10Induction and restriction

II Representation Theory



10 Induction and restriction
This is the last chapter on methods of calculating characters. Afterwards, we
shall start applying these to do something useful.
Throughout the chapter, we will take
F
=
C
. Suppose
H G
. How do
characters of
G
and
H
relate? It is easy to see that every representation of
G
can be restricted to give a representation of
H
. More surprisingly, given a
representation of H, we can induce a representation of G.
We first deal with the easy case of restriction.
Definition
(Restriction)
.
Let
ρ
:
G GL
(
V
) be a representation affording
χ
.
We can think of
V
as an
H
-space by restricting
ρ
’s attention to
h H
. We get
a representation
Res
G
H
ρ
=
ρ
H
=
ρ
H
, the restriction of
ρ
to
H
. It affords the
character Res
G
H
χ = χ
H
= χ
H
.
It is a fact of life that the restriction of an irreducible character is in general
not irreducible. For example, restricting any non-linear irreducible character to
the trivial subgroup will clearly produce a reducible character.
Lemma.
Let
H G
. If
ψ
is any non-zero irreducible character of
H
, then there
exists an irreducible character
χ
of
G
such that
ψ
is a constituent of
Res
G
H
χ
, i.e.
hRes
G
H
χ, ψi 6= 0.
Proof.
We list the irreducible characters of
G
as
χ
1
, ··· , χ
k
. Recall the regular
character π
reg
. Consider
hRes
G
H
π
reg
, ψi =
|G|
|H|
ψ(1) 6= 0.
On the other hand, we also have
hRes
G
H
π
reg
, ψi
H
=
k
X
1
deg χ
i
hRes
G
H
χ
i
, ψi.
If this sum has to be non-zero, then there must be some
i
such that
hRes
G
H
χ
i
, ψi 6
=
0.
Lemma. Let χ be an irreducible character of G, and let
Res
G
H
χ =
X
i
c
i
χ
i
,
with χ
i
irreducible characters of H, and c
i
non-negative integers. Then
X
c
2
i
|G : H|,
with equality iff χ(g) = 0 for all g G \ H.
This is useful only if the index is small, e.g. 2 or 3.
Proof. We have
hRes
G
H
χ, Res
G
H
χi
H
=
X
c
2
i
.
However, by definition, we also have
hRes
G
H
χ, Res
G
H
χi
H
=
1
|H|
X
hH
|χ(h)|
2
.
On the other hand, since χ is irreducible, we have
1 = hχ, χi
G
=
1
|G|
X
gG
|χ(g)|
2
=
1
|G|
X
hH
|χ(h)|
2
+
X
gG\H
|χ(g)|
2
=
|H|
|G|
X
c
2
i
+
1
|G|
X
gG\H
|χ(g)|
2
|H|
|G|
X
c
2
i
.
So the result follows.
Now we shall go in the other direction given a character of
H
, we want to
induce a character of the larger group
G
. Perhaps rather counter-intuitively, it is
easier for us to first construct the character, and then later find a representation
that affords this character.
We start with a slightly more general notion of an induced class function.
Definition
(Induced class function)
.
Let
ψ C
(
H
). We define the induced
class function Ind
G
H
ψ = ψ
G
= ψ
G
by
Ind
G
H
ψ(g) =
1
|H|
X
xG
˚
ψ(x
1
gx),
where
˚
ψ(y) =
(
ψ(y) y H
0 y 6∈ H
.
The first thing to check is, of course, that Ind
G
H
is indeed a class function.
Lemma. Let ψ C
H
. Then Ind
G
H
ψ C(G), and Ind
G
H
ψ(1) = |G : H|ψ(1).
Proof.
The fact that
Ind
G
H
ψ
is a class function follows from direct inspection of
the formula. Then we have
Ind
G
H
ψ(1) =
1
|H|
X
xG
˚
ψ(1) =
|G|
|H|
ψ(1) = |G : H|ψ(1).
As it stands, the formula is not terribly useful. So we need to find an
alternative formula for it.
If we have a subset, a sensible thing to do is to look at the cosets of
H
.
We let
n
=
|G
:
H|
, and let 1 =
t
1
, ··· , t
n
be a left transversal of
H
in
G
, i.e.
t
1
H = H, ··· , t
n
H are precisely the n left-cosets of H in G.
Lemma. Given a (left) transversal t
1
, ··· , t
n
of H, we have
Ind
G
H
ψ(g) =
n
X
i=1
˚
ψ(t
1
i
gt
i
).
Proof.
We can express every
x G
as
x
=
t
i
h
for some
h H
and
i
. We then
have
˚
ψ((t
i
h)
1
g(t
i
h)) =
˚
ψ(h
1
(t
1
i
gt
i
)h) =
˚
ψ(t
1
i
gt
i
),
since
ψ
is a class function of
H
, and
h
1
(
t
1
i
gt
i
)
h H
if and only if
t
1
i
gt
i
H
,
as h H. So the result follows.
This is not too useful as well. But we are building up to something useful.
Theorem (Frobenius reciprocity). Let ψ C(H) and ϕ C(G). Then
hRes
G
H
ϕ, ψi
H
= hϕ, Ind
G
H
ψi
G
.
Alternatively, we can write this as
hϕ
H
, ψi = hϕ, ψ
G
i.
If you are a category theorist, and view restriction and induction as functors,
then this says restriction is a right adjoint to induction, and thsi is just a special
case of the tensor-Hom adjunction. If you neither understand nor care about
this, then this is a good sign [sic].
Proof. We have
hϕ, ψ
G
i =
1
|G|
X
gG
ϕ(g)ψ
G
(g)
=
1
|G||H|
X
x,gG
ϕ(g)
˚
ψ(x
1
gx)
We now write
y
=
x
1
gx
. Then summing over
g
is the same as summing over
y
.
Since ϕ is a G-class function, this becomes
=
1
|G||H|
X
x,yG
ϕ(y)
˚
ψ(y)
Now note that the sum is independent of x. So this becomes
=
1
|H|
X
yG
ϕ(y)
˚
ψ(y)
Now this only has contributions when y H, by definition of
˚
ψ. So
=
1
|H|
X
yH
ϕ(y)ψ(y)
= hϕ
H
, ψi
H
.
Corollary. Let ψ be a character of H. Then Ind
G
H
ψ is a character of G.
Proof. Let χ be an irreducible character of G. Then
hInd
G
H
ψ, χi = hψ, Res
G
H
χi.
Since
ψ
and
Res
G
H
χ
are characters, the thing on the right is in
Z
0
. Hence
Ind
G
H
is a linear combination of irreducible characters with non-negative coefficients,
and is hence a character.
Recall we denote the conjugacy class of
g G
as
C
G
(
g
), while the centralizer
is
C
G
(
g
). If we take a conjugacy class
C
G
(
g
) in
G
and restrict it to
H
, then
the result
C
G
(
g
)
H
need not be a conjugacy class, since the element needed
to conjugate
x, y C
G
(
g
)
H
need not be present in
H
, as is familiar from the
case of
A
n
S
n
. However, we know that
C
G
(
g
)
H
is a union of conjugacy
classes in H, since elements conjugate in H are also conjugate in G.
Proposition. Let ψ be a character of H G, and let g G. Let
C
G
(g) H =
m
[
i=1
C
H
(x
i
),
where the
x
i
are the representatives of the
H
conjugacy classes of elements of
H
conjugate to g. If m = 0, then Ind
G
H
ψ(g) = 0. Otherwise,
Ind
G
H
ψ(g) = |C
G
(g)|
m
X
i=1
ψ(x
i
)
|C
H
(x
i
)|
.
This is all just group theory. Note that some people will think this proof
is excessive everything shown is “obvious”. In some sense it is. Some steps
seem very obvious to certain people, but we are spelling out all the details so
that everyone is happy.
Proof.
If
m
= 0, then
{x G
:
x
1
gx H}
=
. So
˚
ψ
(
x
1
gx
) = 0 for all
x
. So
Ind
G
H
ψ(g) = 0 by definition.
Now assume m > 0. We let
X
i
= {x G : x
1
gx H and is conjugate in H to x
i
}.
By definition of
x
i
, we know the
X
i
’s are pairwise disjoint, and their union is
{x G : x
1
gx H}. Hence by definition,
Ind
G
H
ψ(g) =
1
|H|
X
xG
˚
ψ(x
1
gx)
=
1
|H|
m
X
i=1
X
xX
i
ψ(x
1
gx)
=
1
|H|
m
X
i=1
X
xX
i
ψ(x
i
)
=
m
X
i=1
|X
i
|
|H|
ψ(x
i
).
So we now have to show that in fact
|X
i
|
|H|
=
|C
G
(g)|
|C
H
(x
i
)|
.
We fix some 1
i m
. Choose some
g
i
G
such that
g
1
i
gg
i
=
x
i
. This exists
by definition of x
i
. So for every c C
G
(g) and h H, we have
(cg
i
h)
1
g(cg
i
h) = h
1
g
1
i
c
1
gcg
i
h
We now use the fact that c commutes with g, since c C
G
(g), to get
= h
1
g
1
i
c
1
cgg
i
h
= h
1
g
1
i
gg
i
h
= h
1
x
i
h.
Hence by definition of X
i
, we know cg
i
h X
i
. Hence
C
G
(g)g
i
H X
i
.
Conversely, if
x X
i
, then
x
1
gx
=
h
1
x
i
h
=
h
1
(
g
1
i
gg
i
)
h
for some
h
. Thus
xh
1
g
1
i
C
G
(g), and so x C
G
(g)g
i
h. So
x C
G
(g)g
i
H.
So we conclude
X
i
= C
G
(g)g
i
H.
Thus, using some group theory magic, which we shall not prove, we get
|X
i
| = |C
G
(g)g
i
H| =
|C
G
(g)||H|
|H g
1
i
C
G
(g)g
i
|
Finally, we note
g
1
i
C
G
(g)g
i
= C
G
(g
1
i
gg
i
) = C
G
(x
i
).
Thus
|X
i
| =
|H||C
G
(g)|
|H C
G
(x
i
)|
=
|H||C
G
(g)|
|C
H
(x
i
)|
.
Dividing, we get
|X
i
|
|H|
=
|C
G
(g)|
|C
H
(x
i
)|
.
So done.
To clarify matters, if H, K G, then a double coset of H, K in G is a set
HxK = {hxk : h H, k K}
for some x G. Facts about them include
(i) Two double cosets are either disjoint or equal.
(ii) The sizes are given by
|H||K|
|H xKx
1
|
=
|H||K|
|x
1
Hx K|
.
Lemma.
Let
ψ
= 1
H
, the trivial character of
H
. Then
Ind
G
H
1
H
=
π
X
, the
permutation character of
G
on the set
X
, where
X
=
G/H
is the set of left
cosets of H.
Proof.
We let
n
=
|G
:
H|
, and
t
1
, ··· , t
n
be representatives of the cosets. By
definition, we know
Ind
G
H
1
H
(g) =
n
X
i=1
˚
1
H
(t
1
i
gt
i
)
= |{i : t
1
i
gt
i
H}|
= |{i : g t
i
Ht
1
i
}|
But t
i
Ht
1
i
is the stabilizer in G of the coset t
i
H X. So this is equal to
= |fix
X
(g)|
= π
X
(g).
By Frobenius reciprocity, we know
hπ
X
, 1
G
i
G
= hInd
G
H
1
H
, 1
G
i
G
= h1
H
, 1
H
i
H
= 1.
So the multiplicity of 1
G
in π
X
is indeed 1, as we have shown before.
Example.
Let
H
=
C
4
=
h
(1 2 3 4)
i G
=
S
4
. The index is
|S
4
:
C
4
|
= 6. We
consider the character of the induced representation
Ind
G
H
(
α
), where
α
is the
faithful 1-dimensional representation of C
4
given by
α((1 2 3 4)) = i.
Then the character of α is
1 (1 2 3 4) (1 3)(2 4) (1 4 3 2)
χ
α
1 i 1 i
What is the induced character in S
4
? We know that
Ind
G
H
(α)(1) = |G : H|α(1) = |G : H| = 6.
So the degree of
Ind
G
H
(
α
) is 6. Also, the elements (1 2) and (3 4) are not
conjugate to anything in C
4
. So the character vanishes.
For (1 2)(3 4), only one of the three conjugates in
S
4
lie in
H
(namely
(1 3)(2 4)). So, using the sizes of the centralizers, we obtain
Ind
G
H
((1 3)(2 4)) = 8
1
4
= 2.
For (1 2 3 4), it is conjugate to 6 elements of
S
4
, two of which are in
C
4
, namely
(1 2 3 4) and (1 4 3 2). So
Ind
G
H
(1 2 3 4) = 4
i
4
+
i
4
= 0.
So we get the following character table:
1 6 8 3 6
1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)
Ind
G
H
(α) 6 0 0 2 0
Induced representations
We have constructed a class function, and showed it is indeed the character of
some representation. However, we currently have no idea what this corresponding
representation is. Here we will try to construct such a representation explicitly.
This is not too enlightening, but its nice to know it can be done. We will also
need this explicit construction when proving Mackey’s theorem later.
Let
H G
have index
n
, and 1 =
t
1
, ··· , t
n
be a left transversal. Let
W
be
a H-space. Define the vector space
V = Ind
G
H
W = W (t
2
W ) ··· (t
n
W ),
where
t
i
W = {t
i
w : w W }.
So dim V = n dim W .
To define the
G
-action on
V
, let
g G
. Then for every
i
, there is a unique
j
such that t
1
j
gt
i
H, i.e. gt
i
H = t
j
H. We define
g(t
i
w) = t
j
((t
1
j
gt
i
)w).
We tend to drop the tensor signs, and just write
g(t
i
w) = t
j
(t
1
j
gt
i
w).
We claim this is a G-action. We have
g
1
(g
2
t
i
w) = g
1
(t
j
(t
1
j
g
2
t
i
)w),
where j is the unique index such that g
2
t
i
H = t
j
H. This is then equal to
t
`
((t
1
`
gt
j
)(t
1
j
g
2
t
i
)w) = t
`
((t
1
`
(g
1
g
2
)t
i
)w) = (g
1
g
2
)(t
i
w),
where
`
is the unique index such that
g
1
t
j
H
=
t
`
H
(and hence
g
1
g
2
t
i
H
=
g
1
t
j
H = t
`
H).
Definition
(Induced representation)
.
Let
H G
have index
n
, and 1 =
t
1
, ··· , t
n
be a left transversal. Let
W
be a
H
-space. Define the induced
representation to be the vector space
Ind
G
H
W = W t
2
W ··· t
n
W,
with the G-action
g : t
i
w 7→ t
j
(t
1
j
gt
i
)W,
where t
j
is the unique element (among t
1
, ··· , t
n
) such that t
1
j
gt
i
H.
This has the “right” character. Suppose
W
has character
ψ
. Since
g
:
t
i
w 7→
t
j
(
t
1
j
gt
i
), the contribution to the character is 0 unless
j
=
i
, i.e. if
t
1
i
gt
i
H
.
Then it contributes
ψ(t
1
i
gt
i
).
Thus we get
Ind
G
H
ψ(g) =
n
X
i=1
˚
ψ(t
1
i
gt
i
).
Note that this construction is rather ugly. It could be made much nicer if we
knew a bit more algebra we can write the induced module simply as
Ind
G
H
W = FG
FH
W,
where we view
W
as a left-
FH
module, and
FG
as a (
FG, FH
)-bimodule. Alter-
natively, this is the extension of scalars from FH to FG.