II Representation Theory - Induction and restriction

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

. Suppose

H ≤ G

. How do

characters of

and

relate? It is easy to see that every representation of

can be restricted to give a representation of

. 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

G → GL

(

) be a representation affording

We can think of

as an

-space by restricting

’s attention to

h ∈ H

. We get

a representation

Res

ρ ↓

, the restriction of

. It affords the

character Res

χ = χ

= χ ↓

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

is any non-zero irreducible character of

, then there

exists an irreducible character

such that

is a constituent of

Res

, i.e.

hRes

χ, ψi 6= 0.

Proof.

We list the irreducible characters of

, ··· , χ

. Recall the regular

character π

reg

. Consider

hRes

reg

, ψi =

|G|

|H|

ψ(1) 6= 0.

On the other hand, we also have

hRes

reg

, ψi

deg χ

hRes

, ψi.

If this sum has to be non-zero, then there must be some

such that

hRes

, ψi 6

Lemma. Let χ be an irreducible character of G, and let

Res

χ =

with χ

irreducible characters of H, and c

non-negative integers. Then

≤ |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

χ, Res

χi

However, by definition, we also have

hRes

χ, Res

χi

|H|

h∈H

|χ(h)|

On the other hand, since χ is irreducible, we have

1 = hχ, χi

|G|

g∈G

|χ(g)|

|G|





h∈H

|χ(h)|

g∈G\H

|χ(g)|





|H|

|G|

g∈G\H

|χ(g)|

≥

|H|

|G|

So the result follows.

Now we shall go in the other direction — given a character of

, we want to

induce a character of the larger group

. 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

(

). We define the induced

class function Ind

ψ = ψ ↑

= ψ

Ind

ψ(g) =

|H|

x∈G

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

is indeed a class function.

Lemma. Let ψ ∈ C

. Then Ind

ψ ∈ C(G), and Ind

ψ(1) = |G : H|ψ(1).

Proof.

The fact that

Ind

is a class function follows from direct inspection of

the formula. Then we have

Ind

ψ(1) =

|H|

x∈G

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

We let

, and let 1 =

, ··· , t

be a left transversal of

, i.e.

H = H, ··· , t

H are precisely the n left-cosets of H in G.

Lemma. Given a (left) transversal t

, ··· , t

of H, we have

Ind

ψ(g) =

i=1

ψ(t

−1

Proof.

We can express every

x ∈ G

for some

h ∈ H

and

. We then

have

ψ((t

−1

g(t

h)) =

ψ(h

−1

)h) =

ψ(t

−1

since

is a class function of

, and

−1

(

−1

)

h ∈ H

if and only if

−1

∈ 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

ϕ, ψi

= hϕ, Ind

ψi

Alternatively, we can write this as

hϕ

, ψi = hϕ, ψ

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

i =

|G|

g∈G

ϕ(g)ψ

(g)

|G||H|

x,g∈G

ϕ(g)

ψ(x

−1

gx)

We now write

−1

. Then summing over

is the same as summing over

Since ϕ is a G-class function, this becomes

|G||H|

x,y∈G

ϕ(y)

ψ(y)

Now note that the sum is independent of x. So this becomes

|H|

y∈G

ϕ(y)

ψ(y)

Now this only has contributions when y ∈ H, by definition of

ψ. So

|H|

y∈H

ϕ(y)ψ(y)

= hϕ

, ψi

Corollary. Let ψ be a character of H. Then Ind

ψ is a character of G.

Proof. Let χ be an irreducible character of G. Then

hInd

ψ, χi = hψ, Res

χi.

Since

and

Res

are characters, the thing on the right is in

≥0

. Hence

Ind

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

(

), while the centralizer

(

). If we take a conjugacy class

(

) in

and restrict it to

, then

the result

(

)

∩ H

need not be a conjugacy class, since the element needed

to conjugate

x, y ∈ C

(

)

∩ H

need not be present in

, as is familiar from the

case of

≤ S

. However, we know that

(

)

∩ 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

(g) ∩H =

[

i=1

where the

are the representatives of the

conjugacy classes of elements of

conjugate to g. If m = 0, then Ind

ψ(g) = 0. Otherwise,

Ind

ψ(g) = |C

(g)|

i=1

ψ(x

)

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.

= 0, then

{x ∈ G

−1

gx ∈ H}

∅

. So

(

−1

) = 0 for all

. So

Ind

ψ(g) = 0 by definition.

Now assume m > 0. We let

= {x ∈ G : x

−1

gx ∈ H and is conjugate in H to x

By definition of

, we know the

’s are pairwise disjoint, and their union is

{x ∈ G : x

−1

gx ∈ H}. Hence by definition,

Ind

ψ(g) =

|H|

x∈G

ψ(x

−1

gx)

|H|

i=1

x∈X

ψ(x

−1

gx)

|H|

i=1

x∈X

ψ(x

)

i=1

|H|

ψ(x

So we now have to show that in fact

|H|

(g)|

We fix some 1

≤ i ≤ m

. Choose some

∈ G

such that

−1

. This exists

by definition of x

. So for every c ∈ C

(g) and h ∈ H, we have

(cg

−1

g(cg

h) = h

−1

gcg

We now use the fact that c commutes with g, since c ∈ C

(g), to get

= h

−1

cgg

= h

−1

= h

−1

Hence by definition of X

, we know cg

h ∈ X

. Hence

(g)g

H ⊆ X

Conversely, if

x ∈ X

, then

−1

(

−1

)

for some

. Thus

−1

∈ C

(g), and so x ∈ C

(g)g

h. So

x ∈ C

(g)g

So we conclude

= C

(g)g

Thus, using some group theory magic, which we shall not prove, we get

| = |C

(g)g

H| =

(g)||H|

|H ∩ g

−1

(g)g

Finally, we note

−1

(g)g

= C

−1

) = C

Thus

| =

|H||C

(g)|

|H ∩ C

|H||C

(g)|

Dividing, we get

|H|

(g)|

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|

−1

Hx ∩K|

Lemma.

Let

= 1

, the trivial character of

. Then

Ind

, the

permutation character of

on the set

, where

G/H

is the set of left

cosets of H.

Proof.

We let

, and

, ··· , t

be representatives of the cosets. By

definition, we know

Ind

(g) =

i=1

−1

)

= |{i : t

−1

∈ H}|

= |{i : g ∈ t

−1

But t

−1

is the stabilizer in G of the coset t

H ∈ X. So this is equal to

= |fix

(g)|

= π

(g).

By Frobenius reciprocity, we know

hπ

, 1

= hInd

, 1

= h1

, 1

= 1.

So the multiplicity of 1

in π

is indeed 1, as we have shown before.

Example.

Let

(1 2 3 4)

i ≤ G

. The index is

= 6. We

consider the character of the induced representation

Ind

(

), where

is the

faithful 1-dimensional representation of C

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)

1 i −1 −i

What is the induced character in S

? We know that

Ind

(α)(1) = |G : H|α(1) = |G : H| = 6.

So the degree of

Ind

(

) is 6. Also, the elements (1 2) and (3 4) are not

conjugate to anything in C

. So the character vanishes.

For (1 2)(3 4), only one of the three conjugates in

lie in

(namely

(1 3)(2 4)). So, using the sizes of the centralizers, we obtain

Ind

((1 3)(2 4)) = 8



−1



= −2.

For (1 2 3 4), it is conjugate to 6 elements of

, two of which are in

, namely

(1 2 3 4) and (1 4 3 2). So

Ind

(1 2 3 4) = 4



−i



= 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

(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

, and 1 =

, ··· , t

be a left transversal. Let

a H-space. Define the vector space

V = Ind

W = W ⊕ (t

⊗ W ) ⊕ ··· ⊕ (t

⊗ W ),

where

⊗ W = {t

⊗ w : w ∈ W }.

So dim V = n dim W .

To define the

-action on

, let

g ∈ G

. Then for every

, there is a unique

such that t

−1

∈ H, i.e. gt

H = t

H. We define

g(t

⊗ w) = t

⊗ ((t

−1

)w).

We tend to drop the tensor signs, and just write

g(t

w) = t

−1

w).

We claim this is a G-action. We have

w) = g

−1

)w),

where j is the unique index such that g

H = t

H. This is then equal to

((t

−1

)(t

−1

)w) = t

((t

−1

)w) = (g

)(t

w),

where

is the unique index such that

(and hence

H = t

H).

Definition

(Induced representation)

Let

H ≤ G

have index

, and 1 =

, ··· , t

be a left transversal. Let

be a

-space. Define the induced

representation to be the vector space

Ind

W = W ⊕ t

⊗ W ⊕ ··· ⊕ t

⊗ W,

with the G-action

g : t

w 7→ t

−1

)W,

where t

is the unique element (among t

, ··· , t

) such that t

−1

∈ H.

This has the “right” character. Suppose

has character

. Since

w 7→

(

−1

), the contribution to the character is 0 unless

, i.e. if

−1

∈ H

Then it contributes

ψ(t

−1

Thus we get

Ind

ψ(g) =

i=1

ψ(t

−1

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

W = FG ⊗

where we view

as a left-

module, and

as a (

FG, FH

)-bimodule. Alter-

natively, this is the extension of scalars from FH to FG.