12Mackey theory
II Representation Theory
12 Mackey theory
We work over
C
. We wish to describe the restriction to a subgroup
K ≤ G
of an
induced representation
Ind
G
H
W
, i.e. the character
Res
G
K
Ind
G
H
W
. In general,
K
and
H
can be unrelated, but in many applications, we have
K
=
H
, in which
case we can characterize when Ind
G
H
W is irreducible.
It is quite helpful to first look at a special case, where
W
= 1
H
is the trivial
representation. Thus
Ind
G
H
1
H
is the permutation representation of
G
on
G/H
(coset action on the left cosets of H in G).
Recall that by the orbit-stabilizer theorem, if
G
is transitive on the set
X
,
and
H
=
G
α
for some
α ∈ X
, then the action of
G
on
X
is isomorphic the action
on G/H, namely the correspondence
gα ↔ gH
is a well-defined bijection, and commutes with with the
g
-action (i.e.
x
(
gα
) =
(xg)α ↔ x(gH) = (xg)H).
We now consider the action of
G
on
G/H
and let
K ≤ G
. Then
K
also acts
on
G/H
, and
G/H
splits into
K
-orbits. The
K
-orbit of
gH
contains precisely
kgH for k ∈ K. So it is the double coset
KgH = {kgh : k ∈ K, h ∈ H}.
The set of double cosets
K\G/H
partition
G
, and the number of double cosets
is
|K\G/H| = hπ
G/K
, π
G/H
i.
We don’t need this, but this is true.
If it happens that
H
=
K
, and
H
is normal, then we just have
K\G/H
=
G/H.
What about stabilizers? Clearly,
G
gH
=
gHg
−1
. Thus, restricting to the
action of K, we have K
gH
= gHg
−1
∩ K. We call H
g
= K
gH
.
So by our correspondence above, the action of
K
on the orbit containing
gH
is isomorphic to the action of
K
on
K/
(
gHg
−1
∩ K
) =
K/H
g
. From this,
and using the fact that
Ind
G
H
1
H
=
C
(
G/H
), we get the special case of Mackey’s
theorem:
Proposition.
Let
G
be a finite group and
H, K ≤ G
. Let
g
1
, ··· , g
k
be the
representatives of the double cosets K\G/H. Then
Res
G
K
Ind
G
H
1
H
∼
=
k
M
i=1
Ind
K
g
i
Hg
−1
i
∩K
1.
The general form of Mackey’s theorem is the following:
Theorem
(Mackey’s restriction formula)
.
In general, for
K, H ≤ G
, we let
S = {1, g
1
, ··· , g
r
} be a set of double coset representatives, so that
G =
[
Kg
i
H.
We write
H
g
=
gHg
−1
∩ K ≤ G
. We let (
ρ, W
) be a representation of
H
. For
each
g ∈ G
, we define (
ρ
g
, W
g
) to be a representation of
H
g
, with the same
underlying vector space W , but now the action of H
g
is
ρ
g
(x) = ρ(g
−1
xg),
where h = g
−1
xg ∈ H by construction.
This is clearly well-defined. Since
H
g
≤ K
, we obtain an induced representa-
tion Ind
K
H
g
W
g
.
Let G be finite, H, K ≤ G, and W be a H-space. Then
Res
G
K
Ind
G
H
W =
M
g∈S
Ind
K
H
g
W
g
.
We will defer the proof of this for a minute or two. We will first derive some
corollaries of this, starting with the character version of the theorem.
Corollary. Let ψ be a character of a representation of H. Then
Res
G
K
Ind
G
H
ψ =
X
g∈S
Ind
K
H
g
ψ
g
,
where ψ
g
is the class function (and a character) on H
g
given by
ψ
g
(x) = ψ(g
−1
xg).
These characters
ψ
g
are sometimes known as the conjugate characters. This
obviously follows from Mackey’s restriction formula.
Corollary
(Mackey’s irreducibility criterion)
.
Let
H ≤ G
and
W
be a
H
-space.
Then V = Ind
G
H
W is irreducible if and only if
(i) W is irreducible; and
(ii)
For each
g ∈ S\H
, the two
H
g
spaces
W
g
and
Res
H
H
g
W
have no irreducible
constituents in common, where H
g
= gHg
−1
∩ H.
Note that the set
S
of representatives was chosen arbitrarily. So we may
require (ii) to hold for all
g ∈ G \ H
, instead of
g ∈ S \ H
. While there are
more things to check in
G
, it is sometimes convenient not have to pick out an
S
explicitly.
Note that the only
g ∈ S ∩ H
is the identity, since for any
h ∈ H
,
KhH
=
KH = K1H.
Proof.
We use characters, and let
W
afford the character
ψ
. We take
K
=
H
in
Mackey’s restriction formula. Then we have H
g
= gHg
−1
∩ H.
Using Frobenius reciprocity, we can compute the inner product as
hInd
G
H
ψ, Ind
G
H
ψi
G
= hψ, Res
G
H
Ind
G
H
ψi
H
=
X
g∈S
hψ, Ind
H
H
g
ψ
g
i
H
=
X
g∈S
hRes
H
H
g
ψ, ψ
g
i
H
g
= hψ, ψi +
X
g∈S\H
hRes
H
H
g
ψ, ψ
g
i
H
g
We can write this because if g = 1, then H
g
= H, and ψ
g
= ψ.
This is a sum of non-negative integers, since the inner products of characters
always are. So
Ind
G
H
ψ
is irreducible if and only if
hψ, ψi
= 1, and all the other
terms in the sum are 0. In other words,
W
is an irreducible representation of
H
,
and for all g 6∈ H, W and W
g
are disjoint representations of H
g
.
Corollary.
Let
H C G
, and suppose
ψ
is an irreducible character of
H
. Then
Ind
G
H
ψ
is irreducible if and only if
ψ
is distinct from all its conjugates
ψ
g
for
g ∈ G \ H (where ψ
g
(h) = ψ(g
−1
hg) as before).
Proof.
We take
K
=
H C G
. So the double cosets are just left cosets. Also,
H
g
= H for all g. Moreover, W
g
is irreducible since W is irreducible.
So, by Mackey’s irreducible criterion,
Ind
G
H
W
is irreducible precisely if
W 6
∼
=
W
g
for all g ∈ G \ H. This is equivalent to ψ 6= ψ
g
.
Note that again we could check the conditions on a set of representatives of
(double) cosets. In fact, the isomorphism class of
W
g
(for
g ∈ G
) depends only
on the coset gH.
We now prove Mackey’s theorem.
Theorem
(Mackey’s restriction formula)
.
In general, for
K, H ≤ G
, we let
S = {1, g
1
, ··· , g
r
} be a set of double coset representatives, so that
G =
[
Kg
i
H.
We write
H
g
=
gHg
−1
∩ K ≤ G
. We let (
ρ, W
) be a representation of
H
. For
each
g ∈ G
, we define (
ρ
g
, W
g
) to be a representation of
H
g
, with the same
underlying vector space W , but now the action of H
g
is
ρ
g
(x) = ρ(g
−1
xg),
where h = g
−1
xg ∈ H by construction.
This is clearly well-defined. Since
H
g
≤ K
, we obtain an induced representa-
tion Ind
K
H
g
W
g
.
Let G be finite, H, K ≤ G, and W be a H-space. Then
Res
G
K
Ind
G
H
W =
M
g∈S
Ind
K
H
g
W
g
.
This is possibly the hardest and most sophisticated proof in the course.
Proof.
Write
V
=
Ind
G
H
W
. Pick
g ∈ G
, so that
KgH ∈ K\G/H
. Given a left
transversal T of H in G, we can obtain V explicitly as a direct sum
V =
M
t∈T
t ⊗ W.
The idea is to “coarsen” this direct sum decomposition using double coset
representatives, by collecting together the t ⊗ W ’s with t ∈ KgH. We define
V (g) =
M
t∈KgH∩T
t ⊗ W.
Now each
V
(
g
) is a
K
-space — given
k ∈ K
and
t ⊗w ∈ t ⊗W
, since
t ∈ KgH
,
we have
kt ∈ KgH
. So there is some
t
0
∈ T
such that
ktH
=
t
0
H
. Then
t
0
∈ ktH ⊆ KgH. So we can define
k · (t ⊗ w) = t
0
⊗ (ρ(t
0−1
kt)w),
where t
0
kt ∈ H.
Viewing V as a K-space (forgetting its whole G-structure), we have
Res
G
K
V =
M
g∈S
V (g).
The left hand side is what we want, but the right hand side looks absolutely
nothing like Ind
K
H
g
W
g
. So we need to show
V (g) =
M
t∈KgH∩T
t ⊗ W
∼
=
Ind
K
H
g
W
g
,
as K representations, for each g ∈ S.
Now for
g
fixed, each
t ∈ KgH
can be represented by some
kgh
, and by
restricting to elements in the traversal
T
of
H
, we are really summing over cosets
kgH
. Now cosets
kgH
are in bijection with cosets
k
(
gHg
−1
) in the obvious way.
So we are actually summing over elements in
K/
(
gHg
−1
∩ K
) =
K/H
g
. So we
write
V (g) =
M
k∈K/H
g
(kg) ⊗ W.
We claim that there is a isomorphism that sends
k ⊗W
g
∼
=
(
kg
)
⊗W
. We define
k ⊗W
g
→
(
kg
)
⊗W
by
k ⊗w 7→ kg ⊗w
. This is an isomorphism of vector spaces
almost by definition, so we only check that it is compatible with the action. The
action of x ∈ K on the left is given by
ρ
g
(x)(k ⊗ w) = k
0
⊗ (ρ
g
(k
0−1
xk)w) = k
0
⊗ (ρ(g
−1
k
0−1
xkg)w),
where
k
0
∈ K
is such that
k
0−1
xk ∈ H
g
, i.e.
g
−1
k
0−1
xkg ∈ H
. On the other
hand,
ρ(x)(kg ⊗ w) = k
00
⊗ (ρ(k
00
x
−1
(kg))w),
where
k
00
∈ K
is such that
k
00−1
xkg ∈ H
. Since there is a unique choice of
k
00
(after picking a particular transversal), and
k
0
g
works, we know this is equal to
k
0
g ⊗ (ρ(g
−1
k
0−1
xkg)w).
So the actions are the same. So we have an isomorphism.
Then
V (g) =
M
k∈K/H
g
k ⊗ W
g
= Ind
K
H
g
W
g
,
as required.
Example.
Let
g
=
S
n
and
H
=
A
n
. Consider a
σ ∈ S
n
. Then its conjugacy
class in S
n
is determined by its cycle type.
If the class of
σ
splits into two classes in
A
n
, and
χ
is an irreducible character
of
A
n
that takes different values on the two classes, then by the irreducibility
criterion, Ind
S
n
A
n
χ is irreducible.