2024-01-29--notes-RT.md

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title: Schur's Lemma and irreducible representations
date: 2024-01-29

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\newcommand{\trivial}{\mathbf{1}}

Let $G$ be a finite group and $F$ an algebraically closed field of characteristic 0.

Some notational convention(s)

We will sometimes just write $V$ for a representation $(\rho,V)$ of a group $G$, leaving implicit the mapping $$G \to \operatorname{GL}(V).$$

One exception is when $(\rho,V)$ is a representation for which $\dim V = 1$. In this case, we denote this representation by the notation $\rho$. For example, for any group $G$ one has the trivial representation $\mathbb{1}$: the vector space is simply the 1-dimensional space $F$, and the mapping ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 1: \̲t̲r̲i̲v̲i̲a̲l̲:G \to \operato… is given by $g \mapsto 1$ for each $g \in G$.

Thus for example ParseError: KaTeX parse error: Undefined control sequence: \trivial at position 1: \̲t̲r̲i̲v̲i̲a̲l̲ ̲\oplus \trivial is a two-dimensional $G$-representation.

Irreducible Representations

A representation $(\rho,V)$ of $G$ is irreducible (sometimes one says simple) provided that $V \ne 0$ and for any invariant subspace $W$ of $V$, either $W = 0$ or $W = V$.

Theorem: : Let $(\rho,V)$ be a finite dimensional representation of $G$. Then $V$ is isomorphic to a direct sum of irreducible representations: $$V = L_1 \oplus L_2 \oplus \cdots \oplus L_r.$$

Proof: : First of all, we claim that $V$ has an invariant subspace which is irreducible as a representation for $G$.

 Indeed, we proceed by induction on the dimension $\dim V$. If
 $\dim V = 1$, then $V$ is irreducible since the only *linear*
 subspaces of $V$ are $0$ and $V$.

 Now suppose that $\dim V > 1$. If $V$ is irreducible, we are
 done. Otherwise, $V$ has a non-0 invariant subspace $W$ with
 $\dim W < \dim V$. By the inductive hypotheses, $W$ has an
 invariant subspace which is irreducible as $G$-representation.
 This completes the proof of the *claim*.
 
 We now prove the Theorem again by induction on $\dim V$. If $\dim V =1$,
 again $V$ is already irreducible and the proof is complete. [^1]
 
 [^1]: When $V = 0$ the result is still true, since $V$ is the
 "direct sum" of an empty collection of irreducible representations.
 
 Now, suppose that $\dim V > 1$. Choose an irreducible invariant
 subspace $L_1 \subseteq V$ and use complete reducibility to write
 $V = L_1 \oplus V'$ for an invariant subspace $V'$.
 
 If $V' = 0$, then $V = L_1$ is a direct sum of irreducible
 representations. Otherwise, $\dim V' = \dim V - \dim L_1 < \dim
 V$ and so by the induction hypothesis $V'$ is the direct sum $V'
 = L_2 \oplus \cdots \oplus L_r$ for certain irreducible representations
 $L_j$.
 
 Now notice that
 $$V = L_1 \oplus V' = L_1 \oplus L_2 \oplus \cdots \oplus L_r$$
 as required.

Example: : Suppose that $G$ is the cyclic group $\mathbb{Z}/m\mathbb{Z}$.

 If $\zeta$ is an $m$-th root of unity in $F^\times$, then
 $$\rho_\zeta:G \to \operatorname{GL}_1(F) = F^\times$$ defined by
 $\rho_\zeta(i + m \mathbb{Z}) = \zeta^i$ determines an
 irreducible representation of $G$.
 
 Every irreducible representation of $G$ is isomorphic to
 $\rho_\zeta$ for some root $\zeta$ of $T^m - 1 \in F[T]$.

For $G$-representations $V$ and $W$, write $$\operatorname{Hom}_G(V,W)$$ for the space of all $G$-homomorphisms $\Phi:V \to W$. If $V = W$, write $$\operatorname{End}_G(V) = \operatorname{Hom}_G(V,V)$$ for the space of $G$-endomorphisms.

Notice that $\operatorname{End}_G(V)$ is a ring (in fact, an $F$-algebra) under composition of endomorphisms.

Theorem: : Let $L,L'$ be irreducible representations for $G$.

 a. We have $\operatorname{End}_G(L) = F$.

 b. $\dim_F \operatorname{Hom}_G(L,L') = \left \{
    \begin{matrix}
	1 & \text{if} \quad L \simeq L' \\
	0 & \text{else}
	\end{matrix}
	\right.$ 

Proof: : (a). This is essentially the content of Schur's Lemma. We claim first that $\operatorname{End}_G(L)$ is a division algebra.

 For this, it suffices to argue that any non-zero element
 $\phi$ of $\operatorname{End}_G(L)$ has an inverse.
    
 Since $L$ is irreducible and since the kernel of $\phi$ is
 non-zero, $\ker \phi= 0$. Since $V$ is finite dimensional, it
 follows that $\phi$ is bijective and therefore invertible.
	
 To see that $\operatorname{End}_G(L) = F$, it remains to observe
 that when $F$ is *algebraically closed*, any finite dimensional
 division algebra $D$ with $F \subset Z(D)$ satisfies $D = F$.

 Now (b) follows at once from (a).

Permutation representations and homomorphisms

Let $G$ act transitively on the set $\Omega$, fix $\omega \in \Omega$ and let $H = \operatorname{Stab}_G(\omega)$ be the stabilizer of $x$. Since the action of $G$ is transitive, $\Omega = G.\omega$ is the $G$-orbit of $\omega$, and $\Omega$ may be identified with $G/H$.

Proposition: : Let $V$ be a $G$-representation and let $x \in V$ be a non-zero vector such that $hx = x$ for all $h \in H$.

a. There is a unique homomorphism of $G$-representations
   $$\Phi:F[\Omega] \to V$$ with the property that
   $\Phi(\delta_\omega) = x$.

b. If the $G$-representation $V$ is *irreducible*, then $V$ is
   isomorphic to a direct summand of $F[\Omega]$.

Proof: : (a). Every element $\tau$ of $\Omega$ may be written in the form $\tau = g\omega$ for some $g \in G$. Define $\Phi$ by the rule $$\Phi(\delta_{g\omega}) = gx$$ for all $g \in G$.

Notice that $\delta_{g \omega} = \delta_{g'\omega} \iff g^{-1}g'
\implies gx = g'x$, so $\Phi$ is a well-defined linear mapping.

Let's check that $\Phi$ is a $G$-homomorphism. Let $\gamma \in G$.  We
must argue that $\Phi(\gamma v) = \gamma\Phi(v)$, and it suffices to prove
this identity when $v = \delta_{g \omega}$ is a basis vector in
$F[\Omega]$.

Now, 
$$\Phi(\gamma \delta_{g \omega}) = \Phi(\delta_{\gamma g\omega}) = \gamma g.
= \gamma (g.x) = \gamma \Phi(\delta_{g \omega});$$
this shows that $\Phi$ is indeed a $G$-homomorphism.

Finally, suppose that $\Psi:F[\Omega] \to V$ is any
$G$-homomorphism with $\Psi(\delta_\omega) = x$. Then for $g \in
G$, $$g x = g \Psi(\delta_\omega) = \Psi(g \delta_\omega) =
\Psi(\delta_{g\omega}).$$ which shows that $\Psi$ is given by
precisely the same formula as $\Phi$; this proves the uniqueness.

(b). The homomorphism constructed in (a) is nonzero since $x$ is
contained in its image. Since $V$ is irreducible, this
homomorphism is *surjective*. Let $K \subset F[\Omega]$ be the
*kernel* of this homomorphism.  By complete reducibity, there is a
subrepresentation $W$ of $F[\Omega]$ such that $F[\Omega] = K
\oplus W$.

On the one hand, $F[\Omega]/K = (W \oplus K)/K \simeq W$, and on
the other hand, the homomorphism $F[\Omega] \to V$ induces an
isomorphism $F[\Omega]/K \simeq V$. Thus $W \simeq V$, so indeed
the irreducible representation $W$ is a direct summand of
$F[\Omega]$.

Remark: Let $\Phi:F[\Omega] \to V$ be the mapping of the proposition, so $x \in V$ is fixed by $V$. If $f \in F[\Omega]$, then $$\Phi(f) = \sum_{g \in G} f(g)gx.$$

The Regular Representation

Note that the group $G$ acts on the set $\Omega = G$ by left multiplication. The resulting permutation representation $F[\Omega] = F[G]$ is called the regular representation.

Note that the action of $G$ on itself is transitive, and the stabilizer $H$ of an element (say, $1 \in G$) is the trivial subgroup.

Theorem: : Every irreducible representation is isomorphic to a subrepresentation of the regular representation $F[G]$.

Proof: : The Theorem follows at once from the Proposition in the previous section.

Corollary: : Up to isomorphism, $G$ has only finitely many irreducible representations.

Proof: : Write the regular representation as a direct sum $$F[\Omega] = L_1 \oplus L_2 \oplus \cdots \oplus L_r$$ of irreducible representations $L_i$.

 For each $i$, let $\pi_i:F[\Omega] \to L_i$ be the *projection*
 onto $L_i$ for this direct sum decomposition, and notice that
 $$\operatorname{id}_V = \sum_{i=1}^r \pi_i.$$
 
 If $L \subset F[\Omega]$ is an irreducible invariant subspace, it
 follows that for some $i$, $\pi_i(L) \ne 0$. Since $L$ and $L_i$
 are irreducible, $\pi_i$ induces an isomorphism $L_i
 \xrightarrow{\sim} L$.

Characters and class functions

We are now going to assume $F = \mathbb{C}$

Let $V$ be a representation of $G$ and consider the $\mathbb{C}$-valued function $$\chi = \chi_V:G \to \mathbb{C}$$ defined by the rule $$\chi(g) = \operatorname{tr}(g:V \to V)$$ where $\operatorname{tr}(g)$ denotes the trace of the linear endomorphism of $V$ determined by $g$.

If $\rho:G\to\operatorname{GL}(V)$ denotes the homomorphism determining the representation, $\chi_V(g) = \operatorname{tr}(\rho(g))$.

Proposition: : The character $\chi$ of a representation of $G$ is constant on the conjugacy classes of the group $G$.

Recall that a conjugacy class $C \subseteq G$ is an equivalence class for the relation $$g \sim h \iff g = xhx^{-1} \quad \text{for some $x \in GParseError: KaTeX parse error: Expected 'EOF', got '}' at position 2: .}̲$Thus, a conjugacy class has the form$$C = \{xyx^{-1} \mid x \in G\}$$for some$y \in G$.

Proof of Proposition: : If $g \sim h$ we must show that $\chi(g) = \chi(h)$. But we have $g = xhx^{-1}$ so that $\rho(g) = \rho(x) \rho(h) \rho(x)^{-1}$.

 Now the result follows since for any $m \times m$ matrices
 $M,P$ with $P$ invertible we have
 $$\operatorname{tr}(PMP^{-1}) = \operatorname{tr}(M).$$

Let us write $\operatorname{Cl}(G)$ for the space of $\mathbb{C}$-valued class functions on $G$.

Thus for any representation $V$ of $G$, we have $\chi = \chi_V \in \operatorname{Cl}(G)$.

We introduce a hermitian inner product $\langle \cdot , \cdot \rangle$ on $\operatorname{Cl}(G)$ by the rule $$\langle\phi,\psi\rangle = \dfrac{1}{|G|} \sum_{x \in G} \phi(x) \overline{\psi(x)}.$$

Thus $$\langle \cdot,\cdot\rangle:\operatorname{Cl}(G) \times \operatorname{Cl}(G) \to \mathbb{C}$$ is linear in the first variable and conjugate linear in the second variable.

Proposition: : a. $\dim \operatorname{Cl}(G)$ is equal to the number of conjugacy classes in $G$. b. The hermitian inner product $\langle \cdot,\cdot\rangle$ is positive definite on $\operatorname{Cl}(G)$.

Sketch: : For a conjugacy class $C$, let $\theta_C$ denote the characteristic function of $C$; thus $\theta_C \in \operatorname{Cl}(G)$ and it is clear that the functions $\{\theta_C\}$ form a basis for $\operatorname{Cl}(G)$. This proves (a).

 For (b), consider two conjugacy classes $C,C'$ and compute:
 $$\langle \theta_C,\theta_{C'} \rangle = \dfrac{1}{|G|} \sum_{x
 \in G} \theta_C(g) \overline{\theta_{C'}(g)} = \delta_{C,C'}
 \dfrac{|C|}{|G|}$$ where $\delta_{C,C'}$ denotes the "Kronecker
 delta". Since $\dfrac{|C|}{|G|}$ is a positive real number, this
 suffices to confirm that the inner product is positive definite.