Chapter 1

Group Representations

Definition 1.1 A representation α of a group G in a vector space V over k is defined by a homomorphism

α: G → GL(V).

The degree of the representation is the dimension of the vector space:

degα = dim_{k}V.

Remarks:

1. Recall that GL(V)—the general linear group on V —is the group of invertible linear maps t : V → V.

2. We shall be concerned almost exclusively with representations of finite degree, that is, in finite-dimensional vector spaces; and these will almost always be vector spaces over ℝ or ℂ. Therefore, to avoid repetition, let us agree to use the term ‘representation’ to mean representation of finite degree over ℝ or ℂ, unless the contrary is explicitly stated.

Furthermore, in this first Part we shall be concerned almost exclusively with finite groups; so let us also agree that the term ‘group’ will mean finite group, unless the contrary is stated.

3. Suppose {e_{1},...,e_{n}} is a basis for V. Then each linear map t : V → V is defined (with respect to this basis) by an n × n-matrix T.

Explicitly,

te_{j} = ∑_{i} T_{j}^{i}e_{i};

\underline{GpReps-I–2}

or in terms of coordinates,

( x_{1} ; ⋮ ; x_{n} ) ↦ T ( x_{1} ; ⋮ ; x_{n} )

Thus a representation in V can be defined by a homomorphism

α: G → GL(n,k),

where GL(n,k) denotes the group of non-singular n × n-matrices over k. In other words, α is defined by giving matrices A(g) for each g ∈ G, satisfying the conditions

A(gh) = A(g)A(h)

for all g,h ∈ G; and also

A(e) = I.

4. There is another way of looking at representations which perhaps has greater intuitive content.

Recall that a group is said to act on the set X if we have a map

G × X → X : (g,x) ↦ gx

satisfying

(a) (gh)x) = g(hx),

(b) ex = x.

Now suppose X = V is a vector space. Then we can say that G acts linearly on V if in addition

(c) g(u + v) = gu + gv,

(d) g(ρv) = ρ(gv).

Each representation α of G in V defines a linear action of G on V, by

gv = α(g)v;

and every such action arises from a representation in this way.

Thus the notions of representation and linear action are completely equivalent. We can use whichever we find more convenient in a given case.

5. There are 2 other ways of looking at group representations, completely equivalent to the definition but expressing slightly different points of view.

Firstly, we may speak of the vector space V, with the action of G on it. as a G-space. For those familiar with category theory, this would be the categorical approach. Representation theory, from this point of view, is the study of the category of G-spaces and G-maps, where a G-map

t : U → V

from one G-space to another is a linear map preserving the action of G, ie satisfying

t(gu) = g(tu) (g ∈ G,u ∈ U).

6. Secondly, and finally, mathematical physicists often speak—strikingly—of the vector space V carrying the representation α.

Examples:

1. Recall that the dihedral group D_{4} is the symmetry group of a square ABCD

Figure 1.1: The natural representation of D_{4}

(Figure 1.1). Let us take coordinates Ox,Oy as shown through the centre O of the square. Then

D_{4} = {e,r,r^{2},r^{3},c,d,h,v},

where r is the rotation about O through π/2 (sending A to B), while c,d,h,v are the reflections in AC,BD,Ox,Oy respectively.

By definition a symmetry g ∈ D_{4} is an isometry of the plane E^{2} sending the square into itself. Evidently g must send O into itself, and so gives rise to a linear map

A(g) : ℝ^{2} → ℝ^{2}.

The map

g ↦ A(g) ∈ GL(2,ℝ)

defines a 2-dimensional representation ρ of D_{4} over ℝ. We may describe this as the natural 2-dimensional representation of D_{4}.

(Evidently the symmetry group G of any bounded subset S ⊂ E^{n} will have a similar ‘natural’ representation in ℝ^{n}.) The representation ρ is given in matrix terms by

\begin{matrix}e ↦ & ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) , r ↦ & ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) , r^{2} ↦ & ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) , r^{3} ↦ & ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix}) , \\ c ↦ & ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) , d ↦ & ( \begin{matrix}0 & −1 \\ −1 & 0 \end{matrix} ) , h ↦ & ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) , v ↦ & ( \begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix}) . \end{matrix}

For example, the rotation r is given by

( \begin{matrix}x \\ y \end{matrix}) ↦ ( \begin{matrix}−y \\ x \end{matrix}) = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ( \begin{matrix}x \\ y \end{matrix}) .

Each group relation is represented in a corresponding matrix equation, eg

cd = r^{2} ⟹ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) ( \begin{matrix}0 & −1 \\ −1 & 0 \end{matrix} ) = ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) .

The representation ρ is faithful, ie the homomorphism defining it is injective. Thus a relation holds in D_{4} if and only if the corresponding matrix equation is true. However, representations are not necessarily faithful, and the implication is only one way.

Every finite-dimensional representation can be expressed in matrix form in this way, after choosing a basis for the vector space carrying the representation. However, while such matrix representations are reassuringly concrete, they are impractical except in the lowest dimensions. Better just to keep at the back of one’s mind that a representation could be expressed in this way.

2. Suppose G acts on the set X:

(g,x) ↦ gx.

Let

C(X) = C(X,k)

denote the space of maps

f : X → k.

Then G acts linearly on C(X)—and so defines a representation ρ of G—by

gf(x) = f(g^{−1} x).

(We need g^{−1} rather than g on the right to satisfy the rule

g(hf) = (gh)f.

It is fortunate that the relation

(gh)^{−1} = h^{−1} g^{−1}

enables us to correct the order reversal. We shall often have occasion to take advantage of this, particularly when dealing—as here—with spaces of functions.)

Now suppose that X is finite; say

X = {x_{1},...,x_{n}}.

Then

degρ = n = ∥X∥,

the number of elements in X. For the functions

e_{y}(x) = \{ \begin{matrix}1 if x = y, \\ 0 otherwise. \end{matrix}

(ie the characteristic functions of the 1-point subsets) form a basis for C(X). Also

ge_{x} = e_{gx},

since

ge_{y}(x)= e_{y}(g^{−1} x)

= \{ \begin{matrix}1 & if g^{−1} x = y \\ 0 & if g^{−1} x ≠ y \end{matrix}

= \{ \begin{matrix}1 & if x = gy \\ 0 & if x ≠ gy \end{matrix}

Thus

g ↦ P(g)

where P = P(g) is the matrix with entries

P_{x}^{y} = \{ \begin{matrix}1 if y = gx, \\ 0 otherwise. \end{matrix}

Notice that P is a permutation matrix, ie there is just one 1 in each row and column, all other entries being 0. We call a representation that arises from the action of a group on a set in this way a permutational representation.

As an illustration, consider the natural action of S(3) on the set {a,b,c}. This yields a 3-dimensional representation ρ of S(3), under which

(abc) ↦ ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix}) , (ab) ↦ ( \begin{matrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}) .

(These 2 instances actually define the representation, since (abc) and (ab) generate S(3).)

3. A 1-dimensional representation α of a group G over k = ℝ or ℂ is just a

homomorphism

α: G → k^{×} ,

where k^{×} denotes the multiplicative group on the set k \ {0}. For

GL(1,k) = k^{×} ,

since we can identify the 1 × 1-matrix [x] with its single entry x.

We call the 1-dimensional representation defined by the identity homomorphism

g ↦ 1

(for all g ∈ G) the trivial representation of G, and denote it by 1.

In a 1-dimensional representation, each group element is represented by a number. Since these numbers commute, the study of 1-dimensional representations is much simpler than those of higher dimension.

In general, when investigating the representations of a group G, we start by determining all its 1-dimensional representations.

Recall that 2 elements g,h ∈ G are said to be conjugate if

h = xgx^{−1}

for some third element x ∈ G. Suppose α is a 1-dimensional representation of G. Then

α(h) = α(x)α(g)α(x^{−1} )
= α(x)α(g)α(x)^{−1}
= α(g)α(x)α(x)^{−1}
= α(g),

since the numbers α(x),α(g) commute. It follows that a 1-dimensional representation is constant on each conjugacy class of G.

Consider the group S_{3}. This has 3 classes (we shall usually abbreviate ‘conjugacy class’ to class):

{1}, {(abc),(acb)}, {(bc),(ca),(ab)}.

Let us write

s = (abc), t = (bc).

Then (assuming k = ℂ)

s^{3} = 1 ⟹ α(s)^{3} = 1 ⟹ α(s) = 1,ω or ω^{2},
t^{2} = 1 ⟹ α(s)^{2} = 1 ⟹ α(t) = ±1.

But

tst^{−1} = s^{2}.

It follows that

α(t)α(s)α(t)^{−1} = α(s)^{2},

from which we deduce that

α(s) = 1.

It follows that S_{3} has just 2 1-dimensional representations: the trivial representation

1 : g ↦ 1,

and the parity representation

ϵ: g ↦ \{ \begin{matrix}1 & if g is even, \\ −1 & if g is odd. \end{matrix}

4. The corresponding result is true for all the symmetric groups S_{n} (for n ≥ 2); S_{n} has just 2 1-dimensional representations, the trivial representation 1 and the parity representation ϵ.

To see this, let us recall 2 facts about S_{n}.

(a) The transpositions τ = (xy) generate S_{n}, ie each permutation g ∈ S_{n} is expressible (not uniquely) as a product of transpositions

g = τ_{1}··· τ_{r}.

(b) The transpositions are all conjugate. (This is a particular case of the general fact that 2 permutations in S_{n} are conjugate if and only if they are of the same cyclic type, ie they have the same number of cycles of each length.)

It follows from (1) that a 1-dimensional representation of S_{n} is completely determined by its values on the transpositions. It follows from (2) that the representation is constant on the transpositions. Finally, since each transposition τ satisfies τ^{2} = 1 it follows that this constant value is ±1. Thus there can only be 2 1-dimensional representations of S_{n}; the first takes the value 1 on the transpositions, and so is 1 everywhere; the second takes the value -1 on the transpositions, and takes the value (−1)^{r} on the permutation

g = τ_{1}··· τ_{r}.

Thus S_{n} has just 2 1-dimensional representations; the trivial representation 1 and the parity representation ϵ.

5. Let’s look again at the dihedral group D_{4}, ie the symmetry group of the square ABCD. Let r denote the rotation through π/2, taking A into B; and let c denote the reflection in AC.

It is readily verified that r and c generate D_{4}, ie each element g ∈ D_{4} is expressible as a word in r and c (eg g = r^{2}cr). This follows for example from Lagrange’s Theorem. The subgroup generated by r and c contains at least the 5 elements 1,r,r^{2},r^{3},c, and so must be the whole group. (We shall sometimes denote the identity element in a group by 1, while at other times we shall use e or I.)

It is also easy to see that r and c satisfy the relations

r^{4} = 1, c^{2} = 1, rc = cr^{3}.

In fact these are defining relations for D_{4}, ie every relation between r and c can be derived from these 3.

We can express this in the form

D_{4} = ⟨r,c: r^{4} = c^{2} = 1, rc = cr^{3}⟩.

Now suppose α is a 1-dimensional representation of D_{4}. Then we must have

α(r)^{4} = α(c)^{2} = 1, α(r)α(c) = α(c)α(r)^{3}.

From the last relation

α(r)^{2} = 1.

Thus there are just 4 possibilities

α(r) = ±1, α(c) = ±1.

It is readily verified that all 4 of these satisfy the 3 defining relations for s and t. It follows that each defines a homomorphism

α: D_{4} → k^{×} .

We conclude that D_{4} has just 4 1-dimensional representations.

6. We look now at some examples from chemistry and physics. It should be emphasized, firstly that the theory is completely independent of these examples, which can safely be ignored; and secondly, that we are not on oath when speaking of physics. It would be inappropriate to delve too deeply here into the physical basis for the examples we give.

Figure 1.2: The methane molecule

First let us look at the methane molecule CH_{4}. In its stable state the 4 hydrogen atoms are situated at the vertices of a regular tetrahedron, with the single carbon atom at its centroid (Figure 1.2).

The molecule evidently has symmetry group S_{4}, being invariant under permutations of the 4 hydrogen atoms.

Now suppose the molecule is vibrating about this stable position. We suppose that the carbon atom at the centroid remains fixed. (We shall return to this point later.) Thus the configuration of the molecule at any moment is defined by the displacement of the 4 hydrogen atoms, say

X_{i} = (x_{i1},x_{i2},x_{i3}) (i = 1,2,3,4).

Since the centroid remains fixed,

∑_{i} x_{ij} = 0 (j = 1,2,3).

This reduces the original 12 degrees of freedom to 9.

Now let us assume further that the angular momentum also remains 0, ie the molecule is not slowly rotating. This imposes a further 3 conditions on the x_{ij}, leaving 6 degrees of freedom for the 12 ‘coordinates’ x_{ij}. Mathematically, the coordinates are constrained to lie in a 6-dimensional space. In other words we can find 6 ‘generalized coordinates’ q_{1},...,q_{6} — chosen so that q_{1} = q_{2} = ···= q_{6} = 0 at the point of equilibrium — such that each of the x_{ij} is expressible in terms of the q_{k}:

x_{ij} = x_{ij}(q_{1},...,q_{6}).

The motion of the molecule is governed by the Euler-Lagrange equations

\frac{d}{dt}( \frac{∂K}{∂q̇_{k}}) = −\frac{∂V}{∂q_{k}}

where K is the kinetic energy of the system, and V its potential energy. (These equations were developed for precisely this purpose, to express the motion of a system whose configuration is defined by generalized coordinates.)

The kinetic energy of the system is given in terms of the mass m of the hydrogen atom by

K = \frac{1}{2}m∑_{i,j} ẋ_{ij}^{2}

On substituting

ẋ_{ij} = \frac{∂x_{ij}}{∂q_{1}} q̇_{1}+ ··· + \frac{∂x_{ij}}{∂q_{6}} q̇_{6},

we see that

K = K(q̇_{1},...,q̇_{6}),

where K is a positive-definite quadratic form. Although the coefficients of this quadratic form are actually functions of q_{1},...,q_{6}, we may suppose them constant since we are dealing with small vibrations.

The potential energy of the system, which we may take to have minimal value 0 at the stable position, is given to second order by some positivedefinite quadratic form Q in the q_{k}:

V = Q(q_{1},...,q_{6}) + ....

While we could explicitly choose the coordinates q_{k}, and determine the kinetic energy K, the potential energy form Q evidently depends on the forces holding the molecule together. Fortunately, we can say a great deal about the vibrational modes of the molecule without knowing anything about these forces.

Since these 2 forms are positive-definite, we can simultaneously diagonalize them, ie we can find new generalized coordinates z_{1},...,z_{6} such that

K= ż_{1}^{2}+ ··· +ż_{6}^{2},
V= ω_{1}^{2}z_{1}^{2}+ ··· + ω_{6}^{2}z_{6}^{2}.

The Euler-Lagrange equations now give

z̈_{i} = −ω_{i}^{2}z_{i} (i = 1,...,6).

Thus the motion is made up of 6 independent harmonic oscillations, with frequencies ω_{1},...,ω_{6}.

As usual when studying harmonic or wave motion, life is easier if we allow complex solutions (of which the ‘real’ solutions will be the real part). Each harmonic oscillation then has 1 degree of freedom:

z_{j} = C_{j}e^{iω_{j}t}.

The set of all solutions of these equations (ie all possible vibrations of the system) thus forms a 6-dimensional solution-space V.

So far we have made no use of the S_{4}-symmetry of the CH_{4} molecule. But now we see that this symmetry group acts on the solution space V, which thus carries a representation, ρ say, of S_{4}. Explicitly, suppose π ∈ S_{4} is a permutation of the 4 H atoms. This permutation is ‘implemented’ by a unique spatial isometry Π. (For example, the permutation (123)(4) is effected by rotation through 1/3 of a revolution about the axis joining the C atom to the 4th H atom.)

But now if we apply this isometry Π to any vibration v(t) we obtain a new vibration Πv(t). In this way the permutation π acts on the solution-space V.

In general, the symmetry group G of the physical configuration will act on the solution-space V.

The fundamental result in the representation theory of a finite group G (as we shall establish) is that every representation ρ of G splits into parts, each corresponding to a ‘simple’ representation of G. Each finite group has a finite number of such simple representations, which thus serve as the ‘atoms’ out of which every representation of G is constructed. (There is a close analogy with the Fundamental Theorem of Arithmetic, that every natural number is uniquely expressible as a product of primes.)

The group S_{4} (as we shall find) has 5 simple representations, of dimensions 1,1,2,3,3. Our 6-dimensional representation must be the ‘sum’ of some of these.

It is not hard to see that there is just one 1-dimensional mode (up to a scalar multiple) corresponding to a ‘pulsing’ of the molecule in which the 4 H atoms move in and out (in ‘sync’) along the axes joining them to the central C atom. (Recall that S_{4} has just 2 1-dimensional representations: the trivial representation, under which each permutation leaves everything unchanged, and the parity representation, in which even permutations leave things unchanged, while odd permutations reverse them. In our case, the 4 atoms must move in the same way under the trivial representation, while their motion is reversed under an odd permutation. The latter is impossible. For by considering the odd permutation (12)(3)(4) we deduce that the first atom is moving out while the second moves in; while under the action of the even permutation (12)(34) the first and second atoms must move in and out together.)

We conclude (not rigorously, it should be emphasized!) that

ρ = 1 + α + β

where 1 denotes the trivial representation of S_{4}, α is the unique 2-dimensional representation, and β is one of the two 3-dimensional representations.

Thus without any real work we’ve deduced quite a lot about the vibrations of CH_{4}.

Each of these 3 modes has a distinct frequency. To see that, note that our system — and in fact any similar non-relativistic system — has a time symmetry corresponding to the additive group ℝ. For if (z_{j}(t) : 1 ≤ j ≤ 6) is one solution then (z_{j}(t + c)) is also a solution for any constant c ∈ ℝ.

The simple representations of ℝ are just the 1-dimensional representations

t ↦ e^{iωt}.

(We shall see that the simple representations of an abelian group are always 1-dimensional.) In effect, Fourier analysis — the splitting of a function or motion into parts corresponding to different frequencies — is just the representation theory of ℝ.

The actions of S_{4} and ℝ on the solution space commute, giving a representation of the product group S_{4}× ℝ.

As we shall see, the simple representations of a product group G × H arise from simple representations of G and H: ρ = σ × τ. In the present case we must have

ρ = 1 × E(ω_{1}) + α × E(ω_{2}) + β × E(ω_{3}),

where ω_{1},ω_{2},ω_{3} are the frequencies of the 3 modes.

If the symmetry is slightly broken, eg by placing the vibrating molecule in a magnetic field, these ‘degenerate’ frequencies will split, so that 6 frequencies will be seen: ω′_{1},ω′_{2},ω′′_{2},ω′_{3},ω′′_{3},ω′′′_{3}, where eg ω′_{2} and ω′′_{2}are close to ω. This is the origin of ‘multiple lines’ in spectroscopy.

The concept of broken symmetry has become one of the corner-stones of mathematical physics. In ‘grand unified theories’ distinct particles are seen as identical (like our 4 H atoms) under some large symmetry group, whose action is ‘broken’ in our actual universe.

7. Vibrations of a circular drum. [?]. Consider a circular elastic membrane. The motion of the membrane is determined by the function

z(x,y,t) (x^{2}+ y^{2} ≤ r^{2})

where z is the height of the point of the drum at position (x,y).

It is not hard to establish that under small vibrations this function will satisfy the wave equation

T ( \frac{∂^{2}z}{∂x^{2}}+ \frac{∂^{2}z}{∂y^{2}}) = ρ\frac{∂^{2}z}{∂t^{2}},

where T is the tension of the membrane and ρ its mass per unit area. This may be written

( \frac{∂^{2}z}{∂x^{2}}+ \frac{∂^{2}z}{∂y^{2}}) = \frac{1}{c^{2}} \frac{∂^{2}z}{∂t^{2}},

where c = (T /ρ)^{1/2} is the speed of the wave motion.

The configuration has O(2) symmetry, where O(2) is the group of 2-dimensional isometries leaving the centre O fixed, consisting of the rotations about O and the reflections in lines through O.

Although this group is not finite, it is compact. As we shall see, the representation theory of compact groups is essentially identical to the finite theory; the main difference being that a compact group has a countable infinity of simple representations.

For example, the group O(2) has the trivial representation 1, and an infinity of representations R(1),R(2),..., each of dimension 2.

The circular drum has corresponding modes M(0),M(1),M(2),..., each with its characteristic frequency. As in our last example, after taking time symmetry into account, the solution-space carries a representation ρ of the product group O(2) × ℝ, which splits into

1 × E(ω_{0}) + R(1) × E(ω_{1}) + R(2) × E(ω_{2}) + ···.

8. In the last example but one, we considered the 4 hydrogen atoms in the methane molecule as particles, or solid balls. But now let us consider a single hydrogen atom, consisting of an electron moving in the field of a massive central proton.

According to classical non-relativistic quantum mechanics [?], the state of the electron (and so of the atom) is determined by a wave function ψ(x,y,z,t), whose evolution is determined by Schrödinger’s equation

iℏ\frac{∂ψ}{∂t} = Hψ.

Here H is the hamiltonian operator, given by

Hψ = −\frac{ℏ^{2}}{2m}∇^{2}ψ + V (r)ψ,

where m is the mass of the electron, V (t) is its potential energy, ℏ is Planck’s constant, and

∇^{2}ψ = \frac{∂^{2}ψ}{∂x^{2}} + \frac{∂^{2}ψ}{∂y^{2}} + \frac{∂^{2}ψ}{∂z^{2}} .

Thus Schrodinger’s equation reads, in full,

iℏ\frac{∂ψ}{∂t} = −\frac{ℏ^{2}}{2m}( \frac{∂^{2}ψ}{∂x^{2}} + \frac{∂^{2}ψ}{∂y^{2}} + \frac{∂^{2}ψ}{∂z^{2}} ) − \frac{e^{2}}{r}ψ.

The essential point is that this is a linear differential equation, whose solutions therefore form a vector space, the solution-space.

We regard the central proton as fixed at O. (A more accurate account might take O to be the centre of mass of the system.) The system is invariant under the orthogonal group O(3), consisting of all isometries — that is, distancepreserving transformations — which leave O fixed. Thus the solution space carries a representation of the compact group O(3).

This group is a product-group:

O(3) = SO(3) × C_{2},

where C_{2} = {I,J} (J denoting reflection in O), while SO(3) is the subgroup of orientation-preserving isometries. In fact, each such isometry is a rotation about some axis, so SO(3) is group of rotations in 3 dimensions.

The rotation group SO(3) has simple representations D_{0},D_{1},D_{2},... of dimensions 1,3,5,.... To each of these corresponds a mode of the hydrogen atom, with a particular frequency ω and corresponding energy level E = ℏω.

These energy levels are seen in the spectroscope, although the spectral lines of hydrogen actually correspond to differences between energy levels, since they arise from photons given off when the energy level changes.

This idea — considering the space of atomic wave functions as a representation of SO(3) gave the first explanation of the periodic table of the elements, proposed many years before by Mendeleev on purely empirical grounds [?].

The discussion above ignores the spin of the electron. In fact representation theory hints strongly at the existence of spin, since the ‘double-covering’ SU(2) of SO(3) adds the ‘spin representations’ D_{1/2},D_{3/2},... of dimensions 2,4,... to the sequence above, as we shall see.

Finally, it is worth noting that quantum theory (as also electrodynamics) are linear theories, where the Principle of Superposition rules. Thus the application of representation theory is exact, and not an approximation restricted to small vibrations, as in classical mechanical systems like the methane molecule, or the drum.

9. The classification of elementary particles. [?]. Consider an elementary particle E, eg an electron, in relativistic quantum theory. The possible states of E again correspond to the points of a vector space V. More precisely, they correspond to the points of the projective space P(V) formed by the rays, or 1-dimensional subspaces, of V. For the wave functions ψ and ρψ correspond to the same state of E.

The state space V is now acted on by the Poincaré group E(1,3) formed by the isometries of Minkowski space-time. It follows that V carries a representation of E(1,3).

Each elementary particle corresponds to a simple representation of the Poincaré group E(1,3). This group is not compact. It is however a Lie group; and — as we shall see — a different approach to representation theory, based on Lie algebras, allows much of the theory to be extended to this case.

A last remark. One might suppose, from its reliance on linearity, that representation theory would have no rôle to play in curved space-time. But that is far from true. Even if the underlying topological space is curved, the vector and tensor fields on such a space preserve their linear structure. (So one can, for example, superpose vector fields on a sphere.) Thus representation theory can still be applied; and in fact, the so-called gauge theories introduced in the search for a unified ‘theory of everything’ are of precisely this kind.

Exercises

All representations are over ℂ, unless the contrary is stated.

In Exercises 01–11 determine all 1-dimensional representations of the given group.

1 ∗ C_{2}  2 ∗∗ C_{3}  3 ∗∗ C_{n}  4 ∗∗ D_{2}  5 ∗∗ D_{3}
6 ∗∗∗ D_{n}  7 ∗∗∗ Q_{8}  8 ∗∗∗ A_{4}  9 ∗∗∗∗ A_{n}  10 ∗∗ ℤ

11 ∗∗∗∗ D_{∞} = ⟨r,s: s^{2} = 1,rsr = s⟩

Suppose G is a group; and suppose g,h ∈ G. The element [g,h] = ghg^{−1} h^{−1} is called the commutator of g and h. The subgroup G′≡ [G,G] is generated by all commutators in G is called the commutator subgroup, or derived group of G.

12 ∗∗∗ Show that G′lies in the kernel of any 1-dimensional representation ρ of G, ie ρ(g) acts trivially if g ∈ G′.

13 ∗∗∗ Show that G′is a normal subgroup of G, and that G/G′is abelian. Show moreover that if K is a normal subgroup of G then G/K is abelian if and only if G′⊂ K. [In other words, G′is the smallest normal subgroup such that G/G′is abelian.)

14 ∗∗ Show that the 1-dimensional representations of G form an abelian group G^{∗} under multiplication. [Nb: this notation G^{∗} is normally only used when G is abelian.]

15 ∗∗ Show that C_{n}^{∗} ≅ C_{n}.

16 ∗∗∗ Show that for any 2 groups G,H

(G × H)^{∗} = G^{∗}× H^{∗}.

17 ∗∗∗∗ By using the Structure Theorem on Finite Abelian Groups (stating that each such group is expressible as a product of cyclic groups) or otherwise, show that

A^{∗} ≅ A

for any finite abelian group A.

18 ∗∗Suppose Θ : G → H is a homomorphism of groups. Then each representation α of H defines a representation Θα of G.

19 ∗∗∗ Show that the 1-dimensional representations of G and of G/G′are in oneone correspondence.

In Exercises 20–24 determine the derived group G′of the given group G.

\begin{matrix}20 ∗∗∗ C_{n} & 21 ∗∗∗∗ D_{n} & 22 ∗∗ ℤ & 23 ∗∗∗∗ D_{∞} \\ 24 ∗∗∗ Q_{8} & 25 ∗∗∗ S_{n} & 26 ∗∗∗ A_{4} & 27 ∗∗∗∗ A_{n} \end{matrix}

Equivalent Representations

Every mathematical theory starts from some notion of equivalence—an agreement not to distinguish between objects that ‘look the same’ in some sense.

Definition 2.1 Suppose α,β are two representations of G in the vector spaces U,V over k. We say that α and β are equivalent, and we write α = β, if U and V are isomorphic G-spaces.

In other words, we can find a linear map

t : U → V

which preserves the action of G, ie

t(gu) = g(tu) for all g ∈ G,u ∈ U .

Remarks:

1. Suppose α and β are given in matrix form:

α: g ↦ A(g), β: g ↦ B(g).

If α = β, then U and V are isomorphic, and so in particular dimα = dimβ, ie the matrices A(g) and B(g) are of the same size.

Suppose the linear map t : U → V is given by the matrix P. Then the condition t(gu) = g(tu) gives

B(g) = PA(g)P^{−1}

for each g ∈ G. This is the condition in matrix terms for two representations to be equivalent.

2. Recall that two n × n matrices S,T are said to be similar if there exists a non-singular (invertible) matrix P such that

T = PSP^{−1} .

A necessary condition for this is that A,B have the same eigenvalues. For the characteristic equations of two similar matrices are identical:

det(PSP^{−1} − λI) = detP det(S − λI)detP^{−1}
= det(S − λI).

3. In general this condition is necessary but not sufficient. For example, the

matrices

( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) , ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} )

have the same eigenvalues 1,1, but are not similar. (No matrix is similar to the identity matrix I except I itself.)

However, there is one important case, or particular relevance to us, where the converse is true. Let us recall a result from linear algebra.

An n × n complex matrix A is diagonalisable if and only if it satisfies a separable polynomial equation, ie one without repeated roots.

It is easy to see that if A is diagonalisable then it satisfies a separable equation. For if

A ∼ ( \begin{matrix}λ_{1} \\ & . & . & . \\ & & & & λ_{1} \\ & & & & & λ_{2} \\ & & & & & & . & . & . \end{matrix} )

then A satisfies the separable equation

m(x) ≡ (x − λ_{1})(x − λ_{2})···= 0.

The converse is less obvious. Suppose A satisfies the polynomial equation

p(x) ≡ (x − λ_{1})··· (x − λ_{r}) = 0

with λ_{1},...,λ_{r} distinct. Consider the expression of 1/p(x) as a sum of partial fractions:

\frac{1}{p(x)} = \frac{a_{1}}{x − λ_{1}} + ··· + \frac{a_{r}}{x − λ_{r}} .

Multiplying across,

1 = a_{1}Q_{1}(x) + ··· + a_{r}Q_{r}(x),

where

Q_{i}(x) = ∏_{j≠i} (x − λ_{j}) = \frac{p(x)}{x − λ_{i}} .

Substituting x = A,

I = a_{1}Q_{1}(A) + ··· + a_{r}Q_{r}(A).

Applying each side to the vector v ∈ V,

v = a_{1}Q_{1}(A)v + ··· + a_{r}Q_{r}(A)v
= v_{1}+ ··· + v_{r},

say. The vector v_{i} is an eigenvector of A with eigenvalue λ_{i}, since

(A − λ_{i})v_{i} = a_{i}p(A)v = 0.

Thus every vector is expressible as a sum of eigenvectors. In other words the eigenvectors of A span the space.

But that is precisely the condition for A to be diagonalisable. For we can find a basis for V consisting of eigenvectors, and with respect to this basis A will be diagonal.

4. It is important to note that while each matrix A(g) is diagonalisable separately, we cannot in general diagonalise all the A(g) simultaneously. That would imply that the A(g) commuted, which is certainly not the case in general.

5. However, we can show that if A_{1},A_{2},... is a set of commuting matrices then they can be diagonalised simultaneously if and only if they can be diagonalised separately.

To see this, suppose λ is an eigenvalue of A_{1}. Let

E = {v: A_{1}v = λv}

be the corresponding eigenspace. Then E is stable under all the A_{i}, since

v ∈ E ⟹ A_{1}(A_{i}v) = A_{i}A_{1}v = λA_{i}v ⟹ A_{i}v ∈ E.

Thus we have reduced the problem to the simultaneous diagonalisation of the restrictions of A_{2},A_{3},... to the eigenspaces of A_{1}. A simple inductive argument on the degree of the A_{i} yields the result.

In our case, this means that we can diagonalise some (or all) of our representation matrices

A(g_{1}),A(g_{2}),...

if and only it these matrices commute.

This is perhaps best seen as a result on the representations of abelian groups, which we shall meet later.

6. To summarise, two representations α,β are certainly not equivalent if A(g),B(g) have different eigenvalues for some g ∈ G.

Suppose to the contrary that A(g),B(g) have the same eigenvalues for all g ∈ G. Then as we have seen

A(g) ∼ B(g)

for all g, ie

B(g) = P(g)A(g)P(g)^{−1}

for some invertible matrix P(g).

Remarkably, we shall see that if this is so for all g ∈ G, then in fact α and β are equivalent. In other words, if such a matrix P(g) exists for all g then we can find a matrix P independent of g such that

B(g) = PA(g)P^{−1}

for all g ∈ G.

7. Suppose A ∼ B, ie

B = PAP^{−1} .

We can interpret this as meaning that A and B represent the same linear transformation, under the change of basis defined by P.

Thus we can think of two equivalent representations as being, if effect, the same representation looked at from two points of view, that is, taking two different bases for the representation-space.

Example: Let us look again at the natural 2-dimensional real representation ρ of the symmetry group D_{4} of the square ABCD. Recall that when we took coordinates with respect to axes Ox,Oy bisecting DA, AB, ρ took the matrix form

s ↦ ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) c ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) ,

where s is the rotation through a right-angle (sending A to B), and c is the reflection in AC.

Now suppose we choose instead the axes OA, OB. Then we obtain the equivalent representation

s ↦ ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) c ↦ ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) .

We observe that c has the same eigenvalues, ±1, in both cases.

Since we have identified equivalent representations, it makes sense to ask for all the representations of a given group G of dimension d, say. What we have to do in such a case is to give a list of d-dimensional representations, prove that every d-dimensional representation is equivalent to one of them, and show also that no two of the representations are equivalent.

It isn’t at all obvious that the number of such representations is finite, even after we have identified equivalent representations. We shall see later that this is so: a finite group G has only a finite number of representations of each dimension.

Example: Let us find all the 2-dimensional representations over ℂ of

S_{3} = ⟨s,t : s^{3} = t^{2} = 1,st = ts^{2}⟩,

that is, all 2-dimensional representations up to equivalence.

Suppose α is a representation of S(3) in the 2-dimensional vector space V. Consider the eigenvectors of s. There are 2 possibilities:

1. s has an eigenvector e with eigenvalue λ ≠ 1. Since s^{3} = 1, it follows that λ^{3} = 1, ie λ = ω or ω^{2}.

Now let f = te. Then

sf = ste = ts^{2}e = λ^{2}te = λ^{2}f.

Thus f is also an eigenvector of s, although now with eigenvector λ^{2}.

Since e and f are eigenvectors corresponding to different eigenvalues, they must be linearly independent, and therefore span (and in fact form a basis for) V:

V = ⟨e,f⟩.

Since se = λe, sf = λ^{2}f, we see that s is represented with respect to this basis by the matrix

s ↦ ( \begin{matrix}λ & 0 \\ 0 & λ^{2} \end{matrix} ) .

On the other hand, te = f, tf = t^{2}e = e, and so

t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) .

The 2 cases λ = ω,ω^{2} give the representations

α: s ↦ ( \begin{matrix}ω & 0 \\ 0 & ω^{2} \end{matrix} ) , t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) ;
β: s ↦ ( \begin{matrix}ω^{2} & 0 \\ 0 & ω \end{matrix} ) , t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) ;

In fact these 2 representations are equivalent,

α = β,

since one is got from the other by the swapping the basis elements: e,f ↦ f,e.

2. The alternative possibility is that both eigenvalues of s are equal to 1. In that case, since s is diagonalisable, it follows that

s ↦ I = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )

with respect to some basis. But then it follows that this remains the case with respect to every basis: s is always represented by the matrix I.

In particular, s is always diagonal. So if we diagonalise c—as we know we can—then we will simultaneously diagonalise s and c, and so too all the elements of D_{4}.

Suppose

s ↦ ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) , t ↦ ( \begin{matrix}λ & 0 \\ 0 & μ \end{matrix} ) .

Then it is evident that

s ↦ 1, t ↦ λ

and

s ↦ 1, t ↦ μ

will define two 1-dimensional representations of S_{3}. But we know these representations; there are just 2 of them. In combination, these will give 4 2-dimensional representations of S_{3}. However, two of these will be equivalent. The 1-dimensional representations 1 and ϵ give the 2-dimensional representation

s ↦ ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) , t ↦ ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) .

(Later we shall denote this representation by 1 + ϵ, and call it the sum of 1 and ϵ.)

On the other hand, ϵ and 1 in the opposite order give the representation

s ↦ ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) , t ↦ ( \begin{matrix}−1 & 0 \\ 0 & 1 \end{matrix} ) .

This is equivalent to the previous case, one being taken into the other by the change of coordinates (x,y) ↦ (y,x). (In other words, ϵ + 1 = 1 + ϵ.)

We see from this that we obtain just 3 2-dimensional representations of S_{3} in this way (in the notation above they will be 1 + 1, 1 + ϵ and ϵ + ϵ).

Adding the single 2-dimensional representation from the first case, we conclude that S_{3} has just 4 2-dimensional representations.

It is easy to see that no 2 of these 4 representations are equivalent, by considering the eigenvalues of s and c in the 4 cases.

Exercises

All representations are over ℂ, unless the contrary is stated. In Exercises 01–15 determine all 2-dimensional representations (up to equivalence) of the given group.

\begin{matrix}1 ∗∗ C_{2} & 2 ∗∗ C_{3} & 3 ∗∗ C_{n} & 4 ∗∗∗ D_{2} & 5 ∗∗∗ D_{4} \\ 6 ∗∗∗ D_{5} & 7 ∗∗∗∗ D_{n} & 8 ∗∗∗ S_{3} & 9 ∗∗∗∗ S_{4} & 10 ∗∗∗∗∗ S_{n} \\ 11 ∗∗∗∗ A_{4} & 12 ∗∗∗∗∗ A_{n} & 13 ∗∗∗ Q_{8} & 14 ∗∗ ℤ & 15 ∗∗∗∗ D_{∞} \end{matrix}

16 ∗∗∗ Show that a real matrix A ∈ Mat(n,ℝ) is diagonalisable over ℝ if and only if its minimal polynomial has distinct roots, all of which are real. 17 ∗∗∗ Show that a rational matrix A ∈ Mat(n,ℚ) is diagonalisable over ℚ if and only if its minimal polynomial has distinct roots, all of which are rational. 18 ∗∗∗∗ If 2 real matrices A,B ∈ Mat(n,ℝ) are similar over ℂ, are they necessarily similar over ℝ, ie can we find a matrix P ∈ GL(n,ℝ) such that B = PAP^{−1} ? 19 ∗∗∗∗ If 2 rational matrices A,B ∈ Mat(n,ℚ) are similar over ℂ, are they necessarily similar over ℚ? 20 ∗∗∗∗∗ If 2 integral matrices A,B ∈ Mat(n,ℤ) are similar over ℂ, are they necessarily similar over ℤ, ie can we find an integral matrix P ∈ GL(n,ℤ) with integral inverse, such that B = PAP^{−1} ? The matrix A ∈ Mat(n,k) is said to be semisimple if its minimal polynomial has distinct roots. It is said to be nilpotent if A^{r} = 0 for some r > 0. 21 ∗∗∗ Show that a matrix A ∈ Mat(n,k) cannot be both semisimple and nilpotent, unless A = 0. 22 ∗∗∗ Show that a polynomial p(x) has distinct roots if and only if

gcd(p(x),p′(x))= 1.

23 ∗∗∗∗ Show that every matrix A ∈ Mat(n,ℂ) is uniquely expressible in the form

A = S + N,

where S is semisimple, N is nilpotent, and

SN = NS.

(We call S and N the semisimple and nilpotent parts of A.) 24 ∗∗∗∗ Show that S and N are expressible as polynomials in A.

25 ∗∗∗∗ Suppose the matrix B ∈ Mat(n,ℂ) commutes with all matrices that commute with A, ie

AX = XA ⟹ BX = XB.

Show that B is expressible as a polynomial in A.

Simple Representations

Definition 3.1 The representation α of G in the vector space V over k is said to be simple if no proper subspace of V is stable under G.

In other words, α is simple if it has the following property: if U is a subspace of V such that

g ∈ G,u ∈ U ⟹ gu ∈ U

then either U = 0 or U = V.

Proposition 3.1 1. A 1-dimensional representation over k is necessarily simple.

2. If α is a simple representation of G over k then

dimα ≤ ∥G∥.

Proof ▶ (1) is evident since a 1-dimensional space has no proper subspaces, stable or otherwise.

For (2), suppose α is a simple representation of G in V. Take any v ≠ 0 in V, and consider the set of all transforms gv of V. Let U be the subspace spanned by these:

U = ⟨gv: g ∈ G⟩.

Each g ∈ G permutes the transforms of v, since

g(hv) = (gh)v.

It follows that g sends U into itself. Thus U is stable under G. Since α is simple, by hypothesis,

V = U.

But since U is spanned by the ∥G∥ transforms of v,

dimV = dimU ≤ ∥G∥.

◀

Remark: This result can be greatly improved, as we shall see. If k = ℂ—the case of greatest interest to us—then we shall prove that

dimα ≤ ∥G∥^{\frac{1}{2}}

for any simple representation α.

We may as well announce now the full result. Suppose G is a finite group. Then we shall show (in due course) that

1. The number of simple representations of G over ℂ is equal to the number s of conjugacy classes in G;

2. The dimensions of the simple representations σ_{1},...,σ_{s} of G over ℂ satisfy

the relation

dim^{2}σ_{1}+ ··· + dim^{2}σ_{s} = ∥G∥.

3. The dimension each simple representation σ_{i} divides the order of the group:

dimσ_{i} | ∥G∥.

Of course we cannot use these results in any proof; and in fact we will not even use them in examples. But at least they provide a useful check on our work.

Examples:

1. The first stage in studying the representation theory of a group G is to determine the simple representations of G.

Let us agree henceforth to adopt the convention that if the scalar field k is not explicitly mentioned, then we may take it that k = ℂ.

We normally start our search for simple representations by listing the 1dimensional representations. In this case we know that S_{3} has just 2 1dimensional representations, the trivial representation 1, and the parity representation ϵ.

Now suppose that α is a simple representation of S_{3} of dimension > 1. Recall that

S_{3} = ⟨s,t : s^{3} = t^{2} = 1,/;st = ts^{2}⟩,

where s = (abc),/;t = (ab).

Let e be an eigenvector of s. Thus

se = λe,

where

s^{3} = 1 ⟹ λ^{3} = 1 ⟹ λ = 1, ω, or ω^{2}.

Let

f = te.

Then

sf = ste = ts^{2}e = λ^{2}te = λ^{2}f.

Thus f is also an eigenvector of s, but with eigenvalue λ^{2}. Now consider the subspace

U = ⟨e,f⟩

spanned by e and f. Then U is stable under s and t, and so under S_{3}. For

se = λe, sf = λ^{2}f, te = f, tf = t^{2}e = e.

It follows, since α is simple, that

V = U.

So we have shown, in particular, that the simple representations of S_{3} can only have dimension 1 or 2. Let us consider the 3 possible values for λ:

(a) λ = ω. In this case the representation takes the matrix form

s ↦ ( \begin{matrix}ω & 0 \\ 0 & ω^{2} \end{matrix} ) , t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) .

(b) λ = ω^{2}. In this case the representation takes the matrix form

s ↦ ( \begin{matrix}ω^{2} & 0 \\ 0 & ω \end{matrix} ) , t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) .

But this is the same representation as the first, since the coordinate swap (x,y) ↦ (y,x) takes one into the other.

(c) λ = 1. In this case

se = e, sf = f ⟹ sv = v for all v ∈ V.

In other words s acts as the identity on V. It follows that s is represented by the matrix I with respect to any basis of V. (More generally, is g ∈ G is represented by a scalar multiple ρI of the identity with respect to one basis, then it is represented by ρI with respect to every basis; because

P(ρI)P^{−1} = ρI,

if you like.) So in this case we can turn to t, leaving s to ‘look after itself’. Let e be an eigenvector of t. Then the 1-dimensional space

U = ⟨e⟩

is stable under S_{3}, since

se = e,/;te = ±e.

Since α is simple, it follows that V = U, ie V is 1-dimensional, contrary to hypothesis.

We conclude that S_{3} has just 3 simple representations

1, ϵ and α,

of dimensions 1, 1 and 2, given by

1 : s ↦ 1,/;t ↦ 1
ϵ: s ↦ 1,/;t ↦ −1
α: s ↦ ( \begin{matrix}ω & 0 \\ 0 & ω^{2} \end{matrix} ) , t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) .

2. Now let us determine the simple representations (over ℂ) of the quaternion

group

Q_{8} = ⟨s,t : s^{4} = 1,s^{2} = t^{2},st = ts^{3}⟩,

where s = i,/;t = j. (It is best to forget at this point that one of the elements of Q_{8} is called −1, and another i, since otherwise we shall fall into endless confusion.)

We know that Q_{8} has four 1-dimensional representations, given by

s ↦ ±1, t ↦ ±1.

Suppose α is a simple representation of Q_{8} in V, of dimension > 1. Let e be an eigenvector of s:

se = λe,

where

s^{4} = 1 ⟹ λ = ±1,±i.

Let

te = f.

Then

sf = ste = ts^{3}e = λ^{3}te = λ^{3}f.

So as in the previous example, f is also an eigenvector of s, but with eigenvalue λ^{3}.

Again, as in that example, the subspace

U = ⟨e,f⟩

is stable under Q_{8}, since

se = λe, sf = λ^{3}f, te = f, tf = t^{2}e = s^{2}e = λ^{2}e.

So V = U, and {e,f} is a basis for V. With respect to this basis our representation takes the form

s ↦ ( \begin{matrix}λ & 0 \\ 0 & λ^{3} \end{matrix} ) , t ↦ ( \begin{matrix}0 & λ^{2} \\ 1 & 0 \end{matrix} ) ,

where λ = ±1,±i.

If λ = 1 this representation is not simple, since the 1-dimensional subspace

⟨(1,1)⟩

is stable under Q_{8}. (This is the same argument as before. Every vector is an eigenvector of s, so we can find a simultaneous eigenvector by taking any eigenvector of t.)

The same argument holds if λ = −1, since s is represented by −I with respect to one basis, and so also with respect to any basis. Again, the subspace

⟨(1,1)⟩

is stable under Q_{8}, contradicting our assumption that the representation is simple, and of dimension > 1.

We are left with the cases λ = ±i. In fact these are equivalent. For if λ = −i, then f is an s-eigenvector with eigenvalue λ^{3} = i. So taking f in place of e we may assume that λ = i.

We conclude that Q_{8} has just 5 simple representations, of dimensions 1,1,1,1,2, given by

1 : s ↦ 1,/;t ↦ 1
μ: s ↦ 1,/;t ↦ −1
ν: s ↦ −1,/;t ↦ 1
ρ: s ↦ −1,/;t ↦ −1
α: s ↦ ( \begin{matrix}i & 0 \\ 0 & −i \end{matrix} ) , t ↦ ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) .

We end by considering a very important case: abelian (or commutative) groups.

Proposition 3.2 A simple representation of a finite abelian group over ℂ is necessarily 1-dimensional.

Proof ▶ Suppose a ∈ A. Let λ be an eigenvalue of a, and let

E(λ) = {v ∈ V : av = λv}.

be the corresponding eigenspace.

Then E(λ) is stable under A. For

b ∈ A,v ∈ E(λ) ⟹ a(bv) = (ab)v = (ba)v = b(av) = b(λv) = λ(bv)
⟹ bv ∈ E(λ).

Thus E(λ) is stable under b, and so under A. But since V is simple, by hypothesis, it follows that

E(λ) = V.

In other words a acts as a scalar multiple of the identity:

a = λI.

It follows that every subspace of V is stable under a. Since that is true for each a ∈ A, we conclude that every subspace of V is stable under A. Therefore, since α is simple, V has no proper subspaces. But that is only true if dimV = 1. ◀

\underline{GpReps-I–7}

Example: Consider the group

D_{2} = {1,a,b,c: a^{2} = b^{2} = c^{2} = 1, bc = cb = a, ca = ac = b, ab = ca = c}. This has just four 1-dimensional representations, as shown in the following table.

1 a b c
1 1 1 1 1
μ 1 1 −1 −1
ν 1 −1 1 −1
ρ 1 −1 −1 1

Exercises

In Exercises 01–10 determine all simple representations of the given group over ℂ.

\begin{matrix}1 ∗∗ C_{2} & 2 ∗∗ C_{3} & 3 ∗∗ C_{n} & 4 ∗∗∗ D_{2} & 5 ∗∗∗ D_{4} \\ 6 ∗∗∗ D_{5} & 7 ∗∗∗∗ D_{n} & 8 ∗∗∗ S_{3} & 9 ∗∗∗∗ A_{4} & 10 ∗∗∗∗∗ Q_{8} \end{matrix}

In Exercises 11–20 determine all simple representations of the given group over ℝ.

\begin{matrix}11 ∗∗ C_{2} & 12 ∗∗∗ C_{3} & 13 ∗∗∗ C_{n} & 14 ∗∗∗ D_{2} & 15 ∗∗∗∗ D_{4} \\ 16 ∗∗∗∗ D_{5} & 17 ∗∗∗∗ D_{n} & 18 ∗∗∗∗ S_{3} & 19 ∗∗∗∗∗ A_{4} & 20 ∗∗∗∗∗ Q_{8} \end{matrix}

In Exercises 21–25 determine all simple representations of the given group over the rationalsℚ.

21 ∗∗∗∗ C_{n} 22 ∗∗∗∗ D_{n}  23 ∗∗∗∗ S_{3}  24 ∗∗∗ Q_{8}  25 ∗∗∗∗∗ A_{4}

The Arithmetic of Representations

4.1 Addition

Representations can be added and multiplied, like numbers; and the usual laws of arithmetic hold. There is even a conjugacy operation, analogous to complex conjugation.

Definition 4.1 Suppose α,β are representations of G in the vector spaces U,V over k. Then α + β is the representation of G in U ⨁ V defined by the action

g(u ⊕ v) = gu ⊕ gv.

Remarks:

1. Recall that U ⨁ V is the cartesian product of U and V, where however we write u ⊕ v rather than (u,v). The structure of a vector space is defined on this set in the natural way.

2. Note that α + β is only defined when α,β are representations of the same group G over the same scalar field k.

3. Suppose α,β are given in matrix form

α: g ↦ A(g), β: g ↦ B(g).

Then α + β is the representation

g ↦ ( \begin{matrix}A(g) & 0 \\ 0 & B(g) \end{matrix} )

4.2. MULTIPLICATION  GpReps-I–2

Example: Let us look again at the 2-dimensional representations γ_{1},γ_{2},γ_{3} of S_{3} over ℂ defined in Chapter 2

γ_{1} : s ↦ ( \begin{matrix}ω & 0 \\ 0 & ω^{2} \end{matrix} ) ,t ↦ ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) , γ_{2} : s ↦ ( \begin{matrix}ω & 0 \\ 0 & ω^{2} \end{matrix} ) ,t ↦ ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} ) ,

γ_{3} : s ↦ ( \begin{matrix}ω & 0 \\ 0 & ω^{2} \end{matrix} ) ,t ↦ ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) .

We see now that

γ_{1} = 1 + 1, γ_{2} = 1 + ϵ, γ_{3} = ϵ + ϵ,

where 1 is the trivial 1-dimensional representation of S_{3}, and ϵ is the 1-dimensional parity representation

s ↦ 1, t ↦ −1.

(We can safely write 1 + 1 = 2,ϵ + ϵ = 2ϵ.)

Proposition 4.1 1. dim(α + β) = dimα + dimβ;

2. β + α = α + β;

3. α + (β + γ) = (α + β) + γ.

Proof ▶ These are all immediate. For example, the second part follows from the natural isomorphism

V ⨁ U → U ⨁ V : v ⊕ u ↦ u ⊕ v.

◀

4.2 Multiplication

Definition 4.2 Suppose α,β are representations of G in the vector spaces U,V over k. Then αβ is the representation of G in U ⨂ V defined by the action

g(u_{1}⊗ v_{1}+ ··· + u_{r}⊗ v_{r}) = gu_{1}⊗ gv_{1}+ ··· gu_{r}⊗ gv_{r}.

Remarks:

1. The tensor product U ⨂ V of 2 vector spaces U and V may be unfamiliar. Each element of U ⨂ V is expressible as a finite sum

u_{1}⊗ v_{1}+ ··· + u_{r}⊗ v_{r}.

If U has basis {e_{1},...,e_{m}} and V has basis {f_{1},...,f_{n}} then the mn elements

e_{i}⊗ f_{j} (i = 1,...,m; j = 1,...,n)

form a basis for U ⨂ V. In particular

dim(U ⊗ V) = dimU dimV.

(It is a common mistake to suppose that every element of U ⨂ V is expressible in the form u ⊗ v. That is not so; the general element requires a finite sum.)

Formally, the tensor product is defined as the set of formal sums

u_{1}⊗ v_{1}+ ··· + u_{r}⊗ v_{r},

where 2 sums define the same element if one can be derived from the other by applying the rules

(u_{1}+u_{2})⊗v = u_{1}⊗v+u_{2}⊗v, u⊗(v_{1}+v_{2}) = u⊗v_{1}+u⊗v_{2}, (ρu)⊗v = u⊗(ρv).

The structure of a vector space is defined on this set in the natural way.

2. As with α+β, αβ is only defined when α,β are representations of the same group G over the same scalar field k.

3. It is important not to write α×β for αβ, as we shall attach a different meaning to α × β later.

4. Suppose α,β are given in matrix form

α: g ↦ A(g), β: g ↦ B(g).

Then αβ is the representation

αβ: g ↦ A(g) ⊗ B(g).

But what do we mean by the tensor product S⊗T of 2 square matrices S,T? If S = s_{ij} is an m × m-matrix, and T = t_{kl} is an n × n-matrix, then S ⊗ T is the mn × mn-matrix whose rows and columns are indexed by the pairs (i,k) where 1 ≤ i ≤ m,1 ≤ k ≤ n, with matrix entries

(S ⊗ T)_{(i,k)(j,l)} = S_{ij}T_{kl}.

To write out this matrix S ⊗ T we must order the index-pairs. Let us settle for the ‘lexicographic order’

(1,1),(1,2),...,(1,n),(2,1),...,(2,n),...,(m,1),...,(m,n).

(In fact the ordering does not matter for our purposes. For if we choose a different ordering of the rows, then we shall have to make the same change in the ordering of the columns; and this double change simply corresponds to a change of basis in the underlying vector space, leading to a similar matrix to S ⊗ T.)

Example: Consider the 2-dimensional representation α of S_{3} over ℂ

α: s ↦ ( \begin{matrix}ω & 0 \\ 0 & ω^{2} \end{matrix} ) , t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )

We shall determine the 4-dimensional representation α^{2} = αα. (The notation α^{2} causes no problems.) We have

α^{2} : s ↦ ( \begin{matrix}ω & 0 \\ 0 & ω^{2} \end{matrix} ) ⨂ ( \begin{matrix}ω & 0 \\ 0 & ω^{2} \end{matrix} ) , t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) ⨂ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )

It is simply (!) a matter of working out these 2 tensor products. In fact

( \begin{matrix}ω & 0 \\ 0 & ω^{2} \end{matrix} ) ⨂ ( \begin{matrix}ω & 0 \\ 0 & ω^{2} \end{matrix} ) = ( \begin{matrix}ω · ω & ω · 0 & 0 · ω & 0 · 0 \\ ω · 0 & ω · ω^{2} & 0 · 0 & 0 · ω^{2} \\ 0 · ω & 0 · 0 & ω^{2}· ω & ω^{2}· 0 \\ 0 · 0 & 0 · ω^{2} & ω^{2}· 0 & ω^{2}· ω^{2} \end{matrix} )

= ( \begin{matrix}ω^{2} & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & ω \end{matrix} ) ,

while

( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) ⨂ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) = ( \begin{matrix}0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{matrix} ) ,

We can simplify this by the change of coordinates (x,y,z,t) ↦ (y,z,t,x). This will give the equivalent representation (which we may still denote by αβ):

αβ: s ↦ ( \begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & ω & 0 \\ 0 & 0 & 0 & ω^{2} \end{matrix} ) , t ↦ ( \begin{matrix}0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{matrix} )

But now we see that this splits into 2 2-dimensional representations, the second of which is α itself:

α^{2} = β + α,

where β is the representation

β: s ↦ ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) , t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )

The representation β can be split further. That is evident if we note that since s is represented by I, we can diagonalise t without affecting s. Since t has eigenvalues ±1, this must yield the representation

β: s ↦ ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ) , t ↦ ( \begin{matrix}1 & 0 \\ 0 & −1 \end{matrix} )

Concretely, the change of coordinates (x,y) ↦ (x + y,x − y) brings this about.) Thus

β = 1 + ϵ,

and so

α^{2} = 1 + ϵ + α.

(We hasten to add that this kind of matrix manipulation is not an essential part of representation theory! We shall rapidly develop techniques which will enable us to dispense with matrices altogether.)

Proposition 4.2 1. dim(αβ) = dimαdimβ;

2. βα = αβ;

3. α(βγ) = (αβ)γ;

4. α(β + γ) = αβ + αγ;

5. 1α = α.

4.3. DUALITY  GpReps-I–6

All these results, again, are immediate consequences of ‘canonical isomorphisms’ which it would be tedious to write out explicitly.

We have seen that the representations of G over k can be added and multiplied. They almost form a ring—only subtraction is missing. In fact if we introduce ‘virtual representations’ α−β (where α,β are representations) then we will indeed obtain a ring

R(G) = R(G,k),

the representation-ring of G over k. (By convention if k is omitted then we assume that k = ℂ.)

We shall see later that

α + β = α + γ ⟹ β = γ.

It follows that nothing is lost in passing from representations to R(G); if α = β in R(G) then α = β in ‘real life’.

4.3 Duality

Definition 4.3 Suppose α = is a representation of G in the vector space V over k. Then α^{∗} is the representation of G in the dual vector space U^{∗} defined by the action

(gπ)(v) = π(g^{−1} v) (g ∈ G, π ∈ V^{∗}, v ∈ V)

Remarks:

1. Recall that the dual vector space V^{∗} is the space of linear functionals

π: V → k.

To any basis {e_{1},...,e_{n}} of V there corresponds a dual basis {π_{1},...,π_{n}} of V^{∗}, where

π_{j}(e_{i}) = \{ \begin{matrix}1 & if i = j \\ 0 & otherwise \end{matrix}

2. Suppose α is given in matrix form

α: g ↦ A(g).

Then α^{∗} is the representation

g ↦ (A(g)^{−1} )′,

where T′denotes the transpose of T. Notice the mysterious way in which the inverse and transpose, each of which is ‘contravariant’, ie

(RS)^{−1} = S^{−1} T^{−1} , (RS)′= S′R′,

combine to give the required property ((RS)^{−1} )′= (R^{−1} )′(S^{−1} )′.

Example: Consider α^{∗}, where α is the 2-dimensional representation of S_{3} over ℂ considered above. By the rule above, α^{∗} is given by

α^{∗} : s ↦ ( \begin{matrix}ω^{2} & 0 \\ 0 & ω \end{matrix} ) , t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) .

It is easy to see that swapping the coordinates, (x,y) ↦ (y,x), gives

α^{∗} = α

Many of the representations we shall meet will share this property of self-conjugacy.

Proposition 4.3 1. dim(α^{∗}) = dimα; 2. (α^{∗})^{∗} = α;

3. (α + β)^{∗} = α^{∗}+ β^{∗}.

4. (αβ)^{∗} = α^{∗}β^{∗}.

5. 1^{∗} = 1.

Summary: We have defined the representation ring R(G) of a group G, and shown that it carries an operation α ↦ α^{∗} analogous to complex conjugation.

Semisimple Representations

Definition 5.1 The represenation α of G is said to be semisimple if it is expressible as a sum of simple representations:

α = σ_{1}+ ··· + σ_{r}.

Example: Consider the permutation representation ρ of S_{3} in k^{3}. (It doesn’t matter for the following argument if k = ℝ or ℂ.)

Recall that

g(x_{1},x_{2},x_{3}) = (x_{g^{−1} 1},x_{g^{−1} 2},x_{g^{−1} 3}).

We have seen that k^{3} has 2 proper stable subspaces:

U = {(x,x,x) : x ∈ k}, W = {(x_{1},x_{2},x_{3}) : x_{1}+ x_{2}+ x_{3} = 0}.

U has dimension 1, with basis {(1,1,1)}; W has dimension 2, with basis {(1,−1,0),(−1,0,1)}. Evidently

U ∩ V = 0.

Recall that a sum U + V of vector subspaces is direct,

U + V = U ⊕ V,

if (and only if) U ∩ V = 0. So it follows here, by considering dimensions, that

k^{3} = U ⨁ W.

The representation on U is the trivial representation 1. Thus

ρ = 1 + α,

where α is the representation of S_{3} in W.

We can see that α is simple as follows. Suppose V ⊂ W is stable under S_{3}, where V ≠ 0. Take any element v ≠ 0 in V: say

v = (x,y,z) (x + y + z = 0).

The coefficients x,y,z cannot all be equal. Suppose x ≠ y. Then

(12)v = (y,x,z) ∈ V;

and so

v − (12)v = (x − y,y − x,0) = (x − y)(1,−1,0) ∈ V.

Hence

(1,−1,0) ∈ V.

It follows that

(−1,0,1) = (132)(1,−1,0) ∈ V

also. But these 2 elements generate W; hence

V = W.

So we have shown that W is a simple S_{3}-space, whence the corresponding representation α is simple.

We conclude that the representation

ρ = 1 + α

is a sum of simple representations, and so is semisimple.

It is easy to see that U and W are the only subspaces of k^{3} stable under S_{3}, apart from 0 and the whole space. So it is evident that the splitting U ⊕ V is unique. In general this is not so; in fact we shall show later that there is a unique split into simple subspaces if and only if the representations corresponding to these subspaces are distinct. (So in this case the split is unique because 1 ≠ α.) However the simple representations that appear are unique. This fact, which we shall prove in the next chapter, is the foundation stone of representation theory.

Most of the time we do not need to look behind a representation at the underlying representation-space. But sometimes we do; and the following results should help to clarify the structure of semisimple representation-spaces.

Proposition 5.1 Suppose V is a sum (not necessarily direct) of simple subspaces:

V = S_{1}+ ··· + S_{r}.

Then V is semisimple.

Proof ▶ Since S_{2} is simple,

S_{1}∩ S_{2} = 0 or S_{2}.

In the former case

S_{1}+ S_{2} = S_{1}⨁ S_{2};

in the latter case S_{2} ⊂ S_{1} and so

S_{1}+ S_{2} = S_{1}.

Repeating the argument with S_{1}+ S_{2} in place of S_{1}, and S_{3} in place of S_{2},

(S_{1}+ S_{2}) ∩ S_{3} = 0 or S_{3},

since S_{3} is simple. In the former case

S_{1}+ S_{2}+ S_{3} = (S_{1}+ S_{2})⨁ S_{3};

in the latter case S_{3} ⊂ S_{1}+ S_{2} and so

S_{1}+ S_{2}+ S_{3} = S_{1}+ S_{2}.

Combining this with the previous step

S_{1}+ S_{2}+ S_{3} = S_{1}⨁ S_{2}⨁ S_{3} or S_{1}⨁ S_{3} or S_{1}⨁ S_{2} or S_{1}.

Continuing in this style, at the ith step, since S_{i} is simple,

S_{1}+ ··· + S_{i} = (S_{1}+ ··· + S_{i−1} )⨁ S_{i} or S_{1}+ ··· + S_{i−1} .

We conclude, finally, that

V = S_{1}+ ··· + S_{r} = S_{i_{1}}⨁ ··· S_{i_{s}},

where {S_{i_{1}},...,S_{i_{s}}} is a subset of {S_{1},...,S_{r}}. ◀

Remark: The subset {S_{i_{1}},...,S_{i_{s}}} depends in general on the order in which we take S_{1},...,S_{r}. In particular, since S_{i_{1}} = S_{1}, we can always specify that any one of S_{1},...,S_{n} appears in the direct sum.

Proposition 5.2 The following 2 properties of the G-space V are equivalent:

1. V is semisimple; 2. each stable subspace U ⊂ V has at least one complementary stable sub-

space W, ie

V = U ⨁ W.

Proof ▶ Suppose first that V is semisimple, say

V = S_{1}⨁ ··· ⨁ S_{r}.

Let us follow the proof of the preceding proposition, but starting with U rather than S_{1}. Thus our first step is to note that since S_{1} is simple,

U + S_{1} = U ⨁ S_{1} or U.

Continuing as before, we conclude that

V = U ⨁ S_{i_{1}}⨁ ··· ⨁ S_{i_{s}},

from which the result follows, with

W = S_{i_{1}}⨁ ··· ⨁ S_{i_{s}}.

Now suppose that condition (2) holds. Since V is finite-dimensional, we can find a stable subspace S_{1} of minimal dimension. Evidently S_{1} is simple; and by our hypothesis

V = S_{1}⨁ W_{1}.

Now let us find a stable subspace S_{2} of W_{1} of minimal dimension. As before, this subspace is simple; and

S_{1}∩ S_{2} ⊂ S_{1}∩ W = 0,

so that

S_{1}+ S_{2} = S_{1}⨁ S_{2}.

Applying the hypothesis again to this space, we can find a stable complement W_{2}:

V = S_{1}⨁ S_{2}⨁ W_{2}.

Continuing in this way, since V is finite-dimensional we must conclude with an expression for V as a direct sum of simple subspaces:

V = S_{1}⨁ ··· ⨁ S_{r}.

Hence V is semisimple. ◀

Remark: This Proposition gives an alternative definition of semisimplicity: V is semisimple if every stable subspace U ⊂ V posseses a complementary stable subspace W. This alternative definition allows us to extend the concept of semisimplicity to infinite-dimensional representations.

Exercises

In Exercises 01–15 calculate e^{X} for the given matrix X:

1. Show that any commuting set of diagonalisable matrices can be simultaneously diagonalised. Hence show that any representation of a finite abelian group

2. Show that for all n the natural representation ρ of S_{n} in k^{n} is semisimple.

3. If T ∈ GL(n,k) then the map

ℤ → GL(n,k) : m ↦ T^{m}

defines a representation τ of the infinite abelian group Z. Show that if k = ℂ then τ is semisimple if and only if T is semisimple.

4. Prove the same result when k = ℝ.

5. Suppose k = GF(2) = {0,1}, the finite field with 2 elements. Show that the representation of C_{2} = {e,g} given by

g ↦ ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} )

is not semisimple.

Every Representation of a Finite Group is Semisimple

Theorem 6.1 (Maschke’s Theorem) Suppose α is a representation of the finite group G over k, where k = ℝ or ℂ. Then α is semisimple.

Proof ▶ Suppose α is a representation on V. We take the alternative definition of semisimplicity: every stable subspace U ⊂ V must have a stable complement W.

Our idea is to construct an invariant positive-definite form P on V. (By ‘form’ we mean here quadratic form if k = ℝ, or hermitian form if k = ℂ.) Then we can take W to be the orthogonal complement of U with respect to this form:

W = U^{⊥} = {v ∈ V : P(u,v) = 0 for all u ∈ U}.

We can construct such a form by taking any positive-definite form Q, and averaging it over the group:

P(u,v) = \frac{1}{∥G∥} ∑_{g∈G}Q(gu,gv).

(It’s not really necessary to divide by the order of the group; we do it because the idea of ‘averaging over the group’ occurs in other contexts.)

It is easy to see that the resulting form is invariant:

P(gu,gv) = \frac{1}{∥G∥} ∑_{h∈G}Q(hgu,hgv)
= \frac{1}{∥G∥} ∑_{h∈G}Q(hu,hv)
= P(u,v)

since hg runs over the group as h does so.

It is a straightforward matter to verify that if P is invariant and U is stable then so is U^{⊥}. Writing ⟨u,v⟩ for P(u,v),

g ∈ G,w ∈ U^{⊥} ⟹ ⟨u,w⟩ = 0 ∀u ∈ U
⟹ ⟨gu,gw⟩ = ⟨u,w⟩ = 0 ∀u ∈ U
⟹ ⟨u,gw⟩ = ⟨g(g^{−1} u),w⟩ = 0 ∀u ∈ U
⟹ gw ∈ U^{⊥} .

◀

Examples:

1. Consider the representation of S_{3} in ℝ^{3}. There is an obvious invariant quadratic form—as is often the case—namely

x_{1}^{2}+ x_{2}^{2}+ x_{3}^{2}.

But as an exercise in averaging, let us take the positive-definite form

Q(x_{1},x_{2},x_{3}) = 2x_{1}^{2}− 2x_{1}x_{2}+ 3x_{2}^{2}+ x_{3}^{2}.

Then

Q(e(x_{1},x_{2},x_{3})) = Q(x_{1},x_{2},x_{3}) = 2x_{1}^{2}− 2x_{1}x_{2}+ 3x_{2}^{2}+ x_{3}^{2}
Q((23)(x_{1},x_{2},x_{3})) = Q(x_{1},x_{3},x_{2}) = 2x_{1}^{2}− 2x_{1}x_{3}+ 3x_{3}^{2}+ x_{2}^{2}
Q((13)(x_{1},x_{2},x_{3})) = Q(x_{3},x_{2},x_{1}) = 2x_{3}^{2}− 2x_{3}x_{2}+ 3x_{2}^{2}+ x_{1}^{2}
Q((12)(x_{1},x_{2},x_{3})) = Q(x_{2},x_{1},x_{3}) = 2x_{2}^{2}− 2x_{2}x_{1}+ 3x_{1}^{2}+ x_{3}^{2}
Q((123)(x_{1},x_{2},x_{3})) = Q(x_{3},x_{1},x_{2}) = 2x_{3}^{2}− 2x_{3}x_{1}+ 3x_{1}^{2}+ x_{2}^{2}
Q((132)(x_{1},x_{2},x_{3})) = Q(x_{2},x_{3},x_{1}) = 2x_{2}^{2}− 2x_{2}x_{3}+ 3x_{3}^{2}+ x_{1}^{2}

Adding, and dividing by 6,

P(x_{1},x_{2},x_{3}) = 2(x_{1}^{2}+ x_{2}^{2}+ x_{3}^{2}) − \frac{2}{3} (x_{2}x_{3}+ x_{1}x_{3}+ x_{1}x_{2})
= \frac{7}{3} (x_{1}^{2}+ x_{2}^{2}+ x_{3}^{2}) − \frac{1}{3} (x_{1}+ x_{2}+ x_{3}) ^{2}.

The corresponding inner product is given by

⟨(x_{1},x_{2},x_{3}),(y_{1},y_{2},y_{3})⟩ = 2(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3})− \frac{1}{3}(x_{2}y_{3}+x_{3}y_{2}+x_{3}y_{1}+x_{1}y_{3}+x_{1}y_{2}+x_{2}y_{1})

To see how this is used, let

U = {(x,x,x) : x ∈ ℝ}.

Evidently U is stable. Its orthogonal complement with respect to the form above is

U^{⊥} = {(x_{1},x_{2},x_{3}) : ⟨(1,1,1),(x_{1},x_{2},x_{3})⟩ = 0}
= \{(x_{1},x_{2},x_{3}) : \frac{4}{3}(x_{1}+ x_{2}+ x_{3}) = 0\}
= {(x_{1},x_{2},x_{3}) : x_{1}+ x_{2}+ x_{3} = 0},

which is just the complement we found before. This is not surprising since— as we observed earlier—U and U^{⊥} are the only proper stable subspaces of ℝ^{3}.

2. For an example using hermitian forms, consider the simple representation of D_{4} over ℂ defined by

s ↦ ( \begin{matrix}i & 0 \\ 0 & −i \end{matrix} ) , t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) .

Again, there is an obvious invariant hermitian form, namely

|x_{1}^{2}+ x_{2}^{2}| = \overline{x_{1}}x_{1}+ \overline{x_{2}}x_{2}.

But this will not give us much exercise.

The general hermitian form on ℂ^{2} is

ax̄x + bȳy + cx̄y + c̄ȳx (a,b ∈ ℝ, c ∈ ℂ)

Let us take

Q(x,y) = 2x̄x + ȳy − ix̄y + iȳx.

Note that

D_{4} = {e,s,s^{2},s^{3},t,ts,ts^{2},ts^{3}}.

For these 8 elements are certainly distinct, eg

s^{2} = ts^{3} ⟹ ts = 1 ⟹ s = t.

Now

Q(e(x,y)) = Q(x,y) = 2x̄x + ȳy − ix̄y + iȳx,
Q(s(x,y)) = Q(ix,−iy) = 2x̄x + ȳy + ix̄y − iȳx,
Q(s^{2}(x,y)) = Q(−x,−y) = 2x̄x + ȳy − ix̄y + iȳx,
Q(s^{3}(x,y)) = Q(−ix,iy) = 2x̄x + ȳy + ix̄y − iȳx,
Q(t(x,y)) = Q(y,x) = x̄x + 2ȳy + ix̄y − iȳx,
Q(ts(x,y)) = Q(iy,−ix) = x̄x + 2ȳy − ix̄y + iȳx,
Q(ts^{2}(x,y)) = Q(−y,−x) = x̄x + 2ȳy + ix̄y − iȳx,
Q(ts^{3}(x,y)) = Q(−iy,ix) = x̄x + 2ȳy − ix̄y + iȳx.

Averaging,

P(x,y) = \frac{1}{8} ∑ g ∈ D_{4}Q(g(x,y))
= \frac{3}{2}(x̄x + ȳy)

It is no coincidence that we have ended up with a scalar multiple of |x|^{2}+ |y|^{2}. For it is easy to see that a simple G-space carries a unique invariant hermitian form, up to a scalar multiple. Suppose P,Q were 2 such forms. Let λ be an eigenvalue of Q with respect to P, ie a solution of

det(A − λB) = 0,

where A,B are the matrices of P,Q. Then the corresponding eigenspace

E = {v: Av = λBv}

would be stable under G.

The alternative proof of Maschke’s Theory below may be preferred by the algebraically-minded. It has the advantage of extending to scalar fields other than ℝ and ℂ. Against that, it lacks the intuitive appeal of the earlier proof.

Alternative proof ▶ Recall that a projection p: V → V is a linear map satisfying the relation

p^{2} = p

(ie p is idempotent).

If p is a projection then so is 1 − p:

(1 − p)^{2} = 1 − 2p + p^{2} = 1 − 2p + p = 1 − p.

The projections (p,1 − p) define a splitting of V into a direct sum

V = imp⨁ im(1 − p).

Note that

v ∈ imp ⟺ pv = v.

Note also that

im(1 − p) = kerp,

since

v = (1 − p)w ⟹ pv = (p − p^{2})w = 0,

while

pv = 0 ⟹ v = (1 − p)v.

Thus the splitting can equally well be written

V = imp⨁ kerp.

Conversely, every splitting

V = U ⨁ W

arises from a projection p in this way: if

v = u + w (u ∈ U,w ∈ W)

then we set

pv = u.

(Although the projection p is often referred to as ‘the projection onto U’ it depends on W as well as U. In general there are an infinity of projections onto U, corresponding to the infinity of complements W. When there is a positive-definite form on V —quadratic or hermitian, according as k = ℝ or ℂ—then one of these projections is distinguished: namely the ‘orthogonal projection’ corresponding to the splitting

V = U ⨁ U^{⊥} .

But we are not assuming the existence of such a form at the moment.)

Now suppose U is a stable subspace of V. Choose any complementary subspace W:

V = U ⨁ W.

In general W will not be stable under G. Our task is to find a stable complementary subspace W_{0}:

V = U ⨁ W = U ⨁ W_{0}.

Let p be the projection onto U with complement W. We know that U is stable under G, ie

g ∈ G, u ∈ U ⟹ gu ∈ U.

Thus

g ∈ G,u = pv ⟹ pgu = gu ⟹ pgpv = gpv.

Since this holds for all v ∈ V,

pgp = gp

for all g ∈ G. Conversely, if this is so then U = imp is stable.

By the same argument, W = im(1 − p) will be stable if and only if

(1 − p)g(1 − p) = g(1 − p)

for all g ∈ G. This reduces to

pgp = pg.

Both U and W are stable if and only if

gp = pg.

For in that case

pgp = p(gp) = p(pg) = p^{2}g = pg = gp.

Now

gp = pg ⟺ g^{−1} pg = p.

In other words, p defines a splitting into stable subspaces if and only if it is invariant under G.

In general, we can construct an invariant element by averaging over G. Let us therefore set

P = \frac{1}{∥G∥} ∑_{g∈G}g^{−1} pg.

This will certainly be invariant under G:

g^{−1} Pg = \frac{1}{∥G∥} ∑_{h∈G}g^{−1} h^{−1} phg
= \frac{1}{∥G∥} ∑_{h∈G}(hg)^{−1} p(hg)^{−1}
= \frac{1}{∥G∥} ∑_{h∈G}h^{−1} ph^{−1}

since hg runs over G as h does so.

What is less obvious is that P is a projection, and in fact a projection onto U. To see that, note that

u ∈ U ⟹ gu ∈ U ⟹ p(gu) = gu.

Hence by addition

u ∈ U ⟹ Pu = u.

Conversely,

v ∈ V ⟹ pgv ∈ U ⟹ gpgv ∈ U.

So by addition

v ∈ V ⟹ Pv ∈ U.

These 2 results imply that P^{2} = P, and that P projects onto U. ◀

Remarks:

1. We can show directly that P is a projection, as follows:

P^{2} = \frac{1}{∥G∥^{2}} ∑_{g,h}g^{−1} pgh^{−1} ph
= \frac{1}{∥G∥^{2}} ∑_{g,h}g^{−1} gh^{−1} ph
= \frac{1}{∥G∥^{2}} ∑_{g,h}h^{−1} ph
= \frac{1}{∥G∥} ∑_{h} h^{−1} ph
= P.

Two projections p,q project onto the same (first) subspace if

qp = p, pq = q.

So to prove that P projects onto the same subspace U as p, we must show that Pp = p and pP = P. These follow in much the same way:

Pp = \frac{1}{∥G∥} ∑_{g} g^{−1} pgp
= \frac{1}{∥G∥} ∑_{g} g^{−1} gp

pP = \frac{1}{∥G∥} ∑_{g} pg^{−1} pg
= \frac{1}{∥G∥} ∑_{g} g^{−1} pg
= P.

2. Both proofs of Maschke’s Theorem rely on the same idea: obtaining an invariant element (in the first proof, an invariant form; in the second, and invariant projection) by averaging over transforms of a non-invariant element.

In general, if V is a G-space (in other words, we have a representation of G in V) then the invariant elements form a subspace

V^{G} = {v ∈ V : gv = v∀g ∈ G}.

The averaging operation defines a projection of V onto V^{G}:

v ↦ \frac{1}{∥G∥} ∑_{g} gv.

Clearly V^{G} is a stable subspace of V. Thus if V is simple, either V^{G} = 0 or V^{G} = V. In the first case, all averages vanish. In the second case, the representation in V is trivial, and so V must be 1-dimensional.

3. It is worth noting that our alternative proof works in any scalar field k, provided ∥G∥ ≠ 0 in k. Thus it even works over the finite field GF(p^{n}), unless p | ∥G∥.

Of course we are not considering such modular representations (as representations over finite fields are known); but our argument shows that semisimplicity still holds unless the characteristic p if the scalar field divides the order of the group.

Uniqueness and the Intertwining Number

Definition 7.1 Suppose α,β are representations of G over k in the vector spaces U,V respectively. The intertwining number I(α,β) is defined to be the dimension of the space of G-maps t : U → V,

I(α,β) = dimhom^{G}(U,V).

Remarks:

1. A G-map t : U → V is a linear map which preserves the action of G:

t(gu) = g(tu) (g ∈ G,u ∈ G).

These G-maps evidently form a vector space over k.

2. The intertwining number will remain somewhat abstract until we give a formula for it (in terms of characters) in Chapter . But intuitively I(α,β) measures how much the representations α,β have in common.

3. The intertwining number of finite-dimensional representations is certainly finite, as the following result shows.

Proposition 7.1 We have

I(α,β) ≤ dimαdimβ.

Proof ▶ The space hom(U,V) of all linear maps t: U → V has dimension dimU dimV, since we can represent each such map by an m × n-matrix, where m = dimU,n = dimV.

The result follows, since

hom^{G}(U,V) ⊂ hom(U,V).

◀

Proposition 7.2 Suppose α,β are simple representations over k. Then

I(α,β) = \{ \begin{matrix}0 & if α ≠ β, \\ ≥ 1 & if α = β. \end{matrix}

Proof ▶ Suppose α,β are representations in U,V, respectively; and suppose

t : U → V

is a G-map. Then the subspaces

kert = {u ∈ U : tu = 0} and imt = {v ∈ V : ∃u ∈ U, tu = v}

are both stable under G. Thus

u ∈ kert ⟹ tu = 0
⟹ t(gu) = g(tu) = 0
⟹ gu ∈ kert,

while

v ∈ imt ⟹ v = tu
⟹ t(gu) = g(tu) = gv
⟹ gv ∈ imt.

But since U and V are both simple, by hypothesis, it follows that

kert = 0 or U, imt = 0 or V.

Now kert = U ⟹ t = 0, and imt = 0 ⟹ t = 0. So if t ≠ 0,

kert = 0, imt = V.

But in this case t is an isomorphism of G-spaces, and so α = β.

On the other hand, if α = β then (by the definition of equivalent representations) there exists a G-isomorphis t : U → V, and so I(α,β) ≥ 1. ◀

When k = ℂ we can be more precise.

Proposition 7.3 If α is a simple representation over ℂ then

I(α,α) = 1.

Proof ▶ Suppose V carries the representation α. We have to show that

dimhom^{G}(V,V) = 1.

Since the identity map 1 : V → V is certainly a G-map, we have to show that every G-map t : V → V is a scalar multiple ρ1 of the identity.

Let λ be an eigenvector of t. Then the corresponding eigenspace

E = E(λ) = {v ∈ V : tv = λv}

is stable under G. For

g ∈ G, v ∈ E ⟹ t(gv) = g(tv) = λgv ⟹ gv ∈ E.

Since α is simple, this implies that E = V, ie

t = λ1.

◀

Proposition 7.4 Suppose α,β,γ are representations over k. Then

1. I(α + β,γ) = I(α,γ) + I(β,γ);

2. I(α,β + γ) = I(α,β) + I(α,γ);

3. I(αβ,γ) = I(α,β^{∗}γ).

Proof ▶ Suppose α,β,γ are representations in U,V,W respectively. The first 2 results are immediate, arising from the more-or-less self-evident isomorphisms

hom(U ⨁ V,W) ≅ hom(U,W)⨁ hom(V,W)
hom(U,V ⨁ W) ≅ hom(U,V)⨁ hom(U,W).

Take the first. This expresses the fact that a linear map

t : U ⨁ V → W

can be defined by giving 2 linear maps

t_{1} : U → W, t_{2} : V → W.

In fact t_{1} is the restriction of t to U ⊂ U ⨁ V, and t_{2} the restriction of t to V ⊂ U ⨁ V; and

t(u ⊕ v) = t_{1}u ⊕ t_{2}v.

In much the same way, the second result expresses the fact that a linear map

t : U → V ⨁ W

can be defined by giving 2 linear maps

t_{1} : U → V, t_{2} : U → W.

In fact

t_{1} = π_{1}t, t_{2} = π_{2}t,

where π_{1},π_{2} are the projections of U ⨁ V onto V,W respectively; and

tu = t_{1}u ⊕ t_{2}u.

The third result, although following from a similar ‘natural equivalence’

hom(U ⨂ V,W) ≅ hom(U,V^{∗}⨂ W),

where

V^{∗} = hom(V,k),

is rather more difficult to establish.

We can divide the task in two. First, there is a natural equivalence

hom(U,hom(V,W)) ≅ hom(U ⨂ V,W).

For this, note that there is a 1–1 correspondence between linear maps b: U⨂ V → W and bilinear maps

B: U × V → W.

(This is sometimes taken as the definition of U ⨂ V.) So we have to show how such a bilinear map B(u,v) gives rise to a linear map

t : U → hom(V,W).

But that is evident:

t(u)(v) = B(u,v).

It is a straightforward matter to verify that every such linear map t arises in this way from a unique bilinear map B.

It remains to show that

hom(V,W) ≅ V^{∗}⨂ W.

For this, note first that both sides are ‘additive functors’ in W, ie

hom(V,W_{1}⨁ W_{2})= hom(V,W_{1})⨁ hom(V,W_{2}),
V^{∗}⨂ (W_{1}⨁ W_{2})= (V^{∗}⨂ W_{1})⨁ (V^{∗}⨂ W_{2}).

This allows us to reduce the problem, by expressing W as a sum of 1-dimensional subspaces, to the case where W is 1-dimensional. In that case, we may take W = k; so the result to be proved is

hom(V,k) ≅ V^{∗}⨂ k.

But there is a natural isomorphism

U ⨂ k ≅ U

for every vector space U. So our result reduces to the tautology V^{∗} ≅ V^{∗}.

It’s a straightforward (if tedious) matter to verify that these isomorphisms are all compatible with the actions of the group G. In particular the G-invariant elements on each side correspond:

hom^{G}(U ⨁ V,W) ≅ hom^{G}(U,W)⨁ hom^{G}(V,W),
hom^{G}(U,V ⨁ W) ≅ hom^{G}(U,V)⨁ hom^{G}(U,W),
hom^{G}(U ⨂ V,W) ≅ hom^{G}(U,V^{∗}⨂ W).

The 3 results follow on taking the dimensions of each side. ◀

Theorem 7.1 The expression for a semisimple representation α as a sum of simple parts

α = σ_{1}+ ··· + σ_{r}

is unique up to order.

Proof ▶ Suppose σ is a simple representation of G over k. We can use the intertwining number to compute the number of times, m say, that σ occurs amongst the σ_{i}. For

I(σ,α) = I(σ,σ_{1}) + ··· + I(σ,σ_{r})
= mI(σ,σ),

since only those summands for which σ_{i} = σ will contribute to the sum. Thus

m = \frac{I(σ,α)}{I(σ,σ)} .

It follows that σ will occur the same number m times in every expression for α as a sum of simple parts. Hence two such expressions can only differ in the order of their summands. ◀

Although the expression

α = σ_{1}+ ··· + σ_{r}

for the representation α is unique, the corresponding splitting

V = U_{1}⨁ ··· ⨁ U_{r}

of the representation-space is not in general unique. It’s perfectly possible for 2 different expressions for V as a direct sum of simple G-subspaces to give rise to the same expression for α: say

V = U_{1}⨁ ··· ⨁ U_{r}, V = W_{1}⨁ ··· ⨁ W_{r}

where U_{i} and W_{i} both carry the representation σ_{i}.

For example, consider the trivial representation α = 1 + 1 of a group G in the 2-dimensional space V = k^{2}. Every subspace of V is stable under G; so if we choose any 2 different 1-dimensional subspaces U,W ⊂ V, we will have

V = U ⨁ W.

However, the splitting of V into isotypic components is unique, as we shall see.

Definition 7.2 The representation α, and the underlying representation-space V, are said to be isotypic of type σ, where σ is a simple representation, if

α = eσ = σ + ··· + σ.

In other words, σ is the only simple representation appearing in α.

Proposition 7.5 Suppose V is a G-space.

1. If V is isotypic of type σ then so is every G-subspace U ⊂ V.

2. If U,W ⊂ V are isotypic of type σ then so is U + W.

Proof ▶ These results follow easily from the Uniqueness Theorem. But it is useful to give an independent proof, since we can use this to construct an alternative proof of the Uniqueness Theorem.

Lemma 7.1 Suppose

V = U_{1}+ ··· + U_{r}

is an expression for the G-space V as a sum of simple spaces; and suppose the subspace U ⊂ V is also simple. Then U is isomorphic (as a G-space) to one of the summands:

U ≅ U_{i}

for some i.

Proof of Lemma ▷ We know that

V = U_{i_{1}}⨁ ··· ⨁ U_{i_{t}}

for some subset {U_{i_{1}},...,U_{i_{t}}} ⊂ {U_{1},...,U_{r}}. Thus we may assume that the sum is direct:

V = U_{1}⨁ ··· ⨁ U_{r}.

For each i, consider the composition

U → V → U_{i},

where the second map is the projection of V onto its component U_{i}. Since U and U_{i} are both simple, this map is either an isomorphism, or else 0.

But it cannot be 0 for all i. For suppose u ∈ U,u ≠ 0. We can express u as a sum

u = u_{1}⊕ ··· ⊕ u_{r} (u_{i} ∈ U_{i}).

Not all the u_{i} vanish. Now u ↦ u_{i} under the composition U → V → U_{i}. Thus one (at least) of these compositions is ≠ 0. Hence U ≅ U_{i} for some i. ◁

Turning to the first part of the Proposition, if U ⊂ V, where V is σ-isotypic, then each simple summand of U must be of type σ, by the Lemma. It follows that U is also σ-isotypic.

For the second part, if U and W are both σ-isotypic, then U + W is a sum (not necessarily direct) of simple subspaces X_{i} of type σ:

U + W = X_{1}+ ··· + X_{r}.

But then

U + W = X_{i_{1}}⨁ ··· ⨁ X_{i_{t}},

where {X_{i_{1}},...,X_{i_{t}}} are some of the X_{1},...,X_{r}. In particular U + W is σ-isotypic.

◀

Corollary 7.1 Suppose σ is a simple representation of G over k, Then each Gspace V over k possesses a maximal σ-isotypic subspace V_{σ}, which contains every other σ-isotypic subspace.

Definition 7.3 This subspace V_{σ} is called the σ-component of V.

Proposition 7.6 Every semimsimple G-space V is the direct sum of its isotypic components:

V = V_{σ_{1}} ⨁ ··· ⨁ V_{σ_{r}}.

Proof ▶ If we take an expression for V as a direct sum of simple subspaces, and combine those that are isomorphic, we will obtain an expression for V as a direct sum of isotypic spaces of different types, each of which will be contained in the corresponding isotypic component. It follows that

V = V_{σ_{1}} + ··· + V_{σ_{r}}.

We have to show that this sum is direct.

It is sufficient to show that

(V_{σ_{1}} + ··· + V_{σ_{i−1}} )⋂ V_{σ_{i}} = 0

for i = 2,...,r.

Suppose not. Then we can find a simple subspace

U ⊂ V_{σ_{i}}, U ⊂ V_{σ_{1}} + ··· + V_{σ_{i−1}} .

By the Lemma to the last Proposition, U must be of type σ_{i}, as a subspace of V_{σ_{i}}. On the other hand, as a subspace of V_{σ_{1}} +··· +V_{σ_{i−1}} it must be of one of the types σ_{1},...,σ_{i−1} , by the same Lemma.

This is a contradiction. Hence the sum is direct:

V = V_{σ_{1}} ⨁ ··· ⨁ V_{σ_{r}}.

◀

Corollary 7.2 If the G-space V carries a multiple-free representation

α = σ_{1}+ ··· + σ_{r}

(where the σ_{i} are distinct) then V has a unique expression as a direct sum of simple subspaces.

Remark: It is easy to see that multiplicity does give rise to non-uniqueness. For suppose

V = U ⨁ U,

where U is simple. For each λ ∈ k consider the map

u ↦ u ⊕ λu: U → U ⨁ U = V.

The image of this map is a subspace

U(λ) = {u ⊕ λu: u ∈ U}.

This subspace is isomorphic to U, since U is simple.

It is readily verified that

U(λ) ≠ U(μ) ⟺ λ = μ.

It follows that

V = U(λ)⨁ U(μ)

for any λ,μ with λ ≠ μ.

The Character of a Representation

Amazingly, all the information about a representation of a group G can be encoded in a single function on G, the character of the representation.

Definition 8.1 Suppose α is a representation of G over k. The character χ = χ_{α} of α is the function χ: G → k defined by

χ(g) = tr(α(g)) .

Remarks:

1. Recall that the trace of an n × n-matrix A is the sum of the diagonal ele-

ments:

trA = ∑_{1≤i≤n} A_{ii}.

The trace has the following properties:

(a) tr(A + B) = trA + trB;

(b) tr(λA) = λtrA.

(c) trAB = trBA;

(d) trA′= trA;

(e) trA^{∗} = \overline{trA}. Here A′denotes the transpose of A, and A^{∗} the conjugate transpose:

A′_{ij} = A_{ji}, A_{ij}^{∗} = \overline{A_{ji}}.

The third property is the only one that is not immediate:

trAB = ∑_{i} (AB)_{ii} = ∑_{i} ∑_{j} A_{ij}B_{ji} = ∑_{j} ∑_{i} B_{ji}A_{ij} = trBA.

Note that

trABC ≠ trBAC

in general. However the trace is invariant under cyclic permutations, eg

trABC = trBCA = trCAB.

In particular, if P is invertible (non-singular) then

trPAP^{−1} = trP^{−1} PA = trA:

similar matrices have the same trace.

It follows from this that we can speak without ambiguity of the trace trt of a linear transformation t : V → V; for the matrix T representing t with respect to one basis will be changed to PT P^{−1} with respect to another basis, where P is the matrix of the change of basis.

Example: Consider the 2-dimensional representation α of D_{4} over ℂ given by

s ↦ ( \begin{matrix}i & 0 \\ 0 & −i \end{matrix} ) t ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) .

Writing χ for χ_{α}

χ(e) = dimα = 2
χ(s) = i − i = 0
χ(s^{2}) = tr( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) = −1 − 1 = −2
χ(s^{3}) = tr( \begin{matrix}−i & 0 \\ 0 & i \end{matrix} ) = −i + i = 0
χ(t) = i − i = 0
χ(st) = tr( \begin{matrix}0 & i \\ −i & 0 \end{matrix} ) = 0
χ(s^{2}t) = tr( \begin{matrix}0 & −1 \\ −1 & 0 \end{matrix} ) = 0
χ(s^{3}t) = tr( \begin{matrix}0 & −i \\ i & 0 \end{matrix} ) = 0

In summary

χ(e) = 2,χ(s^{2}) = −2,χ(g) = 0 if g ≠ e,s^{2} .

Proposition 8.1 1. χ_{α+β} (g) = χ_{α}(g) + χ_{β}(g)

2. χ_{αβ}(g) = χ_{α}(g)χ_{β}(g)

3. χ_{α^{∗}}(g) = χ_{α}(g^{−1} )

4. χ_{1}(g) = 1

5. χ_{α}(e) = dimα Proof ▶ (1) follows from the matrix form

g ↦ ( \begin{matrix}A(g) & 0 \\ 0 & B(g) \end{matrix} )

for α + β.

(2) follows from the fact that if A is an m × m-matrix and B is an n × n-matrix then the diagonal elements of the tensor product A ⊗ B are just the products

A_{ii}B_{jj} (1 ≤ i ≤ m, 1 ≤ j ≤ n)

Thus

tr(A ⊗ B) = trAtrB.

(3) If α takes the matrix form

g ↦ A(g)

then its dual is given (with respect to the dual basis) by

g ↦ A(g)′^{−1} = A(g^{−1} )′.

Hence

χ_{α^{∗}}(g) = trA(g^{−1} )′= trA(g^{−1} ) = χ_{α}(g^{−1} ).

(4) and (5) are immediate. ◀ Remark: In effect the character defines a ring-homomorphism

χ: R(G,k) → C(G,k)

from the representation-ring R(G) = R(G,k) to the ring C(G,k) of functions on G (with values in k). Theorem 8.1 Suppose α,β are representations of G over k. Then

I(α,β) = \frac{1}{∥G∥} ∑_{g∈G}χ_{α}(g^{−1} )χ_{β}(g).

Proof ▶ It is sufficient to prove the result when α = 1. For on the left-hand side

I(α,β) = I(1,α^{∗}β);

while on the right-hand side

∑_{g∈G}χ_{α}(g^{−1} )χ_{β}(g) = ∑_{g} χ_{α^{∗}}(g)χ_{β}(g)
= ∑_{g} χα^{∗}β(g)
= ∑_{g} χχ_{1}(g)α^{∗}β (g).

Thus the result for α,β follows from that for 1,α^{∗}β.

We have to show therefore that

I(1,α) = \frac{1}{∥G∥} ∑_{g∈G}χ_{α}(g).

By definition, if α is a representation in V,

I(1,α) = dimhom^{G}(k,V).

Now

hom(k,V) = V,

with the vector v ∈ V corresponding to the map

λ ↦ λv: k → V.

Moreover, the action of G is preserved under this identification; so we may write

hom^{G}(k,V) = V^{G},

where V^{G} denotes the space of G-invariant elements of V:

V^{G} = {v ∈ V : gv = v ∀g ∈ G}

Thus we have to prove that

dimV^{G} = \frac{1}{∥G∥} ∑_{g∈G}χ_{α}(g).

Consider the ‘averaging map’ π: V → V defined by

v ↦ \frac{1}{∥G∥} ∑_{g∈G}gv,

that is,

π = \frac{1}{∥G∥} ∑_{g∈G}α(g).

It is evident that πv ∈ V^{G} for all v ∈ V, ie πv is invariant under G. For

gπv = \frac{1}{∥G∥} ∑_{h∈G}ghv
= \frac{1}{∥G∥} ∑_{h∈G}hv
= πv,

since gh runs over G as h does.

On the other hand, if v ∈ V^{G} then gv = v for all g and so

πv = \frac{1}{∥G∥} ∑_{g∈G}gv = v.

It follws that π is a projection onto V^{G}.

Lemma 8.1 Suppose p: V → V is a projection onto the subspace U ⊂ V. Then

trp = dimU.

Proof of Lemma ▷ We know that

V = imp ⊕ kerp.

Let e_{1},...,e_{m} be a basis for imp = U, and let e_{m+1},...,e_{n} be a basis for kerp. Then

pe_{i} = \{ \begin{matrix}e_{i} & 1 ≤ i ≤ m, \\ 0 & m + 1 ≤ ilen. \end{matrix}

It follows that the matrix of p with respect to the basis e_{1},...,e_{n} is

P = ( \begin{matrix}1 \\ & . & . & . \\ & & & & 1 \\ & & & & & 0 \\ & & & & & & . & . & . \\ & & & & & & & & & 0 \end{matrix} )

with m 1’s down the diagonal and 0’s elsewhere. Hence

trp = trP = m = dimU.

◁

Applying this to the averaging map π,

trπ = dimV^{G}.

On the other hand, by the linearity of the trace,

trπ = \frac{1}{∥G∥} ∑_{g} trα(g)
= \frac{1}{∥G∥} ∑_{g} χ_{α}(g)

Thus

dimV^{G} = \frac{1}{∥G∥} ∑_{g} χ_{α}(g),

as we had to show. ◀

Proposition 8.2 If k = ℝ,

χ_{α^{∗}}(g) = χ_{α}(g^{−1} ) = χ_{α}(g).

If k = ℂ,

χ_{α^{∗}}(g) = χ_{α}(g^{−1} ) = \overline{χ_{α}(g)}.

Proof ▶ First suppose k = ℂ. Let λ_{1},...,λ_{n} be the eigenvalues of α(g). Then

χ_{α}(g) = trα(g) = λ_{1}+ ··· + λ_{n}.

In fact, we can diagonalise α(g), ie we can find a basis with respect to which

g ↦ A(g) = ( \begin{matrix}λ_{1} & & & & 0 \\ & . & . & . \\ 0 & & & & λ_{n} \end{matrix} )

Now

A(g^{−1} ) = A(g)^{−1} = ( \begin{matrix}λ_{1}^{−1} & & & & 0 \\ & . & . & . \\ 0 & & & & λ_{n}^{−1} \end{matrix} )

and so

χ_{α}(g^{−1} ) = trA(g^{−1} ) = λ_{1}^{−1} + ··· + λ_{n}^{−1} .

But since G is a finite group, g^{n} = e for some n (eg for n = ∥G∥), and so

λ_{i}^{n} = 1 ⟹ |λ_{i}| = 1 ⟹ λ_{i}^{−1} = \overline{λ_{i}}

for each eigenvalue λ_{i}. Hence

χ_{α}(g^{−1} ) = \overline{λ_{1}} + ··· + \overline{λ_{n}} = \overline{χ_{α}(g)}.

The result for k = ℝ follows from this. For if A is a real matrix satisfying A^{n} = I then we may regard A as a complex matrix, and so deduce by the argument above that

tr(A^{−1} ) = \overline{trA}.

But since A is real, so is trA, and thereforeHence

tr(A^{−1} ) = trA.

◀

Corollary 8.1 Suppose α,β are representations of G over k. Then

I(α,β) = \{ \begin{matrix}\frac{1}{∥G∥}∑_{g∈G} χ_{α}(g)χ_{β}(g) & if k = ℝ \\ \frac{1}{∥G∥}∑_{g∈G} \overline{χ_{α}(g)}χ_{β}(g) & if k = ℂ \end{matrix}

Definition 8.2 We define the inner product

⟨u,v⟩ (u(g),v(g) ∈ C(G,k))

by

⟨u,v⟩ = \{ \begin{matrix}\frac{1}{∥G∥}∑_{g∈G} \overline{u(g)}v(g) & if k = ℂ \\ \frac{1}{∥G∥}∑_{g∈G} u(g)v(g) & if k = ℝ \end{matrix}

Proposition 8.3 1. The inner product ⟨u,v⟩ is positive-definite.

2. I(α,β) = ⟨χ_{α},χ_{β}⟩.

Proposition 8.4 Two representations are equivalent if and only if their characters are equal:

α = β ⟺ χ_{α}(g) = χ_{β}(g) for all g ∈ G.

Proof ▶ If α = β then

B(g) = PA(g)P^{−1}

for some P. Hence

χ_{β}(g) = trB(g) = trA(g) = χ_{α}(g).

On the other hand, suppose χ_{α}(g) = χ_{β}(g) for all g ∈ G. Then for each simple representation σ of G over k,

I(σ,α) = \frac{1}{∥G∥} ∑_{g∈G}χ_{σ}(g^{−1} )χ_{α}(g)
= \frac{1}{∥G∥} ∑_{g∈G}χ_{σ}(g^{−1} )χ_{β}(g)
= I(σ,β).

It follows that σ occurs the same number of times in α and β. Since this is true for all simple representations σ,

α = β.

◀

Proposition 8.5 Characters are class functions, ie

g′∼ g ⟹ χ_{α}(g′) = χ_{α}(g).

Remark: Recall that we write g′∼ g to mean that g′,g are conjugate, ie there exists an x ∈ G such that

g′= xgx^{−1} .

Proof ▶ If

g′= xgx^{−1}

then (since a representation g ↦ A(g) is a homomorphism)

A(g′) = A(x)A(g)A(x^{−1} )
= A(x)A(g)A(x)^{−1} .

It follows from the basic property of the trace that

χ_{α}(g′) = trA(g′) = trA(g) = χ_{α}(g).

◀

Proposition 8.6 Simple characters are orthogonal, ie if α,β are distinct simple representations of G over k then

⟨χ_{α},χ_{β}⟩ = 0.

Proof ▶ This is just a restatement of the fact that

I(α,β) = 0.

◀

When k = ℂ we can be a little more precise.

Proposition 8.7 The simple characters of G over ℂ form an orthonormal set, ie

⟨χ_{α},χ_{β}⟩ = \{ \begin{matrix}1 & if α = β, \\ 0 & otherwise. \end{matrix}

Proof ▶ Again, this is simply a restatement of the result for the intertwining number. ◀

Theorem 8.2 The group G has at most s simple represenations over k, where s is the number of classes in G.

Proof ▶ The class functions on G form a vector space

X ⊂ C(G,k).

Lemma 8.2 dimX = s.

Proof of Lemma ▷ Suppose the conjugacy classes are C_{1},...,C_{n}. Let c_{i}(g) denote

the characteristic function of C_{i}, ie

c_{i}(g) = \{ \begin{matrix}1 & if g ∈ C_{i}, \\ 0 & otherwise \end{matrix}

Then the functions

c_{i}(g) (1 ≤ i ≤ s)

form a basis for the class functions on G. ◁

Lemma 8.3 Mutually orthogonal vectors (with respect to a positive-definite form) are necessarily linearly independent.

Proof of Lemma ▷ Suppose v_{1},...,v_{r} are mutually orthogonal:

⟨v_{i},v_{j}⟩ = 0 if i ≠ j .

Suppose

λ_{1}v_{1}+ ··· + λ_{r}v_{r} = 0.

Taking the inner product of v_{i} with this relation,

λ_{1}⟨v_{i},v_{1}⟩ + ··· + λ_{r}⟨v_{i},v_{r}⟩ = 0 ⟹ λ_{i} = 0.

Since this is true for all i, the vectors v_{1},...,v_{r} must be linearly independent. ◁

Now consider the simple characters of G over k. They are mutually orthogonal, by the last Proposition; and so they are linearly independent, by the Lemma. But they belong to the space X of class functions. Hence their number cannot exceed the dimension of this space, which by Lemma 1 is s. ◀ Remark: We shall see that when k = ℂ, the number of simple representations is actually equal to the number of classes. This is equivalent, by the reasoning above, to the statement that the characters span the space of class functions.

We shall establish this result (in later Chapters) in two different ways — one using induced representations, and one based on the representation theory of product groups. Example: Since characters are class functions, it is only necessary to compute their values for 1 representative from each class. The character table of a group G over k tabulates the values of the simple representations on the various classes. By convention, if the scalar field k is not specified it is understood that we are speaking of representations over ℂ.

As an illustration, let us take the group S_{3}. The 6 elements divide into 3 classes, corresponding to the 3 cylic types:

1^{3} e

21 (bc),(ac),(ab)

3 (abc),(acb)

It follows that S_{3} has at most 3 simple characters over ℂ. Since we already know 3, namely the 2 1-dimensional representations 1,ϵ and the 2-dimensional representation α, we have the full panoply.

We draw up the character table as follows:

class [1^{3}] [21] [3]
size 1 3 2
1  1 1 1
ϵ  1 −1 1
α  2 0 −1

Let us verify that the simple characters form an orthonormal set:

I(1,1) = \frac{1}{6}(1 · 1 · 1 + 3 · 1 · 1 + 2 · 1 · 1) = 1
I(1,ϵ) = \frac{1}{6}(1 · 1 · 1 + 3 · 1 · −1 + 2 · 1 · 1) = 0
I(1,α) = \frac{1}{6}(1 · 1 · 2 + 3 · 1 · 0 + 2 · 1 · −1) = 0
I(ϵ,ϵ) = \frac{1}{6}(1 · 1 · 1 + 3 · −1 · −1 + 2 · 1 · 1) = 1
I(ϵ,α) = \frac{1}{6}(1 · 1 · 2 + 3 · −1 · 0 + 2 · 1 · −1) = 0
I(α,α) = \frac{1}{6}(1 · 2 · 2 + 3 · 0 · 0 + 2 · −1 · −1) = 1

It is very easy to compute the character of a permutational representation, that is, a representation arising from the action of the group G on the finite set X. Recall that this is the representation in the function-space C(X,k) given by

(gf)(x) = f(g^{−1} x).

Proposition 8.8 Suppose α is the permutational representation of G arising from the action of G on the finite set X. Then

χ_{a}lpha(g) = ∥{x: gx = x}∥,

ie χ(g) is equal to the number of elements of X left fixed by g.

Proof ▶ Let c_{x}(t) denote the characteristic function of the 1-point subset {x}, ie

c_{x}(t) = \{ \begin{matrix}1 & if t = x, \\ 0 & otherwise. \end{matrix}

The ∥X∥ functions c_{x}(t) form a basis for the vector space C(X,k); and the action of g ∈ G on this basis is given by

gc_{x} = c_{gx},

since

gc_{x}(t) = c_{x}(g^{−1} t) = 1 ⟺ g^{−1} t = x ⟺ t = gx.

It follows that with respect to this basis

g ↦ A(g),

where A = A(g) is the matrix with entries

A_{xy}= \{ \begin{matrix}1 & if x = gy, \\ 0 & otherwise. \end{matrix}

In particular

A_{xx}= \{ \begin{matrix}1 & if x = gx, \\ 0 & otherwise. \end{matrix}

Hence

χ_{a}lpha(g) = trA = ∑_{x} A_{xx}= ∥{x: gx = x}∥.

◀

Example: Consider the action of the group S_{3} on X = {a,b,c}, Let us denote the resulting representation by ρ. We only need to compute χ_{ρ}(g) for 3 values of g, namely 1 representative of each class.

We know that

χ_{ρ}(e) = dimρ = ∥X∥ = 3.

The transposition (bc) (for example) has just 1 fixed point, namely a. Hence

χ_{ρ}(bc) = 1.

On the other hand, the 3-cycle (abc) has no fixed points, so

χ_{ρ}(abc) = 0.

Let us add this character to our table:

class [1^{3}] [21] [3]
size 1 3 2
1  1 1 1
ϵ  1 −1 1
α  2 0 −1
ρ  3 1 0

We know that ρ is some integral combination of the simple characters, say

ρ = r · 1 + s · ϵ + t · α,

where r,s,t ∈ ℕ. These ‘coefficients’ r,s,t are unique, since the simple characters are linearly independent.

It would be easy to determine them by observation. But let us compute them from the character of ρ. Thus

r = I(1,ρ) = \frac{1}{6}(1 · 1 · 3 + 3 · 1 · 1 + 2 · 1 · 0) = 1

s = I(ϵ,ρ) = \frac{1}{6}(1 · 1 · 3 + 3 · −1 · 1 + 2 · 1 · 0) = 0

t = I(α,ρ) = \frac{1}{6}(1 · 2 · 3 + 3 · 0 · 1 + 2 · −1 · 0) = 0

Thus

ρ = 1 + α.

The regular representation

The group G acts on itself in 3 ways:

• By left multiplication: (g,x) ↦ gx

• By right multiplication: (g,x) ↦ xg^{−1}

• By inner automorphism: (g,x) ↦ gxg^{−1}

The first action leads to the regular representation defined below. The second action also leads to the regular representation, as we shall see. The third action leads to the adjoint representation, which we shall consider later.

Definition 9.1 The regular representation reg of the group G over k is the permutational representation defined by the action

(g,x) ↦ gx

of G on itself.

Proposition 9.1 The character of the regular representation is given by

χ_{reg}(g) = \{ \begin{matrix}1 & if g = e , \\ 0 & otherwise. \end{matrix}

Proof ▶ We have to determine, for each g ∈ G, the number of elements x ∈ G left fixed by g, ie satisfying

gx = x.

But

gx = x ⟹ g = e.

Thus no element g ≠ e leaves any element fixed; while g = e leaves every element fixed. ◀ Proposition 9.2 The permutational representation defined by right multiplication

(g,x) ↦ xg^{−1}

is equivalent to the regular representation.

Proof ▶ No element g ≠ e leaves any element fixed; while g = e leaves every element fixed:

xg^{−1} = x ⟺ g = e.

Thus this representation has the same character as the regular representation; and so it is equal (that is, equivalent) to it. ◀

Alternative proof ▶ In fact it is readily verified that the representation defined by right multiplication is the dual reg^{∗} of the regular representation. But the regular representation is self-dual, since its character is real. ◀

Proposition 9.3 Suppose α is a representation of G over k. Then

I(α,reg) = dimα.

Proof ▶ Plugging the result for the character of reg above into the formula for the intertwining number,

I(α,reg) = \frac{1}{∥G∥} ∑_{g∈G}χ_{α}(g^{−1} )χ_{reg}(g)
= \frac{1}{∥G∥}∥G∥χ_{α}(e)
= dimα.

◀

This result shows that every simple representation occurs in the regular representation, since I(σ,reg) > 0. When k = ℂ we can be more precise.

Proposition 9.4 Each simple representation σ of G over ℂ occurs just dimσ times in the regular representation reg of G over ℂ:

reg = ∑_{σ} (dimσ)sigma,

where the sum extends over all simple representations σ of G over ℂ.

Proof ▶ We know that reg, as a semisimple representation, is expressible in the form

reg = ∑_{σ} e_{σ}σ (e_{σ} ∈ ℕ).

Taking the intertwining number of a particular simple representation σ with each side,

I(σ,reg) = e_{σ}I(σ,σ) = e_{σ}
= dimσ,

by the Proposition. ◀

Theorem 9.1 The dimensions of the simple representations σ_{1},...,σ_{r} of G over ℂ satisfy the relation

dim^{2}σ_{1}+ ··· + dim^{2}σ_{r} = ∥G∥.

Proof ▶ This follows at once on taking the dimensions on each side of the identity

reg = ∑_{σ} (dimσ)sigma.

◀

Example: Consider S_{5}. We have

∥S_{5}∥ = 120;

while S_{5} has 7 classes:

[1^{5}], [21^{3}], [2^{2}1], [31^{2}], [32], [41], [5].

Thus S_{5} has at most 7 simple representations over ℂ.

Let us review the information on these representations that we already have:

1. S_{5} has just 2 1-dimensional representations, 1 and ϵ;

2. The natural 5-dimensional representation ρ of S_{5} splits into 2 parts:

ρ = 1 + α,

where α is a simple 4-dimensional representation of S_{5};

3. If σ is a simple representation of S_{5} of odd dimension then ϵσ ≠ σ; 4. More generally, if σ is a simple representation of S_{5} with σ([21^{3}]) ≠ 0 then

ϵσ ≠ σ;

We can apply this last result to α. For

χ_{α}([21^{3}]) = χ_{ρ}([21^{3}]) − 1
= 3 − 1
= 2.

Hence

ϵα ≠ α.

Thus we have found 4 of the 7 (or fewer) simple representations of S_{5}: 1,ϵ,α,ϵα. Our dimensional equation reads

120 = 1^{2}+ 1^{2}+ 4^{2}+ 4^{2}+ a^{2}+ b^{2}+ c^{2},

where a,b,c ∈ ℕ, with a,b,c ≠ 1. (We are allowing for the fact that S_{5} might have < 7 simple representations.) In other words,

a^{2}+ b^{2}+ c^{2} = 86.

It follows that

a^{2}+ b^{2}+ c^{2} ≡ 6 (mod 8).

Now

n^{2} ≡ 0,1, or 4 (mod 8)

according as n ≡ 0 (mod 4), or n is odd, or n ≡ 2 (mod 4). The only way to get 6 is as 4 + 1 + 1. In other words, 2 of a,b,c must be odd, and the other must be ≡ 2 (mod 4). (In particular a,b,c ≠ 0. So S_{5} must in fact have 7 simple representations.)

By (3) above, the 2 odd dimensions must be equal: say a = b. Thus

2a^{2}+ c^{2} = 86.

Evidently a = 3 or 5. Checking, the only solution is

a = b = 5,c = 6.

We conclude that S_{5} has 7 simple representations, of dimensions

1,1,4,4,5,5,6.

Induced Representations

Each representation of a group defines a representation of a subgroup, by restriction; that much is obvious. More subtly, each representation of the subgroup defines a representation of the full group, by a process called induction. This provides the most powerful tool we have for constructing group representations.

Definition 10.1 Suppose H is a subgroup of G; and suppose α is a representation of G in V. Then we denote by α_{H} the representation of H in the same space V defined by restricting the group action from G to H. We call α_{H} the restriction of α to H.

Proposition 10.1 1. (α + β)_{H} = α_{H} + β_{H}

2. (αβ)_{H} = α_{H}β_{H}

3. (α^{∗})_{H} = (α_{H})^{∗}

4. 1_{H} = 1

5. dimα_{H} = dimα

6. χ_{α_{H}} (h) = χ_{α}(h)

Example: We can learn much about the representations of G by considering their restrictions to subgroups H ⊂ G. But induced representations give us the same information—and more—much more easily, as we shall see; so the following example is of more intellectual interest than practical value.

Let us see what we can discover about the simple characters of S_{4} (over ℂ) from the character table for S_{3}. Let’s assume we know—as we shall prove later in this chapter—that the number of simple characters of S_{4} is equal to the number of classes, 5. Let’s suppose too that we know S_{4} has just 2 1-dimensional representations, 1 and ϵ. Let γ be one of the 3 other simple representations of S_{4}.

Let

γ_{S_{3}} = a1 + bϵ + cα (a,b,c ∈ ℕ).

By the Proposition above, if h̄⊂ ḡ(where h̄is a class in H and ḡa class in G) then

χ_{γ}(ḡ) = χ_{γ_{H}}(h̄).

So we know some of the values of χ_{γ}:

Class  [1^{4}] [21^{2}] [2^{2}] [31] [4]
size  1  6  3  8  6
1  1  1  1  1  1
ϵ  1  −1 1  1  −1
γ a + b + 2c a − b x a + b − c y

We have found nothing about χ([2^{2}]) and χ([4]), since these 2 classes don’t intersect S_{3}. However, if we call the values x and y as shown, then the 2 equations

I(1,γ) = 0, I(ϵ,γ) = 1

give

15a + 3b − 6c + 3x + 6y = 0
3a + 15b − 6c + 3x + 6y = 0

Setting

s = a + b, t = a − b,

for simplicity, these yield

x = −3s + 2t, y = −t.

The table now reads

Class [1^{4}] [21^{2}] [2^{2}] [31] [4]
size  1  6  3  8 6
1  1  1  1  1 1
ϵ  1  −1  1  1 −1
γ s + 2c t −3s + 2c s − c −t

Since γ is—by hypothesis—simple,

I(γ,γ) = 1.

Thus

24 = (s + 2c)^{2}+ 6t^{2}+ 3(−3s + 2c)^{2}+ 8(s − c)^{2}+ 6t^{2}.

On simplification this becomes

2 = 3s^{2}− 4sc + 2c^{2}+ t^{2}
= s^{2}+ 2(s − c)^{2}+ t^{2}.

Noting that s,t,c are all integral, and that s,c ≥ 0, we see that there are just 3 solutions to this diophantine equation:

(a,b,c) = (1,0,1), (0,1,1), (0,0,1).

These must yield the 3 missing characters.

We have determined the character table of S_{4} without constructing—even implicitly—the corresponding representations. This has an interesting parallel in recent mathematical history. One of the great achievements of the last 25 years has been the determination of all finite simple groups, ie groups possessing no proper normal (or self-conjugate) subgroups. The last link in the chain was the determination of the exceptional simple groups, ie those not belonging to the known infinite families (such as the family of alternating groups A_{n} for n ≥ 5). Finally, all was known except for the largest exceptional group—the so-called mammoth group. The character table of this group had been determined several years before it was established that a group did indeed exist with this table.

As we remarked earlier, the technique above is not recommended for serious character hunting. The method of choice must be induced representations, our next topic.

Suppose V is a vector space. Then we denote by C(G,V) the G-space of maps f : G → V, with the action of G defined by

(gf)(x) = f(g^{−1} x)

(This extends our earlier definition of C(G,k).)

Definition 10.2 Suppose H is a subgroup of G; and suppose α is a representation of H in U. Then we define the induced representation α^{G} of G as follows. Let

V = {F ∈ C(G,U) : F (gh) = h^{−1} F (g) for all g ∈ G,h ∈ H}.

Then V is a G-subspace of C(G,U); and α^{G} is the representation of G in this subspace.

Remark: That V is a G-subspace follows from the fact that we are acting on G with G and H from opposite sides (G on the left, H on the right); and their actions therefore commute:

(gx)h = g(xh).

Thus if F ∈ V then

(gF)(xh) = F (g^{−1} xh)
= h^{−1} F (g^{−1} x)
= h^{−1} ((gF)(x)),

ie gF ∈ V.

This definition is too cumbersome to be of much practical use. The following result offers an alternative, and usually more convenient, starting point.

Lemma 10.1 Suppose e = g_{1},g_{2},...,g_{r} are representatives of the cosets of H in G, ie

G = g_{1}H ∪ g_{2}H ∪ ... ∪ g_{r}H.

Then there exists an H-subspace U′⊂ V such that

(a) U′is isomorphic to U as an H-space,

(b) V = g_{1}U′⨁ g_{2}U′⨁ ...⨁ g_{r}U′.

Moreover the induced representation α^{G} is uniquely characterised by the existence of such a subspace.

Remarks:

1. After the lemma, we may write

V = g_{1}U ⨁ g_{2}U ⨁ ...⨁ g_{r}U.

2. The action of G on V is implicit in this description of V. For suppose v is in

the ith summand, say

v = g_{i}u;

and suppose gg_{i} is in the jth coset, say

gg_{i} = g_{j}h.

Then gv is in the jth summand:

gv = gg_{i}u = g_{j}(hu).

3. The difficulty of taking this result as the definition of α^{G} lies in the awkwardness of showing that the resulting representation does not depend on the choice of coset-representatives.

Proof ▶ To each u ∈ U let us associate the function u′= u′(g) ∈ C(G,U) by

u′(g) = \{ \begin{matrix}gu & if g ∈ H \\ 0 & otherwise \end{matrix}

Then it is readily verified that

(a) u′∈ V, ie u′(gh) = h^{−1} u′(g) for all h ∈ H.

(b) If u ↦ u′then hu ↦ hu′.

Thus the map u ↦ u′sets up an H-isomorphism between U and an H-subspace U′⊂ V.

Suppose F ∈ V. From the definition of V,

F (gh) = h^{−1} F (g).

It follows that the values of F on any coset g_{i}H are completely determined by its value at one point g_{i}. Thus F is completely determined by its r values

u_{1} = F (e),u_{2} = F (g_{2}),...,u_{r} = F (g_{r}).

Let us write

F ⟷ (u_{1},u_{2},...,u_{r}).

Then it is readily verified that

u′⟷ (u,0,...,0);

and more generally

g_{i}u′⟷ (0,..,u,..,0),

ie the function g_{i}u′vanishes on all except the ith coset g_{i}H, and takes the value u at g_{i}.

It follows that

F = g_{1}u′_{1}+ g_{2}u′_{2}+ ...+ g_{r}u′_{r}

since the 2 functions take the same values at the r points g_{i}. Moreover the argument shows that this expression for F ∈ V as a sum of functions in the subspaces U′= g_{1}U′,g_{2}U′,...,g_{r}U′, respectively, is unique: so that

V = g_{1}U′⨁ g_{2}U′⨁ ...⨁ g_{r}U′.

Finally this uniquely characterises the representation α^{G}, since the action of G on V is completely determined by the action of H on U, as we saw in Remark 1 above. ◀

Example: Suppose α is the representation of S_{3} in U = ℂ^{2} given by

(abc) ↦ ( \begin{matrix}ω & 0 \\ 0 & ω^{−1} \end{matrix} ) (ab) ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )

Let us consider the representation of S_{4} induced by α (where we identify S_{3} with the subgroup of S_{4} leaving d fixed).

First we must choose representatives of the S_{3}-cosets in S_{4}. The nicest way to choose coset representatives of H in G is to find—if we can—a subgroup T ⊂ G transverse to H, ie such that

1. T ∩ H = {e}

2. ∥T ∥ ∥H∥ = ∥G∥.

It is readily verified that these 2 conditions imply that each element g ∈ G is uniquely expressible in the form

g = th (t ∈ T,h ∈ H)

It follows that the elements of T represent the cosets gH of H in G.

In the present case we could take T to be the subgroup generated by a 4-cycle: say

{e,(abcd),(ac)(bd),(adcb)}.

Or we could take

T = V_{4} = {e,(ab)(cd),(ac)(bd),(ad)(bc)}

(the Viergruppe). Let’s make the latter choice; the fact that T is normal (selfconjugate) in G should simplify the calculations. We have

S_{4} = S_{3}∪ (ab)(cd)S_{3}∪ (ac)(bd)S_{3}∪ (ad)(bc)S_{3};

and so α^{G} is the represention in the 8-dimensional vector space

V = U ⨁ (ab)(cd)U ⨁ (ac)(bd)U ⨁ (ad)(bc)U.

As basis for this space we may take

\begin{matrix}e_{1} = e, & e_{2} = f, & e_{3} = (ab)(cd)e, & e_{4} = (ab)(cd)f, \\ e_{5} = (ac)(bd)e, & e_{6} = (ac)(bd)f, & e_{7} = (ad)(cb)e, & e_{8} = (ad)(bc)f, \end{matrix}

where e = (1,0),f = (0,1).

To simplify our calculations, recall that if g,x ∈ S_{n}, and

x = (a_{1}a_{2}...a_{r})(b_{1}b_{2}...b_{s})...

in cyclic notation, then

gxg^{−1} = (ga_{1},ga_{2},...,ga_{r})(gb_{1},gb_{2},...,gb_{s})...,

since, for example,

(gxg^{−1} )(ga_{1}) = gxa_{1} = ga_{2}.

(This is how we show that 2 elements of S_{n} are conjugate if and only if they are of the same type.) In our case, suppose h ∈ S_{3},t ∈ V_{4}. Then

hth^{−1} ∈ V_{4}

since V_{4} is normal. In other words,

ht = sh,

where s ∈ V_{4}.

Now let’s determine the matrix representing (ab). By the result above, we have

(ab) · (ab)(cd) = (ab)(cd) · (ab)
(ab) · (ac)(bd) = (bc)(ad) · (ab)
(ab) · (ad)(bc) = (bd)(ac) · (ab).

Thus

(ab)e_{6} = (ab) · (ac)(bd)f
= (ad)(bc) · (ab)f
= (ad)(bc)e
= e_{7}.

In fact

(ab)e_{1} = e_{2},  (ab)e_{2} = e_{1},
(ab)e_{3} = e_{4},  (ab)e_{4} = e_{3},
(ab)e_{5} = e_{8},  (ab)e_{6} = e_{7},
(ab)e_{7} = e_{6},  (ab)e_{8} = e_{5}.

Hence

(ab) ↦ ( \begin{matrix}0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \end{matrix} )

It is not hard to see that (ab) and (abcd) generate S_{4}. So the representation α^{S_{4}} will be completely determined—in principle, at least—if we establish the matrix representing (abcd). We see now that it was easier to detemine the matrix representing (ab), because (ab) ∈ S_{3}. But the general case is not difficult. Notice that

(abcd) = (ad)(bc) · (ac)

It follows that (for example)

(abcd) · (ac)(bd) = (ad)(bc) · (ac) · (ac)(bd)
= (ad)(bc) · (ac)(bd) · (ac)
= (ab)(cd) · (ac).

Now (ac) = (abc)(ab); so under α,

(ac) ↦ ( \begin{matrix}0 & ω \\ ω^{−1} & 0 \end{matrix} )

We see that, for example,

(abcd)e_{5} = (abcd) · (ac)(bd)e
= (ab)(cd) · (ac)e
= (ab)(cd)ω^{−1} f
= ω^{−1} e_{2}.

We leave it to the reader to complete this calculation of the matrix representing (abcd).

Clearly this is too time-consuming a hobby to pursue.

It is evident that

h ∼ h′in H ⟹ h ∼ h′in G

In other words, each class h̄in H lies in a unique class ḡin G:

h̄⊂ ḡ.

Or, to put it the other way round, each class ḡin G splits into classes \overline{h_{1}},...,\overline{h_{r}} in H:

ḡ∩ H = \overline{h_{1}}∪ ··· ∪ \overline{h_{r}}.

Theorem 10.1 Suppose H is a subgroup of G; and suppose β is a representation of H. Then

χ_{β^{G}}(ḡ) = \frac{∥G∥}{∥H∥ ∥ḡ∥} ∑_{h̄⊂ḡ} ∥h̄∥χ_{β}(h̄),

where the sum runs over those H-classes h̄contained in ḡ.

Proof ▶ Let g_{1},...,g_{r} be representatives of the cosets of H in G, so that β^{G} is the representation in

V = g_{1}U ⨁ ··· ⨁ g_{r}U.

Lemma 10.2 With the notation above

χ_{β^{G}}(g) = ∑_{i:g_{i}^{−1} gg_{i}=h∈H} χ_{β}(h),

where the sum extends over those coset-representatives g_{i} for which g_{i}^{−1} gg_{i} ∈ H.

Proof ▶ Let us extend the function χ_{β} (which is of course defined on H) to G by setting

χ_{β}(g) = 0 if g ∉ H,

then our formula can be written:

χ_{β^{G}}(g) = ∑_{i} χ_{β}(g_{i}^{−1} gg_{i}),

with the sum now extending over all coset-representatives.

Suppose e_{1},...,e_{m} is a basis for U. Then g_{i}e_{j} (1 ≤ i ≤ r,1 ≤ j ≤ m) is a basis for V.

Suppose v belongs to the ith summand of V, say

v = g_{i}u;

and suppose gg_{i} belongs to the jth coset, say

gg_{i} = g_{j}h.

Then

gv = gg_{i}u = g_{j}(hu).

So

g(g_{i}U) ⊂ g_{j}U.

Thus the basis elements in g_{i}U cannot contribute to χ_{β^{G}}(g) unless i = j, that is, unless gg_{i} = g_{i}h, ie

g_{i}^{−1} gg_{i} = h ∈ H.

Moreover if this is so then

g(g_{i}e_{j}) = g_{i}(he_{j}),

ie the m × m matrix defining the action of g on g_{i}U with respect to the basis g_{i}e_{1},...,g_{i}e_{m} is just B(h); so that its contribution to χ_{β^{G}}(g) is

χ_{β}(h).

The result follows on adding the contributions from all those summands sent into themselves by g. ◀

Lemma 10.3 For each g ∈ G,

χ_{β^{G}}(g) = \frac{1}{∥H∥} ∑_{g′∈G:g′^{−1} gg′=h∈H} χ_{β}(h)

Proof ▶ Suppose we take a different representative of the ith coset, say

g′_{i}= g_{i}h.

This will make the same contribution to the sum, since

g′_{i}^{−1} gg′_{i}= h^{−1} (g_{i}^{−1} gg_{i})h;

and

χ_{β}(h^{−1} h′h) = χ_{β}(h′).

Thus if we sum over all the elements of G, we shall get each coset-contribution just ∥H∥ times. ◀

To return to the proof of the Proposition, we compute how many times each element h ∈ H occurs in the sum above.

Two elements g′,g′′define the same conjugate of g in G, ie

g′^{−1} gg′= g′′^{−1} gg′′,

if and only if g′′g′^{−1} and g commute, ie if and only

g′′N(g) = g′N(g),

where

N(g) = {x ∈ G: gx = xg}.

It follows that each G-conjugate h of g in H will occur just ∥N(g)∥ times in the sum of Corollary 1. Thus if we sum over these elements h we must multiply by ∥N(g)∥.

The result follows, since

|N(g)∥ = \frac{∥G∥}{∥ḡ∥}

by the same argument, each conjugate x^{−1} gx of g arising from ∥N(g)∥ elements x. ◀

Examples:

1. Let us look again at α^{S_{3}→S_{4}} . The classes of S_{4} and S_{3} are related as follows:

[1^{4}] ∩ S_{3} = [1^{3}]
[21^{2}] ∩ S_{3} = [21]
[2^{2}] ∩ S_{3} = ∅
[31] ∩ S_{3} = [3]
[4] ∩ S_{3} = ∅

Hence

χ_{α^{S_{4}}}([1^{4}]) = \frac{24}{6 · 1}χ_{α}(1^{3}) = 8
χ_{α^{S_{4}}}([21^{2}]) = \frac{24}{6 · 6}3χ_{α}(21) = 0
χ_{α^{S_{4}}}([2^{2}]) = 0
χ_{α^{S_{4}}}([31]) = \frac{24}{6 · 8}2χ_{α}(3) = −1
χ_{α^{S_{4}}}(4) = 0

Class [1^{4}] [21^{2}] [2^{2}] [31] [4]
size 1  6  3 8 6
α^{S_{4}} 8  0  0 −1 0

Since

I(α^{S_{4}},α^{S_{4}}) = \frac{1}{24} (8^{2}+ 8 · 1^{2}) = 3,

α^{S_{4}} has just 3 distinct simple parts. whose determination is left to the reader.

The relation between S_{4} and S_{3} is unusual, in that classes never split. If ḡis a class in S_{4} then ḡ∩S_{3} is either a whole class h̄in H, or else is empty. This is true more generally for S_{n} and S_{m}(m < n), where S_{m}is identified with the subgroup of S_{n} leaving the last n − m elements fixed. If ḡis a class in S_{n}, then

ḡ∩ S_{m}= h̄or ∅.

2. Now let’s look at the cyclic subgroup

C_{4} = ⟨(abcd)⟩ = {e,(abcd),(ac)(bd),(adcb)}

of S_{4}. Since C_{4} is abelian, each element is in a class by itself. Let θ be the 1-dimensional representation of C_{4} defined by

(abcd) ↦ i

We have

[1^{4}] ∩ C_{4} = {e}
[21^{2}] ∩ C_{4} = ∅
[2^{2}] ∩ C_{4} = {(ac)(bd)}
[31] ∩ C_{4} = ∅
[4] ∩ C_{4} = {(abcd),(adcb)}

Hence

χ_{θ^{S_{4}}}([1^{4}]) = \frac{24}{4 · 1}χ_{θ}(e) = 6
χ_{θ^{S_{4}}}([21^{2}]) = 0
χ_{θ^{S_{4}}}([2^{2}]) = \frac{24}{4 · 3}χ_{θ}((ac)(bd)) = −2
χ_{θ^{S_{4}}}([31]) = 0
χ_{θ^{S_{4}}}([4]) = \frac{24}{4 · 6}(χ_{θ}((abcd)) + χ_{θ}((adcb)))
= i + (−i) = 0

Class [1^{4}] [21^{2}] [2^{2}] [31] [4]
size 1 6 3 8 6
θ^{S_{4}} 6 −2 0 0 0

Since

I(θ^{S_{4}},θ^{S_{4}}) = \frac{1}{24} (6^{2}+ 6 · 2^{2}) = 3,

θ^{S_{4}} has just 3 distinct simple parts. whose elucidation we again leave to the reader.

Proposition 10.2 1. (α + α′)^{G} = α^{G}+ α′^{G};

2. (α^{∗})^{G} = (α^{G})^{∗};

3. dimα^{G} = [G: H]dimα.

It is worth noting that permutation representations are a particular case of induced representations.

Lemma 10.4 Suppose G acts transitively on the finite set X. Let α be the corresponding representation of G. Take x ∈ X; and let

S_{x} = {g ∈ G: gx = x}

be the corresponding stabilizer subgroup. Then

α = 1^{S_{x}→G} ,

ie α is the representation of G obtained by induction from the trivial representation of S_{x}.

Remark: The result is easily extended to non-transitive actions. For in that case the set splits into a number of orbits, on each of which G acts transitively. On applying the Proposition to each orbit, we conclude that any permutation representation can be expressed as a sum of representations, each of which arises by induction from the trivial representation of some subgroup of G.

Proof ▶ By Definition 1,

α′= 1^{S_{x}→G}

is the representation in the subspace

V ⊂ C(G)

consisting of those functions F : G → k satisfying

F (gh) = h^{−1} F (g) ∀h ∈ S_{x}.

But since S_{x} acts trivially on k this condition reduces to

F (gh) = F (g),

ie F is constant on each coset gS_{x}. Thus V can be identified with the space C(G/S_{x},k) of functions on the set G/S_{x} of S_{x}-cosets in G.

On the other hand, the G-sets X and G/S_{x} can be identified, with the element gx ∈ X corresponding to the coset gS_{x}. Thus

C(G/S_{x},k) = C(X,k).

Since α is by definition the representation of G in C_{X} the result follows. ◀

Proof ▶ (Alternative) By Proposition 2,

χ_{α′}(g) = ∥{i: g_{i}^{−1} gg_{i} ∈ S_{x}}∥
= ∥{i: gg_{i}x = g_{i}x}∥
= ∥{y ∈ X : gy = y}∥,

since each y ∈ X is uniquely expressible in the form y = g_{i}x. But by Proposition ???,

χ_{α}(g) = ∥{y ∈ X : gy = y}∥.

Thus

χ_{α} = χ_{α′},

and so

α = α′= 1^{S_{x}→G} .

◀

Induced representations are of great practical value. But we end with an extremely important theoretical application.

Proposition 10.3 The number of simple representations of a finite group G is equal to the number of conjugacy classes in G.

Proof ▶ Let s denote the number of classes in G. We already know that

• The characters of G are class functions.

• The simple characters are linearly independent.

Thus G has at most s simple characters; and the result will follow if we can show that every class function is a linear combination of characters.

It suffices for the latter to show that we can find a linear combination of characters taking the value 1 on a given class ḡ, and vanishing on all other classes.

We can extend the formula in the Theorem above to define a map

f(h̄) ↦ f^{G}(ḡ) : X(H,k) → X(G,k)

from the space X(H,k) of class functions on the subgroup H ⊂ G to the space X(G,k) of class functions on G, by

f^{G}(ḡ) = \frac{∥G∥}{∥H∥∥ḡ∥} ∑_{h̄⊂ḡ} ∥h̄∥f(h̄).

Evidently this map is linear:

F (h) = af(h) + bg(h) ⟹ F^{G}(g) = af^{G}(g) + bf^{G}(g).

Choose any g ∈ ḡ. Let H be the subgroup generated by g. Thus if g is of order d,

H = C_{d} = ⟨g⟩ = {e,g,g^{2},...,g^{d−1} }.

Let θ denote the 1-dimensional character on H defined by

θ(g) = ω = e^{2πi/d}.

Since H is abelian, each element is in a class by itself, so all functions on H are class functions. The d characters on H are

1,θ,θ^{2},...,θ^{d−1} .

Let f(h) denote the linear combination

f = 1 + ω^{−1} θ + ω^{−2} θ^{2}+ ··· + ω^{−(d−1)} θ^{d−1} .

Then

f(h^{i}) = \{ \begin{matrix}d & if i = 1 , \\ 0 & if i = 0 . \end{matrix} ,

ie f vanishes off the H-class {g}, but is not identically 0.

It follows that the induced function f^{G}(g) has the required property; it vanishes off ḡ, while

f^{G}(ḡ) = \frac{∥G∥d}{∥H∥∥ḡ∥} ≠ 0.

This proves the result, since f^{G} is a linear combination of characters:

f^{G} = 1^{G}+ ωθ^{G}+ ω^{2}(θ^{2})^{G}+ ··· + ω^{−(d−1)} (θ^{d−1} )^{G}.

◀

Examples:

1. S_{3} has 3 classes: 1^{3}, 21 and 3. So it has 3 simple representations over ℂ, as of course we already knew: namely 1, ϵ and α.

2. D(4) has 5 classes: {e}, {s^{2}}, {s,s^{3}}, {c,d} and {h,v}. So it has 5 simple representations over ℂ. We already know of 4 1-dimensional representations. In addition the matrices defining the natural 2-dimensional representation in ℝ^{2} also define a 2-dimensional complex representation. (We shall consider this process of complexification more carefully in Chapter ???.) This representation must be simple, since the matrices do not commute, as they would if it were the sum of 2 1-dimensional representations. Thus all the representations of D_{4} are accounted for.

Proposition 10.4 (Frobenius’ Reciprocity Theorem) Suppose α is a representation of G, and β a representation of H ⊂ G. Then

I_{G}(α,β^{G}) = I_{H}(α_{H},β).

Proof ▶ We have

I_{G}(α,β^{G}) = \frac{1}{∥G∥} ∑_{ḡ} ∥ḡ∥\overline{χ_{α}(ḡ)} \frac{∥G∥}{∥H∥∥ḡ∥} ∑_{h̄⊂ḡ} ∥h̄∥χ_{β}(h̄)

= \frac{1}{∥H∥} ∑_{h̄} ∥h̄∥\overline{χ_{α}(h̄)}χ_{β}(h̄)
= I_{H}(α_{H},β).

◀

This short proof does not explain why Frobenius’ Reciprocity Theorem holds. For that we must take a brief excursion into category theory.

Let C_{G} denote the category of G-spaces and G-maps. Then restriction and induction define functors

S: C_{G} → C_{H}, ,I : C_{H} → C_{G}.

Now 2 functors

E : C_{1} → C_{2}, F : C_{2} → C_{1}

are said to be adjoint if for any 2 objects X ∈ C_{1},Y ∈ C_{2} there are bijections

M_{C_{1}}(X,F Y) = M_{C_{2}}(EX,Y)

which are natural in the sense that given any morphism

f : X → X′

in C_{1} the diagram

M (X,F Y) ⟵ M (X′,F Y)
∥ ∥
M (EX,Y) ⟵ M (EX′,Y)

is commutative, and similarly given any morphism

e: Y → Y′

in C_{2} the diagram

M (X,F Y) −→ M (X,F Y′)
∥ ∥
M (EX,Y) −→ M (EX,Y′)

is commutative.

It’s not difficult to establish—but would take us too far out of our way—that the induction and restriction functors are adjoint in this sense: if V is a G-space, and U a H-space, then

hom^{H}(V_{H},U) = hom^{G}(V,U^{G}).

On taking dimensions, this gives Frobenius’ Theorem:

I_{H}(α_{H},β) = I_{G}(α,β^{G}).

Representations of Product Groups

The representations of a product group G × H can be expressed—in as neat a way as one could wish—in terms of the representations of G and H.

Definition 11.1 Suppose α is a representation of G in the vector space U over k, and β a representation of H in the vector space V over k. Then we denote by α×β the representation of the product group G×H in the tensor product U ⊗V defined by

(g,h)∑ u ⊗ v = ∑ gu ⊗ hv.

Lemma 11.1 1. χ_{α×β} (g,h) = χ_{α}(g)χ_{β}(h)

2. dim(α × β) = dimαdimβ

3. if α and β are both representations of G then

(α × β)_{G} = αβ,

where the restriction is to the diagonal subgroup

G = {(g,g) : g ∈ G} ⊂ G × G.

Proposition 11.1 The representation α × β of G × H over ℂ is simple if and only if α and β are both simple. Moreover, every simple representation of G × H is of this form.

Proof ▶

Lemma 11.2 If α_{1},α_{2} are representations of G, and β_{1},β_{2} are representations of H, all over k, then

I(α_{1}× β_{1},α_{2}× β_{2}) = I(α_{1},β_{1})I(α_{2},β_{2})

Proof ▶ We have

I(α_{1}× β_{1},α_{2}× β_{2}) = \frac{1}{|G||H|} ∑_{(g,h)∈G×H} \overline{χ_{α_{1}×β_{1}} (g,h)}χ_{α_{2}×β_{2}} (g,h)

= \frac{1}{|G||H|} ∑_{(g,h)∈G×H} \overline{χ_{α_{1}}(g)χ_{β_{1}}(h)}χ_{α_{2}}(g)χ_{β_{2}}(h)

= \frac{1}{|G|} ∑_{g∈G} \overline{χ_{α_{1}}(g)}χ_{α_{2}}(g)\frac{1}{|G|} ∑_{h∈H} \overline{χ_{β_{1}}(h)}χ_{β_{2}}(h)
= I(α_{1},β_{1})I(α_{2},β_{2})

◀

Recall that a representation α over ℂ is simple if and only if

I(α,α) = 1.

Thus if α is a simple representation of G and β is a simple representation of H (both over ℂ) then

I(α × β,α × β) = I(α,α)I(β,β) = 1;

and therefore α × β is simple.

Now suppose G has r classes and H has s classes. Then G × H has rs classes, since

(g,h) ∼ (g′,h′) ⟺ g ∼ g′and h ∼ h′.

But we have just produced rs simple representations α × β of G × G; so these are in fact the full complement.

(The lemma shows that these representations are distinct; for

I(α_{1}× β_{1},α_{2}× β_{1}) = I(α_{1},α_{2})I(β_{1},β_{2}) = 0

unless α_{1} = α_{2} and β_{1} = β_{2}.) ◀

It is useful to give a proof of the last part of the Proposition not using the fundamental result that the number of simple representations is equal to the number of classes; for we can give an alternative proof of this result using product groups.

Proof ▶ (of last part of Proposition). Suppose γ is a representation of G × H in W over ℂ.

Consider the restriction γ_{H} of γ to the subgroup H = e×H ⊂ G×H. Let V be a simple part of W_{H}:

W_{H} = V ⊕ ···

Let

X = hom^{H}(V,W)

be the vector space formed by the H-maps t : V → W. This is non-trivial since V is a subspace of W.

Now X is a G-space, under the action

(gt)(v) = g(tv)

Let U be a simple G-subspace of X. Then

hom^{G}(U,X) = hom^{G}(U,hom^{H}(V,W))
= hom^{G×H} (U ⊗ V,W).

Since this space is non-trivial, there exists a G × H map

θ: U ⊗ V → W.

But since both U ⊗ V and W are simple, we must have

kerθ = 0, imθ = W.

Hence θ is an isomorphism, ie

W = U ⊗ V.

Thus

γ = α × β,

where α is the representation of G in U, and β is the representation of H in V.

◀

Theorem 11.1 Suppose G has n elements and s classes. Then

1. G has s simple representations over ℂ;

2. if these are σ_{1},...,σ_{s} then

dim^{2}σ_{1}+ ··· + dim^{2}σ_{s} = n.

Proof ▶ Let τ be the permutation representation of G × G in C(G,k) induced by the action

(g,h)x = gxh^{−1}

of G × G on G.

Lemma 11.3 The character of τ is given by

χ_{τ}(g,h) = \{ \begin{matrix}|G|/|ḡ| & ifg ∼ h \\ 0 & otherwise \end{matrix}

Proof ▶ Since τ is a permutational representation,

χ_{τ}(g,h) = |{x: (g,h)x = x}|
= |{x: gxh^{−1} = x}|
= |{x: x^{−1} gx = h}|.

If g ≁ h then clearly no such x exists.

Suppose g ∼ h. Then there exists at least one x, say x_{0}, such that

h = x_{0}^{−1} gx_{0}.

Now

x^{−1} gx = h ⟺ x^{−1} gx = x_{0}^{−1} gx_{0}
⟺ (xx_{0}^{−1} )g = g(xx_{0}^{−1} )
⟺ xx_{0}^{−1} ∈ Z(g)
⟺ x ∈ Z(g)x_{0}.

Thus

χ_{τ}(g,h) = |{x: gxh^{−1} = x|
= |Z(g)|
= |G|/|ḡ|.

◀

Lemma 11.4 Suppose G has simple representations σ_{1},...σ_{s}. Then

τ = σ_{1}^{∗}× σ_{1}+ ··· + σ_{s}^{∗}× σ_{s}.

Proof ▶ We know that the simple representations of G × G are σ_{i}× σ_{j}. Thus

τ = ∑_{i,j} e(i,j)σ_{i}× σ_{j},

where e(i.j) ∈ ℕ.

To determine e(i,j) we must compute the intertwining number

I(τ,σ_{i}× σ_{j}) = \frac{1}{|G|^{2}} ∑_{g,h} \overline{χ_{τ}(g,h)}χ_{σ_{i}×σ_{j}} (g,h)
= \frac{1}{|G|^{2}} ∑_{g,h} \overline{χ_{τ}(g,h)}χ_{σ_{i}}(g)χ_{σ_{j}} (h)

= \frac{1}{|G|^{2}} ∑_{h∼g} \frac{|G|}{|ḡ|}χ_{σ_{i}}(g)χ_{σ_{j}} (h)
= \frac{1}{|G|}χ_{σ_{i}}(g)χ_{σ_{j}} (g)
= \frac{1}{|G|}\overline{χ_{σ_{i}^{∗}}(g)}χ_{σ_{j}} (g)
= I(σ_{i}^{∗},σ_{j}).

Thus

I(τ,σ_{i}× σ_{j}) = \{ \begin{matrix}1 & if σ_{i}^{∗} = σ_{j} \\ 0 & otherwise \end{matrix}

In other words, σ_{i} × σ_{j} occurs in τ if and only if σ_{i}^{∗} = σ_{j}, and then occurs just once. ◀

It follows from this result in particular that the number of simple representations is equal to I(τ,τ).

Lemma 11.5 I(τ,τ) is equal to the number of classes in G.

Proof ▶ We have

I(τ,τ) = \frac{1}{|G|^{2}} ∑_{g,h}|χ_{τ}(g,h)|^{2}

= \frac{1}{|G|^{2}} ∑_{g} ∑_{h∼g} ( \frac{|G|}{|ḡ|}) ^{2}

= \frac{1}{|G|^{2}} ∑_{g} |ḡ|\frac{|G|^{2}}{|ḡ|^{2}}
= ∑_{g} \frac{1}{|ḡ|}.

Since each class contributes |ḡ| terms to this sum, each equal to 1/|ḡ|, the sum is equal to the number of classes. ◀

That proves the first part of the Theorem; the number of simple representations is equal to I(τ,τ), which in turn is equal to the number of classes.

The second part follows at once on taking dimensions in

τ = σ_{1}^{∗}× σ_{1}+ ··· + σ_{s}^{∗}× σ_{s}.

◀

Example: We can think of product groups in 2 ways—as a method of constructing new groups, or as a way of splitting up a given group into factors.

We say that G = H × K, where H,K are subgroups of G, if the map

H × K → G: (h,k) ↦ hk

is an isomorphism.

A necessary and sufficient condition for this— supposing G finite—is that

1. elements of H and K commute, ie

hk = kh

for all h ∈ H,k ∈ K; and

2. |G| = |H||K|.

Now consider the symmetry group G of a cube. This has 48 elements; for there are 8 vertices, and 6 symmetries leaving a given vertex fixed.

Of these 48 symmetries, half are proper and half improper. The proper symmetries form a subgroup P ⊂ G.

Let Z = {I,J}, where J denotes reflection in the centre of the cube. In fact Z is the centre of G:

Z = ZG = {z ∈ G: zg = gz for all g ∈ G}.

By the criterion above,

G = Z × P.

Moreover,

P = S_{4},

as we can see by considering the action of symmetries on the 4 diagonals of the cube. This defines a homomorphism

Θ : P → S_{4}.

Since no symmetry send every diagonal into itself,

kerΘ = {I}.

Thus Θ is injective; and so it is bijective, since

|P| = 24 = |S_{4}|.

Hence Θ is an isomorphism.

Thus

G = C_{2}× S_{4}.

In theory this allows us to dtermine the character table of G from that of S_{4}. However, to make use of this table we must know how the classes of C_{2}× S_{4} are to be interpreted geometrically. This is described in the following table.

class in C_{2}× P size order  geometricaldescription
{I} × 1^{4}  1  1 identity
{J} × 1^{4}  1  1 reflection in centre
{I} × 21^{2}  6  2 half-turn about axis joining centres of op-

posite edges

{J} × 21^{2} 6 2 relflection in plane through opposite

edges

{I} × 2^{2} 3 2 rotation about axis parallel to edge

through π

{J} × 2^{2}  3  2 relflection in central plane parallel to face
\{I\} × 31  8  3 rotation about diagonal through ±\frac{π}{3}
{J} × 31  8  6 screw reflection about diagonal
{I} × 4  6  4 rotation about axis parallel to edge
through ±\frac{π}{2}

{J} × 4 6 4 screw reflection about axis parallel to edge

The character table of C_{2}× S_{4} id readily derived from that of S_{4}. We denote the non-trivial character of C_{2} (J ↦ −1) by η.

\begin{matrix}Class & I × 1^{4} & J × 1^{4} & I × 21^{2} & J × 21^{2} & I × 2^{2} & J × 2^{2} & I × 31 & J × 31 & I × 4 & J × 4 \\ 1 × 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ η × 1 & 1 & −1 & 1 & −1 & 1 & −1 & 1 & −1 & 1 & −1 \\ 1 × ϵ & 1 & 1 & −1 & −1 & 1 & 1 & 1 & 1 & −1 & −1 \\ η × ϵ & 1 & −1 & −1 & 1 & 1 & −1 & 1 & −1 & −1 & 1 \\ 1 × α & 2 & 2 & 0 & 0 & 2 & 2 & −1 & −1 & 0 & 0 \\ η × α & 2 & −2 & 0 & 0 & 2 & −2 & −1 & 1 & 0 & 0 \\ 1 × β & 3 & 3 & 1 & 1 & −1 & −1 & 0 & 0 & −1 & −1 \\ η × β & 3 & −3 & 1 & −1 & −1 & 1 & 0 & 0 & −1 & 1 \\ 1 × ϵβ & 3 & 3 & −1 & −1 & −1 & −1 & 0 & 0 & 1 & 1 \\ η × ϵβ & 3 & −3 & −1 & 1 & −1 & 1 & 0 & 0 & 1 & −1 \end{matrix}

Now suppose π is the 6-dimensional permutational representation of G induced by its action on the 6 faces of the cube. Its character is readily determined:

\begin{matrix}Class & I × 1^{4} & J × 1^{4} & I × 21^{2} & J × 21^{2} & I × 2^{2} & J × 2^{2} & I × 31 & J × 31 & I × 4 & J × 4 \\ π & 6 & 0 & 0 & 2 & 2 & 4 & 0 & 0 & 2 & 0 \end{matrix}

For example, to determine χ_{π}({J} × 4) we note that an element of this class is a rotation about an axis parallel to an edge followed by reflection in the centre. This will send each of the 4 faces parallel to the edge into an adjacent face, and will swap the other 2 faces. Thus it will leave no face fixed; and so

χ_{π}({J} × 4) = 0.

We have

I(π,1 × 1) = \frac{1}{48}(1 · 1 · 6 + 6 · 2 · 1 + 3 · 2 · 1 + 3 · 4 · 1 + 6 · 2 · 1) = 1

(as we knew it would be, since the action is transitive). Similarly,

I(π,η × 1) = \frac{1}{48}(1 · 1 · 6 − 6 · 2 · 1 + 3 · 2 · 1 − 3 · 4 · 1 + 6 · 2 · 1) = 0,
I(π,1 × ϵ) = \frac{1}{48}(1 · 1 · 6 − 6 · 2 · 1 + 3 · 2 · 1 + 3 · 4 · 1 − 6 · 2 · 1) = 0,
I(π,η × ϵ) = \frac{1}{48}(1 · 1 · 6 + 6 · 2 · 1 + 3 · 2 · 1 − 3 · 4 · 1 − 6 · 2 · 1) = 0.

It is clear at this point that the remaining simple parts of π must be of dimensions 2 and 3. Thus π contains either 1 × α or η × α. In fact

I(π,1 × α) = \frac{1}{48}(1 · 6 · 2 + 3 · 2 · 2 + 3 · 4 · 2) = 1.

The remaining part drops out by subtraction; and we find that

π = 1 × 1 + 1 × α + η × ϵβ.

Exterior Products

12.1 The exterior products of a vector space

Suppose V is a vector space. Recall that its rth exterior product ∧^{r}V is a vector space, spanned by elements of the form

v_{1}∧ ··· ∧ v_{r} (v_{1},...,v_{r} ∈ V),

where

v_{π1} ∧ ··· ∧ v_{π_{r}} = ϵ(π)v_{1}∧ ··· ∧ v_{r}

for any permutation π ∈ S_{r}.

This implies in particular that any product containing a repeated element vanishes:

··· ∧ v ∧ ··· ∧ v ∧ ···= 0.

(We are assuming here that the characteristic of the scalar field k is not 2. In fact we shall only be concerned with the cases k = ℝ or ℂ.)

The exterior product ∧^{r}V could be defined rigorously as the quotient-space

∧^{r}V = V^{⊗r} /X,

where X is the subspace of V^{⊗r} spanned by all elements of the form

v_{π1} ∧ ...v_{π_{r}} − ϵ(π)v_{1}∧ ··· ∧ v_{r},

where v_{1},...,v_{r} ∈ V,π ∈ S_{r}, and ϵ denotes the parity representation of S_{r}. Suppose e_{1},...,e_{n} is a basis for V. Then

e_{i_{1}} ∧ e_{i_{2}} ∧ ··· ∧ e_{i_{r}} (i_{1} < i_{2} < ···< i_{r})

12.1. THE EXTERIOR PRODUCTS OF A VECTOR SPACE  GpReps-I–2

is a basis for ∧^{r}V. (Note that there is one basis element corresponding to each subset of {e_{1},...,e_{n}} containing r elements.) It follows that if dimV = n then

∧^{r}V = 0 if r > n;

while if r ≤ n then

dim∧^{r}V = \binom{n}{r} .

Now suppose T : V → V is a linear map. Then we can define a linear map

∧^{r}T : ∧^{r}V → ∧^{r}V

by

(∧^{r}T)(v_{1}∧ ··· ∧ v_{r}) = (T v_{1}) ∧ ··· ∧ (T v_{r}).

(To see that this action is properly defined, it is sufficient to see that it sends the subspace X ⊂ V^{⊗n} described above into itself; and that follows at once since

(∧^{r}T)(v_{π1} ∧ ··· ∧ v_{π_{r}}) −ϵ(π)v_{1}∧...v_{r} = (T v_{π1})∧··· ∧(T v_{π_{r}})−ϵ(π)(T v_{1})∧··· ∧(T v_{r})

is again one of the spanning elements of X.)

In the case r = n, ∧^{n}V is 1-dimensional, with the basis element

e_{1}∧ ··· ∧ e_{n};

and

∧^{n}T = (detT)I.

This is in fact the “true” definition of the determinant.

Although we shall not make use of this, the spaces ∧^{r}V can be combined to form the exterior algebra ∧V of V

∧V = ⨁ ∧^{r}V,

with the “wedge multiplication”

∧: ∧^{r}V × ∧^{s}V → ∧^{r+s} V

defined by

(u_{1}∧ ··· ∧ u_{r}) ∧ (v_{1}∧ ··· ∧ v_{s}) = u_{1}∧ ··· ∧ u_{r}∧ v_{1}∧ ··· ∧ v_{s},

extended to ∧V by linearity.

Observe that if a ∈ ∧^{r}V,b ∈ ∧^{s}V then

b ∧ a = (−1)^{rs}a ∧ b.

In particular the elements of even order form a commutative subalgebra of ∧V.

12.2. THE EXTERIOR PRODUCTS OF A GROUP REPRESENTATION

GpReps-I–3

12.2 The exterior products of a group representation

Definition 12.1 Suppose α is a representation of G in V. Then we denote by ∧^{r}α the representation of G in ∧^{r}V defined by

g(v_{1}∧ ··· ∧ v_{r}) = (gv_{1}) ∧ ··· ∧ (gv_{r}).

In other words, g acts through the linear map ∧^{r}(α(g)).

Proposition 12.1 Suppose g ∈ G has eigenvalues λ_{1},...,λ_{n} in the representation α. Then the character of ∧^{r}α is the rth symmetric sum of the λ’s, ie

χ_{∧^{r}α} (g) = ∑_{i_{1}<i_{2}<··· <i_{r}} λ_{i_{1}}λ_{i_{2}}...λ_{i_{r}}.

Proof ▶ Let us suppose that k = ℂ. We know that α(g) can be diagonalised, ie we can find a basis e_{1},...,e_{n} of the representation-space V such that

ge_{i} = λ_{i}e_{i} (i = 1,...,n).

But now

ge_{i_{1}} ∧ e_{i_{2}} ∧ ··· ∧ e_{i_{r}} = λ_{i_{1}}λ_{i_{2}}...λ_{i_{1}}e_{i_{1}} ∧ e_{i_{2}} ∧ ··· ∧ e_{i_{r}},

from which the result follows, since these products form a basis for ∧^{r}V. ◀

12.3 Symmetric polynomials

We usually denote the symmetric product in the Proposition above by

∑ λ_{1}...λ_{r}.

It is an example of a symmetric polynomial in λ_{1},...,λ_{n}.

More generally, suppose A is a commutative ring, with 1. (In fact we shall only be interested in the rings ℤ and ℚ.) As usual, A[x_{1},...,x_{n}] denotes the ring of polynomials in x_{1},...,x_{n} with coefficients in A.

The symmetric group S_{n} acts on this ring, by permutation of the variables:

(πP)(x_{1},...,x_{n}) = P ( x_{π^{−1} (1)},...,x_{π^{−1} (n)}) (π ∈ S_{n}).

12.3. SYMMETRIC POLYNOMIALS  GpReps-I–4

The polynomial P(x_{1},...,x_{n}) is said to be symmetric if it is left invariant by this action of S_{n}. The symmetric polynomials evidently form a sub-ring of A[x_{1},...,x_{n}], which we shall denote by Σ_{n}(A).

The n polynomials

a_{1} = ∑ x_{i}, a_{2} = ∑_{i_{1}<i_{2}} x_{i_{1}}x_{i_{2}}, ...,a_{n} = x_{1}··· x_{n}

are symmetric; as are

s_{1} = ∑ x_{i}, s_{2} = ∑ x_{i}^{2},;s_{3} = ∑ x_{i}^{3},....

Proposition 12.2 The ring Σ_{ℤ}(n) is freely generated over ℤ by a_{1},...,a_{n}, ie the map

p(x_{1},...,x_{n}) ↦ p(a_{1},...,a_{n}) : ℤ[x_{1},...,x_{n}] → Σ_{n}(ℤ)

is a ring-isomorphism.

Proof ▶ We have to show that

1. Every symmetric polynomial P(x_{1},...,x_{n}) over ℤ (ie with integer coefficients) is expressible as a polynomial in a_{1},...,a_{n} over ℤ:

P(x_{1},...,x_{n}) = p(a_{1},...,a_{n}).

This will show that the map is surjective.

2. The map is injective, ie

p(a_{1},...,a_{n}) ≡ 0 ⟹ p ≡ 0.

1. Any polynomial is a linear combination of monomials x_{1}^{e_{1}}··· x_{n}^{e_{n}}. We order the monomials first by degree, with higher degree first, and then within each degree lexicographically, eg if n = 2 then

1 < x_{2} < x_{1} < x_{2}^{2} < x_{1}x_{2} < x_{1}^{2} < x_{2}^{3} < ···.

The leading term in p(x_{1},...,x_{n}) is the non-zero term cx_{1}^{e_{1}}··· x_{n}^{e_{n}} containing the greatest monomial in this ordering. Now suppose the polynomial P(x_{1},...,x_{n}) is symmetric. Evidently e_{1} ≥ e_{2} ≥ ··· ≥ e_{n}in the leading term. For if say e_{1} < e_{2} then the term cx_{1}^{e_{2}}x_{2}^{e_{1}}··· x_{n}^{e_{n}} — which must also appear in P(x_{1},...,x_{n}).

◀

Corollary 12.1 The ring Σ_{ℚ}(n) is freely generated over ℚ by a_{1},...,a_{n},

Proposition 12.3 The ring Σ_{ℚ}(n) is freely generated over ℚ by s_{1},...,s_{n},

Proof ▶ ◀

12.4. NEWTON’S FORMULA  GpReps-I–5

12.4 Newton’s formula

It follows from the Propositions above that the power-sums s_{n} are expressible in terms of the symmetric products a_{n}, and vice versa. More precisely, there exist polynomials S_{n}(x_{1},...,x_{n}) and A_{n}(x_{1},...,x_{n}) such that

s_{n} = S_{n}(a_{1},..., n), a_{n} = A_{n}(a_{1},..., n),

with the coefficients of S_{n} integral and those of A_{n} rational. Newton’s formula allows these polynomials to be determined recursively.

Let

f(t) = (1 − x_{1}t)··· (1 − x_{n}t)
= 1 − a_{1}t + a_{2}t^{2}− ··· + (−1)^{n}a_{n}t^{n}.

Then

\frac{f′(t)}{f(t)} = frac−x_{1} 1 − x_{1}t + ··· + frac−x_{n}1 − x_{n}t
= −s_{1}− s_{2}t − s_{3}t^{2}− ···.

Thus

−a_{1}+2a_{2}t −3a_{3}t^{2}+··· +(−1)^{n}na_{n}t^{n−1} = (1−a_{1}t +a_{2}t^{2}−··· +(−1)^{n}a_{n}t^{n})(−s_{1}−s_{2}t −s_{3}t^{2}−··· ).

Equating coefficients,

a_{1} = s_{1}
2a_{2} = s_{1}a_{1}− s_{2}
3a_{3} = s_{1}a_{2}− s_{2}a_{1}+ s_{3}
...
ra_{r} = s_{1}a_{r}− s_{2}a_{r−1} + ··· + (−1)^{r−1} s_{r}
...

12.5. PLETHYSM  GpReps-I–6

Evidently these equations allow us to express s_{1},s_{2},s_{3},... successively in terms of a_{1},a_{2},a_{3},..., or vice versa:

s_{1} = a_{1}
s_{2} = a_{1}^{2}− 2a_{2}
s_{3} = a_{1}^{3}− 3a_{1}a_{2}+ 3a_{3}
...
a_{1} = s_{1}
2a_{2} = s_{1}^{2}− s_{2}
6a_{3} = s_{1}^{3}− 3s_{1}s_{2}+ 2s_{3}
...

12.5 Plethysm

There is another way of looking at the exterior product — as a particular case of the plethysm operator on the representation-ring R(G).

Suppose V is a vector space over a field k of characteristic 0. (We shall only be interested in the cases k = R or ℂ.) Then the symmetric group S_{n} acts on the tensor product V^{⊗n} by permutation of the factors:

π(v_{1}⊗ ··· ⊗ v_{n}) = v_{π^{−1} 1} ⊗ ··· ⊗ v_{π^{−1} n}.

Thus V^{⊗n} carries a representation of S_{n}. As we know this splits into components V^{Σ} corresponding to the simple representations Σ of S_{n}:

V^{⊗n} = V^{Σ_{1}} ⊕ ··· ⊕ V^{Σ_{s}},

where Σ_{1},...,Σ_{s} are the simple representations of S_{n}. (We shall find it convenient to use superfixes rather than suffixes for objects corresponding to representations of S_{n}.)

We are particularly interested in the components corresponding to the 2 1dimensional representations of S_{n}: the trivial representation 1_{n} and the parity representation ϵ_{n}, and we shall write

V^{P} = V^{1_{n}}, V^{N} = V^{ϵ_{n}}.

We also use P and N to denote the operations of symmetrisation and skewsymmetrisation on V^{⊗n}; that is, the linear maps

P,N : V^{⊗n} → V^{⊗n}

defined by

P(v_{1}⊗ ··· ⊗ v_{n}) = \frac{1}{n!} ∑_{π∈S_{n}} π(v_{1}⊗ ··· ⊗ v_{n}),
N(v_{1}⊗ ··· ⊗ v_{n}) = \frac{1}{n!} ∑_{π∈S_{n}} ϵ(π)π(v_{1}⊗ ··· ⊗ v_{n}).

Suppose π ∈ S_{n}. Regarding π as a map V^{⊗n} → V^{⊗n}, we have

πP = P = Pπ, πN = ϵ(π)N = Nπ.

It follows that

{P^{2} = P, N^{2} = N,

ie P and N are both projections onto subspaces of V^{⊗n}.

We say that x ∈ V^{⊗n} is symmetric if

πx = x

for all π ∈ S_{n}; and we say that x is skew-symmetric if

πx = ϵ(π)x

for all π. It follows at once from the relations πP = P,πN = ϵN that x is symmetric if and only if

Px = x;

while x is skew-symmetric if and only if

Nx = x.

Thus P is a projection onto the symmetric elements in V^{⊗n}, and N is a projection onto the skew-symmetric elements.

To see the connection with the exterior product ∧^{n}V, recall that we could define the latter by

∧^{n}V = V^{⊗n} /X,

where X ⊂ V^{⊗n} is the subspace spanned by elements of the form

πx − ϵ(π)x.

It is easy to see that Nx = 0 for such an element x; while conversely, for any x ∈ V^{⊗n}

x − Nx = \frac{1}{n!} ∑_{π∈S_{n}} ϵ(π)(ϵ(π)x − πxs)
∈ X

It follows that

X = kerN;

and so (since N is a projection)

∧^{n}V = V /X ≅ imN = V^{N};

that is, the nth exterior product of V can be identified with the ϵ-component of V^{⊗n}.

Now suppose that V carries a representation α of some group G. Then G acts on V^{⊗n} through the representation α^{n}.

Proposition 12.4 Suppose α is a represenation of G in V. Then the actions of G and S_{n} on V^{⊗n} commute.

For each simple representation Σ of S_{n}, the component V^{Σ} of V^{⊗n} is stable under G, and so carries a representation α^{Σ} of G. Thus

α^{n} = α^{Σ_{1}} + ··· + α^{Σ_{s}},

where Σ_{1},...,Σ_{s} are the simple representations of S_{n}.

Proof ▶ We have

πg(v_{1}⊗ ··· ⊗ v_{n}) = π(gv_{1}⊗ ··· ⊗ gv_{n})
= (gv_{π^{−1} 1} ⊗ ··· ⊗ gv_{π^{−1} n})
= gπ(v_{1}⊗ ··· ⊗ v_{n}).

◀

Since the actions of G and S_{n} on V^{⊗n} commute, they combine to define a representation of the product group G × S_{n} on this space.

Corollary 12.2 The representation of G × S_{n} on V^{⊗n} is given by

α^{Σ_{1}} × Σ_{1}+ ··· + α^{Σ_{s}} × Σ_{s}.

Suppose g ∈ G (or more accurately, α(g)) has eigenvalues λ_{1},...,λ_{d}. We know that the character of

∧^{n}α = α^{ϵ_{n}}

is the nth symmetric product of the λ_{i}:

χ_{∧^{n}α} (g) = a_{n}(λ_{1},...,λ_{d}).

Proposition 12.5 To each simple representation Σ of S_{n}there corresponds a unique symmetric function S_{Σ} of degree n such that for any representation α of G, and for any g ∈ G with eigenvalues λ_{1},...,λ_{d},

χ_{∧^{n}α} (g) = S_{Σ}(λ_{1},...,λ_{d}).

Proof ▶ We begin by establishing an important result which should perhaps have been proved when we discussed the splitting of a G-space V into components

V = V_{σ_{1}} ⊕ ··· ⊕ V_{σ_{s}}

corresponding to the simple representations σ_{1},...,σ_{s} of G.

Lemma 12.1 The projection P_{σ} onto the σ-component of V is given by

P_{σ} = \frac{dimσ}{∥G∥} ∑_{g∈G}χ_{σ}(g^{−1} )g.

Proof of Lemma ▷ Suppose α is a representation of G in V. Then the formula

above defines a linear map

P : V → V.

Suppose h ∈ G. Then (writing d for dimσ)

hPh^{−1}= \frac{d}{∥G∥} ∑_{g} χ_{σ}(g^{−1} )hgh^{−1}

= \frac{d}{∥G∥} ∑_{g′} χ_{σ}(h^{−1} g′^{−1} h)g′

= \frac{d}{∥G∥} ∑_{g′} χ_{σ}(g′^{−1} )g′

= P.

Now suppose α is simple. By Schur’s Lemma, the only linear transformations commuting with all α(g) are multiples of the identity. Thus

P = ρI

for some ρ ∈ ℂ. Taking traces,

\frac{d}{∥G∥} ∑_{g} χ_{σ}(g^{−1} )χ_{α}(g) = ρd.

It follows that

ρ = \{ \begin{matrix}1 & α = σ \\ 0 & α ≠ σ \end{matrix}

◁ In other words,

P = \{ \begin{matrix}I & α = σ \\ 0 & α ≠ σ \end{matrix}

It follows that P acts as the identity on all simple G-subspaces carrying the representation σ, and as 0 on all simple subspaces carrying a representation σ′≠ σ. In particular, P = I on V_{σ} and P = 0 on V_{σ′} for all σ′≠ σ. In other words, P is the projection onto the component V_{σ}.

◀

Real Representations

Representation theory over ℂ is much simpler than representation theory over ℝ. For that reason, we usually complexify real representations—extend the scalars from ℝ to ℂ—just as we do with polynomial equations. But at the end of the day we must determine if the representations—or solutions—that we have obtained are in fact real.

Suppose U is a vector space over ℝ. Then we can define a vector space V = ℂU over ℂ by “extension of scalars”. More precisely,

V = ℂ ⊗_{ℝ}U.

In practical terms,

V = U ⊕ iU,

ie each element v ∈ V is uniquely expressible in the form

v = u_{1}+ iu_{2} (u_{1},u_{2} ∈ U).

If e_{1},...,e_{n}is a basis for U over ℝ, then it is also a basis for V over ℂ. In particular,

dim_{ℂ}V = dim_{ℝ}U.

On the other hand, suppose V is a vector space over ℂ. Then we can define a vector space U = ℝV over ℝ by “forgetting” scalar multiplication by non-reals. Thus the elements of U are precisely the same as those of V. If e_{1},...,e_{n} is a basis for V over V, then e_{1},ie_{1},e_{2},ie_{2},...,e_{n},ie_{n} is a basis for U over ℝ. In particular,

dim_{ℝ}U = 2dim_{ℂ}V.

Now suppose G acts on the vector space U over ℝ. Then G also acts on ℂU, by

g(u_{1}+ iu_{2}) = (gu_{1}) + i(gu_{2}).

On the other hand, suppose G acts on the vector space V over ℂ. Then G also acts on ℝV by the same rule

(g,v) ↦ gv.

Definition 13.1 1. Suppose β is a real representation of G in U. Then we denote by ℂβ the complex representation of G in the vector space

ℂU = U ⊕ iU

derived from U by extending the scalars from ℝ to ℂ.

2. Suppose α is a complex representation of G in V. Then we denote by ℝα the real representation of G in the vector space ℝV derived from V by “forgetting” scalar multiplication by non-reals.

Remarks:

1. Suppose β is described in matrix terms, by choosing a basis for U and giving the matrices B(g) representing β(g) with respect to this basis. Then we can take the same basis for ℂU, and the same matrices to represent ℂβ(g). Thus from the matrix point of view, β and ℂβ appear the same. The essential difference is that ℂβ may split even if β is simple, ie we may be able to find a complex matrix P such that

PB(g)P^{−1} = ( \begin{matrix}C(g) & 0 \\ 0 & D(g) \end{matrix} )

for all g ∈ G, although no real matrix P has this property.

2. Suppose α is described in matrix form, by choosing a basis e_{1},e_{2},...,e_{n} for V, and giving the matrices A(g) representing α(g) with respect to this basis. Then we can take the 2n elements e_{1},ie_{1},e_{2},ie_{2},...,e_{n},ie_{n}as a basis for ℝV; and the matrix representing ℝα(g) with respect to this basis can be derived from the matrix A = A(g) representing α(g) as follows. By definition,

ge_{r} = ∑_{s} A_{sr}e_{s}.

Let

A_{r,s} = X_{r,s}+ iY_{r,s},

where X_{r,s},Y_{r,s} ∈ ℝ. Then

ge_{r} = X_{sr}e_{s}+ Y_{sr}ie_{s}
g(ie_{r}) = −Y_{sr}e_{s}+ X_{sr}ie_{s}

Thus the entry A_{rs} is replaced in ℝα(g) by the 2 × 2-matrix

( \begin{matrix}X_{r,s} & −Y_{r,s} \\ Y_{r,s} & X_{r,s} \end{matrix} )

Proposition 13.1 1. ℂ(β + β′) = ℂβ + ℂβ′ 2. ℂ(ββ′) = (ℂβ)(ℂβ′)

3. ℂ(β^{∗}) = (ℂβ)^{∗}

4. ℂ1 = 1

5. dimℂβ = dimβ

6. χ_{ℂβ}(g) = χ_{β}(g)

7. I(ℂβ,ℂβ′) = I(β,β′)

8. ℝ(α + α′) = ℝα + ℝα′

9. ℝ(α^{∗}) = (ℝα)^{∗}

10. dimℝα = 2dimα

11. χ_{ℝα}(g) = 2ℜχ_{α}(g) = χ_{α}(g) + χ_{α}(g^{−1} )

12. ℝℂβ = 2β

13. ℂℝα = α + α^{∗}

Proof ▶ All is immediate except (perhaps) parts (11) and (13).

11. Suppose α(g) is represented by the n × n matrix

A = X + iY,

where X,Y are real. Then—as we saw above—the entry A_{rs} is replaced in ℝα(g) by the 2 × 2 matrix

( \begin{matrix}X_{r,s} & −Y_{r,s} \\ Y_{r,s} & X_{r,s} \end{matrix} )

\underline{GpReps-I–4}

Thus

trℝα(g) = 2∑_{r} X_{rr}
= 2ℜ(∑_{r} A_{rr})
= 2ℜtrα(g)
= 2ℜχ_{α}(g)
= χ_{α}(g) + χ_{α}(g^{−1} )

since

\overline{χ(g)} = χ(g^{−1} ).

13. This now follows on taking characters, since

χ_{ℂℝα}(g) = χ_{ℝα}(g)
= χ_{α}(g) + χ_{α}(g^{−1} )
= χ_{α}(g) + χ_{α^{∗}}(g)

Since this holds for all g,

ℂℝα = α + α^{∗}.

◀

Lemma 13.1 Given a representation α of G over ℂ there exists at most one real representation β of G over ℝ such that

α = ℂβ.

Proof ▶ By Proposition 1,

ℂβ = ℂβ′ ⟹ χ_{ℂβ}(g) = χ_{ℂβ′}(g)
⟹ χ_{β}(g) = χ_{β′}(g)
⟹ β = β′.

◀

Definition 13.2 A representation α of G over ℂ is said to be real if α = ℂβ for some representation β over ℝ. Remarks:

1. In matrix terms α is real if we can find a complex matrix P such that the

matrices

PB(g)P^{−1}

are real for all g ∈ G.

2. Since β is uniquely determined by α in this case, one can to some extent confuse the two (as indeed in speaking of α as real), although eg if discussing simplicity it must be made clear whether the reference is to α or to β.

Lemma 13.2 Consider the following 3 properties of the representation α over ℂ:

1. α is real

2. χ_{α} is real, ie χ_{α}(g) ∈ ℝ for all g ∈ G

3. α = α^{∗}

We have

(1) ⟹ (2) ⟺ (3).

Proof ▶ (1) ⟹ (2): If α = ℂβ then

χ_{α}(g) = χ_{β}(g).

But the trace of a real matrix is necessarily real.

(2) ⟺ (3): If χ_{α} is real then

χ_{α}(g) = \overline{χ_{α}(g)} = χ_{α^{∗}}(g)

for all g ∈ G. Hence

α = α^{∗}.

◀

Problems involving representations over ℝ often arise in classical physics, since the spaces there are normally real, eg those given by the electric and magnetic fields, or the vibrations of a system. The best way of tackling such a problem is usually to complexify, ie to extend the scalars from ℝ to ℂ. This allows the powerful techniques developed in the earlier chapters to be applied. But at the end of the day it may be necessary to determine whether or not the representations that arise are real. The Lemma above gives a necessary condition: if α is real then its character must be real. But this condition is not sufficient; and our aim in the rest of the Chapter is to find necessary and sufficient conditions for reality, of as practical a nature as possible.

Definition 13.3 Suppose α is a simple representation over ℂ. Then we say that α is strictly complex if χ_{α} is not real; and we say that α is quaternionic if χ_{α} is real, but α itself is not real.

Thus the simple representations of G over ℂ fall into 3 mutually exclusive classes:

ℝ real: α = ℂβ

ℂ strictly complex: χ_{α} not real

ℍ quaternionic: χ_{α} real but α not real

Lemma 13.3 Suppose α is a simple representation over ℂ. Then

1. If α is real, ℝα = 2β, where β is a simple representation over ℝ;

2. if α is strictly complex or quaternionic, ℝα = β is a simple representation over ℝ.

In particular, if χ_{α} is not real then ℝα must be simple.

Proof ▶ If α is real, say α = ℂβ, then by Proposition 1

ℝα = ℝℂβ = 2β.

Conversely, suppose ℝα splits, say

ℝα = β + β′.

Then by Proposition 1,

α + α^{∗} = ℂℝα = ℂβ + ℂβ′.

But since α and α^{∗} are simple, this implies (by the unique factorisation theorem) that

α = ℂβ or α = ℂβ′.

In either case α is real. ◀ This gives a (not very practical) way of distinguishing between the 3 classes:

ℝ: α real ⟺ χ_{α} real and ℝα splits

ℂ: α quaternionic ⟺ χ_{α} real and ℝα simple

ℍ: α strictly complex ⟺ χ_{α} not real (⟹ ℝα simple)

The next Proposition shows that the classification of simple representations over ℂ into 3 classes leads to a similar classification of simple representations over ℝ. Proposition 13.2 Suppose β is a simple representation over ℝ. Then there are 3 (mutually exclusive) possibilities: ℝ: ℂβ = α is simple ℂ: ℂβ = α + α^{∗}, ℍ: ℂβ = 2α, with α simple with α (and α^{∗}) simple, and α ≠ α^{∗}

In case (ℝ), α is real and

I(β,β) = 1.

In case (ℂ), α is strictly complex and

I(β,β) = 2.

In case (ℍ), α is quaternionic and

I(β,β) = 4.

Proof ▶ Since

ℝℂβ = 2β,

ℂβ cannot split into more than 2 parts. Thus there are 3 possibilities:

1. ℂβ = α is simple

2. ℂβ = 2α, with α simple

3. ℂβ = α + α′, with α, α′simple and α ≠ α′ Since

I(β,β) = I(ℂβ,ℂβ)

by Proposition 1, the values of I(β,β) in the 3 cases follow at once. Thus it only remains to show that α is in the class specified in each case, and that α′= α^{∗} in case (3).

In case (1), α is real by definition.

In case (2),

2χ_{α}(g) = χ_{2α}(g) = χ_{β}(g)

is real for all g ∈ G. Hence χ_{α}(g) is real, and so α is either real or quaternionic. If α were real, say α = ℂβ′, we should have

ℂβ = 2ℂβ′

which would imply that

β = 2β′

by Proposition 2. Hence α is quaternionic.

In case (3)

2β = ℝℂβ = ℝα + ℝα′.

Hence

ℝα = β = ℝα′.

But then

α + α′= ℂβ = α + α^{∗}.

Hence

α′= α^{∗}.

Finally, since α^{∗} = α′≠ α, α is strictly complex. ◀

Proposition 5 gives a practical criterion for determining which of the 3 classes a simple representation β over ℝ belongs to, namely by computing I(β,β) from χ_{β}. Unfortunately, the question that more often arises is: which class does a given simple representation α over ℂ belong to? and this is more difficult to determine.

Lemma 13.4 Suppose α is a simple representation of G over ℂ in V. Then

ℝ: if α is real,there exists an invariant symmetric (quadratic) form on V, unique up to a scalar multiple—but there is no invariant skew-symmetric form on V;

ℂ: if α is complex, there is no invariant bilinear form on V.

ℍ: if α is quaternionic, there exists an invariant skew-symmetric form on V, unique up to a scalar multiple—but there is no invariant symmetric form on V;

Proof ▶ A bilinear form on V is a linear map

V ⊗ V → ℂ,

ie an element of

(V ⊗ V)^{∗} = V^{∗}⊗ V^{∗}.

Thus the space of bilinear maps carries the representation (α^{∗})^{2} of G. Hence the invariant bilinear maps form a space of dimension

I(1,(α^{∗})^{2}) = I(1,α^{∗}α^{∗}) = I(α,α^{∗})

Since α and α^{∗} are simple, this is 0 or 1 according as α = α^{∗} or not, ie according as α is either real or quaternionic, or strictly complex. In other words, if α is complex there is no invariant bilinear form; while if α is real or quaternionic there is an invariant bilinear form on V, say F (u,v), unique up to a scalar multiple.

Now any bilinear form can be split into a symmetric (or quadratic) part and a skew-symmetric part; say

F (u,v) = Q(u,v) + S(u,v),

where

Q(u,v) = \frac{1}{2} (F (u,v) + F (v,u)), S(u,v) = \frac{1}{2} (F (u,v) − F (v,u))

But it is easy to see that if F is invariant then so are Q and S. Since F is the only invariant bilinear form on V, it follows that either

F = Q or F = S,

ie F is either symmetric or skew-symmetric. It remains to show that the former occurs in the real case, the latter in the quaternionic case.

Suppose α is real, say α = ℂβ, where β is a representation in the real vector space U. We know that U carries an invariant symmetric form (in fact a positivedefinite one), say Q(u,u′). But this defines an invariant symmetric form ℂQ on V = ℂU by extension of scalars. So if α is real, V carries an invariant symmetric form.

Finally, suppose α is quaternionic. Then V carries either a symmetric or a skew-symmetric invariant form (but not both). Suppose the former; say Q(v,v′) is invariant. By Proposition 3, β = ℝα is simple. Hence there exists a real invariant positive-definite symmetric form on ℝV; and this is the only invariant symmetric form on ℝV, up to a scalar multiple. But the real part of Q(v,v′) is also an invariant form on ℝV; and it is certainly not positive-definite, since

ℜQ(iv,iv) = −ℜQ(v,v).

This contradiction shows that V cannot carry an invariant symmetric form. We conclude that it must carry an invariant skew-symmetric form. ◀

We deduce from this Proposition the following more practical criterion for reality.

Proposition 13.3 Suppose α is a simple representation over ℂ. Then

\frac{1}{|G|} ∑_{g∈G}χ_{α}(g^{2}) = \{ \begin{matrix}1 & if α is real \\ 0 & if α is strictly complex \\ −1 & if α is quaternionic \end{matrix}

Proof ▶ Every bilinear form has a unique expression as the sum of its symmetric and skew-symmetric parts. In other words, the space of bilinear forms is the direct sum of the spaces of symmetric and of skew-symmetric forms; say

V^{∗}⊗ V^{∗} = V^{Q}⊕ V^{S}.

Moreover, each of these subspaces is stable under G; so the representation (α^{∗})^{2} in the space of bilinear forms splits in the same way; say

(α^{∗})^{2} = α^{Q}+ α^{S},

where α^{Q} is the representation of G in the space V^{Q} of symmetric forms on V, and α^{S} is the representation in the space V^{S} of skew-symmetric forms.

Now the dimensions of the spaces of invariant symmetric and skew-symmetric space are

I(1,α^{Q}) and I(1,α^{S}),

respectively. Thus Proposition 6 can be reworded as follows:

ℝ: If α is real then

I(1,α^{Q}) = 1 and I(1,α^{S}) = 0.

ℂ: If α is complex then

I(1,α^{Q}) = 0 and I(1,α^{S}) = 0.

ℍ: If α is quaternionic then

I(1,α^{Q}) = 0 and I(1,α^{S}) = 1.

Thus all (!) we have to do is to compute these 2 intertwining numbers. In fact it suffices to find one of them, since

I(1,α^{Q}) + I(1,α^{S}) = I(1,(α^{∗})^{2}) = I(α,α^{∗})

which we already know to be 1 if α is real or quaternionic, and 0 if α is complex.

To compute I(1,α^{Q}), choose a basis e_{1},...,e_{n} for V; and let the corresponding coordinates be x_{1},...,x_{n}. Then the n(n + 1)/2 quadratic forms

x_{i}^{2} (1 ≤ i ≤ n), 2x_{i}x_{j} (1 ≤ i < j ≤ n)

form a basis for V^{Q}. Let g_{ij} denote the matrix defined by α(g). Thus if v = (x_{1},...,x_{n}) ∈ V, then the coordinates of gv are

(gv)_{i} = ∑_{j} g_{ij}x_{j}.

Hence

g(x_{i}^{2}) = ∑_{j,k} g_{ij}x_{j}g_{ik}x_{k}.

In particular, the coefficient of x_{i}^{2} in this (which is all we need to know for the trace) is g_{ii}^{2}. Similarly, the coefficient of 2x_{i}x_{j} in g(2x_{i}x_{j}) is

g_{ii}g_{jj} + g_{ij}g_{ji}.

We conclude that

χ_{α^{Q}}(g) = ∑_{i} g_{ii}^{2}+ ∑_{i,j:i<j} (g_{ii}g_{jj} + g_{ij}g_{ji}).

But

χ_{α}(g) = ∑_{i} g_{ii}, χ_{α}(g^{2}) = ∑_{i,j} g_{ij}g_{ji}.

Thus

χ_{α^{Q}}(g) = \frac{1}{2} (χ_{α}(g)^{2}+ χ_{α}(g^{2})) .

Since

I(1,α^{Q}) = \frac{1}{|G|} ∑_{g∈G}χ_{α^{[2]}}(g),

it follows that

2I(1,α^{Q}) = \frac{1}{|G|} ∑_{g∈G}(χ_{α}(g))^{2}+ χ_{α}(g^{2})) .

But

\frac{1}{|G|} ∑_{g} χ_{α}(g)^{2} = I(α,α^{∗}).

Thus

2I(1,α^{Q}) = I(α,α^{∗}) + \frac{1}{|G|} ∑_{g} χ_{α}(g^{2}).

The result follows, since I(α,α^{∗}) = 1 in the real and quaternionic cases, and 0 in the complex case. ◀ 13.1. THE ENDOMORPHISM ALGEBRA OF A REPRESENTATION

GpReps-I–12

13.1 The endomorphism algebra of a representation

There is another way of looking at the classification of simple representations over ℂ into real, quaternionic and essentially complex — and one that justifies that term quaternionic.

Recall that an associative algebra over a field k is defined by giving a vector space A over k, together with a map

A × A → A,

which we denote by (a,b) ↦ ab, and which is both bilinear and associative. The element 1 ∈ A is said to be a unity element if 1 · a = a · 1 = a for all a ∈ A. We shall use the term algebra to mean ‘associative algebra with 1’.

Proposition 13.4 If α is a representation of the group G in the vector space V over k then the G-maps t : V → V, form an algebra over k of dimension I(α,α).

Proof ▶ The G-maps certainly form a vector space over k of dimension

I(α,α) = dimhom^{G}(V,V).

Moreover, composition of maps defines a multiplication which is both bilinear and associative. ◀

A G-map V → V is often called an endomorphism of the G-space V.

Definition 13.4 We denote the algebra of G-maps by End(()α).

An algebra A is said to be a division-algebra or skew-field (the two terms are used interchangeably) if each non-zero element a ∈ A has an inverse b:

ab = 1 = ba.

Proposition 13.5 If α is a simple representation of G then End(α) is a divisionalgebra.

Proof ▶ Suppose t : V → V is a non-zero G-map. It is readily verified that kert and imt are both stable under G. Since α is simple, it follows that

kert = 0 or V, imt = 0 or V.

Thus if t ≠ 0,

kert = 0, imt = V,

13.1. THE ENDOMORPHISM ALGEBRA OF A REPRESENTATION

GpReps-I–13

ie t is injective and surjective. It follows that t is bijective and so invertible; and it is a straightforward matter to verify that t^{−1} is also a G-map. ◀

The most familiar example of a non-commutative division-algebra is furnished by the algebra of quaternions ℍ Recall that ℍ is a 4-dimensional algebra over ℝ with basis 1,i,j,k, so that each quaternion q ∈ ℍ is uniquely expressible in the form

q = t + xi + yj + zk,

where t,x,y,z ∈ ℝ. Multiplication of quaternions is defined by the familiar rules

i^{2} = j^{2} = k^{2} = −1,
ij = k = −ji, jk = i = −kj, ki = j = −ik,

extended to ℍ by bilinearity, eg

(i + 2j)(1 − k) = i − ik + 2j − 2jk = −i + 3j.

It is a straightforward matter to verify that this multiplication is associative — in effect one has to verify that

x(yz) = (xy)z

in the 3^{3} = 27 cases x,y,z ∈ {i,j,k}. If we observe that the multiplication rules are invariant under the cyclic permutation i ↦ j ↦ k ↦ i then the number of cases is reduced to 9, since we may assume eg that x = i.

There is a conjugacy operation on ℍ analogous to complex conjugation, with

q̄= t − xi − yj − zk.

It is readily verified that

\overline{q_{1}q_{2}} = \overline{q_{2}}\overline{q_{1}}

(eg by considering the 9 cases q_{1},q_{2} ∈ {i,j,k} and applying bilinearity).

Also

q̄q = t^{2}+ x^{2}+ y^{2}+ z^{2}.

We define |q| to be the positive square-root of this, so that

|q|^{2} = t^{2}+ x^{2}+ y^{2}+ z^{2}.

Evidently

|q| = 0 ⟺ q = 0.

It follows that every non-zero quaternion q has an inverse namely

q^{−1} = \frac{q̄}{|q|^{2}}.

13.1. THE ENDOMORPHISM ALGEBRA OF A REPRESENTATION

GpReps-I–14

Thus the quaternions form a skew-field or division algebra. (The two terms are interchangeable.)

The quaternions have a certain uniqueness, as we shall see.

Proposition 13.6 The only finite-dimensional division algebra over ℂ is ℂ itself.

Proof ▶ If A is an algebra over k then we can identify λ ∈ k with the element λ · 1 ∈ A. Thus k can be identified with a subalgebra of A: k ⊂ A.

Suppose A is a finite-dimensional division-algebra over ℂ; and suppose A ≠ ℂ. Let a ∈ A \ ℂ.

If dimA = n then then n + 1 elements 1,a,a^{2},...,a^{n} must be linearly dependent, ie a satisfies a polynomial equation of degree ≤ n.

Let m(x) be the minimal polynomial of a, ie the monic polynomial of minimal degree satisfied by a. Then m(x) must be irreducible; for if m(x) = f(x)g(x) then

m(a) = 0 ⟹ f(a)g(a) = 0 ⟹ f(a) = 0 or g(a) = 0,

contradicting the minimality of m(x).

But every polynomial over ℂ factorises completely into linear factors, by the Fundamental Theorem of Algebra. It follows that m(x) = x−c, and so a = c ∈ ℂ. We conclude that A = ℂ. ◀

Proposition 13.7 The only finite-dimensional division algebras over ℝ are ℝ, ℂ and the quaternions ℍ.

Proof ▶ Suppose A is a finite-dimensional division algebra over ℝ; and suppose A ≠ ℝ. Let a ∈ A \ ℝ.

As above, a satisfies a polynomial equation over ℝ; and if m(x) is the minimal such polynomial, then m(x) is irreducible. But an irreducible polynomial m(x) over ℝ is necessarily linear or quadratic. In effect, if α ∈ ℂ is a root of m(x) then so is ᾱ; and

(x − α)(x − ᾱ)

is a polynomial over ℝ satisfied by a.

If m(x) is linear then a ∈ ℝ. So

m(x) = x^{2}+ 2bx + c,

where

d = b^{2}− ac < 0,

since otherwise m(x) could be factorised over ℝ. Let d = −e^{2}. Then

i = \frac{a + b}{e}

13.1. THE ENDOMORPHISM ALGEBRA OF A REPRESENTATION

GpReps-I–15

satisfies

i^{2} = −1.

It follows that the elements

{x + yi: x,y ∈ ℝ}

form a subalgebra of A which we can identify with ℂ.

If A = ℂ we are done; otherwise consider the linear map

t : x ↦ ixi^{−1} = −ixi: A → A.

Evidently

t^{2} = 1.

It follows that

A = E_{1}⊕ E_{−1} ,

where E_{1},E_{−1} are the 1− and −1−eigenspaces of t, ie

E_{1} = {a ∈ A: ai = ia}, E_{−1} = {a ∈ A: ai = −ia}.

(Explicitly,

a = \frac{1}{2}(a + ta) + \frac{1}{2}(a − ta),

and a + ta ∈ E_{1},a − ta ∈ E_{−1} .)

Evidently ℂ ∈ E_{1}. We show that in fact E_{1} = ℂ. For suppose a ∈ E_{1}\ℂ. Since ai = ia it follows that az = za for all z ∈ ℂ. Consider the subalgebra B generated by ℂ and a, ie the set of elements expressible as polynomials

c_{0}+ c_{1}a + ··· + c_{r}a^{r} (c_{i} ∈ ℂ).

Evidently B is a commutative sub-algebra of A. Moreover it is a divisionalgebra, ie the inverse of every non-zero element b ∈ B lies in B. For suppose the minimal polynomial of b is

m(x) = x^{r}+ a_{1}x^{r−1} + ··· + a_{r}.

Then a_{r} ≠ 0 since m(x) is irreducible; and so

b^{−1} = −\frac{1}{a_{r}} (b^{r−1} + a_{1}b^{r−2} + ··· + a_{r−1} ) ∈ B.

Since ℂ ⊂ B, we can regard B as a division-algebra over ℂ; and so from above B = ℂ. Hence

E_{1} = ℂ.

13.1. THE ENDOMORPHISM ALGEBRA OF A REPRESENTATION

GpReps-I–16

If A = ℂ we are done; otherwise E_{−1} ≠ 0. Let a ∈ E_{−1} ,a ≠ 0.

Then

E_{−1} = aE_{1} = aℂ.

For if z commutes with i then az anti-commutes with i; hence aE_{1} ⊂ E_{−1} . Conversely, a^{−1} anti-commutes with i; so if a′ also anti-commutes with i then z = a_{−1} a′commutes with i, ie a′= az ∈ aE_{1}.

From the definition of E_{−1} ,

ai = −ia.

It follows that

az = z̄a

for all z = x + yi ∈ ℂ.

In particular, if z ∈ ℂ then

z ∈ ℝ ⟺ az = za.

Evidently a^{2} commutes with i, and so

a^{2} ∈ ℂ.

In fact, since

a · a^{2} = a^{3} = a^{2}· a,

it follows that

a^{2} ∈ ℝ.

Thus

a^{2} = ±d^{2},

where d ∈ ℝ.

If a^{2} = d^{2} then (a − d)(a + d) = 0 and so (since we are in a division-algebra) a = ±d, contradicting our assumption that a ≠ ℂ. Hence a^{2} = −d^{2}. Thus if we set

j = \frac{a}{d}

then

j^{2} = −1;

while since j ∈ E_{−1} ,

ji = −ij.

Finally, on setting

k = ij = −ji

it is readily verified that i,j,k satisfy Hamilton’s identities, so that

A = ℍ.

◀

13.2. APPLICATION TO SIMPLE REPRESENTATIONS  GpReps-I–17

13.2 Application to simple representations

If α is a simple representation over ℂ then I(α,α) = 1 and so End(α) = ℂ. Thus Proposition 13.6 tells us nothing new; in effect it is a re-statement of Schur’s Lemma.

But we do learn something new if we consider simple representations over ℝ. First we note that the intertwining number is invariant under extension of the scalars.

Proposition 13.8 Suppose β_{1},β_{2} are representations of G over ℝ. Then

I(ℂβ_{1},ℂβ_{2}) = I(β_{1},β_{2}).

Proof ▶ Since

χ_{ℂβ}(g) = χ_{β}(g),

this follows at once from our formula for the intertwining number:

I(β_{1},β_{2}) = \frac{1}{∥G∥} ∑_{g∈G} \overline{χ_{β_{1}}(g)}χ_{β_{2}}(g).

But it is also easy to establish directly. If we think in terms of matrices, I(β_{1},β_{2}) is the dimension of the space of matrices X satisfying the linear equations

XB_{1}(g) = B_{2}(g)X

for all g ∈ G. Now the Proposition follows from the fact that if we have a set of equations over the field k, and if k ⊂ K (ie k is a subfield of K) then the dimension of the solution-space over K is equal to the dimension of the solution-space over k. ◀

Now we can complete our classification of simple representations of G over ℝ.

Proposition 13.9 Suppose β is a simple representation of the finite group G over ℝ. Then

1. If β is essentially real, ie ℂβ is simple, then

I(β,β) = 1

and

End(β) = ℝ;

13.3. REPRESENTATIONS OVER THE QUATERNIONS ℍ  GpReps-I–18

2. If β is quaternionic, ie ℂβ = 2α, then

I(β,β) = 4

and

End(β) = ℍ;

3. If β is essentially complex, ie ℂβ = α + α^{∗}, where α^{∗} ≠ α, then

I(β,β) = 2

and

End(β) = ℂ;

Proof ▶ If ℂβ is simple then

I(β,β) = I(ℂβ,ℂβ) = 1 ⟹ End(β) = ℝ.

Similarly, if ℂβ = 2α then

I(β,β) = I(2α,2α) = 4 ⟹ End(β) = ℍ.

Finally, if ℂβ = α + α^{∗} then

I(β,β) = I(α + α^{∗},α + α^{∗}) = 2 ⟹ End(β) = ℂ.

In effect, End(β) is completely determined by its dimension, since there is just one division-algebra of each dimension 1,2,4. ◀

These results refer to representations over ℝ; but we shall see next that they have a bearing on representations over ℂ.

13.3 Representations over the quaternions ℍ

Although vector spaces are normally defined over a commutative field k, the basic concepts of linear algebra carry over unchanged if we allow k to be a noncommutative or skew field, The most familiar skew-field is provided by the quaternions ℍ.

If we look at the definitions (and basic properties) of vector space and subspace, linear independence and dependence, spanning set, basis and dimension, we see that none of them make use of the commutativity of k; so all extend to the non-commutative case.

Thus a vector space over ℍ is defined by giving an additive group W together with a map

ℍ × W → W,

which we denote by (q,w) ↦ qw, and which satisfies the rules

1. q(w_{1}+ w_{2}) = qw_{1}+ qw_{2},

2. (q_{1}+ q_{2}w) = q_{1}w + q_{2}w,

3. q_{1}(q_{2}w) = (q_{1}q_{2})w,

4. 1w = w.

An n-dimensional vector space over ℍ can be identified in the usual way with the space of n − tuples of quaternions ℍ^{n}.

The concept of a linear map t : W_{1} → W_{2} of vector spaces over ℍ also goes over unchanged: t is linear if

t(qw) = qt(w)

for all q ∈ ℍ,w ∈ W.

But we need to take a little care when representing such maps by matrices. For simplicity we restrict the discussion to linear maps t : W → W, since this is the only kind we shall meet. Note that we cannot define a linear map t : ℍ^{n} → ℍ^{n} in the usual way, by

w ↦ Tw,

ie

w′_{r}= ∑_{s} T_{rs}w_{s},

where T is an n × n matrix with quaternionic entries. For in general

T_{rs}(qw_{s}) ≠ q(T_{rs}w_{s}),

so the map is not linear.

In effect, the matrix must act on the right:

w′= (q′_{1},...,q′_{n}) = (q_{1},...,q_{n})T.

Suppose e_{1},...,e_{n} is a basis for W. In the commutative case we define the associated matrix T by

te_{s} = ∑_{r} T_{rs}e_{r}.

Appendix A

Quaternions

The quaternions are the only non-commutative finite-dimensional skew-field over ℝ.

Definition A.1 An associative algebra A over a field k is defined by giving a vector space over k (which we also denote by A) together with a multiplication on A given by a map

A × A → A: (a,b) ↦ ab,

which is bilinear and associative, ie

1. for each a ∈ A the maps

x ↦ ax, x ↦ xa: A → A

are linear; and

2. for all a,b,c ∈ A

a(bc) = (ab)c.

The dimension of A is the dimension of the vector space over k.

The element 1 ∈ A is said to be a unity if

a · 1 = a = 1 · 1

for all a ∈ A.

By convention, the unadorned term ‘algebra’ is reserved for ‘associative algebra’. Non-associative algebras (such as Lie algebras and Jordan algebras) should be explicitly labelled as such.

If there is a unity 1 the algebra is said to be ‘unitary’ or ‘with 1’. Since we shall be concerned exclusively with associative algebras with 1, we shall use the term ‘algebra’ in this sense.

Examples:

1. The complex numbers ℂ form a 2-dimensional algebra over the reals ℝ.

2. The n × n matrices over a field k form an n^{2}-dimensional algebra over k.

If e_{1},...,e_{n} is a basis for the n-dimensional algebra A then the multiplication on A is completely determined by the n^{2} products e_{i}e_{j} since

(∑_{i} λ_{i}e_{i})(∑_{j} μ_{j}e_{j}) = ∑_{i,j} (λ_{i}μ_{j})e_{i}e_{j}.

By the same token, the multiplication is associative if the n^{3} identities

(e_{i}e_{j})e_{k} = e_{i}(e_{j}e_{k})

all hold. If 1 is one of the basis elements the identities involving 1 are trivially true, leaving (n − 1)^{3} identities to verify.

Definition A.2 The quaternion algebra ℍ is the 4-dimensional algebra over ℝ with basis 1,i,j,k and multiplication defined by

i^{2} = j^{2} = k^{2} = −1,
ij = k = −ji, jk = i = −kj, ki = j = −ik.

We should verify that this multiplication is associative. In principle this means verifying 3^{3} = 27 identities. However, the rules are evidently invariant under cyclic permutation of i,j,k. Thus it is sufficient to consider the 9 triple products with i in first place, ie

i^{3},i^{2}j,i^{2}k,iji,ij^{2},ijk,iki,ikj,ik^{2}.

For example,

(ij)k = k^{2} = −1, i(jk) = i^{2} = −1.

We leave the other 8 identities to the reader.

The letter ℍ (for Hamilton) is used for the quaternions because ℚ is already in use for the rationals.

Definition A.3 Each quaternion

q = t + xi + yj + zk

has conjugate

q̄= t − xi − yj − zk.

Proposition A.1 If

q = t + xi + yj + zk

then

qq̄= q̄q = t^{2}+ x^{2}+ y^{2}+ z^{2}.

Definition A.4 The norm q ∈ ℍ is the positive square root

|q| = \sqrt{qq̄}

Evidently

|q| = 0 ⟺ q = 0.

Proposition A.2 Every quaternion q ≠ 0 has an inverse,

Proof ▶ Let

q^{∗} = q̄/|q|^{2}.

Then

qq^{∗} = 1 = q^{∗}q;

and it follows easily that q^{∗} is the only quaternion with this property, since

qr = 1 ⟹ q^{∗}(qr) = q^{a}st ⟹ (q^{∗}q)r = q^{∗} ⟹ r = q^{∗}.

◀

Definition A.5 The algebra A is said to be a division-algebra or skew-field if each non-zero a ∈ A has an inverse b:

ab = 1 = ba.

We use the terms ‘division algebra’ and ‘skew field’ interchangeably. There is an almost universal convention that the unadorned word ‘field’ means commutative field. On the other hand, the term skew-field or division-algebra includes commutative field, unless that is explicitly excluded.

Proposition A.3 The only finite-dimensional division-algebras (or skew-fields) over ℝ are ℝ, ℂ and ℍ.

Proof ▶ Suppose A is a finite-dimensional skew-field over ℝ.

If A ≠ ℝ then we can find an element a ∈ A \ ℝ.

Suppose dimA = n. Then the elements 1,a,a^{2},...,a^{n} are linearly dependent over ℝ, ie a satisfies a polynomial equation

p(a) = 0,

where p(x) is a polynomial of degree ≤ n over ℝ.

Let m(x) be a polynomial satisfied by a of minimal degree. Note that in a division-algebra

ab = 0 ⟹ a = 0 or b = 0.

For if a ≠ 0 then we can multiply by its inverse:

ab = 0 ⟹ a^{−1} (ab) = 0 ⟹ (a^{−1} a)b = 0 ⟹ b = 0.

It follows that the minimal polynomial m(x) must be irreducible over ℝ, ie it cannot be expressed as the product m(x) = f(x)g(x) of two polynomials over ℝ of lower degree; for then f(a)g(a) = 0 and so f(a) = 0 or g(a) = 0.

But an irreducible polynomial m(x) over ℝ is necessarily of degree 1 or 2. For consider the factorisation of m(x) over ℂ, say

m(x) = (x − α_{1})··· (x − α_{r}).

Each root α_{i} is either real, or else one of a conjugate pair α,ᾱ of roots. In the first case (x − α) is a factor of m(x), and so is m(x) itself; in the second case the real quadratic (x − α)(x − ᾱ) is a factor of m(x), and so again is m(x) itself.

If m(x) is linear then a ∈ ℝ, which we have excluded. Hence each element a ∈ A \ ℝ satisfies a quadratic equation without real roots.

Suppose a has minimal equation

x^{2}+ 2Bx + C = 0 (B,C ∈ ℝ),

where B^{2}− C < 0, say

B^{2}− C = −D^{2}.

Then

(a + B)^{2} = −D^{2},

ie

( \frac{a + B}{D} ) ^{2} = −1.

Setting

i = \frac{a + B}{D}

we see that

i^{2} = −1,

and A has a subalgebra ⟨1,i⟩ > isomorphic to ℂ.

Two observations: Firstly, this subalgebra is not in general unique; A may have many subalgebras isomorphic to ℂ. Secondly, A is not in general an algebra over ℂ; for if z ∈ ℂ, a ∈ A then in general

az ≠ za,

ie

a(z · 1) ≠ z(a · 1).

Lemma A.1 The only finite-dimensional commutative algebras over ℝ are ℝ itself and ℂ.

Proof of Lemma ▷ If A ≠ ℝ then A has a subalgebra which we can identify with

ℂ. Since A is commutative, it can be regarded as an algebra over ℂ. Then each a ∈ A satisfies a polynomial equation m(a) = 0 where m(x) is irreducible over ℂ. But an irreducible polynomial over ℂ is necessarily linear. Hence a ∈ ℂ, and so A = ℂ. ◁

Returning to the general case, we have seen that if A ≠ ℝ then A has a subalgebra which we can identify with ℂ. If A = ℂ we are done. If not, let a ∈ A \ ℂ.

Now consider the map α: A → A under which

x ↦ ixi^{−1} = −ixi.

It is easy to see that

α^{2} = 1.

It follows that A is the direct sum of the 1- and −1-eigenspaces of A:

A = E_{1}⊕ E_{−1} .

(Explicitly,

x = \frac{1}{2}(x + αx) + \frac{1}{2}(x − αx)

for each x ∈ A.)

We have

E_{1} = {x ∈ A: xi = ix}, E_{−1} = {x ∈ A: xi = −ix},

It follows easily that

E_{1}E_{1} ⊂ E_{1}, E_{1}E_{−1} ⊂ E_{−1} , E_{−1} E_{1} ⊂ E_{−1} , E_{−1} E_{−1} ⊂ E_{1}.

In particular E_{1} is a subalgebra of A, which evidently contains ℂ. We claim that E_{1} = ℂ. For suppose b ∈ E_{1}\ ℂ. Consider the sub-algebra B generated by i and a. (Since b satisfies a quadratic equation over ℝ, B consists of the elements z + wb, where z,w ∈ ℂ.) Evidently B is commutative, since i,a commute.

Now every subalgebra S ⊂ A is in fact a division-algebra. For suppose s ∈ S has minimal equation

m(x) = x^{r}+ c_{1}x^{r−1} + ··· c_{r}

over ℝ. (We know of course that the degree r of m(x) is either 1 or 2, but that is not relevant here.) Then c_{r} ≠ 0 since m(x) is irreducible, and so

b^{−1} = −c_{r}^{−1} (b^{r−1} + c_{1}b^{r−2} + ··· + c_{r−1} ).

Thus b^{−1} is expressible as a polynomial in b, and so belongs to S.

In particular B is a commutative division-algebra, and so by the Lemma B = ℂ. Hence E_{1} = ℂ; the only elements in A that commute with i are the elements of ℂ.

It follows that if A ≠ ℂ then E_{−1} ≠ 0. Let a ∈ E_{−1} ,a ≠ 0.

If b ∈ E_{−1} then ib = −bi, and so

zb = bz̄

for all zinℂ. In particular, if z ∈ ℂ then

z ∈ ℝ ⟺ az = za.

Consider the element a^{2}. We know that a ∈ E_{1} = ℂ. But

a(a^{2}) = (a^{2})a.

It follows that a^{2} ∈ ℝ.

Furthermore a^{2} < 0; for if a^{2} = λ^{2} (with λ ∈ ℝ) then a = ±λ, and in particular a ∈ ℝ, contrary to hypothesis.

Hence

a^{2} = −λ^{2}

for some λ ∈ ℝ. Let

j = a/λ.

A.1. LINEAR ALGEBRA OVER ℍ  GpReps-I–6

Then

j^{2} = −1, ji = −ij.

Set

k = ij.

Then — we leave the verification to the reader —

k^{2} = −1, ki = −ik = j, jk = −kj = i.

Thus A is isomorphic to ℍ. ◀

A.1 Linear algebra over ℍ

It is perhaps surprising that the basic concepts of linear algebra all go over to vector spaces over ℍ.

A vector space V over ℍ is defined in the usual way, by giving an additive group V together with a scalar multiplication

ℍ × V → V : (q,v) ↦ qv,

such that

Appendix A

My favourite group — A_{5}

Recall that S_{n} is the group of permutations of n objects, and A_{n} is the subgroup formed by the even permutations. Just half the permutations are even (if n ≥ 2); for if g_{0} is odd then the odd permutations are those of the form gg_{0} with g even.

In particular,

∥A_{5}∥= 5!/2 = 60.

A.1 Dodecahedron and icosahedron

Each of the 5 regular solids S has a dual S̄whose vertices are the centres of the faces of S. Thus the dual of the cube is the octahedron, and the dual of the dodecahedron is the icosahedron, while the regular tetrahedron is self-dual.

Evidentl S and S̄have the same symmetry group.

The symmetry group of the regular tetrahedron is S_{4}, since every permutation of its vertices can be effected by a symmetry. Just half of these symmetries are proper. (For if g_{0} is an improper symmetry, then the improper symmetries are those of the form gg_{0} with g proper.) It follows that the proper symmetry group is A_{4}, since this is the only subgroup of S_{4} of index 2.

It is easy to identify the proper symmetries of the regular tetrahedron ABCD:

• the identity element I;

• 4 rotations through angle 2π/3 about the axes OA,OB,OC,OD (where O is the centroid of the tetrahedron);

• 4 rotations through angle −2π/3 about the same axes;

• 3 rotations through angle π about the axes joining mid-points of opposite edges.

A.1. DODECAHEDRON AND ICOSAHEDRON  GpReps-I–1

Each of these is a conjugacy class in the proper symmetry group; For example, if g is the rotation about OA through 2π/3 then xgx^{−1} is the rotation through the same angle about OX, where X = x(A).

It is no coincidence that the proper symmetries are all rotations; every proper symmetry of a bounded figure F in 3 dimensions is a rotation about some axis through the centroid O of F.

To see this, note that every isometry in n dimensions is expressible as the product of at most n + 1 reflections in hyperplanes. Very briefly, suppose the isometry α takes the simplex P_{0}P_{1}...P_{n} into Q_{0}Q_{1}...Q_{n}. Then we can take P_{0},...,P_{n} into Q_{0},...,Q_{n} by successive reflections. Thus if P_{0} ≠ Q_{0} let ρ_{0} be the reflection in the perpendicular bisector of P_{0}Q_{0}. (If P_{0} = Q_{0} let ρ_{0} = 1.) Suppose P′_{1}= ρ_{0}(P_{1}). If P′_{1} ≠ Q_{1} let ρ_{1} be the reflection in the perpendicular bisector of P′_{1}Q_{1}. The essential point is that this reflection leaves Q_{0} fixed; for since α and ρ_{0} are isometries,

d(Q_{0},Q_{1}) = d(P_{0},P_{1}) = d(Q_{0},P′_{1}),

ie Q_{0} is equidistant from P′_{1},Q_{1}, and so lies on their perpendicular bisector.

By the same argument, once P_{0},...,P_{r−1} have been brought to Q_{0},...,Q_{r−1} (by at most r reflections) the next reflection bringing P_{r+1} to its final position Q_{r+1} will leave the first r points fixed.

The original simplex is taken into its final position by a sequence of at most n + 1 reflections; and it is easy to see that the isometry α. is the product of these reflections.

Two further observations. Firstly, if the isometry α sends a point O into itself then we can take this as the first point of the simplex; so in this case α is expressible as the product of at most n reflections, all in hyperplanes through O.

Secondly, reflection in a hyperplane is improper. It follows that a proper isometry can only be the product of an even number of reflections.

Putting these ideas together, we deduce that a proper isometry in 3 dimensions which leaves a point O fixed is the product of 2 reflections in planes through O, ie a rotation about an axis through O.

Turning to the symmetry group G of the cube (and the octahedron), consider the action of G on the set of 6 faces. Let F = ABCD be one face. By Lagrange’s theorem,

∥G∥= ∥O(F)∥· ∥S(F)∥,

where O(F) is the orbit of F and S(F) is its stabilizer subgroup. The orbit consists of all 6 faces, while the stabilizer is the symmetry group of the square ABCD, since every symmetry of the square extends to a symmetry of the cube. Hence

S(F) = D_{4},

and so

∥G∥= 6 · 8 = 48.

Thus the proper symmetry group P of the cube has order 24. It is easy to identify these symmetries, knowing that they are all rotations:

• the identity element I;

• 8 rotations through angle ±2π/3 about the 4 diagonals of the cube;

• 6 rotations through angle ±π/2 about the axes joining the centres of opposite faces;

• 3 rotations through angle π about the same axes;

• 6 rotations through angle π about the axes joining mid-points of opposite edges.

Each of these is a conjugacy class in P. (Note that rotation through −2π/3 about the diagonal AC′, for example, is the same as rotation through 2π/3 about C′A.)

We can identify P by considering its action on the set of 4 diagonals of the cube. This gives a homomorphism

θ: P → S_{4}.

It is easy to see that kerθ = 1, so that θ is injective. Since ∥P∥= ∥S_{4}∥= 24, it is bijective, ie

P = S_{4}.

It is left to the reader to identify the conjugacy classes of P described above with the conjugacy classes of S_{4} defined by permutation type.

Reflection J in the centre of the cube is improper, and J^{2} = I. Also J ∈ Z(G); for if X is any vertex of the cube, then X and JX are diametrically opposite. Hence so are gX and gJX for any symmetry g, since g preserves distances. Thus gJX = JgX; and since this is true for all vertices X,

gJ = Jg.

It follows that

G= P × {I,J}
= S_{4}× C_{2}.

A.2. SIMPLICITY OF A_{N}  GpReps-I–3

Turning to the dodecahedron (and icosahedron), let us (as before) denote the symmetry group by G and the proper symmetry group by P. By considering the action of G on the 12 faces we deduce that

∥G∥= 12 · 10 = 120,

and so

∥P∥= 60.

Again, it is easy to identify these symmetries, knowing that they are all rotations:

• the identity element I;

• 6·2 = 12 rotations through angle ±2π/5 about the 6 axes joining the centres of opposite faces;

• 12 rotations through angle ±4π/5 about the same axes;

• 15 rotations through angle π about the axes joining mid-points of opposite edges;

• 10 ·2 = 20 rotations through angle ±2π/3 about the 10 diagonals (joining opposite vertices).

Each of these is a conjugacy class in P. It is easy to deduce from this that P is simple. For any normal subgroup N ◁ P must be a union of classes, and must include the class {I}. Thus ∥N∥ must be a sum of numbers chosen from {12,12,15,20}, plus 1. But no such sum divides 60 (except for 1 and 60 itself).

A.2 Simplicity of A_{n}

Proposition A.1 A_{n} is generated by 3-cycles.

Proof ▶ We argue by induction on n. Suppose

g = (a_{1}...a_{r})(b_{1}...b_{s})···∈ A_{n}.

We may suppose by the inductive hypothesis that all n elements occur in these cycles.

Case 1: There are more than 2 cycles in g. In this case we can divide these cycles into two parts, each of which is even, and so is expressible as a product of 3-cycles by the inductive hypothesis.

A.3. A_{5} IS THE SMALLEST NON-ABELIAN SIMPLE GROUP GpReps-I–4

Case 2: There are 2 cycles in g of odd length. In this case each cycle is expressible as a product of 3-cycles.

Case 3: ¿ ¿ David Ames ¿ ¿ David Ames

◀

A.3 A_{5} is the smallest non-abelian simple group

To round off this diversion, let us show that there is no non-abelian group with order < 60.

We know of course that a group of prime order is cyclic. This, together with a couple of Lemmas, will eliminate most cases.

Lemma A.1 A group of order pq, where p,q are distinct primes, cannot be simple

Proof of Lemma ▷ Suppose p > q. Then

n_{p} ≡ 1 modp,n_{p} | q ⟹ n_{p} = 1.

Thus there is one Sylow p-subgroup, which must be normal. ◁

Lemma A.2 A finite p-group has non-trivial centre.

Proof of Lemma ▷ Suppose

∥G∥= p^{e}.

The element g ∈ G lies in a conjugacy class by itself if and only if g ∈ Z(G), since

xgx^{−1} = g ⟺ xg = gx.

On the other hand, the number of elements in a conjugacy class divides ∥G∥, since

∥[g]∥· ∥S(g)∥= ∥G∥.

In this case, therefore, each class contains p^{f} elements for some f < e. In particular, the number of elements in each class [g] is divisible by p, unless g ∈ Z(G). It follows that the number of elements in Z(G) is divisible by p, and so Z(G) ≠ 1.

◁

In particular, the only simple p-group is C_{p}.

Although we shall not use this, one consequence of the last Lemma is of some interest.

Corollary A.1 A finite p-group contains subgroups of all poossible orders; that is, if ∥G∥= p^{e} then G contains subgroups of order p^{f} for 0 ≤ f ≤ e.

Proof of Lemma ▷ Any subgroup of Z(G) is normal in G. Take an element of

order p in Z(G). Then ⟨g⟩= C_{p} ◁ G. Consider G/⟨g⟩. Any normal subgroup of this is of the form N/⟨g⟩where N ◁ G. Thus the result follows by induction on e.

◁

Now let us go through the possible orders from 1–59, omitting those excluded by the above lemmas.

12 We have

n_{3} ≡ 1 mod3, n_{3} | 4.

Hence

n_{3} = 4.

The 4 Sylow 3-subgroups contain 4·2 = 8 elements of order 3, leaving only 4 elements for the Sylow 2-subgroups, each containing 4 elements. Hence there is only one such subgroup, which must therefore be normal.

18 We have

n_{3} ≡ 1 mod3, n_{3} | 2 ⟹ n_{3} = 1.

Thus the Sylow 3-subgroup is normal.

(Alternatively, a subgroup of index 2 is necessarily normal.)

20 We have

n_{5} ≡ 1 mod5, n_{5} | 4 ⟹ n_{5} = 1.

24 We have

n_{3} ≡ 1 mod3, n_{3} | 8 ⟹ n_{3} = 4.

The action of G on the 4 Sylow 3-subgroups gives a homomorphism

θ: G → S_{4},

which must be injective if G is simple. Hence

G = S_{4},

which is not simple.

28 We have

n_{7} ≡ 1 mod7, n_{7} | 4 ⟹ n_{7} = 1.

30 We have

n_{3} ≡ 1 mod3, n_{3} | 10 ⟹ n_{3} = 10.

Thus the Sylow 3-subgroups contain 10·2 = 20 elements of order 3. At the same time,

n_{5} ≡ 1 mod5, n_{5} | 6 ⟹ n_{5} = 6.

Thus the Sylow 5-subgroups contain 6·4 = 24 elements of order 5. We have too many elements!

36 We have

n_{3} ≡ 1 mod3, n_{3} | 4 ⟹ n_{3} = 4.

Thus we have a homomorphism

θ: G → S_{4},

which must be injective if G is simple. But that is a contradiction since

∥G∥> ∥S_{4}∥.

40 We have

n_{5} ≡ 1 mod5, n_{5} | 8 ⟹ n_{5} = 1.

42 We have

n_{7} ≡ 1 mod7, n_{7} | 6 ⟹ n_{7} = 1.

44 We have

n_{11} ≡ 1 mod11, n_{11} | 4 ⟹ n_{11} = 1.

45 We have

n_{3} ≡ 1 mod3, n_{3} | 5 ⟹ n_{3} = 1.

48 We have

n_{3} ≡ 1 mod3, n_{3} | 16 ⟹ n_{3} = 4 or 16.

If n_{3} = 4 then we have a homomorphism

θ: G → S_{4},

which must be injective if G is simple. But that is impossible since ∥G∥> ∥S_{4}∥.

If n_{3} = 16 then the 16 Sylow 3-subgroups contain 16 · 2 = 32 elements of order 3. This leaves only 16 elements for the Sylow 2-subgroups, each of which contains 16 elements. Hence there is only one Sylow 2-subgroup, which is therefore normal.

50 We have

n_{5} ≡ 1 mod5, n_{5} | 2 ⟹ n_{5} = 1.

52 We have

n_{13} ≡ 1 mod13, n_{13} | 4 ⟹ n_{13} = 1.

54 We have

n_{3} ≡ 1 mod3, n_{3} | 2 ⟹ n_{3} = 1.

56 We have

n_{7} ≡ 1 mod7, n_{7} | 8 ⟹ n_{7} = 8.

The 8 Sylow 7-subgroups contain 8 · 6 = 48 elements of order 7, leaving just 8 elements. These are enough for just one Sylow 2-subgroup, which is therefore normal.

Appendix D

The semisimple and nilpotent parts of a matrix

D.1 Semisimple matrices

Definition D.1 An n × n-matrix A over a field k is said to be semisimple if it satisfies a separable polynomial equation m(x) = 0, ie an equation with distinct roots.

Recall that a necessary and sufficient condition for m(x) to be separable is that

gcd(m(x),m′(x)) = 1,

where m′(x) is the derivative of m(x).

Clearly any factor of a separable polynomial is separable. The following result follows at once from this.

Proposition D.1 The matrix A is semisimple if and only if its minimal polynomial is separable.

Proposition D.2 The matrix A over k is semisimple if and only if it is diagonalisable over the algebraic closure k̄of k.

Proof ▶ If a matrix over k is semisimple then it will remain semisimple over any extension K of k. For the minimal polynomial over K will be a factor of the minimal polynomail over k.

It is sufficient therefore to consider matrices over an algebraically closed field K = k̄; and we already saw (in Chapter 2) that such a matrix is diagonalisable.

◀

D.2. NILPOTENT MATRICES  GpReps-I–2

D.2 Nilpotent matrices

Definition D.2 An n × n-matrix N is said to be nilpotent if

N^{r} = 0

for some r ≥ 1.

Proposition D.3 The matrix A is nilpotent if and only if

trA^{r} = 0

for all r ≥ 1.

D.3 The canonical splitting

Theorem D.1 Suppose A is an n × n-matrix over a field k of characteristic 0. Then A is uniquely expressible in the form

A = S + N,

where S is semisimple, N is nilpotent, and NS = SN. Moreover, S and N are expressible as polynomials in A.

Proof ▶ We show first that there are polynomials s(x),n(x) ∈ k[x] such that S = s(A) is semisimple, N = n(A) is nilpotent, and A = S + N.

Suppose m(x) is the minimal polynomial of A. We know that there is unique factorisation of polynomials over k, for any field k. Let

m(x) = p_{1}(x)^{e_{1}}··· p_{r}(x)^{e_{r}},

where p_{1}(x),...,p_{r}(x) are distinct irreducible polynomials.

Set

p(x) = p_{1}(x)··· p_{r}(x).

(In other words, each irreducible factor of m(x) is taken just once in p(x).) Evidently p(x) is separable; and

p(x) | m(x) | p(x)^{e},

where e = maxe_{i}.

In particular,

p(A)^{e} = 0,

D.3. THE CANONICAL SPLITTING  GpReps-I–3

ie p(A) is nilpotent. Conversely, suppose f(A) is nilpotent for some polynomial f(x) ∈ k[x]. Then

f(A)^{r} = 0 ⟹ m(x) | f(x)^{r} ⟹ p(x) | f(x).

Thus f(A) is nilpotent if and only if f(x) is a multiple of p(x). In particular,

n(x) = p(x)r(x)

for some polynomial r(x).

Suppose

B = A + N,

where NA = AN; and suppose the minimal polynomials of A,B are m(x),m_{1}(x). Then

m(B) = m(A + N) = m(A) + Nf(A,N),

since all terms in the expansion except those in m(A) will contain N as a factor. But m(A) = 0. Hence

m(B) = Nf(A,N);

and so

m(B)^{e} = 0

for some e. It follows that

m(x) | m_{1}(x)^{e};

and similarly

m_{1}(x) | m(x)^{f} .

In other words, m(x) and m_{1}(x) have the same separable part.

In particular, if A = S + N as in the theorem then S must have minimal polynomial p(x). Thus

p(S) = p(A − N) = 0.

In other words,

p(x − n(x)) ≡ 0 modm(x).

But as we have seem, n(x) = p(x)r(x). Thus we are looking for a polynomial r(x) such that

p(x − p(x)r(x)) ≡ 0 modm(x).

We shall show that we can satisfy the stronger congruence

p(x − p(x)r(x)) ≡ 0 modp(x)^{e},

by succesively finding solutions to higher powers of p(x).

Evidently r(x) = 0 is a solution for e = 1. Suppose we have a solution r_{f} (x) for e = f, ie

p(x − p(x)r_{f} (x)) = p(x)^{f} u(x).

Set

r(x) = r_{f} (x) + p(x)^{f−1} t(x).

Then we have to solve

p(x − p(x)r_{f} (x) − p(x)^{f} t(x)) ≡ 0 modp(x)^{f+1} ,

ie

p(x)^{f} u(x) − p′(x)p(x)^{f} t(x) modp(x)^{f+1} ,

ie

u(x) ≡ p′(x)t(x) modp(x).

But since gcd(p(x),p′(x)) = 1 we can find polynomials v(x),w(x) such that

v(x)p(x) + w(x)p′(x) = 1;

and then

t(x) = −u(x)w(x)

will satisfy our congruence.

Accordingly, we have found polynomials n(x) = p(x)r(x) and s(x) = x − n(x) such that S = s(A) is semisimple, N = n(A) is nilpotent, and A = S + N.

To prove uniqueness, suppose we had a second solution

A = S′+ N′,

where N′S′= S′N′. Then

S′= S + N′′,

where N′′= N − N′.

Now S′ and N′ commute with A = S′+ N′; and so they both commute with S = s(A) and N = n(A). In particular, N and N′commute. Hence N′′= N − N′is nilpotent; for if N^{e} = 0 and N^{′e′}= 0 then

N^{′′e+e′}= 0.

From above, S′has the same minimal polynomial p(x) as S. Thus

p(S + N′′) = p′(S)N′′+ N^{′′2}f(S,N′′) = 0.

D.4. THE RESULT OVER FINITE FIELDS  GpReps-I–5

But p′(S) — again from above — is invertible. For since gcd(p(x),p′(x)) = 1 we can find polynomials v(x),w(x) such that

v(x)p(x) + w(x)p′(x) = 1;

and then

w(S)p′(S) = I.

Accordingly,

N′′= N^{′′2}g(S,N),

and then N′′is a multiple of N^{′′4}, of N^{′′8}, etc. Hence

N′′= 0,

ie N = N′and S = S′. ◀

D.4 The result over finite fields

What goes wrong with the above argument if k is of finite characteristic? There is only one point where the argument fails; the irreducible factors p_{i}(x) of m(x) may not be separable.

For example, if k is a field of characteristic p, and a ∈ k does not have a pth root in k, then

f(x) = x^{p}− a

is irreducible, but it is not separable since it factorises into

f(x) = (x − b)^{p}

in an extension field of k.

Actually, the result still holds if k is perfect, ie if every element a ∈ k has a pth root in k. For in that case an irreducible polynomial f(x) over k is necessarily separable.

To see this, note that f′(x) cannot vanish identically; for suppose

f(x) = x^{n}+ a_{1}x^{n−1} + ··· + a_{n−1} x + a_{n}.

Then

f′(x) = nx^{n}+ (n − 1)a_{1}x^{n−1} + ··· + a_{n−1} ;

and this can only vanish identically if the only non-xero terms are those involving powers x^{pr}, ie

f(x) = x^{mp}+ a_{p}x^{(m−1)p} + ··· + a_{0}.

But then, since k is perfect, there are elements b_{1},b_{2},···∈ k such that

a_{p} = b_{1}^{p},a_{2p} = b_{2}^{p},...,a_{mp}= b_{m}^{p};

and so

f(x) = g(x)^{p},

where

g(x) = x^{m}+ b_{1}x^{m−1} + ··· + b_{m}.

This is particularly significant, since all finite fields are perfect. of perfect fields are the finite fields 𝔽_{p^{n}}. For if k is a field of characteristic p the map

x ↦ x^{p} : k → k

is a ring-homomorphism, since

(x + y)^{p} = x^{p}+ y^{p}, (xy)^{p} = x^{p}y^{p};

and this homomorphism is injective since x^{p} = 0 ⟹ x = 0. Thus if the field is finite, the homomorphism is surjective, ie every element is a pth power.

Thus our result still holds over the finite field k = F_{q} (where q is a prime-power, q = p^{n}): Every n × n-matrix A over 𝔽_{q} is uniquely expressible in the form

A = S + N,

with S semisimple, N nilpotent, and NS = SN. Moreover, S and N are expressible as polynomials in A.