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 \[\mathbf{1}:G \to \operatorname{GL}(V) = \operatorname{GL}_1(F) = F^\times\] is given by \(g \mapsto 1\) for each \(g \in G\).

Thus for example \(\mathbf{1}\oplus \mathbf{1}\) 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

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 \(G\).

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

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

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 \(h \in H\).

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

2. 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 \[g \sim h \iff g = xhx^{-1} \quad \text{for some $x \in G$.}\] 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:

1. \(\dim \operatorname{Cl}(G)\) is equal to the number of conjugacy classes in \(G\).
2. 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.