21. Modules and representations¶

Linear algebra works over a field: you can divide by any nonzero scalar, and every vector space has a basis. What happens if the scalars are only a ring, such as the integers or the polynomials? You get modules, and one theorem about them, the structure theorem over a principal ideal domain, turns out to contain both the classification of finite abelian groups and the Jordan normal form of a matrix. Then we turn the idea around: instead of a ring acting on a vector space, let a group act on it. That is a representation, and its character, a short list of traces, holds almost everything about it. This is the algebra that chemists use to count the ways a molecule can vibrate and that physicists use to sort particles.

In this chapter

You will learn:

what a module over a ring is, with examples that you already know under other names
how modules differ from vector spaces: torsion, and modules with no basis
the structure theorem for finitely generated modules over a PID, the Smith normal form, and its two children: finite abelian groups and the Jordan form
what a representation of a finite group is, and Maschke's theorem (every representation splits into irreducibles)
Schur's lemma, characters and the orthogonality relations, with the full character table of \(S_3\)
how chemistry and physics use all this: molecular vibrations, atomic orbitals, the eightfold way, crystals

You need: Chapter 17 and Chapter 18 (rings, ideals, PIDs), Chapter 15 (quotients, isomorphism theorems, finite abelian groups stated), Chapter 16 (group actions, conjugacy classes), and linear algebra: matrices, eigenvalues, change of basis, as in MAT240 Unit 6 and Unit 10.

Levels: 21.1 is advanced; 21.2 is advanced to expert; 21.3 and 21.4 are advanced; 21.5 is expert; 21.6 is applications for everyone who has read 21.3.

21.1 Modules over a ring¶

Level: advanced. Here are four situations that look different.

A vector space \(V\) over a field \(F\): you can add vectors and multiply them by scalars.
An abelian group \(A\), written additively: you can add, and you can multiply by an integer, \(3a = a + a + a\) and \((-2)a = -(a + a)\).
An ideal \(I\) of a ring \(R\): you can add elements of \(I\) and multiply them by anything in \(R\), staying in \(I\).
A vector space \(V\) with a fixed linear map \(T: V \to V\): you can multiply a vector by a polynomial, by the rule \(p(x) \cdot v = p(T)v\). For instance \((x^2 + 1)\cdot v = T(T(v)) + v\).

All four are the same structure: an abelian group with a ring acting on it.

Definition. Let \(R\) be a ring with 1. A (left) \(R\)-module is an abelian group \((M, +)\) with a multiplication \(R \times M \to M\), \((r, m) \mapsto rm\), such that for all \(r, s \in R\) and \(m, n \in M\):

\[ r(m + n) = rm + rn, \qquad (r + s)m = rm + sm, \qquad (rs)m = r(sm), \qquad 1m = m . \]

These are exactly the axioms of a vector space, with "field" replaced by "ring". So:

an \(F\)-module, for a field \(F\), is a vector space over \(F\);
a \(\mathbb{Z}\)-module is an abelian group (there is only one way to multiply by integers, repeated addition);
an ideal of \(R\), and \(R\) itself, are \(R\)-modules; so is \(R^n\), columns of \(n\) elements of \(R\);
a vector space with a linear map \(T\) is an \(F[x]\)-module, written \(V_T\).

The usual words carry over. A submodule is a subgroup closed under the ring action; a homomorphism of modules \(\varphi: M \to N\) respects addition and \(\varphi(rm) = r\varphi(m)\) (for vector spaces, a linear map); the kernel and image are submodules; the quotient \(M/N\) by a submodule is a module, and the isomorphism theorems hold word for word as for groups (Chapter 15). For \(V_T\), a submodule is a subspace \(W\) with \(T(W) \subseteq W\): an invariant subspace. That observation is why modules will produce the Jordan form.

Generators and bases. \(M\) is generated by \(m_1, \ldots, m_n\) if every element is \(r_1 m_1 + \cdots + r_n m_n\); finitely generated if some finite list works. The list is independent if \(r_1 m_1 + \cdots + r_n m_n = 0\) only when all \(r_i = 0\), and a basis if it is both. A module with a basis is free; then \(M \cong R^n\).

Where modules differ from vector spaces¶

Every vector space has a basis. Modules often do not.

\(\mathbb{Z}/6\) as a \(\mathbb{Z}\)-module is generated by \(1\), but \(6 \cdot 1 = 0\) with \(6 \ne 0\), so \(\{1\}\) is not independent; in fact no nonempty subset is independent, since \(6m = 0\) for every \(m\). It is not free.
\(\mathbb{Q}\) as a \(\mathbb{Z}\)-module is not finitely generated (finitely many fractions have a common denominator \(d\), and \(\tfrac{1}{2d}\) is not a combination of them), and not free (any two elements \(\tfrac{a}{b}, \tfrac{c}{d}\) satisfy \((bc)\tfrac{a}{b} - (ad)\tfrac{c}{d} = 0\)).
A submodule of a free module need not be free. In \(R = \mathbb{Z}[x]\), the ideal \(I = (2, x)\) is a submodule of the free module \(R\). It needs two generators (it is not principal, Chapter 18), but any two elements \(f, g\) of \(I\) satisfy \(g \cdot f - f \cdot g = 0\), so no two are independent. Not free.

The first example shows the main new phenomenon. An element \(m\) is torsion if \(rm = 0\) for some nonzero \(r \in R\). Over an integral domain the torsion elements form a submodule \(M_{\mathrm{tors}}\); \(M\) is torsion-free if it is \(0\), and torsion if it is everything. In a finite abelian group every element is torsion (Lagrange). In \(V_T\), every element is torsion: if \(\dim V = n\), the \(n + 1\) vectors \(v, Tv, \ldots, T^n v\) are dependent, so some nonzero polynomial kills \(v\). That fact, read as module theory, is a version of the Cayley-Hamilton theorem.

21.2 Modules over a PID: the structure theorem¶

Level: advanced, then expert. Over a field, a finitely generated module is \(F^n\), classified by one number. Over a principal ideal domain the answer is almost as clean.

Theorem 21.1 (structure theorem). Let \(R\) be a PID and \(M\) a finitely generated \(R\)-module. Then

\[ M \;\cong\; R^r \oplus R/(d_1) \oplus R/(d_2) \oplus \cdots \oplus R/(d_k), \]

where \(d_1, \ldots, d_k\) are nonzero non-units with \(d_1 \mid d_2 \mid \cdots \mid d_k\). The rank \(r\) and the ideals \((d_i)\), called the invariant factors, are uniquely determined by \(M\).

By the Chinese remainder theorem (Chapter 15), \(R/(ab) \cong R/(a) \oplus R/(b)\) when \(a, b\) are coprime, so each \(R/(d_i)\) splits into pieces \(R/(p^e)\) for prime powers dividing \(d_i\). This gives the equivalent elementary divisor form:

\[ M \;\cong\; R^r \oplus R/(p_1^{e_1}) \oplus \cdots \oplus R/(p_s^{e_s}), \]

with the prime powers \(p_j^{e_j}\) (repetitions allowed) unique up to order and associates.

Idea of the proof: matrices over \(R\)¶

Choose generators \(m_1, \ldots, m_n\) of \(M\). The map \(R^n \to M\), \((r_1, \ldots, r_n) \mapsto \sum r_i m_i\), is onto, so \(M \cong R^n / K\), where \(K\) is the module of relations among the generators. Over a PID a submodule of \(R^n\) is free of rank at most \(n\) (proof by induction on \(n\), using that every ideal of \(R\) is principal: Dummit and Foote, Theorem 12.4). So \(K\) has a finite basis, and we can record it as an integer-like matrix \(A\) whose rows are the relations.

Now change bases. Column operations on \(A\) (add a multiple of one column to another, swap columns, multiply a column by a unit) change the generators of \(M\); row operations change the basis of the relations. Neither changes \(M\). Over a Euclidean domain these operations reach the Smith normal form:

Theorem 21.2 (Smith normal form, 1861). Over a Euclidean domain, every matrix can be brought by invertible row and column operations to a diagonal matrix \(\operatorname{diag}(d_1, \ldots, d_k, 0, \ldots, 0)\) with \(d_1 \mid d_2 \mid \cdots \mid d_k\).

Proof. If \(A = 0\) there is nothing to do. Otherwise move a nonzero entry of smallest size (absolute value in \(\mathbb{Z}\), degree in \(F[x]\)) to the top-left corner. Divide every other entry of the first row and column by it; subtracting the quotient times the first column (or row) leaves remainders. If a remainder is nonzero it is smaller than the corner, so move it to the corner and repeat; sizes cannot decrease forever, so eventually the first row and column are zero apart from the corner \(a\). If some other entry \(b\) is not divisible by \(a\), add its row to the first row and repeat the process, which produces a smaller corner (a remainder of \(b\) by \(a\)). When it stops, \(a\) divides everything else. Now do the same to the smaller matrix left over, whose entries are all multiples of \(a\), and so on. \(\blacksquare\)

(Henry Smith published the theorem for integer matrices in 1861. For a general PID the same is true with a slightly different argument, using gcds instead of division.)

If \(A\) reduces to \(\operatorname{diag}(d_1, \ldots, d_k, 0, \ldots)\), then in the new generators the relations are just \(d_1 m_1' = 0, \ldots, d_k m_k' = 0\), and the remaining \(n - k\) generators are free. That is the structure theorem, with the units \(d_i\) (giving \(R/(1) = 0\)) dropped. Uniqueness is proved by counting: for instance, for a prime \(p\), the dimension of \(M/pM\) over the field \(R/(p)\), and of \(p^jM/p^{j+1}M\), can be read off from either decomposition and pin down the elementary divisors.

Worked example. The abelian group \(A\) with generators \(a, b\) and relations \(4a + 6b = 0\), \(6a + 4b = 0\). The relation matrix and its reduction:

\[ \begin{pmatrix} 4 & 6 \\ 6 & 4 \end{pmatrix} \xrightarrow{C_2 - C_1} \begin{pmatrix} 4 & 2 \\ 6 & -2 \end{pmatrix} \xrightarrow{C_1 \leftrightarrow C_2} \begin{pmatrix} 2 & 4 \\ -2 & 6 \end{pmatrix} \xrightarrow{C_2 - 2C_1} \begin{pmatrix} 2 & 0 \\ -2 & 10 \end{pmatrix} \xrightarrow{R_2 + R_1} \begin{pmatrix} 2 & 0 \\ 0 & 10 \end{pmatrix} . \]

So \(A \cong \mathbb{Z}/2 \oplus \mathbb{Z}/10\), and in elementary divisors \(\mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/5\). Check: \(|A| = |\det| = |16 - 36| = 20 = 2 \cdot 10\). (In general the order of \(\mathbb{Z}^n/A\mathbb{Z}^n\), for a square \(A\) with nonzero determinant, is \(|\det A|\), because row and column operations change the determinant only by a sign.)

First child: finitely generated abelian groups¶

Take \(R = \mathbb{Z}\). Every finitely generated abelian group is

\[ \mathbb{Z}^r \oplus \mathbb{Z}/d_1 \oplus \cdots \oplus \mathbb{Z}/d_k, \qquad d_1 \mid d_2 \mid \cdots \mid d_k, \quad d_i > 1, \]

uniquely. This is the classification stated in Chapter 15, now with its proof. A finite abelian group has \(r = 0\). To list the abelian groups of order \(n = \prod p^{a}\), choose for each prime a partition of its exponent \(a\); the number of groups is the product of the partition numbers.

Example: order \(72 = 2^3 \cdot 3^2\). Partitions of 3: \(3\), \(2+1\), \(1+1+1\); of 2: \(2\), \(1+1\). So \(3 \times 2 = 6\) groups:

2-part	3-part	elementary divisors	invariant factors
\(2^3\)	\(3^2\)	\(\mathbb{Z}/8 \oplus \mathbb{Z}/9\)	\(\mathbb{Z}/72\)
\(2^3\)	\(3, 3\)	\(\mathbb{Z}/8 \oplus \mathbb{Z}/3 \oplus \mathbb{Z}/3\)	\(\mathbb{Z}/3 \oplus \mathbb{Z}/24\)
\(2^2, 2\)	\(3^2\)	\(\mathbb{Z}/4 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/9\)	\(\mathbb{Z}/2 \oplus \mathbb{Z}/36\)
\(2^2, 2\)	\(3, 3\)	\(\mathbb{Z}/4 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/3 \oplus \mathbb{Z}/3\)	\(\mathbb{Z}/6 \oplus \mathbb{Z}/12\)
\(2, 2, 2\)	\(3^2\)	\(\mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/9\)	\(\mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/18\)
\(2, 2, 2\)	\(3, 3\)	\(\mathbb{Z}/2^{\oplus 3} \oplus \mathbb{Z}/3^{\oplus 2}\)	\(\mathbb{Z}/2 \oplus \mathbb{Z}/6 \oplus \mathbb{Z}/6\)

To get invariant factors from elementary divisors, multiply the largest power of each prime to get \(d_k\), then the next largest to get \(d_{k-1}\), and so on.

Second child: the Jordan normal form¶

Take \(R = \mathbb{C}[x]\) and \(M = V_T\) for a linear map \(T\) on \(V = \mathbb{C}^n\). \(M\) is finitely generated (by a basis of \(V\)) and torsion (21.1), so \(r = 0\). The primes of \(\mathbb{C}[x]\) are \(x - \lambda\) (every polynomial splits, by the fundamental theorem of algebra), so the elementary divisor form is

\[ V_T \;\cong\; \mathbb{C}[x]/\big((x - \lambda_1)^{e_1}\big) \oplus \cdots \oplus \mathbb{C}[x]/\big((x - \lambda_s)^{e_s}\big) . \]

Each piece is an invariant subspace. In the piece \(\mathbb{C}[x]/((x - \lambda)^e)\), take the basis

\[ w_1 = (x - \lambda)^{e-1}, \quad w_2 = (x - \lambda)^{e-2}, \quad \ldots, \quad w_e = 1 . \]

Multiplication by \(x\) is \(T\), and \(x = \lambda + (x - \lambda)\), so \(T w_1 = \lambda w_1\) (because \((x-\lambda)^e = 0\) here) and \(T w_j = \lambda w_j + w_{j-1}\). In this basis \(T\) is the Jordan block

\[ J_e(\lambda) = \begin{pmatrix} \lambda & 1 & & \\ & \lambda & \ddots & \\ & & \ddots & 1 \\ & & & \lambda \end{pmatrix} . \]

Theorem 21.3 (Jordan normal form). Every complex square matrix is similar to a block-diagonal matrix of Jordan blocks, unique up to the order of the blocks. The blocks are the elementary divisors of \(V_T\); the largest invariant factor \(d_k\) is the minimal polynomial and the product \(d_1 \cdots d_k\) is the characteristic polynomial.

Over a field that is not algebraically closed (like \(\mathbb{Q}\) or \(\mathbb{R}\)) you use invariant factors instead: each \(F[x]/(d_i)\) gives a companion matrix, and the result is the rational canonical form, which exists over any field.

Worked example. \(T = \begin{pmatrix} 3 & 1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}\).

Characteristic polynomial: the top-left block has trace 4 and determinant \(3 + 1 = 4\), so it contributes \(\lambda^2 - 4\lambda + 4 = (\lambda - 2)^2\); with the \(2\) in the corner, \((\lambda - 2)^3\).
\(T - 2I = \begin{pmatrix} 1 & 1 & 0 \\ -1 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}\) has rank 1, so the eigenspace \(\ker(T - 2I)\) has dimension 2: two Jordan blocks.
\((T - 2I)^2 = 0\) (multiply it out), so the minimal polynomial is \((x - 2)^2\): the largest block has size 2.

So \(V_T \cong \mathbb{C}[x]/(x - 2) \oplus \mathbb{C}[x]/((x - 2)^2)\), invariant factors \(x - 2\) and \((x - 2)^2\), and the Jordan form is \(J_2(2) \oplus J_1(2)\). A Jordan basis: \(v = e_1\), \(w = (T - 2I)e_1 = (1, -1, 0)\), and \(e_3\). Check: \(Te_1 = (3, -1, 0) = 2e_1 + w\) and \(Tw = (2, -2, 0) = 2w\).

The structure theorem has thus done two jobs: abelian groups (scalars \(\mathbb{Z}\)) and matrices (scalars \(\mathbb{C}[x]\)), because \(\mathbb{Z}\) and \(F[x]\) are both Euclidean. MAT247's Unit 9 and Unit 10 develop the Jordan and rational forms by linear algebra alone; it is worth comparing the two routes.

21.3 Representations of finite groups¶

Level: advanced. A group is an abstract thing: a set with a multiplication table. A good way to understand it is to watch it act. In Chapter 16 groups acted on sets. Now let them act on vector spaces, by linear maps, so all of linear algebra becomes available.

Definition. A representation of a group \(G\) on a vector space \(V\) over a field \(F\) is a homomorphism

\[ \rho : G \to GL(V), \]

where \(GL(V)\) is the group of invertible linear maps \(V \to V\) (invertible \(n \times n\) matrices, once a basis is chosen). So each \(g\) gives an invertible matrix \(\rho(g)\), with \(\rho(gh) = \rho(g)\rho(h)\). The dimension (or degree) is \(\dim V\). Throughout, \(G\) is finite and \(F = \mathbb{C}\) unless we say otherwise.

The module view. The group algebra \(\mathbb{C}[G]\) has basis the elements of \(G\) and multiplication from the group. A representation is exactly a \(\mathbb{C}[G]\)-module (let \(\sum a_g g\) act by \(\sum a_g \rho(g)\)). So representations are modules over a non-commutative ring, and everything in 21.1 applies.

Examples.

The trivial representation: \(\rho(g) = 1\) on \(\mathbb{C}\), for all \(g\).
The sign representation of \(S_n\): \(\rho(\sigma) = \operatorname{sgn}(\sigma) = \pm 1\) on \(\mathbb{C}\).
Cyclic groups. For \(G = \mathbb{Z}/n\) and any \(k\), \(\rho_k(j) = \zeta^{jk}\) with \(\zeta = e^{2\pi i/n}\). These are \(n\) one-dimensional representations.
The permutation representation of \(S_3\) on \(\mathbb{C}^3\): \(\rho(\sigma)\) permutes coordinates, \(\rho(\sigma)e_i = e_{\sigma(i)}\). The matrices are permutation matrices.
The symmetries of a triangle. \(S_3\) is the symmetry group of an equilateral triangle centred at the origin (Chapter 13). The rotation \(r\) by \(120^\circ\) and a reflection \(s\) in the \(x\)-axis act on \(\mathbb{R}^2\) by \(\rho(r) = \begin{pmatrix} -1/2 & -\sqrt3/2 \\ \sqrt3/2 & -1/2 \end{pmatrix}\) and \(\rho(s) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\). Check the defining relations: \(\rho(r)^3 = I\) (rotation by \(360^\circ\)), \(\rho(s)^2 = I\), and \(\rho(s)\rho(r)\rho(s) = \rho(r)^{-1}\) (a reflected rotation turns the other way).
The regular representation: \(G\) acts on \(\mathbb{C}[G]\), a space of dimension \(|G|\), by \(\rho(g)h = gh\).

Subrepresentations and irreducibility. A subspace \(W \subseteq V\) is invariant if \(\rho(g)W \subseteq W\) for all \(g\); then \(\rho\) restricts to a subrepresentation on \(W\). A nonzero representation is irreducible if its only invariant subspaces are \(0\) and \(V\). Two representations are isomorphic if there is an invertible linear map \(\varphi: V \to V'\) with \(\varphi \rho(g) = \rho'(g)\varphi\) for all \(g\): the same representation in a different basis.

In example 4, the line spanned by \((1, 1, 1)\) is invariant (permuting the coordinates of \((1,1,1)\) does nothing): a copy of the trivial representation. So is the plane \(W = \{x + y + z = 0\}\) (permuting coordinates does not change their sum). So

\[ \mathbb{C}^3 = \mathbb{C}(1,1,1) \oplus W, \]

and the plane \(W\) turns out to be isomorphic to example 5, the standard representation of \(S_3\). It is irreducible: an invariant line in \(W\) would be a common eigenvector of \(\rho(r)\) and \(\rho(s)\), but \(\rho(r)\), a rotation by \(120^\circ\), has no real eigenvectors, and its complex eigenvectors \((1, \mp i)\) (eigenvalues \(e^{\pm 2\pi i/3}\)) are not eigenvectors of \(\rho(s)\), which swaps them.

21.4 Maschke's theorem¶

Level: advanced. In the example, the invariant line had an invariant complement, and the representation split. Is that always so?

Theorem 21.4 (Maschke, 1898). Let \(G\) be a finite group and \(V\) a representation of \(G\) over a field \(F\) whose characteristic does not divide \(|G|\) (for example \(F = \mathbb{C}\)). Every invariant subspace \(W\) has an invariant complement \(W'\), with \(V = W \oplus W'\). Hence every finite-dimensional representation is a direct sum of irreducible ones.

Proof. Choose any linear projection \(\pi: V \to W\), that is, linear with \(\pi(w) = w\) for \(w \in W\) (extend a basis of \(W\) to a basis of \(V\) and send the extra vectors to 0). It need not respect the group. Average it over \(G\):

\[ \pi_0 = \frac{1}{|G|}\sum_{g \in G} \rho(g)\, \pi\, \rho(g)^{-1} . \]

(We divide by \(|G|\), which is allowed because \(|G| \ne 0\) in \(F\).) Three facts:

\(\pi_0\) maps into \(W\): \(\rho(g)^{-1}v\) is sent by \(\pi\) into \(W\), and \(\rho(g)\) keeps \(W\) in \(W\).
\(\pi_0\) is the identity on \(W\): for \(w \in W\), \(\rho(g)^{-1}w \in W\), so \(\pi\) fixes it, and \(\rho(g)\rho(g)^{-1}w = w\). Each of the \(|G|\) terms gives \(w\); the average is \(w\).
\(\pi_0\) commutes with the group: \(\rho(h)\pi_0\rho(h)^{-1} = \tfrac{1}{|G|}\sum_g \rho(hg)\pi\rho(hg)^{-1} = \pi_0\), since \(hg\) runs over \(G\) as \(g\) does.

By 1 and 2, \(\pi_0\) is a projection onto \(W\), so \(V = W \oplus \ker\pi_0\). By 3, if \(\pi_0 v = 0\) then \(\pi_0\rho(h)v = \rho(h)\pi_0 v = 0\): \(\ker \pi_0\) is invariant. For the last sentence, induct on dimension: split off invariant subspaces until all pieces are irreducible. \(\blacksquare\)

The averaging trick, "make anything symmetric by averaging over the group", is one of the most useful ideas in the subject. Over \(\mathbb{C}\) it gives another proof: average any inner product to get an invariant one, \(\langle u, v \rangle_0 = \tfrac{1}{|G|}\sum_g \langle \rho(g)u, \rho(g)v \rangle\); then the orthogonal complement of an invariant subspace is invariant, and all \(\rho(g)\) are unitary in a suitable basis.

When it fails. Both hypotheses matter.

Characteristic dividing \(|G|\). Let \(G = \mathbb{Z}/p\) act on \(\mathbb{F}_p^2\) by \(\rho(k) = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\). (This is a homomorphism, and \(\rho(p) = I\) since \(p = 0\) in \(\mathbb{F}_p\).) The line spanned by \(e_1\) is invariant. Any invariant line is spanned by an eigenvector of \(\rho(1)\), whose only eigenvalue is \(1\) with eigenspace \(\mathbb{F}_p e_1\). So there is no invariant complement. This is the beginning of modular representation theory, which is much harder.
Infinite groups. The same matrices with \(k \in \mathbb{Z}\) and entries in \(\mathbb{C}\) give a representation of \(\mathbb{Z}\) that does not split, for the same reason.

21.5 Characters and orthogonality¶

Level: expert. A representation is a list of matrices, one per group element, and they depend on a choice of basis. We want invariants. The simplest invariant of a matrix under change of basis is its trace.

Definition. The character of a representation \(\rho\) is the function \(\chi_\rho: G \to \mathbb{C}\), \(\chi_\rho(g) = \operatorname{tr}\rho(g)\).

Basic properties.

\(\chi(1) = \dim V\) (the trace of the identity matrix).
Isomorphic representations have the same character (similar matrices have equal traces).
\(\chi\) is a class function: \(\chi(hgh^{-1}) = \operatorname{tr}(\rho(h)\rho(g)\rho(h)^{-1}) = \chi(g)\). So it is constant on conjugacy classes.
\(\chi_{V \oplus W} = \chi_V + \chi_W\) (block-diagonal matrices).
\(\chi(g^{-1}) = \overline{\chi(g)}\). (If \(g^m = 1\) then \(\rho(g)^m = I\), so \(\rho(g)\) is diagonalizable with eigenvalues that are \(m\)-th roots of unity, \(|\lambda| = 1\), and \(\lambda^{-1} = \bar\lambda\).)

For class functions \(\alpha, \beta\) define the inner product

\[ \langle \alpha, \beta \rangle = \frac{1}{|G|}\sum_{g \in G} \alpha(g)\,\overline{\beta(g)} . \]

Lemma 21.5 (Schur, 1905). Let \(V\), \(W\) be irreducible representations and \(\varphi: V \to W\) a linear map with \(\varphi\rho_V(g) = \rho_W(g)\varphi\) for all \(g\) (an intertwiner). Then \(\varphi\) is \(0\) or an isomorphism. If \(V = W\) (over \(\mathbb{C}\)), then \(\varphi\) is a scalar, \(\varphi = \lambda I\).

Proof. \(\ker\varphi\) is invariant in \(V\) (if \(\varphi v = 0\) then \(\varphi\rho_V(g)v = \rho_W(g)\varphi v = 0\)), so it is \(0\) or \(V\); \(\operatorname{im}\varphi\) is invariant in \(W\), so it is \(0\) or \(W\). Either \(\varphi = 0\), or \(\varphi\) is injective and onto. If \(V = W\), \(\varphi\) has an eigenvalue \(\lambda\) (over \(\mathbb{C}\), by the fundamental theorem of algebra), and \(\varphi - \lambda I\) is an intertwiner with nonzero kernel, so it is \(0\). \(\blacksquare\)

Theorem 21.6 (orthogonality). If \(\chi\) and \(\psi\) are the characters of irreducible representations \(V\) and \(W\), then \(\langle\chi, \psi\rangle = 1\) if \(V \cong W\) and \(0\) otherwise.

Proof. Let \(A: V \to W\) be any linear map and average it into an intertwiner: \(A_0 = \tfrac{1}{|G|}\sum_g \rho_W(g) A \rho_V(g^{-1})\) (the check is the same as fact 3 in Maschke's proof). By Schur, if \(V \not\cong W\) then \(A_0 = 0\) for every \(A\). Choose bases and take \(A = E_{jk}\), the matrix with a single 1 in position \((j, k)\). Entry \((i, l)\) of \(A_0 = 0\) reads

\[ \frac{1}{|G|}\sum_g \rho_W(g)_{ij}\, \rho_V(g^{-1})_{kl} = 0 \quad \text{for all } i, j, k, l . \]

Set \(j = i\) and \(l = k\) and sum over \(i\) and \(k\): the left side becomes \(\tfrac{1}{|G|}\sum_g \chi_W(g)\chi_V(g^{-1}) = \tfrac{1}{|G|}\sum_g \chi_W(g)\overline{\chi_V(g)} = \langle\psi, \chi\rangle\), so it is \(0\).

If \(V = W\), Schur gives \(A_0 = \lambda I\), and taking traces, \(\operatorname{tr}A_0 = \operatorname{tr}A\) (each term \(\rho(g)A\rho(g)^{-1}\) has the trace of \(A\)), so \(\lambda = \operatorname{tr}A/d\) with \(d = \dim V\). For \(A = E_{jk}\), \(\operatorname{tr}A = \delta_{jk}\), so entry \((i, l)\) gives \(\tfrac{1}{|G|}\sum_g \rho(g)_{ij}\rho(g^{-1})_{kl} = \tfrac{1}{d}\delta_{jk}\delta_{il}\). Setting \(j = i\), \(l = k\) and summing over \(i, k\) gives \(\langle \chi, \chi \rangle = \tfrac{1}{d}\sum_{i,k}\delta_{ik}\delta_{ik} = \tfrac{d}{d} = 1\). \(\blacksquare\)

Consequences. Write a representation as \(V \cong U_1^{\oplus a_1} \oplus \cdots \oplus U_s^{\oplus a_s}\) with the \(U_i\) distinct irreducibles (Maschke). Then \(\chi_V = \sum a_i\chi_i\), and orthogonality gives:

the multiplicity of \(U_i\) in \(V\) is \(a_i = \langle \chi_V, \chi_i \rangle\), so a representation is determined by its character;
\(V\) is irreducible exactly when \(\langle \chi_V, \chi_V \rangle = \sum a_i^2 = 1\);
the regular representation has \(\chi_{\mathrm{reg}}(1) = |G|\) and \(\chi_{\mathrm{reg}}(g) = 0\) for \(g \ne 1\) (left multiplication by \(g \ne 1\) moves every basis element), so \(\langle \chi_{\mathrm{reg}}, \chi_i \rangle = \tfrac{1}{|G|}|G|\chi_i(1) = \dim U_i\): every irreducible appears in the regular representation, as often as its dimension.

Comparing the dimensions of both sides of the regular representation gives the famous formula

\[ \sum_i (\dim U_i)^2 = |G| . \]

One more fact, which we state: the number of irreducible representations equals the number of conjugacy classes of \(G\) (the irreducible characters form an orthonormal basis of the space of class functions; Dummit and Foote, section 18.3).

The character table of \(S_3\)¶

\(S_3\) has three conjugacy classes: the identity (1 element), the transpositions (3) and the 3-cycles (2). So it has three irreducible representations, with \(d_1^2 + d_2^2 + d_3^2 = 6\), forcing dimensions \(1, 1, 2\). They are the trivial, sign and standard representations. The standard character: at a transposition, a reflection matrix like \(\rho(s)\), trace \(0\); at a 3-cycle, a rotation by \(120^\circ\), trace \(2\cos 120^\circ = -1\).

	\(e\)	\((12)\) (3 elements)	\((123)\) (2 elements)
trivial \(\chi_1\)	\(1\)	\(1\)	\(1\)
sign \(\chi_2\)	\(1\)	\(-1\)	\(1\)
standard \(\chi_3\)	\(2\)	\(0\)	\(-1\)

Check orthogonality, weighting each column by its class size:

\[ \begin{aligned} \langle \chi_3, \chi_3 \rangle &= \tfrac16\big(1 \cdot 4 + 3 \cdot 0 + 2 \cdot 1\big) = 1, & \langle \chi_3, \chi_1 \rangle &= \tfrac16\big(2 + 0 - 2\big) = 0, \\ \langle \chi_2, \chi_1 \rangle &= \tfrac16\big(1 - 3 + 2\big) = 0, & \langle \chi_3, \chi_2 \rangle &= \tfrac16\big(2 + 0 - 2\big) = 0 . \end{aligned} \]

The columns are orthogonal too ("column orthogonality"): the sum of \(|\chi_i(g)|^2\) down a column is the size of the centralizer of \(g\): \(1 + 1 + 4 = 6\), \(1 + 1 + 0 = 2\), \(1 + 1 + 1 = 3\).

Decomposing with characters. The permutation representation on \(\mathbb{C}^3\) has character "number of fixed points": \((3, 1, 0)\). Then \(\langle \chi, \chi_1 \rangle = \tfrac16(3 + 3 + 0) = 1\), \(\langle \chi, \chi_2 \rangle = \tfrac16(3 - 3 + 0) = 0\), \(\langle \chi, \chi_3 \rangle = \tfrac16(6 + 0 + 0) = 1\). So \(\mathbb{C}^3 \cong\) trivial \(\oplus\) standard, which we found by hand in 21.3, and now without a single invariant subspace in sight.

History. Characters of finite groups were introduced by Georg Frobenius in 1896, prompted by a question of Richard Dedekind about factoring the "group determinant". William Burnside, Frobenius and Issai Schur built the theory in the next decade. Burnside used it in 1904 to prove that every group of order \(p^a q^b\) (\(p\), \(q\) primes) is solvable, a statement about groups for which no character-free proof was found for about 70 years. Characters are also central to the proof of the Feit-Thompson theorem (1963) that every finite group of odd order is solvable, and to the classification of finite simple groups (Chapter 16).

21.6 Uses in physics and chemistry¶

The principle behind all the applications: if a system has a symmetry group \(G\), then every space of states that the symmetry preserves is a representation of \(G\), and it splits into irreducibles. Characters do the splitting by arithmetic.

Molecular vibrations¶

A water molecule, \(\mathrm{H_2O}\), is bent: the oxygen at the apex, the two hydrogens symmetric. Its symmetry group, called \(C_{2v}\) by chemists, has four elements: the identity \(E\); the rotation \(C_2\) by \(180^\circ\) about the axis through the oxygen (the \(z\)-axis); the reflection \(\sigma_v(xz)\) in the plane perpendicular to the molecule; and the reflection \(\sigma_v'(yz)\) in the plane of the molecule. The group is abelian (it is \(\mathbb{Z}/2 \times \mathbb{Z}/2\)), so it has four classes and four one-dimensional irreducible representations:

\(C_{2v}\)	\(E\)	\(C_2\)	\(\sigma_v(xz)\)	\(\sigma_v'(yz)\)	transforms like
\(A_1\)	\(1\)	\(1\)	\(1\)	\(1\)	\(z\)
\(A_2\)	\(1\)	\(1\)	\(-1\)	\(-1\)	rotation \(R_z\)
\(B_1\)	\(1\)	\(-1\)	\(1\)	\(-1\)	\(x\), \(R_y\)
\(B_2\)	\(1\)	\(-1\)	\(-1\)	\(1\)	\(y\), \(R_x\)

The three atoms can each move in three directions: a 9-dimensional space of displacements, on which \(G\) acts. Its character at each element is (number of atoms the operation leaves in place) \(\times\) (trace of the operation on one atom's \(x, y, z\)):

\(E\): 3 atoms fixed, trace 3: \(\chi = 9\).
\(C_2\): only O fixed; a half-turn about \(z\) is \(\operatorname{diag}(-1, -1, 1)\), trace \(-1\): \(\chi = -1\).
\(\sigma_v(xz)\): only O fixed (the hydrogens swap); trace of a reflection is \(1\): \(\chi = 1\).
\(\sigma_v'(yz)\): all 3 fixed; trace 1: \(\chi = 3\).

So \(\Gamma = (9, -1, 1, 3)\), and with the four-element inner product,

\[ \begin{aligned} a_{A_1} &= \tfrac14(9 - 1 + 1 + 3) = 3, & a_{A_2} &= \tfrac14(9 - 1 - 1 - 3) = 1, \\ a_{B_1} &= \tfrac14(9 + 1 + 1 - 3) = 2, & a_{B_2} &= \tfrac14(9 + 1 - 1 + 3) = 3 . \end{aligned} \]

\(\Gamma = 3A_1 + A_2 + 2B_1 + 3B_2\). Of these nine motions, three are translations of the whole molecule (like \(x, y, z\): \(B_1 + B_2 + A_1\)) and three are rotations (\(R_x, R_y, R_z\): \(B_2 + B_1 + A_2\)). What is left is the vibrations:

\[ \Gamma_{\mathrm{vib}} = 2A_1 + B_2 . \]

Three vibrational modes: two symmetric (\(A_1\): the symmetric stretch and the bend) and one antisymmetric (\(B_2\): one O-H bond stretches as the other shrinks). This is what infrared spectra of water show, and the symmetry labels also give selection rules: a mode absorbs infrared light only if it transforms like \(x\), \(y\) or \(z\), which all three of these do. The same computation, done by every chemistry student, works for any molecule.

Atoms and particles¶

Atomic orbitals. The hydrogen atom's energy does not change if you rotate it, so each space of states with a given energy is a representation of the rotation group \(SO(3)\). Its irreducible representations have dimensions \(1, 3, 5, 7, \ldots\) (\(2l + 1\) for \(l = 0, 1, 2, \ldots\)), and these are the numbers of \(s\), \(p\), \(d\), \(f\) orbitals: the shape of the periodic table is in part a list of representation dimensions (doubled by spin). Eugene Wigner's 1931 book Group Theory and its Application to the Quantum Mechanics of Atomic Spectra made this standard. Spin is a representation of \(SU(2)\), a group that covers \(SO(3)\) twice (Chapter 22), and an electron's spin-\(\tfrac12\) is its 2-dimensional representation.

The eightfold way. In 1961 Murray Gell-Mann and, independently, Yuval Ne'eman organized the many particles found in accelerators into families matching irreducible representations of the group \(SU(3)\), of dimensions 8 and 10. One slot of the 10-dimensional family was empty, and Gell-Mann predicted a particle to fill it, the \(\Omega^-\), with its mass and charge. It was found at Brookhaven in 1964. The quark model grew out of this: quarks span the 3-dimensional representation.

Crystals. The symmetries of a crystal that fix a point form one of exactly 32 crystallographic point groups; with translations included there are 230 space groups, enumerated independently by Evgraf Fedorov and Arthur Schoenflies around 1891. X-ray diffraction reads the symmetry and spacing of a crystal from the pattern of reflected rays. On Mars, the CheMin instrument on NASA's Curiosity rover does exactly this inside Gale crater: it identifies minerals in drilled rock powder by their X-ray diffraction patterns, the first such measurements made on another planet.

In number theory. Groups like \(\operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})\), the symmetries of all algebraic numbers, are too big to see directly, so number theorists study their representations, Galois representations. The proof of Fermat's last theorem (Chapter 18) is a theorem about 2-dimensional Galois representations attached to elliptic curves, and the Langlands programme (Chapter 22) is, at heart, a conjectured dictionary between such representations and analysis.

Check yourself¶

1. Is \(\mathbb{Z}/6\) a free \(\mathbb{Z}\)-module? Is it a free \(\mathbb{Z}/6\)-module? What is its torsion submodule as a \(\mathbb{Z}\)-module?

Not free over \(\mathbb{Z}\): every element \(m\) has \(6m = 0\), so no nonempty set is independent (and it is not \(0\)). Over the ring \(\mathbb{Z}/6\) it is free of rank 1 (basis \(\{1\}\)): freeness depends on the ring. As a \(\mathbb{Z}\)-module everything is torsion, so the torsion submodule is all of \(\mathbb{Z}/6\).

2. Using only the module axioms, prove that \(0 \cdot m = 0\) and \((-1)m = -m\).

\(0m = (0 + 0)m = 0m + 0m\); subtract \(0m\) from both sides: \(0m = 0\). Then \(m + (-1)m = 1m + (-1)m = (1 + (-1))m = 0m = 0\), so \((-1)m\) is the additive inverse of \(m\).

3. List the abelian groups of order 36 in both forms.

\(36 = 2^2 \cdot 3^2\); two partitions each, so four groups. Elementary divisors and invariant factors: \(\mathbb{Z}/4 \oplus \mathbb{Z}/9 \cong \mathbb{Z}/36\); \(\mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/9 \cong \mathbb{Z}/2 \oplus \mathbb{Z}/18\); \(\mathbb{Z}/4 \oplus \mathbb{Z}/3 \oplus \mathbb{Z}/3 \cong \mathbb{Z}/3 \oplus \mathbb{Z}/12\); \(\mathbb{Z}/2 \oplus \mathbb{Z}/2 \oplus \mathbb{Z}/3 \oplus \mathbb{Z}/3 \cong \mathbb{Z}/6 \oplus \mathbb{Z}/6\).

4. Identify the abelian group with generators \(a, b\) and relations \(2a + 6b = 0\), \(4a + 8b = 0\).

Relation matrix \(\begin{pmatrix} 2 & 6 \\ 4 & 8 \end{pmatrix}\). \(C_2 - 3C_1\): \(\begin{pmatrix} 2 & 0 \\ 4 & -4 \end{pmatrix}\). \(R_2 - 2R_1\): \(\begin{pmatrix} 2 & 0 \\ 0 & -4 \end{pmatrix}\). Multiply the second column by the unit \(-1\): \(\operatorname{diag}(2, 4)\), and \(2 \mid 4\). So the group is \(\mathbb{Z}/2 \oplus \mathbb{Z}/4\), of order \(8 = |\det| = |16 - 24|\).

5. Find the Jordan form of \(\begin{pmatrix} 5 & 4 & 2 \\ 0 & 1 & 0 \\ -4 & -4 & -1 \end{pmatrix}\).

Expanding along the second row, the characteristic polynomial is \((\lambda - 1)\big((\lambda - 5)(\lambda + 1) + 8\big) = (\lambda - 1)(\lambda^2 - 4\lambda + 3) = (\lambda - 1)^2(\lambda - 3)\). \(T - I = \begin{pmatrix} 4 & 4 & 2 \\ 0 & 0 & 0 \\ -4 & -4 & -2 \end{pmatrix}\) has rank 1, so the eigenvalue 1 has a 2-dimensional eigenspace: two blocks of size 1. The Jordan form is \(\operatorname{diag}(1, 1, 3)\); the matrix is diagonalizable, with minimal polynomial \((x - 1)(x - 3)\).

6. A \(4 \times 4\) complex matrix has characteristic polynomial \((x - 2)^4\) and minimal polynomial \((x - 2)^2\). What can its Jordan form be? How could you tell the cases apart?

The block sizes are a partition of 4 with largest part 2: either \(2 + 2\) or \(2 + 1 + 1\). The number of blocks is \(\dim\ker(T - 2I)\): 2 in the first case, 3 in the second. Invariant factors: \((x-2)^2, (x-2)^2\) or \((x-2), (x-2), (x-2)^2\).

7. Find all one-dimensional representations of \(S_3\) and of \(\mathbb{Z}/4\), with their characters.

A 1-dimensional representation is a homomorphism to \(\mathbb{C}^\times\), an abelian group, so it kills all commutators. For \(S_3\) the commutator subgroup is \(A_3\) and \(S_3/A_3 \cong \mathbb{Z}/2\), so there are two: trivial and sign. For \(\mathbb{Z}/4\) the generator must go to a 4th root of unity: four representations \(\rho_k(j) = i^{jk}\), \(k = 0, 1, 2, 3\), with character table rows \((1,1,1,1)\), \((1, i, -1, -i)\), \((1, -1, 1, -1)\), \((1, -i, -1, i)\) on the elements \(0, 1, 2, 3\).

8. Decompose the regular representation of \(S_3\) using characters.

\(\chi_{\mathrm{reg}} = (6, 0, 0)\). \(\langle \chi_{\mathrm{reg}}, \chi_1 \rangle = 6/6 = 1\), \(\langle \chi_{\mathrm{reg}}, \chi_2 \rangle = 1\), \(\langle \chi_{\mathrm{reg}}, \chi_3 \rangle = 12/6 = 2\). So \(\mathbb{C}[S_3] \cong \text{trivial} \oplus \text{sign} \oplus \text{standard}^{\oplus 2}\), of dimension \(1 + 1 + 4 = 6\).

9. Decompose the tensor square of the standard representation of \(S_3\) (its character is \(\chi_3^2\)).

\(\chi_3^2 = (4, 0, 1)\). \(\langle \chi_3^2, \chi_1 \rangle = \tfrac16(4 + 0 + 2) = 1\), \(\langle \chi_3^2, \chi_2 \rangle = \tfrac16(4 - 0 + 2) = 1\), \(\langle \chi_3^2, \chi_3 \rangle = \tfrac16(8 + 0 - 2) = 1\). So it is trivial \(\oplus\) sign \(\oplus\) standard, dimension \(1 + 1 + 2 = 4\).

10. \(S_4\) has 5 conjugacy classes and exactly two one-dimensional representations. Find the dimensions of its irreducible representations.

Five irreducibles with \(\sum d_i^2 = 24\) and exactly two \(d_i = 1\): the other three satisfy \(a^2 + b^2 + c^2 = 22\) with each at least 2. Squares available: 4, 9, 16. \(16 + 4 + 4 = 24\) too big; \(9 + 9 + 4 = 22\). So the dimensions are \(1, 1, 2, 3, 3\).

11. Show that \(\rho(k) = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\) over \(\mathbb{F}_2\) is a representation of \(\mathbb{Z}/2\) that is not a direct sum of irreducibles.

\(\rho(1)^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = I\) over \(\mathbb{F}_2\), so it is a homomorphism. The invariant lines are spanned by eigenvectors of \(\rho(1)\); \(\rho(1) - I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\) has kernel \(\mathbb{F}_2 e_1\) only. One invariant line, so \(\mathbb{F}_2^2\) is not a sum of two invariant lines, and it is not irreducible (it has that line). Maschke fails because the characteristic 2 divides \(|G| = 2\).

12. Ammonia, \(\mathrm{NH_3}\), has symmetry group \(C_{3v} \cong S_3\). Its displacement representation has character \((12, 0, 2)\) on the classes \(E\), \(C_3\), \(\sigma_v\) (with class sizes 1, 2, 3). Decompose it, and find the vibrational modes, given that translations span \(A_1 + E\) and rotations \(A_2 + E\).

Here \(A_1\), \(A_2\), \(E\) are the trivial, sign and 2-dimensional representations, with characters \((1,1,1)\), \((1,1,-1)\), \((2,-1,0)\) on \((E, C_3, \sigma_v)\). Multiplicities: \(A_1\): \(\tfrac16(12 + 0 + 3 \cdot 2) = 3\); \(A_2\): \(\tfrac16(12 + 0 - 6) = 1\); \(E\): \(\tfrac16(24 + 0 + 0) = 4\). So \(\Gamma = 3A_1 + A_2 + 4E\) (dimension \(3 + 1 + 8 = 12\)). Removing \(A_1 + E\) and \(A_2 + E\) leaves \(\Gamma_{\mathrm{vib}} = 2A_1 + 2E\): six vibrational degrees of freedom (\(3 \cdot 4 - 6\)), in four distinct frequencies, two of them doubly degenerate.

What you learned¶

A module is a vector space over a ring; abelian groups, ideals and "vector space plus linear map" are all modules. Modules can have torsion and need not have a basis.
Over a PID every finitely generated module is \(R^r \oplus R/(d_1) \oplus \cdots \oplus R/(d_k)\) with \(d_1 \mid \cdots \mid d_k\), computed by the Smith normal form of a relation matrix.
With \(R = \mathbb{Z}\) this classifies finitely generated abelian groups; with \(R = \mathbb{C}[x]\) it is the Jordan form, with the minimal and characteristic polynomials read from the invariant factors.
A representation is a homomorphism \(G \to GL(V)\); over \(\mathbb{C}\) (Maschke) every representation of a finite group is a direct sum of irreducibles, by averaging.
Characters are traces; irreducible characters are orthonormal (Schur's lemma), a representation is determined by its character, and \(\sum (\dim U_i)^2 = |G|\). \(S_3\) has the table \((1,1,1)\), \((1,-1,1)\), \((2,0,-1)\).
Chemists count molecular vibrations with characters; physicists read orbitals and particle families as irreducible representations; number theorists study Galois representations.

Next: Chapter 22, a guided tour of where algebra goes after this book.