A primer on group representations

In this post, we will give a brief introduction to the representation theory of general (finite) groups.

Definition: A representation of a group G is a homomorphism G\to\mathrm{GL}_n(\mathbf{C}), where n is called the dimension of the representation.

A representation is faithful if the map G\to\mathrm{GL}_n(\mathbf{C}) is injective. A representation G\to\mathrm{GL}_n(\mathbf{C}) can also be viewed as the action of G on an n-dimensional \mathbf{C}-vector space V because \mathrm{Aut}(V) is isomorphic to \mathrm{GL}_n(\mathbf{C}) via the isomorphism V\cong \mathbf{C}^n. However, this requires fixing a basis. This leads to the following definition.

Definition: Let V be a \mathbf{C}-vector space and G a group. A representation of G on V is a map \rho:G\to\mathrm{Aut}(V). This is equivalent to an operation G\times V\to V defined via (g,v)\mapsto \rho(g)(v).

An isomorphism of representations \rho and \rho^\prime of G on V and V^\prime respectively is an isomorphism V\to V^\prime that is consistent with the action of G on V and V^\prime. We will henceforth denote \mathrm{Aut}(V) by \mathrm{GL}(V). If \rho:G\to\mathrm{GL}_n(\mathbf{C}) is a representation of G, the character of \rho is \mathbf{C}-valued function on G defined as \chi_\rho(g) = \mathrm{tr}(\rho(g)). We claim that characters are constant on conjugate classes. Suppose that g^\prime = hgh^{-1}. Then \rho(g^\prime) = \rho(h)\rho(g)\rho(h^{-1}). Since \rho(h^{-1}) = \rho(h)^{-1}, \rho(g^\prime) and \rho(g) are conjugate, they have the same trace.

Consider a character \chi associated to a representation \rho:G\to\mathrm{GL}(V). Then \chi(1) is the dimension of the representation of G since \rho(1) is the identity on V, whose trace is the dimension of V. Another important statement is the following. Suppose |G|=k, and suppose g\in G. Then the roots of the characteristic polynomial of \rho(g) are powers \zeta^n of the kth roots of unity. Since the trace of a matrix is a sum of its eigenvalues, and if \omega is an eigenvalue of \rho(g), then \omega^k is an eigenvalue of \rho(g)^k = \rho(g^k)= \rho(1), and hence \omega must be a power of e^{2\pi i/k}. This also shows that \rho(g^{-1}) = \overline{\rho(g)}.

Let G act on a \mathbf{C}-vector space V. A subspace W of V is G-invariant if the action of G on V reduces to an action of G on W. If V=W\oplus Z, where W and Z are G-invariant subspaces, then the representation on V is the direct sum of the representations on W and Z. If \rho is the representation on V, then the matrix associated to \rho(g) will be in block diagonal form. This leads to the following definition.

Definition: A representation of G on a \mathbf{C}-vector space V is irreducible if there is no proper G-invariant subspace of V.

Let \rho be a non-irreducible (reducible) representation of G on V, and consider a proper G-invariant subspace W\subset V. The basis of W can be extended to V, and then the matrix associated to \rho(g) will be in block form, where the upper left matrix acts on the basis of W, and the lower right matrix acts on some other representation of G. Maschke’s theorem says:

Theorem: If G is a finite group, every representation of G on a nonzero finite-dimensional \mathbf{C}-vector space is the direct sum of irreducible representations of G.

Proof: The proof is two-part. Say that a representation \rho:G\to\mathrm{GL}(V) on a \mathbf{C}-vector space V is unitary if for all g\in G, the matrix associated to \rho(g) is unitary, i.e., \rho(g)^\ast = \rho(g)^{-1}. Suppose \rho is a reducible unitary representation of G on a Hermitian space V. If W is a G-invariant subspace of V, then W^\perp is also G-invariant, because since \rho is unitary, if v\perp W, then \rho(g)(v)\perp \rho(g)(W); since W is G-invariant, \rho(g)(v)\in W^\perp. Since V=W\oplus W^\perp, it follows that \rho is the direct sum of the restrictions to W and W^\perp. Therefore it follows that every unitary representation on a Hermitian space is an orthogonal sum of irreducible representations.

Now it suffices to prove that if \rho:G\to\mathrm{GL}(V) is a representation of a finite group on a \mathbf{C}-vector space V, then there is a G-invariant positive-definite Hermitian form on V (making it a Hermitian space). Consider some positive-definite Hermitian form [,] on V, and define

\displaystyle\langle v,w \rangle = \frac{1}{|G|}\sum_{g\in G}[gv,gw].

Since G is finite, this is clearly G-invariant. Checking that this new form is positive-definite and Hermitian is easy, and this proves the required result. QED.

We have Schur’s lemma:

Theorem: If \rho,\rho^\prime are irreducible representations of G on V and V^\prime, and if T:V\to V^\prime is a G-invariant linear transformation, then either T is an isomorphism or T=0. In addition, if T:V\to V is a G-invariant linear transformation then T = cI for some scalar c.

Proof: To prove the first statement, consider the subspace \ker T. Clearly this is G-invariant, so \ker T is either V or \{0\}. If T\neq 0, then \ker T cannot be V, so \ker T=\{0\}, i.e., T is injective. In this case, because \mathrm{im}(T) is also G-invariant, it is either V^\prime or \{0\}. Since T\neq 0, the image of T is V^\prime, and T is an isomorphism. To prove the second statement, let \zeta be an eigenvalue of T. Then \ker(T-\zeta I) is not zero since it contains an eigenvector of T. Therefore, T-\zeta I is not an isomorphism, and so T-\zeta I = 0, i.e., T=\zeta I. QED.

There are a number of important consequences. For example, suppose that G is abelian, and consider an irreducible representation \rho:G\to\mathrm{GL}(V). Because G is abelian, \rho(g)\rho(h)v = \rho(h)\rho(g)v, and therefore, for every g\in G, by Schur’s lemma, \rho(g) must be scalar multiplication. In this case, if V is not one-dimensional, then every subspace of V is a G-invariant subspace – but this contradicts irreducibility! Therefore, every irreducible representation of an abelian group G is one-dimensional. We will see another proof below of this when the abelian group is finite.

Definition: Let \chi,\chi^\prime be characters of G. There is a Hermitian product defined via

\displaystyle\langle \chi,\chi^\prime \rangle = \frac{1}{|G|}\sum_g\chi^\prime(g)\overline{\chi(g)}.

Because \chi is constant on conjugacy classes it is clear that if C_j denotes the conjugacy classes of G and g_j are representatives of C_j, then

\displaystyle\langle \chi,\chi^\prime \rangle = \frac{1}{|G|}\sum |C_j|\chi^\prime(g_j)\overline{\chi(g_j)} = \sum\frac{1}{|C_G(g_j)|}\chi^\prime(g_j)\overline{\chi(g_j)}.

The main theorem of representation theory (without proof) is the following.

Theorem: Suppose G is a finite group. Then:

  1. The irreducible characters of G form an orthonormal basis of class functions (functions that are constant on conjugacy classes) of G. If \rho,\rho^\prime are irreducible representations of G, then \langle \chi,\chi^\prime \rangle = 1 if \rho and \rho^\prime are isomorphic, and is 0 otherwise.
  2. The number of isomorphism classes of irreducible representations is finite, and is the equal to the number of conjugacy classes of the group.
  3. Suppose \rho_i represent the irreducible representations of G. Then each \mathrm{dim}(\rho_i) divides |G|, and: |G| = \sum \mathrm{dim}(\rho_i)^2.

There are many corollaries of this. For example, we get another proof that representations of abelian groups are one-dimensional, in the case when the group is finite. If G is finite abelian and \chi:G\to\mathbf{C}^\times is an irreducible character of G, then \chi must be one-dimensional. To see this, note that since G is finite abelian, the number of conjugacy classes of G is |G|, which is also the number of irreducible characters of G. Since |G| = \sum^{|G|} \mathrm{dim}(\chi_i)^2, each \mathrm{dim}(\chi_i) must be 1. We will conclude with the following result.

Corollary: Representations of a finite group G are isomorphic if and only if their characters are equal.

Proof. Since any representation \rho can be written as \bigoplus_i m_i\rho_i, so the associated character \chi can be written as \sum m_i\chi_i. Clearly m_i = \langle \chi,\chi_i \rangle by the orthogonal projection formula, completing the proof.



Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s