Representation Theory IV 3 Character theory 5 Linear Lie groups and their Lie algebras

4 Induced representations

4.1 Definition

Suppose that $H$ is a subgroup of $G$ . Given a representation of $G$ , we may construct a representation of $H$ by restriction. What about the other direction — given a representation of $H$ , is there a natural way to construct a representation of $G$ ? The answer is yes, and this is the induced representation.

More precisely, we start with a representation $(\sigma,W)$ of $H$ and want to construct a representation $(\rho,V)$ of $G$ which contains $W$ as an $H$ -subrepresentation. Suppose we have such a thing. Then, for each $g\in G$ ,

gW=\{\rho(g)w:w\in W\}

is a subspace of $W$ (not necessarily an $H$ -subrepresentation!). We should have that, if $g=g^{\prime}h$ for some $h\in H$ , then

gW=g^{\prime}hW=g^{\prime}W

so that $g W$ only depends on the left coset $g H$ . The representation is induced if there are “no more relations”. Formally:

Definition 4.1.

Let $G$ be a finite group and let $H$ be a subgroup. Let $V$ be a representation of $H$ and let $W$ be a representation of $G$ . We say that $W$ is induced from $V$ if:

1.

There is an $H$ -subrepresentation $V_{0}$ of $W$ that is isomorphic to $V$ ; and
2.

If $g_{1},\ldots,g_{r}$ are a set of left coset representatives for $H$ in $G$ , then

$W=g_{1}V_{0}\oplus\ldots\oplus g_{r}V_{0}.$

The action of $G$ is determined by the direct sum decomposition: if $g\in G$ and $g_{i}v\in g_{i}V_{0}$ then we may write $gg_{i}=g_{j}h$ for some $j$ and some $h\in H$ , and then

g(g_{i}v)=(gg_{i})v=(g_{j}h)v=g_{j}(hv)\in g_{j}V_{0}

and $h v$ is known since we have that $V_{0}$ is isomorphic to $V$ as an $H$ -representation.

One can show that induced representations always exist (we didn’t do this in class). Here is one construction: Let

W=\{\text{$f:G\to V$ such that $f(gh)=hf(g)$ for all $h\in H,g\in g$}\}

with $G$ -action $(gf)(x)=f(g^{-1}x)$ for all $g,x\in G$ , $f\in W$ . For any $f\in W$ we can define its support

\operatorname{supp}(f)=\{g\in G:f(g)\neq 0\}

which will be a union of left cosets of $H$ . We can then take $V_{0}=\{f:\operatorname{supp}(f)\subset H\}$ ; then $g_{i}V_{0}=\{f:\operatorname{supp}(f)\subset g_{i}H\}$ and $W=\bigoplus_{i=1}^{r}g_{i}V_{0}$ .

Lemma 4.2.

If $V$ is a representation of $H$ then any two representation of $G$ induced from $V$ are isomorphic.

Proof.

We defer this proof until after the discussion of Frobenius reciprocity below. ∎

The significance of the lemma is that we can talk about ‘the’ induced representation, since it is unique up to isomorphism.

Example 4.3.

Let $(\sigma,\mathbb{C}^{2})$ be the two-dimensional representation of $G=D_{n}$ such that $\sigma(r)=\begin{pmatrix}\omega&0\\ 0&\omega^{-1}\end{pmatrix}$ and $\sigma(s)=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$ — irreducible if $\omega\neq\pm 1$ . Let $(\chi,\mathbb{C})$ be the one-dimensional representation of the subgroup $H=\left\langle r\right\rangle\cong C_{n}$ with $\chi(r)=\omega$ .

The cosets of $C_{n}$ in $D_{n}$ are $C_{n},sC_{n}$ . If $V_{0}=\left\langle e_{1}\right\rangle\subset\mathbb{C}^{2}$ then $V_{0}$ is a $C_{n}$ -subrepresentation isomorphic to $\chi$ , and $sV_{0}=\left\langle e_{2}\right\rangle$ . We clearly have $\mathbb{C}^{2}=V_{0}\oplus sV_{0}$ , so this shows

\operatorname{Ind}_{C_{n}}^{D_{n}}\chi\cong\sigma.

Example 4.4.

Suppose that $V=\mathbb{C}$ is the trivial representation of $H$ . Then the induced representation $\operatorname{Ind}_{H}^{G}(\mathbb{C})$ coincides with the permutation representation associated to the left action of $G$ on $G/H$ . Indeed, $\operatorname{Ind}_{H}^{G}(\mathbb{C})$ contains a vector $v$ fixed by $H$ (that is, a copy of the trivial representation of $H$ ), and if $g_{1}H,\ldots,g_{r}H$ are the left cosets of $H$ in $G$ then

g_{1}v,\ldots,g_{r}v

are a basis for $\operatorname{Ind}_{H}^{G}(\mathbb{C})$ . Moreover, if $g(g_{i}H)=g_{j}H$ then $gg_{i}=g_{j}h$ for some $h\in H$ and so

gg_{i}v=g_{j}hv=g_{j}v.

This shows that $G$ acts on these basis vectors ‘in the same way’ as it acts on the elements $G/H$ , which is what we have to prove.

Proposition 4.5.

Inducing multiplies the dimension by the index of the subgroup:

\dim\operatorname{Ind}_{H}^{G}V=[G:H]\dim V.

Proof.

If $g_{1}H,\ldots,g_{r}H$ are the cosets of $H$ in $G$ then

\operatorname{Ind}_{H}^{G}V=g_{1}V_{0}\oplus\ldots\oplus g_{r}V_{0}

for some copy $V_{0}$ of $V$ in $G$ . The dimension formula follows. ∎

4.2 Frobenius reciprocity

Theorem 4.6.

(Frobenius reciprocity) Let $H\subset G$ be finite groups, let $V$ be a representation of $H$ , and let $W$ be a representation of $G$ induced from $V$ . Then, for any representation $U$ of $G$ , there is an isomorphism of vector spaces

\operatorname{Hom}_{G}(W,U)\xrightarrow{\sim}\operatorname{Hom}_{H}(V,U).

Proof.

We use the decomposition of $W$ as

\bigoplus_{g_{i}H\in G/H}g_{i}V_{0}.

Since $V_{0}$ is isomorphic to $V$ as an $H$ -representation, and the right hand side only depends on $V$ up to isomorphism, we may as well assume that $V$ is actually a subrepresentation of $\operatorname{Ind}_{H}^{G}V$ .

If $\tilde{\phi}:W\to U$ is a $G$ -homomorphism, then the restriction $\phi$ of $\tilde{\phi}$ to $V$ is an element of $\operatorname{Hom}_{H}(V,U)$ . We show that the map $\tilde{\phi}\mapsto\phi$ is a (linear) bijection, which will prove the theorem.

Given $\phi:V\rightarrow U$ an $H$ -homomorphism, we must show that there is a unique $G$ -homomorphism $\tilde{\phi}:W\rightarrow U$ such that $\tilde{\phi}|_{V}=\phi$ .

Let $g_{1}H,\ldots,g_{r}H$ be the left cosets as usual. We must have

\tilde{\phi}(g_{i}v)=g_{i}\phi(v)

for all $i=1,\ldots,r$ and $v\in V$ , which shows the uniqueness of $\tilde{\phi}$ . To show existence, define $\psi:W\to U$ by $\psi(g_{i}v)=g_{i}\psi(v)$ for all $i=1,\ldots r$ and $v\in V$ ; since $W=\bigoplus_{i=1}^{r}g_{i}V$ this defined a unique linear map. Since we may take $g_{1}=e$ , we see that $\psi$ extends $\phi$ . To show it $\psi$ is a $G$ -homomorphism it is enough to show that $\psi(gg_{i}v)=g\psi(g_{i}v)$ for all $v\in V$ , $i=1,\ldots,r$ , and $g\in G$ . Then $gg_{i}=g_{j}h$ for some $j\in\{1,\ldots,r\}$ and $h\in H$ and we have

	$\displaystyle\psi(gg_{i}v)$	$\displaystyle=\psi(g_{j}hw)$
		$\displaystyle=g_{j}\phi(hw)$
		$\displaystyle=g_{j}h\phi(w)$
as $\phi$ is an $H$ -homomorphism
		$\displaystyle=gg_{i}\phi(w)$
		$\displaystyle=g\psi(g_{i}w).$

This shows the existence of $\tilde{\phi}=\psi$ , and we are done. ∎

Corollary 4.7.

Any two representations induced from isomorphic representations of $H$ are isomorphic.

Proof.

If $W,W^{\prime}$ are both induced from $V$ , and $V_{0},V_{0}^{\prime}$ are their respective $H$ -subrepresentations isomorphic to $V$ , then let $\phi:V_{0}\to V_{0}^{\prime}$ be an $H$ -isomorphism. Then the construction in the previous proof provides a $G$ -homomorphism $\tilde{\phi}:W\to W^{\prime}$ such that $\tilde{\phi}(v)=\phi(v)$ if $v\in V_{0}$ . As $\tilde{\phi}$ then defines an isomorphism of vector spaces $g_{i}V_{0}\to g_{i}V_{0}^{\prime}$ for each left coset representative $g_{i}H$ and each of $W$ and $W^{\prime}$ is the direct sum of these, $\tilde{\phi}$ is an isomorphism. ∎

Corollary 4.8.

Let $(\rho,V)$ be a representation of $H$ with character $\chi$ , and let $\psi$ be any class function on $G$ . Then

\left\langle\operatorname{Ind}_{H}^{G}\chi,\psi\right\rangle_{G}=\left\langle% \chi,\operatorname{Res}^{G}_{H}\psi\right\rangle_{H}.

Proof.

This is true when $\psi$ is the character of a representation of $G$ by Frobenius reciprocity. In general, we can write $\chi$ as a linear combination of characters of representations of $G$ , and deduce the result by linearity. ∎

If $X$ and $Y$ are vector spaces with inner products, then the adjoint of a linear map $T:X\rightarrow Y$ is a linear map $T^{*}:Y\rightarrow X$ such that

\left\langle Tx,y\right\rangle=\left\langle x,T^{*}y\right\rangle

for all $x\in X$ , $y\in Y$ . Thus we say that induction is adjoint to restriction.

Exercise 4.9.

If $W_{1},W_{2}$ are representations of $H$ , then

\operatorname{Ind}_{H}^{G}(W_{1}\oplus W_{2})\cong\operatorname{Ind}_{H}^{G}W_% {1}\oplus\operatorname{Ind}_{H}^{G}W_{2}.

Exercise 4.10.

If $H\subset K\subset G$ are subgroups, and $W$ is a representation of $H$ , then

\operatorname{Ind}_{H}^{G}W\cong\operatorname{Ind}_{K}^{G}(\operatorname{Ind}_% {H}^{K}W).

Exercise 4.11.

If $V$ is a representation of $G$ and $H$ is a subgroup, then

\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H}^{G}V\cong V\otimes% \operatorname{Ind}_{H}^{G}\mathbb{1}.

4.2.1 Example: S₃ to S₄

Table 3 below shows the irreducible characters $\psi_{i}$ of $S_{4}$ and $\chi_{i}$ of $S_{3}$ .

\begin{array}[]{l|rrrrr}&e&(12)&(12)(34)&(123)&(1234)\\ \hline\cr\psi_{0}&1&1&1&1&1\\ \psi_{1}&1&-1&1&1&-1\\ \psi_{2}&2&0&2&-1&-1\\ \psi_{3}&3&1&-1&0&-1\\ \psi_{4}&3&-1&-1&0&1\\ \hline\cr\hline\cr\chi_{0}&1&1&NA&1&NA\\ \chi_{1}&1&-1&NA&1&NA\\ \chi_{2}&2&0&NA&-1&NA\\ \end{array}

Table 3: Characters of

S_{3}

and

S_{4}

We regard $S_{3}$ as the subgroup of $S_{4}$ of elements that fix 4, and use Frobenius reciprocity to compute $\operatorname{Ind}_{S_{3}}^{S_{4}}\chi_{2}$ .

For each $i$ , Frobenius reciprocity implies that

\left\langle\operatorname{Ind}_{S_{3}}^{S_{4}}\chi_{2},\psi_{i}\right\rangle_{% S_{4}}=\left\langle\chi_{2},\psi_{i}|_{S_{3}}\right\rangle_{S_{3}}.

The right hand side is easily seen to be zero for $i=0,1$ and one for $i=2,3,4$ . We therefore have

\operatorname{Ind}_{S_{3}}^{S_{4}}\chi_{2}=\psi_{2}+\psi_{3}+\psi_{4}.

4.3 Characters

Let $H\subset G$ be finite groups, and let $(\rho,V)$ be a representation of $H$ with character $\chi$ . We use Frobenius reciprocity to determine the character $\operatorname{Ind}_{H}^{G}\chi$ of $\operatorname{Ind}_{H}^{G}\rho$ .

Theorem 4.12.

Suppose that we are in the above situation. Suppose that $\mathcal{C}$ is a conjugacy class of $G$ , and let $\mathcal{C}\cap H=\mathcal{D}_{1}\cup\ldots\cup\mathcal{D}_{n}$ where $\mathcal{D}_{1},\ldots,\mathcal{D}_{n}$ are conjugacy classes of $H$ . Then

(\operatorname{Ind}_{H}^{G}\chi)(\mathcal{C})=\frac{|G|}{|H|}\sum_{i=1}^{n}% \frac{|\mathcal{D}_{i}|}{|\mathcal{C}|}\chi(\mathcal{D}_{i}).

Proof.

Let $1_{\mathcal{C}}$ be the indicator function of $\mathcal{C}$ . Then for any class function $\chi$ on $G$ ,

\left\langle\chi,1_{\mathcal{C}}\right\rangle_{G}=\frac{|\mathcal{C}|}{|G|}% \chi(\mathcal{C}).

So, combining this with Frobenius reciprocity,

	$\displaystyle(\operatorname{Ind}_{H}^{G}\chi)(\mathcal{C})$	$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\left\langle\operatorname{Ind}_{H}^{G}% \chi,1_{\mathcal{C}}\right\rangle_{G}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\left\langle\chi,1_{\mathcal{C}}\right% \rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\sum_{i=1}^{n}\left\langle\chi,1_{% \mathcal{D}_{i}}\right\rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\sum_{i=1}^{n}\frac{\|\mathcal{D}_{i}\|}{% \|H\|}\chi(\mathcal{D}_{i})$
		$\displaystyle=\frac{\|G\|}{\|H\|}\sum_{i=1}^{n}\frac{\|\mathcal{D}_{i}\|}{\|\mathcal{% C}\|}\chi(\mathcal{D}_{i})$

which gives the claimed formula. ∎

Remark 4.13.

We derived the character formula from Frobenius reciprocity. It is also possible to go the other way around: prove the character formula directly, then derive Frobenius reciprocity as a consequence. Personally, I find that knowing Frobenius reciprocity is the easiest way to remember the character formula.

Remark 4.14.

If $g\in G$ , we write $C_{G}(g)$ for the centralizer of $g$ :

C_{G}(g)=\{x\in G:x^{-1}gx=g\}.

By the orbit-stabiliser theorem, if $\mathcal{C}(g)$ is the conjugacy class of $g$ , then

\frac{|G|}{|\mathcal{C}(g)|}=|C_{G}(g)|

giving an interpretation for some of the factors in the above formula.

Exercise 4.15.

Let $H\subset G$ be groups and let $\chi$ be a character of $H$ . If $\dot{\chi}(g)=\chi(g)$ if $g\in H$ and $0$ otherwise.

1.

Show that the formula for the induced character may be rewritten

$(\operatorname{Ind}_{H}^{G}\chi)(g)=\frac{1}{|H|}\sum_{x\in G}\dot{\chi}(x^{-1% }gx).$
2.

If $g_{1}H,\ldots,g_{r}H$ are the left cosets of $H$ in $G$ , show that

$(\operatorname{Ind}_{H}^{G}\chi)(g)=\sum_{i=1}^{r}\dot{\chi}(g_{i}^{-1}gg_{i}).$

Example 4.16.

We continue with the example of the dihedral group. Let $H=C_{n}\subset G=D_{n}$ , and let $\psi:H\to\mathbb{C}^{\times}$ be a homomorphism with $\psi(r)=\omega$ . Then:

•

$\operatorname{Ind}_{H}^{G}\psi(e)=[G:H]=2$ .
•

$\operatorname{Ind}_{H}^{G}\psi(s)=\operatorname{Ind}_{H}^{G}\psi(rs)=0$ as the conjugacy class of $s$ or $r s$ does not intersect $H$ .
•

If $0<i<n/2$ , then the conjugacy class $\{r,r^{-1}\}$ of $G$ splits into two conjugacy classes, $\{r\}$ and $\{r^{-1}\}$ , of $H$ . We have

$\operatorname{Ind}_{H}^{G}(\psi)(r^{i})=2(\frac{1}{2}\psi(r^{i})+\frac{1}{2}% \psi(r)^{-i})=\omega^{i}+\omega^{-i}.$
•

If $i=n/2$ , then the conjugacy class $\{r^{n/2}\}$ of $G$ remains as a single conjugacy class of $H$ and

$\operatorname{Ind}_{H}^{G}(\psi)(r^{n/2})=2\psi(r^{n/2})=2\omega^{n/2}.$

Taking $\omega\neq\pm 1$ we again obtain all the irreducible two-dimensional characters of $D_{n}$ .

4.4 Example: D₄ to S₄

Let $H$ be the subgroup of $G=S_{4}$ isomorphic to $D_{4}$ , obtained by labeling the vertices of a square 1, …, 4 and letting $D_{4}$ act on them. In other words, $H$ is the image of the injective homomorphism $D_{4}\rightarrow S_{4}$ sending

r\mapsto(1234),s\mapsto(12)(34).

Recall that $D_{4}$ has five irreducible representations with characters as shown:

Table 4: Character table of

D_{4}

	$e$	$r=(1234)$	$r^{2}=(13)(24)$	$s=(12)(34)$	$rs=(13)$
	1	2	1	2	2
$\mathbbm{1}$	$1$	$1$	$1$	$1$	$1$
$\delta$	$1$	$1$	$1$	$-1$	$-1$
$\phi_{+}$	$1$	$-1$	$1$	$1$	$-1$
$\phi_{-}$	$1$	$-1$	$1$	$-1$	$1$
$\theta$	$2$	$0$	$-2$	$0$	$0$

while recall that $S_{4}$ has character table

\begin{array}[h]{c|rrrrr}&e&(12)&(12)(34)&(123)&(1234)\\ &1&6&3&8&6\\ \hline\cr\mathbbm{1}&1&1&1&1&1\\ \epsilon&1&-1&1&1&-1\\ \chi&3&1&-1&0&-1\\ \chi\epsilon&3&-1&-1&0&1\\ \psi&2&0&2&-1&0\end{array}

We want to first find out how the conjugacy classes of $S_{4}$ intersect with $D_{4}$ . The result is as follows, writing $C(g)$ for the conjugacy class of $C(g)$ in $S_{4}$ or $D_{4}$ .

\begin{array}[]{l|l|l}g\in S_{4}&C(g)\cap D_{4}&\text{sizes}\\ \hline\cr e&C(e)&1\\ (12)&C(rs)&2\\ (12)(34)&C(r^{2})\cup C(s)&1,2\\ (123)&\emptyset&-\\ (1234)&C(r)&2\\ \end{array}

We use this and the character formula to determine $\operatorname{Ind}_{H}^{G}\phi_{+}$ . The factor $|G|/|H|$ is constant, equal to 3. We obtain:

\begin{array}[]{l|r|r}&&\operatorname{Ind}_{H}^{G}\phi_{+}\\ \hline\cr e&1&3\\ (12)&6&3\cdot 2/6\cdot(-1)=-1\\ (12)(34)&3&3(1/3+2/3)=3\\ (123)&8&0\\ (1234)&6&3(-2/6)=-1\\ \end{array}

Decomposing this character, we see that

\operatorname{Ind}_{H}^{G}\phi_{+}=\epsilon+\psi.

But note that we did not need to know the character table of $S_{4}$ to find the induced character.

We check that this is consistent with Frobenius reciprocity: the restriction of $\epsilon$ to $D_{4}$ is $\phi_{+}$ , so

\left\langle\operatorname{Res}^{G}_{H}\epsilon,\phi_{+}\right\rangle=\left% \langle\epsilon,\operatorname{Ind}_{H}^{G}\phi_{+}\right\rangle=1

as required. The restriction of $\psi$ to $D_{4}$ is $\mathbbm{1}+\phi_{+}$ so

\left\langle\operatorname{Res}^{G}_{H}\psi,\phi_{+}\right\rangle=\left\langle% \psi,\operatorname{Ind}_{H}^{G}\phi_{+}\right\rangle=1

(we could also check that the restrictions of the other irreducible characters of $S_{4}$ to $D_{4}$ do not contain $\phi_{+}$ ).

4.5 Mackey’s formula — nonexaminable

We want to come up with a criterion for $\operatorname{Ind}_{H}^{G}\chi$ to be irreducible. This is equivalent to

\left\langle\operatorname{Ind}_{H}^{G}\chi,\operatorname{Ind}_{H}^{G}\chi% \right\rangle_{G}=1

which is in turn equivalent, by Frobenius reciprocity, to

\left\langle\operatorname{Res}^{G}_{H}\operatorname{Ind}_{H}^{G}\chi,\chi% \right\rangle_{H}=1.

So we try to solve the more general problem of decomposing $\operatorname{Res}^{G}_{H}\operatorname{Ind}_{H}^{G}V$ , as a representation of $H$ . In fact, we will solve the more general problem of finding

\operatorname{Res}^{G}_{H}\operatorname{Ind}_{K}^{G}V

for $H$ and $K$ arbitrary subgroups of $G$ .

4.5.1 Double cosets

Let $G$ be a group and let $H, K$ be subgroups. A double coset $H g K$ is a set of the form

HgK=\{hgk:h\in H,k\in K\}.

Like cosets, any two double cosets are either equal or disjoint, so they partition the group. Unlike cosets, they do not have to have the same size. A sequence $g_{1},\ldots,g_{r}$ such that $G$ is the disjoint union

G=\coprod_{i=1}^{r}Hg_{i}K

is called a set of double coset representatives for $H$ and $K$ in $G$ . We write $H\backslash G/K$ for the set of double cosets of $H$ and $K$ in $G$ .

Lemma 4.17.

The double cosets $H g K$ are the in bijection with the orbits of $H$ acting on $G/K$ .

Proof.

Two cosets $g K$ and $g^{\prime}K$ are in the same $H$ -orbit, if and only if $hgK=g^{\prime}K$ for some $h\in H$ , if and only if

HgK=Hg^{\prime}K.

So we have a bijection which takes the double coset $H g K$ to the $H$ -orbit of $g K$ . ∎

Example 4.18.

Let $H=K=S_{n-1}\subset G=S_{n}$ , $n\geq 3$ , thinking of $H$ as the set of elements that fix the number $n$ . Then $H=HeH$ is one double coset. I claim that $G=H\cup H(n-1\,n)H$ . Indeed, a complete set of left coset representatives for $H$ in $G$ is

\{e,(1n),(2n),\ldots,((n-1)n)\}.

As I can write $(in)=(i\,n-1)(n-1\,n)(i\,n-1)\in H(n-1\,n)H$ for $i\leq n-1$ , we see that the left cosets $(1n)H,\ldots,(n-1\,n)H$ are all contained in the same double coset $H(n-1\,n)H$ , as required.

This may be clearer from the orbits point of view. There is a bijection $G/H\to\{1,\ldots,n\}$ which respects the action of $G$ — the bijection takes $g H$ to $g(n)$ . The orbits of $H=S_{n-1}$ acting on $\{1,\ldots,n\}$ are $\{n\}$ and $\{1,\ldots,n-1\}$ , which gives the two double cosets.

If $s K$ is a left coset of $K$ , we write $H_{s}=H\cap sKs^{-1}$ ; note that this is independent of choice of representative $s$ . The notation is motivated by the fact that $H_{s}$ is the stabiliser of $s K$ for the action of $H$ on $G/K$ discussed above. This leads to the following lemma.

Lemma 4.19.

If $s\in G$ , then as a disjoint union of cosets of $K$ we have

HsK=h_{1}sK\cup h_{2}sK\ldots\cup h_{r}sK

where $h_{1},\ldots,h_{r}$ are a complete set of left coset representatives for $H_{s}$ in $H$ .

4.5.2 Mackey’s formula

Suppose that $K\subset G$ , that $s\in G$ , and that $(\rho,V)$ is a representation of $K$ . Then we can define a representation $({}^{s}\rho,V)$ of ${}^{s}H=sHs^{-1}$ by taking ${}^{s}V=V$ (same vector space) and ${}^{s}\rho(h)=\rho(s^{-1}hs)$ for $h\in{}^{s}H$ .

Theorem 4.20.

(Mackey’s formula) Let $H, K$ be subgroups of $G$ and let $(\rho,V)$ be a representation of $K$ . Let $s_{1},\ldots,s_{r}$ be a set of double coset representatives for $H$ and $K$ in $G$ .

Then

\operatorname{Res}^{G}_{H}\operatorname{Ind}_{K}^{G}V\cong\bigoplus_{s\in\{s_{% 1},\ldots,s_{r}\}}\operatorname{Ind}_{H_{s}}^{H}\operatorname{Res}^{{}^{s}K}_{% H_{s}}{}^{s}V.

Proof.

Let $g_{1},\ldots,g_{n}$ be the left coset representatives for $K$ in $G$ . Then

\operatorname{Ind}_{K}^{G}V=g_{1}V\oplus\ldots\oplus g_{n}V.

The action of $H$ permutes the $g_{i}V$ among themselves, with $g_{i}V$ and $g_{j}V$ in the same $H$ -orbit if and only if $Hg_{i}K=Hg_{j}K$ . Therefore for each double coset $H s K$ we have an $H$ -subrepresentation

W_{s}=\bigoplus_{i:g_{i}K\subset HsK}g_{i}V.

Now, by lemma 4.19 above, these $g_{i}$ have the form $h_{i}s$ for $h_{i}$ a set of left coset representatives for $H_{s}$ in $H$ . From the decomposition

W_{s}=\bigoplus_{h_{i}H_{s}\in H/H_{s}}h_{i}sV

we see that

W_{s}\cong\operatorname{Ind}_{H_{s}}^{H}sV.

It remains to show that $s V$ , as a representation of $H_{s}$ , is isomorphic to ${}^{s}V$ . The map ${}^{s}V\rightarrow sV$ taking $v$ to $s v$ is the required isomorphism, because

\rho(h)(sv)=\rho(h)\rho(s)v=\rho(s)\rho(s^{-1}hs)v=s{}^{s}\rho(h)v.

∎

4.5.3 Irreducibility criterion

Theorem 4.21.

Let $\rho$ be an irreducible representation of $H\subset G$ . Then $\operatorname{Ind}_{H}^{G}\rho$ is irreducible if and only if, for every $g$ in a set of double coset representatives of $H\backslash G/H$ , with $g\not\in H$ ,

\operatorname{Hom}_{H_{g}}(\rho,{}^{g}\rho)=0

where $H_{g}=H\cap gHg^{-1}$ .

Proof.

Indeed, let $S=g_{0},\ldots,g_{r}$ be such a set of double coset representatives, with $g_{0}\in H$ . By Mackey’s formula and Frobenius reciprocity,

	$\displaystyle\left\langle\operatorname{Ind}_{H}^{G}\rho,\operatorname{Ind}_{H}% ^{G}\rho\right\rangle_{G}$	$\displaystyle=\left\langle\rho,\operatorname{Res}^{G}_{H}\operatorname{Ind}_{H% }^{G}\rho\right\rangle_{H}$
		$\displaystyle=\left\langle\rho,\bigoplus_{g\in S}\operatorname{Ind}_{H_{g}}^{H% }{}^{g}\rho\right\rangle_{H}$
		$\displaystyle=\left\langle\rho,\rho\right\rangle_{H}+\sum_{g\in\{g_{1},\ldots,% g_{r}\}}\left\langle\rho,\operatorname{Ind}_{H_{g}}^{H}{}^{g}\rho\right\rangle% _{H}$
		$\displaystyle=\left\langle\rho,\rho\right\rangle_{H}+\sum_{g\in\{g_{1},\ldots,% g_{r}\}}\left\langle\rho,{}^{g}\rho\right\rangle_{H_{g}}.$

Since $\rho$ is irreducible, the first term on the right is equal to 1 and all the other terms are at least 0. So

\left\langle\operatorname{Ind}_{H}^{G}\rho,\operatorname{Ind}_{H}^{G}\rho% \right\rangle_{G}\geq 1

with equality if and only if $\operatorname{Hom}_{H_{g}}(\rho,{}^{g}\rho)=0$ for all $g\in S\setminus H$ . The result follows. ∎

Corollary 4.22.

If $H\lhd G$ and $\rho$ is an irreducible representation of $H$ , then

\operatorname{Ind}_{H}^{G}\rho

is irreducible if and only if $\rho\not\cong{}^{g}\rho$ for all $g$ in a set of coset representatives of $H$ in $G$ with $g\not\in H$ .

Proof.

If $H$ is normal, then double cosets, left cosets, and right cosets are all the same. For $g\in G$ , ${}^{g}\rho$ is another irreducible representation of $H$ so

\operatorname{Hom}_{H}(\rho,{}^{g}\rho)\neq 0

if and only if $\rho\not\cong{}^{g}\rho$ . ∎

4.5.4 Mackey examples

Example 4.23.

Let $G=D_{n}\supset H=C_{n}$ . Let $\chi:C_{n}\to\mathbb{C}^{\times}$ with $\chi(r)=\omega$ . The previous corollary shows that

\operatorname{Ind}_{H}^{G}\chi

is reducible if and only if $\chi\cong{}^{s}\chi$ . But

{}^{s}\chi(r)=\chi(s^{-1}rs)=\chi(r^{-1})=\omega^{-1}.

So $\operatorname{Ind}_{H}^{G}\chi$ is irreducible if and only if $\omega\neq\omega^{-1}$ (which we already knew).

Example 4.24.

Let $\rho$ be an irreducible representation of $S_{n-1}\subset S_{n}$ with character $\chi$ . Then $\{e,s\}$ is a set of double coset representatives for $S_{n-1}$ in $S_{n}$ , where $s=(n-1\,n)$ . We have

{}^{s}S_{n-1}\cap S_{n-1}=S_{n-2}

of elements fixing $n-1$ and $n$ .

Moreover, ${}^{s}\chi\cong\chi$ (because elements of $S_{n-2}$ are conjugate in $S_{n-2}$ if and only if they are conjugate in $S_{n-1}$ , so they have the same character!) and so we see

	$\displaystyle\left\langle\operatorname{Ind}_{S_{n-1}}^{S_{n}}\chi,% \operatorname{Ind}_{S_{n-1}}^{S_{n}}\chi\right\rangle$	$\displaystyle=\left\langle\chi,\chi\right\rangle_{S_{n-1}}+\left\langle\chi,% \chi\right\rangle_{S_{n-2}}$
		$\displaystyle=1+\left\langle\chi,\chi\right\rangle_{S_{n-2}}.$

For example, if $\chi$ is the degree 2 irreducible character of $S_{4}$ , then $\chi|_{S_{3}}$ is still irreducible so the right hand side of this formula gives two. And indeed,

\operatorname{Ind}_{S_{4}}^{S_{5}}\chi=\psi+\psi\epsilon

where $\psi$ is one of the irreducible five-dimensional representations of $S_{5}$ .

4.5.5 Index two

Let $G$ be a group and let $H$ be a subgroup of index two (necessarily normal). Let $s\in G$ with $s\not\in H$ . Let $\epsilon:G\to\mathbb{C}^{\times}$ be the unique (order two) character with kernel $H$ .

Of course, we have in mind the example $H=A_{n}$ , $G=S_{n}$ .

Theorem 4.25.

Suppose that $\rho$ is an irreducible representation of $H$ . Then

1.

if $\rho\cong{}^{s}\rho$ , then

$\operatorname{Ind}^{G}_{H}\rho\cong\sigma\oplus\epsilon\sigma$

for some irreducible representation $\sigma$ of $G$ with $\operatorname{Res}^{G}_{H}\sigma\cong\rho$ .
2.

if $\rho\not\cong{}^{s}\rho$ , then

$\sigma=\operatorname{Ind}_{H}^{G}\rho$

is irreducible, with $\sigma\cong\epsilon\sigma$ .

Proof.

Let $\rho$ have character $\chi$ . By the normal subgroup case of Mackey’s theorem,

\left\langle\operatorname{Ind}_{H}^{G}\chi,\operatorname{Ind}_{H}^{G}\chi% \right\rangle_{G}=1+\left\langle\chi,{}^{s}\chi\right\rangle_{H}.

It follows that the induction $\operatorname{Ind}_{H}^{G}\chi$ is irreducible if and only if $\chi\neq{}^{s}\chi$ . This gives the second statement, and most of the first (we leave the proof that, in all cases, $\operatorname{Ind}_{H}^{G}\chi\cong\epsilon\operatorname{Ind}_{H}^{G}\chi$ as an exercise.)

If $\chi={}^{s}\chi$ , then we have $\operatorname{Ind}_{H}^{G}\chi=\psi+\psi^{\prime}$ for some irreducible characters $\psi$ and $\psi^{\prime}$ of $G$ . Restricting to $H$ , we have $\psi|_{H}+\psi^{\prime}|_{H}=2\chi$ so that $\psi$ extends $\chi$ . But then we have

\operatorname{Ind}_{H}^{G}\chi=\operatorname{Ind}_{H}^{G}\operatorname{Res}^{G% }_{H}\psi=\psi(\operatorname{Ind}_{H}^{G}\mathbbm{1})=\psi(\mathbbm{1}+% \epsilon)=\psi+\epsilon\psi

so that $\operatorname{Ind}_{H}^{G}\chi=\psi+\epsilon\psi$ as required. ∎

4.6 Example: GL₂(F_p) — nonexaminable

Let $\mathbb{F}_{p}$ be the field of integers modulo $p$ , with $p$ a prime. We will construct some of the irreducible representation of $G=GL_{2}(\mathbb{F}_{p})$ by induction from the subgroup of upper triangular matrices. We will use Mackey theory to show that these representations are irreducible.

4.6.1 The group GL₂(F_p)

Proposition 4.26.

The order of the group $GL_{2}(\mathbb{F}_{p})$ is

p(p+1)(p-1)^{2}.

We let $B\subset GL_{2}(\mathbb{F}_{p})$ be the subgroup of upper triangular invertible matrices:

B=\left\{\begin{pmatrix}*&*\\ 0&*\end{pmatrix}\in GL_{2}(\mathbb{F}_{p})\right\}.

The group $B$ is sometimes called a Borel subgroup.

Proposition 4.27.

The index of $B$ in $G$ is $p+1$ . A set of left coset representatives for $G/B$ is

\left\{\begin{pmatrix}0&1\\ 1&0\end{pmatrix}\right\}\cup\left\{\begin{pmatrix}1&0\\ i&1\end{pmatrix}:0\leq i\leq p-1\right\}.

We write $w=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$ . We will want to know the double cosets of $B$ in $G$ , and the result is

Proposition 4.28.

We have

B\backslash G/B=B\cup BwB.

For later use, we compute

w\begin{pmatrix}x&z\\ 0&y\end{pmatrix}w^{-1}=\begin{pmatrix}y&0\\ z&x\end{pmatrix}

so that

B_{w}=B\cap{}^{w}B=\left\{\begin{pmatrix}x&0\\ 0&y\end{pmatrix}:x,y\in\mathbb{F}_{p}^{\times}\right\}.

4.6.2 Characters of B and their inductions.

Let $\chi_{1}$ and $\chi_{2}$ be two characters of $\mathbb{F}_{p}^{\times}$ . Then there is a character

\chi_{1}\cdot\chi_{2}:B\to\mathbb{C}^{\times}

defined by

(\chi_{1}\cdot\chi_{2})\left(\begin{pmatrix}x&z\\ 0&y\end{pmatrix}\right)=\chi_{1}(x)\chi_{2}(y).

Define

\pi(\chi_{1},\chi_{2})=\operatorname{Ind}_{B}^{G}(\chi_{1}\cdot\chi_{2}).

If $\chi:\mathbb{F}_{p}^{\times}\to\mathbb{C}^{\times}$ is a character, then we get a character

\chi\circ\det:G\to\mathbb{C}^{\times}

taking $g$ to $\chi(\det(g))$ .

Lemma 4.29.

If $\chi_{1},\chi_{2},\chi:\mathbb{F}_{p}^{\times}\to\mathbb{C}^{\times}$ are characters, then

\pi(\chi_{1}\chi,\chi_{2}\chi)\cong(\chi\circ\det)\otimes\pi(\chi_{1},\chi_{2}).

Lemma 4.30.

On $B_{w}$ , we have

{}^{w}(\chi_{1}\cdot\chi_{2})=\chi_{2}\cdot\chi_{1}.

Theorem 4.31.

If $\chi_{1},\chi_{2}:\mathbb{F}_{p}^{\times}\to\mathbb{C}^{\times}$ are characters, then:

1.

If $\chi_{1}\neq\chi_{2}$ , then

$\pi(\chi_{1},\chi_{2})$

is an irreducible representation of dimension $p+1$ .
2.

There is an irreducible representation $\sigma$ of dimension $p$ such that, for any $\chi:\mathbb{F}_{p}^{\times}\to\mathbb{C}^{\times}$ ,

$\pi(\chi,\chi)\cong(\chi\circ\det)\otimes\left(\mathbbm{1}\oplus\sigma\right).$
3.

There is an isomorphism

$\pi(\chi_{1},\chi_{2})\cong\pi(\chi_{1}^{\prime},\chi_{2}^{\prime})$

if and only if $(\chi_{1},\chi_{2})=(\chi_{1},\chi_{2})$ or $(\chi_{2},\chi_{1})$ .
4.

For $\chi\neq\chi^{\prime}$ , the representations $(\chi\circ\det)\otimes\sigma$ and $(\chi^{\prime}\circ\det)\otimes\sigma$ are not isomorphic.

One can prove the theorem using Mackey theory, but it is also possible to avoid that and directly compute the inner products of the induced characters.

The representations $\pi(\chi_{1},\chi_{2})$ , $\chi\circ\det$ , and $\chi\circ\det\otimes\sigma$ are not all of the irreducible representations of $GL_{2}(\mathbb{F}_{p})$ . There is another family of $p-1$ -dimensional representations called the cuspidal representations, which are more difficult to construct: see [MR1153249] Lecture 5, or [MR1864147] Chapter 28.

4.6.3 Computing the induced character

Finally, we can compute the character of $\pi(\chi_{1},\chi_{2})$ .

We will need to know the conjugacy classes of $G$ and of $B$ . The characteristic polynomial almost determines the conjugacy classes; the result is as follows. Recall that the size of the conjugacy class of $g\in G$ is $\frac{|G|}{|C_{G}(g)|}$ where $C_{G}(g)$ is the centralizer of $g$ in $G$ .

Proposition 4.32.

Every element of $G$ is conjugate to exactly one of the following elements:

1.

$\begin{pmatrix}x&0\\ 0&x\end{pmatrix},x\in\mathbb{F}_{p}^{\times}$
2.

$\begin{pmatrix}x&1\\ 0&x\end{pmatrix},x\in\mathbb{F}_{p}^{\times}$
3.

$\begin{pmatrix}x&0\\ 0&y\end{pmatrix},x,y\in\mathbb{F}_{p}^{\times},x\neq y$
4.

$\begin{pmatrix}0&-d\\ 1&t\end{pmatrix},t\in\mathbb{F}_{p},d\in\mathbb{F}_{p}^{\times}$ such that $X^{2}-tX+d=0$ has no solutions in $\mathbb{F}_{p}$ .

For each type of element $G$ , the order of the centralizer subgroup is:

1.

$|G|=p(p+1)(p-1)^{2}$
2.

$p(p-1)$
3.

$(p-1)^{2}$
4.

$(p-1)(p+1)$ .

Proof.

We sketch the proof. Let $X^{2}-tX+d$ be the characteristic polynomial of an element $g\in G$ . If it has distinct roots in $\mathbb{F}_{p}$ , then $g$ is diagonalizable and we are in case 3. If it has no roots in $\mathbb{F}_{p}$ , then choosing a basis $v, g v$ (for some $V\in\mathbb{F}_{p}^{2}$ ) puts us in case 4. If the roots are identical, both equal to $x$ , there are two possibilities: either the matrix is a scalar matrix, giving case 1, or $v,(g-xI)v$ is a basis for some $v\in\mathbb{F}_{p}^{2}$ and then we are in case 2.

In each case, the centralizer can be computed explicitly and checked to have the order claimed (the most difficult case is the fourth). ∎

Now let $\chi:\mathbb{F}_{p}^{\times}\to\mathbb{C}^{\times}$ be a character. We compute the character of $\pi(\chi,\mathbbm{1})$ ; after twisting by a character of the form $\chi^{\prime}\circ\det$ this gives the general case. In the case $\chi=\mathbbm{1}$ , we also record the character of $\sigma$ .

Theorem 4.33.

The characters are as follows (where $x\neq y$ , and $X^{2}-tX+d$ has no solutions in $\mathbb{F}_{p}$ ):

\begin{array}[]{cc|ccc}&\begin{pmatrix}x&0\\ 0&x\end{pmatrix}&\begin{pmatrix}x&1\\ 0&x\end{pmatrix}&\begin{pmatrix}x&0\\ 0&y\end{pmatrix}&\begin{pmatrix}0&-d\\ 1&t\end{pmatrix}\\ &1&(p-1)(p+1)&p(p+1)&p(p-1)\\ \hline\cr\pi(\chi,\mathbbm{1})&(p+1)\chi(x)&\chi(x)&\chi(x)+\chi(y)&0\\ \sigma&p&0&1&-1\\ \end{array}

	$\displaystyle(\operatorname{Ind}_{H}^{G}\chi)(\mathcal{C})$	$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\left\langle\operatorname{Ind}_{H}^{G}% \chi,1_{\mathcal{C}}\right\rangle_{G}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\left\langle\chi,1_{\mathcal{C}}\right% \rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\sum_{i=1}^{n}\left\langle\chi,1_{% \mathcal{D}_{i}}\right\rangle_{H}$
		$\displaystyle=\frac{\|G\|}{\|\mathcal{C}\|}\sum_{i=1}^{n}\frac{\|\mathcal{D}_{i}\|}{% \|H\|}\chi(\mathcal{D}_{i})$
		$\displaystyle=\frac{\|G\|}{\|H\|}\sum_{i=1}^{n}\frac{\|\mathcal{D}_{i}\|}{\|\mathcal{% C}\|}\chi(\mathcal{D}_{i})$

4 Induced representations

4.1 Definition

Definition 4.1.

Lemma 4.2.

Proof.

Example 4.3.

Example 4.4.

Proposition 4.5.

Proof.

4.2 Frobenius reciprocity

Theorem 4.6.

Proof.

Corollary 4.7.

Proof.

Corollary 4.8.

Proof.

Exercise 4.9.

Exercise 4.10.

Exercise 4.11.

4.2.1 Example: S3 to S4

4.3 Characters

Theorem 4.12.

Proof.

Remark 4.13.

Remark 4.14.

Exercise 4.15.

Example 4.16.

4.4 Example: D4 to S4

4.5 Mackey’s formula — nonexaminable

4.5.1 Double cosets

Lemma 4.17.

Proof.

Example 4.18.

Lemma 4.19.

4.5.2 Mackey’s formula

Theorem 4.20.

Proof.

4.5.3 Irreducibility criterion

Theorem 4.21.

Proof.

Corollary 4.22.

Proof.

4.5.4 Mackey examples

Example 4.23.

Example 4.24.

4.5.5 Index two

Theorem 4.25.

Proof.

4.6 Example: GL2(Fp) — nonexaminable

4.6.1 The group GL2(Fp)

Proposition 4.26.

Proposition 4.27.

Proposition 4.28.

4.6.2 Characters of B and their inductions.

Lemma 4.29.

Lemma 4.30.

Theorem 4.31.

4.6.3 Computing the induced character

Proposition 4.32.

Proof.

Theorem 4.33.

4.2.1 Example: S₃ to S₄

4.4 Example: D₄ to S₄

4.6 Example: GL₂(F_p) — nonexaminable

4.6.1 The group GL₂(F_p)