Another important question to ask of new mathematical objects:
“How do these things talk to each other?”
From another perspective, we can ask: “When are two of these objects the same?” and “What are functions like on these objects?” This subsection is devoted to making sense of (homo)morphisms of representations and, by extension, isomorphisms of representations.
Morphisms of Actions
Given a group \(G\) and a pair of representations over some common field \(F\), \[
\rho: G \to \operatorname{GL}(V) \text{ and } \sigma: G \to \operatorname{GL}(W),
\] what should be a morphism of representations \(\rho \to \sigma\)? Remember that a representation is about a linear action of a group on a specific vector space; the group \(G\) fixes our context and its action on \(V\) plays the central role. This is why, when context is clear, we often casually refer to \(V\) as the representation of \(G\) rather than invoking \(\rho\) explicitly.
Just as how a homomorphism of groups or rings respects their underlying operations, a morphism of representations should be a way of translating the group action on \(V\) to the group action on \(W\). That is, we want a linear map \(\varphi: V \to W\) so that \(\varphi \circ \rho(g) = \sigma(g) \circ \varphi\) for all \(g \in G\). A more modern way to phrase this is to say “the following diagram commutes”:
\[
\begin{CD}
V @>{\rho(g)}>> V \\
@VV{\varphi}V @VV{\varphi}V \\
W @>{\sigma(g)}>> W
\end{CD}
\tag{5.1}\]
which just means that all composable paths through the diagram with the same start and endpoints produce equal functions. Hence we have the following:
Definition 5.1 Fix \(G\) and a pair of representations \(\rho: G \to \operatorname{GL}(V)\) and \(\sigma: G \to \operatorname{GL}(W)\) over \(F\). Then an intertwiner (a.k.a. a morphism of \(G\)-representations, a \(G\)-equivariant map, or simply a \(G\)-linear map) from \(\rho\) to \(\sigma\) is an \(F\)-linear transformation \(\varphi: V \to W\) such that Equation 5.1 commutes. We write \[
\operatorname{Hom}_G(V,W) \coloneqq \{ \varphi \in \operatorname{Hom}(V,W): \varphi \circ \rho(g) = \sigma(g) \circ \varphi \text{ for all }g \in G \}
\] for the set of all intertwiners, a (potentially trivial) subspace of \(\operatorname{Hom}(V,W)\). Moreover, if an intertwiner \(\varphi\) is an isomorphism of the underlying vector spaces, we say \(\varphi\) is an isomorphism of \(G\)-representations and write \(V \cong W\).
We first note that, as we would hope, this notion of morphism satisfies the expected composition rules:
Proposition 5.1 If \(\psi: U \to V\) and \(\varphi: V \to W\) are intertwiners, then so is \(\varphi \circ \psi\) is also an intertwiner. Moreover, if \(\varphi\) is an isomorphism of \(G\)-representations, then \(\varphi^{-1}\) is also an intertwiner.
Example 5.1 The zero map \(V \to W\) is the trivial intertwiner. If two representations only admit the trivial intertwiner between them, they encode highly incompatible actions of \(G\).
Example 5.2 For any representation \(V\) of \(G\), the identity map \(\mathbb{I}: V \to V\) is an intertwiner. Moreover, if \(\lambda \in F\), then the homothety \(\lambda \mathbb{I}\) is also an intertwiner.
Example 5.3 Consider the trivial and alternating representations, \(U\) and \(U'\), respectively, of \(\mathcal{S}_{n}\) over a field \(F\) with \(\operatorname{char}(F) \not = 2\) and let \(\varphi: U \to U'\) be an intertwiner. Since \(U\) and \(U'\) are both 1-dimensional, \(\varphi = \lambda \mathbb{I}\) for some \(\lambda \in F\). But since \[
\lambda = \varphi(1) = \varphi( (1\ 2) \cdot 1) = (1\ 2) \cdot \varphi(1) = (1\ 2) \cdot \lambda = -\lambda,
\] we must have \(\lambda = 0\). Hence the only intertwiner \(U \to U'\) is the trivial map.
Example 5.4 (Brute Force) Recall the action of Example 3.50, where the group \(\mathcal{A}_{4} = \langle (1\ 2\ 3), (1\ 2)(3\ 4) \rangle\) acts on a tetrahedron by sending a permutation \(\sigma\) to the unique rotation that induces \(\sigma\) on the vertices (with some fixed ordering). If we center the tetrahedron at the origin with vertices \[
x_1 = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}, \quad
x_2 = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix}, \quad
x_3 = \begin{pmatrix} -1 \\ -1 \\ 1 \\ \end{pmatrix}, \quad
\text{ and } \quad x_4 = \begin{pmatrix} 1 \\ 1 \\ 1 \\ \end{pmatrix},
\] then this action gives rise to a representation \(A_4 \to \operatorname{GL}_3(\mathbb{R})\) via \[
(1\ 2\ 3) \mapsto A \coloneqq
\begin{pmatrix}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{pmatrix} \quad \text{ and } \quad (1\ 2)(3\ 4) \mapsto B \coloneqq
\begin{pmatrix}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{pmatrix}.
\] Consider an intertwiner \(\mathbb{R}^3 \to \mathbb{R}^3\) of this action with itself, encoded as a \(3 \times 3\) matrix \(M\). Then the intertwiner condition is equivalent to asking that \(M\) commutes with \(A\) and \(B\). If we write \(m_{i,j}\) for the entry of \(M\) in the \(i\)th row and \(j\)th column, then we can directly compute the commutators (see Listing 5.1 for an example with SageMath): \[
\begin{aligned}
\left[ M, A \right] & = \begin{pmatrix}
m_{3,1}-m_{1,2} & m_{3,2}-m_{1,3} & m_{3,3}-m_{1,1} \\
m_{1,1}-m_{2,2} & m_{1,2}-m_{2,3} & m_{1,3}-m_{2,1} \\
m_{2,1}-m_{3,2} & m_{2,2}-m_{3,3} & m_{2,3}-m_{3,1}
\end{pmatrix}, \\
\left[ M, B \right] & = \begin{pmatrix}
0 & 0 & 2 m_{1,3} \\
0 & 0 & 2 m_{2,3} \\
-2 m_{3,1} & -2 m_{3,1} & 0
\end{pmatrix}.
\end{aligned}
\] We can see directly from the requirement \([M,B]=0\) that \[
m_{1,3}=m_{2,3}=m_{3,1}=m_{3,1}=0,
\] while the additional condition \([M,A]=0\) forces \[
m_{1,2}=m_{2,1}=0 \text{ and } m_{1,1}=m_{2,2}=m_{3,3} \text{ is free}.
\] Hence \(M = \lambda \mathbb{I}\) for some \(\lambda \in \mathbb{R}\), i.e., the only intertwiners are homotheties. As we shall see, this means that this representation \(A_4 ⟳\mathbb{R}^3\) is irreducible.
A voracious student might be dissatisfied with these examples, as they seem rather boring. But, as we are beginning to see, asking for a linear transformation to intertwine actions is a tall order!
Proposition 5.2 Let \(V\) and \(W\) be representations of \(G\) and suppose that \(\varphi: V \to W\) is an intertwiner. Then \(\ker \varphi\) and \(\operatorname{im}\varphi\) are subrepresentations of \(V\) and \(W\), respectively.
Proof. We will show that \(\ker \varphi \leq_G V\) is a subrepresentation; \(\operatorname{im}\varphi \leq_G W\) is left for Exercise D.1. Because \(\varphi\) is linear, we know \(\ker \varphi \leq V\); we need to show that \(\ker \varphi\) is closed under the \(G\)-action. To that end, suppose that \(x \in \ker \varphi\) and \(g \in G\), so that we can show \(g \cdot x \in \ker \varphi\). We can evaluate directly: \[
\overbrace{\varphi(g \cdot x) = g \,\cdot}^{\varphi \text{ is $G$-linear}} \underbrace{\varphi(x) = g \cdot 0}_{x \in \ker \varphi} = 0.
\]
Schur’s Lemma
Next we will quantify exactly how hard it is to find intertwiners. Despite being relatively elementary, this result will allow us to deduce extremely useful computational techniques in later sections.
Lemma 5.1 Let \(V\) and \(W\) be irreducible representations of a finite group \(G\) over a field \(F\) and suppose that \(\varphi: V \to W\) is an intertwiner. Then:
- Either \(\varphi\) is an isomorphism or \(\varphi\) is identically zero.
- If \(V=W\) and \(F=\mathbb{C}\), then \(\varphi = \lambda \mathbb{I}\) for some \(\lambda \in \mathbb{C}\).
In particular, if \(V\) and \(W\) are irreducible representations of a finite group \(G\) over \(\mathbb{C}\), then \[
\dim_{\mathbb{C}} \operatorname{Hom}_G(V,W) = \left\{ \begin{array}{ll}
1 & \text{ if } V \cong W \\
0 & \text{ if } V \not \cong W
\end{array} \right.
\tag{5.2}\]
Proof. Part (a) follows from Proposition 5.2. On the one hand, because the kernel of an intertwiner \(\varphi: V \to W\) is a subrepresentation and \(V\) is irreducible, we can only have \(\ker \varphi = 0\) or \(\ker \varphi = V\); so \(\varphi\) is injective or zero. On the other hand, since the image is a subrepresentation and \(W\) is irreducible, we can only have \(\operatorname{im}\varphi = W\) or \(\operatorname{im}\varphi = 0\); so \(\varphi\) is surjective or zero. All told, \(\varphi\) is an isomorphism or is trivial.
Part (b), we can actually make the argument using any algebraically closed field \(F\). The result follows because any linear transformation has an eigenvalue \(\lambda \in F\), since its characteristic polynomial has at least one root. This means that the map \(\varphi-\lambda \mathbb{I}: V \to V\), which is still an intertwiner, has a nontrivial kernel. As we have seen from part (a), \(\varphi - \lambda \mathbb{I}\) is therefore identically zero and the result follows.
So, given two irreducible representations of the same group \(G\), we now see that they are either completely the same or utterly unintelligible to each other. Morally, this result makes concrete the fact that intertwining group actions is very hard—that we should treat non-trivial instances of this phenomena as exceedingly special! In fact, the powerful and precise theory we will develop in this course is in some sense due to the rigidity of this result.
Next, we prove an important lemma:
Lemma 5.2 Let \(G\) be a finite group with \(V\) a finite-dimensional representation over \(\mathbb{C}\). If we suppose further that \(W \leq_G V\) is an irreducible representation, then the inclusion \(W \hookrightarrow V\) and orthogonal projection \(P: V \twoheadrightarrow W\) (with respect to the \(G\)-invariant inner product of Lemma 4.1) are intertwiners.
Proof. That inclusion is an intertwiner is clear, since the action \(G ⟳W\) is a restriction of the action \(G ⟳V\). On the other hand, for any \(x \in V\) and \(y \in W\), we have \[
\langle x, y \rangle = \langle Px, y \rangle,
\] since \(x - Px \in W^\perp\). Now for any \(g \in G\), we have \[
\langle P(g \cdot x), y \rangle
= \langle g \cdot x, y \rangle
= \langle x, g^{-1} \cdot y \rangle
= \langle Px, g^{-1} \cdot y \rangle
= \langle g \cdot Px, y \rangle,
\] where we use that \(g^{-1} \cdot y \in W\). Hence \(P(g \cdot x) = g \cdot Px\), which is what we wanted to show.
Our main result of this secton is the following:
Theorem 5.1 Let \(G\) be a finite group and \(V\) a finite-dimensional representation over \(\mathbb{C}\). Then there is a decomposition \[
V = V_1 \oplus \cdots \oplus V_r,
\] where the \(V_i\) are irreducible representations. Moreover, this decomposition is unique in the sense that, if we have another decomposition \[
V = W_1 \oplus \cdots \oplus W_s,
\] then \(r=s\) and, after re-ordering, we have \(V_i \cong W_i\) for each \(i\).
Before proceeding with the proof, we consider some examples.
Example 5.5 Recall the permutation action of \(\mathcal{S}_{3}\) on \(\mathbb{C}^3\) expressed in the ordered basis \[
(e_1+e_2+e_3,e_1-e_2,e_2-e_3),
\] where we write \(\rho: G \to \operatorname{GL}_3(\mathbb{C})\) to denote the associated matrices, and in the basis \[
(e_1+e_2+e_3,\zeta^2 e_1 + \zeta e_2 + e_3, \zeta e_1 + \zeta^2 e_2 + e_3),
\] where \(\zeta = e^{2 \pi i/3}\), which we denote by \(\sigma: G \to \operatorname{GL}_3(\mathbb{C})\) to emphasize the distinction. An intertwiner of \(\mathbb{C}^3_\rho\) and \(\mathbb{C}^3_\sigma\) is given by \(M \in \operatorname{Mat}_3(\mathbb{C})\) such that \[
\begin{aligned}
M & \underbrace{\begin{pmatrix}
1 & 0 & 0 \\
0 & -1 & 1 \\
0 & 0 & 1
\end{pmatrix}}_{\rho((1\ 2))} = \underbrace{\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}}_{\sigma((1\ 2))} M \\
M & \underbrace{\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & -1 \\
0 & 1 & -1
\end{pmatrix}}_{\rho((1\ 2\ 3))} = \underbrace{\begin{pmatrix}
1 & 0 & 0 \\
0 & \zeta & 0 \\
0 & 0 & \zeta^2
\end{pmatrix}}_{\sigma((1\ 2\ 3))} M.
\end{aligned}
\] If we write \(m_{ij}\) for the entries of \(M\), we can compute the modified commutators via Sage:
\(\displaystyle \verb|(1|\verb| |\verb|2)|\verb| |\verb|condition:| \left(\begin{array}{rrr}
0 & -2 \, m_{12} & m_{12} \\
m_{21} - m_{31} & -m_{22} - m_{32} & m_{22} + m_{23} - m_{33} \\
-m_{21} + m_{31} & -m_{22} - m_{32} & -m_{23} + m_{32} + m_{33}
\end{array}\right)\)
\(\displaystyle \verb|(1|\verb| |\verb|2|\verb| |\verb|3)|\verb| |\verb|condition:| \left(\begin{array}{rrr}
0 & -m_{12} + m_{13} & -m_{12} - 2 \, m_{13} \\
-\frac{1}{2} i \, \sqrt{3} m_{21} + \frac{3}{2} \, m_{21} & -\frac{1}{2} i \, \sqrt{3} m_{22} + \frac{1}{2} \, m_{22} + m_{23} & -\frac{1}{2} i \, \sqrt{3} m_{23} - m_{22} - \frac{1}{2} \, m_{23} \\
\frac{1}{2} i \, \sqrt{3} m_{31} + \frac{3}{2} \, m_{31} & \frac{1}{2} i \, \sqrt{3} m_{32} + \frac{1}{2} \, m_{32} + m_{33} & \frac{1}{2} i \, \sqrt{3} m_{33} - m_{32} - \frac{1}{2} \, m_{33}
\end{array}\right)\)
By inspection, we see that \(m_{12} = m_{13} = m_{21} = m_{31} = 0\) and \(m_{22} = -m_{32} = \zeta^2 m_{23} = -\zeta m_{33}\). Writing our free variables as \(\alpha = m_{11}\) and \(\beta = m_{12}\), we have shown: \[
\begin{aligned}
\operatorname{Hom}_{\mathcal{S}_{3}}(\mathbb{C}^3_\rho,\mathbb{C}^3_\sigma) & =
\left\{ \begin{psmallmatrix}
\alpha & 0 & 0 \\
0 & \beta & \zeta \beta \\
0 & -\beta & - \zeta^2 \beta
\end{psmallmatrix} \mid \alpha, \beta \in \mathbb{C}\right\} \\
& = \operatorname{Span}\left\{
\begin{psmallmatrix}
1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{psmallmatrix}, \begin{psmallmatrix}
0 & 0 & 0 \\
0 & 1 & \zeta \\
0 & -1 & - \zeta^2
\end{psmallmatrix}
\right\}.
\end{aligned}
\] Note the block diagonal form of \(M\), which will be reflected in the proof of Theorem 5.1 to follow; since \[
\begin{gathered}
\mathbb{C}^3_\rho = \underbrace{\operatorname{Span}\{e_1+e_2+e_3\}}_U \oplus \underbrace{\operatorname{Span}\{e_1-e_2,e_2-e_3\}}_V \quad \text{and} \\
\mathbb{C}^3_\sigma = \underbrace{\operatorname{Span}\{e_1+e_2+e_3\}}_U \oplus \underbrace{\operatorname{Span}\{\zeta^2 e_1 + \zeta e_2 + e_3, \zeta e_1 + \zeta^2 e_2 + e_3\}}_V
\end{gathered}
\] give decompositions of the respective representations, the intertwiners can only hope to compare compatible irreducible constituents of \(\rho\) and \(\sigma\). The fact that these representations have \(2\) irreducible subrepresentations in common is reflected by the fact that the space of intertwiners between them is \(2\)-dimensional.
Example 5.6 With the previous example in mind, consider an entirely new action on \(\mathbb{C}^3\): \[
\tau((1\ 2)) = \begin{pmatrix}
0 & 0 & -1 \\
-1 & -1 & 1 \\
-1 & 0 & 0
\end{pmatrix} \quad \text{ and } \quad \tau((1\ 2\ 3)) = \begin{pmatrix}
0 & 0 & 1 \\
1 & 1 & -1 \\
0 & 1 & -1
\end{pmatrix}.
\] If we repeat the previous analysis, attempting to intertwine \(\mathbb{C}^3_\rho \to \mathbb{C}^3_\tau\), we find that \[
\operatorname{Hom}_{\mathcal{S}_{3}}(\mathbb{C}^3_\rho,C^3_\tau) = \operatorname{Span}\left\{ \begin{pmatrix}
0 & 1 & 1 \\
0 & -1 & -1 \\
0 & 1 & -2
\end{pmatrix} \right\}
\] None of these intertwiners can be full-rank, so we can conclude that \(\mathbb{C}^3_\rho\) and \(\mathbb{C}^3_\sigma\) are not isomorphic as \(\mathcal{S}_{3}\)-representations. If we write \(U\) for the trivial representation, \(U'\) for the alternating representation, and \(V\) for the standard representation, all this reflects the (unproven, so far) fact that \[
\mathbb{C}^3_{\rho} \cong U \oplus V \cong \mathbb{C}^3_{\sigma} \quad \text{ but } \quad \mathbb{C}^3_{\tau} \cong U' \oplus V.
\] Since \(U\) and \(U'\) are distinct irreducible representations, Lemma 5.1 tells us that the only intertwiner between them is \(0\). Therefore, any intertwiner \(\mathbb{C}^3_\tau \to \mathbb{C}^3_\sigma\) can only compare the \(V\)-components of their substructure.
We conclude by stating an explicit corollary to Lemma 5.1 in this vein:
Corollary 5.1 Let \(V\) and \(W\) be irreducible \(G\)-representations. Then \[
\dim \operatorname{Hom}_G(V,W) = \left\{ \begin{array}{ll}
1 & V \cong W \\
0 & V \not \cong W.
\end{array} \right.
\]
Proof. If \(V\) and \(W\) are non-isomorphic, then Schur’s Lemma says any intertwiner between them is the zero map, i.e., \(\dim \operatorname{Hom}_G(V,W) = 0\). If instead \(V \cong W\), then there is some \(\varphi: V \to W\) an intertwiner isomorphism. We will show that if \(\psi: V \to W\) is another intertwiner, then \(\psi = \lambda \varphi\) for some \(\lambda \in \mathbb{C}\). Indeed, note that \[
\psi \circ \varphi^{-1}: W \to W
\] is also an intertwiner, and by the second part of Schur’s Lemma we have \(\psi \circ \varphi^{-1} = \lambda \mathbb{I}\). Hence \(\psi = \lambda \varphi\), and so \(\operatorname{Hom}_G(V,W) = \operatorname{Span}\{ \varphi \}\) is \(1\)-dimensional.
Structure of Intertwiners
Definition 5.2 Suppose that \(V\) is a \(G\)-representation which decomposes into irreducibles as \[
V = V_1 \oplus \cdots \oplus V_r.
\] For a given irreducible representation \(U\), its \(U\)-isotypic component in \(V\) is the subrepresentation \[
V_{(U)} \coloneqq \bigoplus_{\substack{i \operatorname{s.t.}\\ V_i \cong U}} V_i.
\] We will see momentarily (in the proof of Theorem 5.1) that the isotypic component are well-defined regardless of how we chose to decompose \(V\) into irreducibles.
Proposition 5.3 Suppose \(V\) and \(W\) are \(G\)-representations, decomposed into irreducibles as \[
V = V_1 \oplus \cdots \oplus V_r \quad \text{and} \quad W = W_1 \oplus \cdots \oplus W_s.
\] Let \(\varphi: V \to W\) be an intertwiner. Then \(\varphi\) sends isotypic components of \(V\) into isotypic components of \(W\), i.e., \[
\varphi\left(V_{(U)}\right) \leq W_{(U)}
\] for all irreducible \(G\)-representations \(U\).
Proof. Given irreducible decompositions as above, an intertwiner \(\varphi: V \to W\) determines a family of intertwiners \(\varphi_{ij}: V_i \to W_j\) by prepending the inclusion \(V_i \hookrightarrow V\) and appending the orthogonal projection \(W \twoheadrightarrow W_j\) (as in Lemma 5.2): \[
\varphi_{ij} : V_i \hookrightarrow V \stackrel{\varphi}{\to} W \twoheadrightarrow W_j
\] From here, Lemma 5.1 tells us that each \(\varphi_{ij}\) is either an isomorphism or the zero map. In particular, if \(\varphi\) is injective, then for each \(i\) there must be some \(j\) such that \(\varphi_{ij}\) is an isomorphism, and hence \(V_i \cong W_j\). This means that the image of \(\varphi\) restricted to a given irreducible \(V_i\) is contained in the direct sum of all the subrepresentations of \(W\) which are isomorphic to \(V_i\). Indeed, if \[
x = x_1 + \cdots + x_r \in V_1 \oplus \cdots \oplus V_r
\] is the unique expression of \(x\) in terms of our decomposition, then \[
\varphi(x) = (\underbrace{\varphi_{11}(x_1) + \cdots + \varphi_{r1}(x_r)}_{\in W_1}) + \cdots + (\underbrace{\varphi_{1s}(x_1) + \varphi_{2s}(x_2) + \cdots + \varphi_{rs}(x_r)}_{\in W_s}) \\
\tag{5.3}\] gives the unique decomposition of \(\varphi(x)\) as an element of \(W_1 \oplus \cdots \oplus W_s\). In particular, when we only input \(x_i \in V_i\), we get \[
\varphi(x_i) = \underbrace{\varphi_{i1}(x_i)}_{\in W_1} + \cdots + \underbrace{\varphi_{is}(x_i)}_{\in W_s}. \\
\tag{5.4}\] The only terms in the above which can be non-zero are those for which \(V_i \cong W_j\), i.e., \(\varphi(x_i)\) lies in the indicated component of \(W\).
Uniqueness of Decompositions
We are now equipped to prove the uniqueness of decompositions:
Proof (Proof of Theorem 5.1). Suppose we have two decompositions of \(V\) into irreducibles: \[
V = V_1 \oplus \cdots \oplus V_r = W_1 \oplus \cdots \oplus W_s.
\] For a given irreducible representation \(U\) of \(G\), we consider the \(U\)-isotypic components (with respect to each decomposition) and their complements under the \(G\)-invariant inner products on \(V\). We set \[
X = \bigoplus_{\substack{i \operatorname{s.t.}\\ V_i \cong U}} V_i \quad \text{and} \quad Y = \bigoplus_{\substack{j \operatorname{s.t.}\\ W_j \cong U}} W_j,
\] so that \(X^{\perp}\) and \(Y^{\perp}\) are the sums of \(V_i\) and \(W_j\), respectively, which are not isomorphic to \(U\). Hence we have \[
V = X \oplus X^{\perp} = Y \oplus Y^{\perp}
\] and, by Proposition 5.3, any intertwiner \(\varphi: V \to V\) must have \(\varphi(X) \leq_G Y\).
In the case when \(\varphi\) is an isomorphism (in particular, \(\varphi = \mathbb{I}_V\)) we must have \(\varphi(X)=Y\) and \(\varphi(X^{\perp})=Y^{\perp}\). Hence the \(U\)-isotypic components in the decompositions are identified: \(X = Y\). It follows that the number of \(V_i\) isomorphic to \(U\) is equal to the number of \(W_j\) isomorphic to \(U\), simply by counting dimensions. This is true for each irreducible \(U\), and hence we have exactly the desired identification of irreducible factors.
We emphasize that two decompositions of \(V\) into irreducibles—literally as individual subspaces—need not agree, but isotypic components are uniquely determined. Indeed, this gives a useful rephrasing of Theorem 5.1, akin to how we factor integers into prime powers.
Now that we know the isotypic components of a representation are well-defined as subspaces regardless of our choice of decomposition, we introduce a bit of notation. For a given subrepresentation \(U \leq V\) which occurs \(a\) times in the irreducible decomposition of \(V\), we write \[
U^{\oplus a} \coloneqq \underbrace{U \oplus \cdots \oplus U}_a,
\] which we identify with the specific \(U\)-isotypic component in \(V\).
First Consequences
With all this in mind, we modify our priorities of representation theory:
The first of these questions can be readily answered by studying a particular representation which will continue to arise many times throughout this course: the group algebra \(\mathbb{C}[G]\).
Proposition 5.4 Let \(G\) be a finite group and let \(V\) be an irreducible representation over \(\mathbb{C}\). Then \(V\) is isomorphic to a subrepresentation of \(\mathbb{C}[G]\).
Since \(\dim \mathbb{C}[G] = |G| < \infty\), we conclude:
Corollary 5.3 Up to isomorphism, \(G\) has only finitely many irreducible representations.
Proof (Proof of Proposition 5.4). Let \(V\) be an irreducible representation of \(G\) and pick any nonzero \(x \in V\). We construct a linear map \(\varphi: \mathbb{C}[G] \to V\) as follows: \[
\varphi\left( \sum_{h\in G} \alpha_h e_h \right) \coloneqq \sum_{h \in G} \alpha_h (h \cdot x).
\] Note that \(\varphi\) is an intertwiner, which we can check on basis vectors: \[
\varphi(g \cdot e_h) = \varphi(e_{gh}) = (gh) \cdot x = g \cdot (h \cdot x) = g \cdot \varphi(e_h)
\] Moreover, \(\operatorname{im}\varphi\) is nonzero since \(x = \varphi(e_e) \in \operatorname{im}\varphi\) and \(V\) is irreducible and so, by Proposition 5.2, \(\varphi\) must be onto. By Proposition 5.3, in the decomposition of \(\mathbb{C}[G]\) into irreducibles, there must be some subrepresentation isomorphic to \(V\).
Moreover, Schur’s Lemma is sufficient to conclude strong constraints on the representations of abelian groups.
Proposition 5.5 Any irreducible representation over \(\mathbb{C}\) of an abelian group is \(1\)-dimensional.
Proof. Let \(\rho: A \to \operatorname{GL}(V)\) be of degree \(n\). The transformations in \(\operatorname{im}\rho\) all commute, so we can choose a basis for \(V\) so that the \(\rho(g)\) are diagonal for every \(g \in G\). Thus \(V\) decomposes into \(n\) subrepresentations, as in Example 2.24, so \(V\) can only be irreducible if \(n=1\).
There is a delightful alternative argument for this claim:
Proof (Alternative proof). Take \(\rho: A \to \operatorname{GL}(V)\) a representation and fix \(h \in G\). Then the transformation \(\rho(h): V \to V\) is itself an intertwiner: \[
\rho(h)(g \cdot x) = h \cdot (g \cdot x) = (hg) \cdot x = (gh) \cdot x = g \cdot \rho(h)(x).
\] If \(V\) is irreducible, then Lemma 5.1 guarantees that \(\rho(h) = \lambda_h \mathbb{I}\) for some \(\lambda_h \in \mathbb{C}^\times\). This holds for all \(h \in G\), so as before we can only have \(\dim V = 1\).
We conclude with another result in a similar spirit, giving us a way of picking out the invariants of a representation by averaging over all the constituent linear transformations.
Theorem 5.2 For any finite group \(G\) and representation \(\rho: G \to \operatorname{GL}(V)\), the map \[
\varphi = \frac{1}{|G|} \sum_{h \in G} \rho(h)
\] is an orthogonal projection \(V \twoheadrightarrow V^G\) with respect to the inner product \(\langle \cdot, \cdot \rangle_G\) of Lemma 4.1.