
0
(g
−1
)
ii
R(g)
jj
.
For any linear map
ϕ
:
V → V
0
, we define a new map by averaging by
ρ
0
and
ρ
.
˜ϕ : V → V
0
v 7→
1
|G|
X
ρ
0
(g
−1
)ϕρ(g)v
We first check ˜ϕ is a G-homomorphism — if h ∈ G, we need to show
ρ
0
(h
−1
) ˜ϕρ(h)(v) = ˜ϕ(v).
We have
ρ
0
(h
−1
) ˜ϕρ(h)(v) =
1
|G|
X
g∈G
ρ
0
((gh)
−1
)ϕρ(gh)v
=
1
|G|
X
g
0
∈G
ρ
0
(g
0−1
)ϕρ(g
0
)v
= ˜ϕ(v).
(i)
Now we first consider the case where
ρ, ρ
0
is not isomorphic. Then by
Schur’s lemma, we must have ˜ϕ = 0 for any linear ϕ : V → V
0
.
We now pick a very nice
ϕ
, where everything disappears. We let
ϕ
=
ε
αβ
,
the operator having matrix
E
αβ
with entries 0 everywhere except 1 in the
(α, β) position.
Then ˜ε
αβ
= 0. So for each i, j, we have
1
|G|
X
g∈G
(R
0
(g
−1
)E
αβ
R(g))
ij
= 0.
Using our choice of ε
αβ
, we get
1
|G|
X
g∈G
R
0
(g
−1
)
iα
R(g)
βj
= 0
for all i, j. We now pick α = i and β = j. Then
1
|G|
X
g∈G
R
0
(g
−1
)
ii
R(g)
jj
= 0.
We can sum this thing over all i and j to get that hχ
0
, χi = 0.
(ii)
Now suppose
ρ, ρ
0
are isomorphic. So we might as well take
χ
=
χ
0
,
V
=
V
0
and ρ = ρ
0
. If ϕ : V → V is linear, then ˜ϕ ∈ End
G
(V ).
We first claim that tr ˜ϕ = tr ϕ. To see this, we have
tr ˜ϕ =
1
|G|
X
g∈G
tr(ρ(g
−1
)ϕρ(g)) =
1
|G|
X
g∈G
tr ϕ = tr ϕ,
using the fact that traces don’t see conjugacy (and
ρ
(
g
−1
) =
ρ
(
g
)
−1
since
ρ is a group homomorphism).
By Schur’s lemma, we know
˜ϕ
=
λι
v
for some
λ ∈ C
(which depends on
ϕ). Then if n = dim V , then
λ =
1
n
tr ϕ.
Let ϕ = ε
αβ
. Then tr ϕ = δ
αβ
. Hence
˜ε
αβ
=
1
|G|
X
g
ρ(g
−1
)ε
αβ
ρ(g) =
1
n
δ
αβ
ι.
In terms of matrices, we take the (i, j)th entry to get
1
|G|
X
R(g
−1
)
iα
R(g)
βj
=
1
n
δ
αβ
δ
ij
.
We now put α = i and β = j. Then we are left with
1
|G|
X
g
R(g
−1
)
ii
R(g)
jj
=
1
n
δ
ij
.
Summing over all i and j, we get hχ, χi = 1.
After learning about tensor products and duals later on, one can provide a
shorter proof of this result as follows:
Alternative proof. Consider two representation spaces V and W . Then
hχ
W
, χ
V
i =
1
|G|
X
χ
W
(g)χ
V
(g) =
1
|G|
X
χ
V ⊗W
∗
(g).
We notice that there is a natural isomorphism
V ⊗ W
∗
∼
=
Hom
(
W, V
), and
the action of
g
on this space is by conjugation. Thus, a
G
-invariant element
is just a
G
-homomorphism
W → V
. Thus, we can decompose
Hom
(
V, W
) =
Hom
G
(
V, W
)
⊕U
for some
G
-space
U
, and
U
has no
G
-invariant element. Hence
in the decomposition of
Hom
(
V, W
) into irreducibles, we know there are exactly
dim Hom
G
(
V, W
) copies of the trivial representation. By Schur’s lemma, this
number is 1 if V
∼
=
W , and 0 if V 6
∼
=
W .
So it suffices to show that if χ is a non-trivial irreducible character, then
X
g∈G
χ(g) = 0.
But if
ρ
affords
χ
, then any element in the image of
P
g∈G
ρ
(
g
) is fixed by
G
.
By irreducibility, the image must be trivial. So
P
g∈G
ρ(g) = 0.
What we have done is show the orthogonality of the rows. There is a similar
one for the columns:
Theorem (Column orthogonality relations). We have
k
X
i=1
χ
i
(g
j
)χ
i
(g
`
) = δ
j`
|C
G
(g
`
)|.
This is analogous to the fact that if a matrix has orthonormal rows, then it
also has orthonormal columns. We get the extra factor of
|C
G
(
g
`
)
|
because of
the way we count.
This has an easy corollary, which we have already shown previously using
the regular representation:
Corollary.
|G| =
k
X
i=1
χ
2
i
(1).
Proof of column orthogonality.
Consider the character table
X
= (
χ
i
(
g
j
)). We
know
δ
ij
= hχ
i
, χ
j
i =
X
`
1
|C
G
(g
`
)|
χ
i
(g
`
)χ
k
(g
`
).
Then
¯
XD
−1
X
T
= I
k×k
,
where
D =
|C
G
(g
1
)| ··· 0
.
.
.
.
.
.
.
.
.
0 ··· |C
G
(g
k
)|
.
Since
X
is square, it follows that
D
−1
¯
X
T
is the inverse of
X
. So
¯
X
T
X
=
D
,
which is exactly the theorem.
The proof requires that
X
is square, i.e. there are
k
many irreducible repre-
sentations. So we need to prove the last part of the completeness of characters.
Theorem.
Each class function of
G
can be expressed as a linear combination
of irreducible characters of G.
Proof.
We list all the irreducible characters
χ
1
, ··· , χ
`
of
G
. Note that we don’t
know the number of irreducibles is
k
. This is essentially what we have to prove
here. We now claim these generate C(G), the ring of class functions.
Now recall that
C
(
G
) has an inner product. So it suffices to show that the
orthogonal complement to the span
hχ
1
, ··· , χ
`
i
in
C
(
G
) is trivial. To see this,
assume f ∈ C(G) satisfies
hf, χ
j
i = 0
for all
χ
j
irreducible. We let
ρ
:
G → GL
(
V
) be an irreducible representation
affording χ ∈ {χ
1
, ··· , χ
`
}. Then hf, χi = 0.
Consider the function
ϕ =
1
|G|
X
g
f(g)ρ(g) : V → V.
For any h ∈ G, we can compute
ρ(h)
−1
ϕρ(h) =
1
|G|
X
g
¯
f(g)ρ(h
−1
gh) =
1
|G|
X
g
¯
f(h
−1
gh)ρ(h
−1
gh) = ϕ,
using the fact that
¯
f
is a class function. So this is a
G
-homomorphism. So as
ρ
is irreducible, Schur’s lemma says it must be of the form λι
V
for some λ ∈ C.
Now we take the trace of this thing. So we have
nλ = tr
1
|G|
X
g
f(g)ρ(g)
!
=
1
|G|
X
g
f(g)χ(g) = hf, χi = 0.
So
λ
= 0, i.e.
P
g
f(g)ρ
(
g
) = 0, the zero endomorphism on
V
. This is valid for
any irreducible representation, and hence for every representation, by complete
reducibility.
In particular, take ρ = ρ
reg
, where ρ
reg
(g) : e
1
7→ e
g
for each g ∈ G. Hence
X
f(g)ρ
reg
(g) : e
1
7→
X
g
f(g)e
g
.
Since this is zero, it follows that we must have
P
f(g)e
g
= 0. Since the
e
g
’s are
linearly independent, we must have f(g) = 0 for all g ∈ G, i.e. f = 0.