Course 424

Group Representations II

Dr Timothy Murphy

EELT3  Tuesday, 13 April 1999  16:00–17:30

Answer as many questions as you can; all carry the same number of marks. All representations are finite-dimensional over ℂ.

1. What is meant by a measure on a compact space X? What is meant by saying that a measure on a compact group G is invariant? Sketch the proof that every compact group G carries such a measure. To what extent is this measure unique?

Answer: A measure μ on X is a continuous linear functional

μ : C(X) → ℂ,

where C(X) = C(X, ℝ) is the space of real-valued continuous functions on X with norm ∥f∥ = sup |f(x)|.

The compact group G acts on C(G) by

(gf)(x) = f(g^{−1}x).

The measure μ is said to be invariant under G if

μ(gf)μ(f)

for all g ∈ G, f ∈ C(G).

By an average F of f ∈ C(G) we mean a function of the form

F = λ_{1}g_{1}f + λ_{2}g_{2}f + ⋯ + λ_{r}g_{r}f,

where 0 ≤ λ_{i} ≤ 1, ∑ λ_{i} = 1 and g_{1}, g_{2}, …, g_{r} ∈ G.

If F is an average of f then

(a) inf f ≤ inf F ≤ sup F ≤ supf;

(b) If μ is an invariant measure then μ(F) = μ(f);

(c) An average of F is an average of f.

Continued overleaf If we set

var(f) = sup f − inf f

then

var(F) ≤ var(f)

for any average F of f. We shall establish a sequence of averages F_{0} = f, F_{1}, F_{2}, … (each an average of its predecessor) such that var(F_{i}) → 0. It follows that

F_{i} → c ∈ ℝ,

ie F_{i}(g) → c for each g ∈ G.

Suppose f ∈ C(G). It is not hard to find an average F of f with var(F) < var(f). Let

V = \{g ∈ G : f(g) < \frac{1}{2}(sup f + inf f),

ie V is the set of points where f is ‘below average’. Since G is compact, we can find g_{1}, …, g_{r} such that

G = g_{1}V ∪ ⋯ ∪ g_{r}V.

Consider the average

F = \frac{1}{r} (g_{1}f + ⋯ + g_{r}f) .

Suppose x ∈ G. Then x ∈ g_{i}V for some i, ie

g_{i}^{−1}x ∈ V.

Hence

(g_{i}f)(x) < \frac{1}{2}(sup f + inf f),

and so

F (x) < \frac{r − 1}{r} sup f + \frac{1}{2r}(sup f + inf f)
= sup f − \frac{1}{2r}sup f − inf f.

Hence sup F < supf and so

var(F) < var(f).

This allows us to construct a sequence of averages F_{0} = f, F_{1}, F_{2}, … such that

var(f) = var(F)_{0} > var(F)_{1} > var(F)_{2} > · · · .

But that is not sufficient to show that var(F)_{i} → 0. For that we must use the fact that any f ∈ C(G) is uniformly continuous.

[I would accept this last remark as sufficient in the exam, and would not insist on the detailed argument that follows.]

In other words, given ϵ > 0 we can find an open set U ∋ e such that

x^{−1}y ∈ U ⟹ |f(x) − f(y)| < ϵ.

Since

(g^{−1}x)^{−1}(g^{−1}y) = x^{−1}y,

the same result also holds for the function gf. Hence the result holds for any average F of f.

Let V be an open neighbourhood of e such that

V V ⊂ U, V^{−1} = V.

(If V satisfies the first condition, then V ∩V^{−1} satisfies both conditions.) Then

xV ∪ yV ≠ ∅ ⟹ |f(x) − f(y)| < ϵ.

For if xv = yv′ then

x^{−1}y = vv′^{−1} ∈ U.

Since G is compact we can find g_{1}, …, g_{r} such that

G = g_{1}V ∪ ⋯ ∪ g_{r}V.

Suppose f attains its minimum inf f at x_{0} ∈ g_{i}V; and suppose x ∈ g_{j}V. Then

g_{i}^{−1}x_{0}, g_{j}^{−1}x ∈ V.

Hence

(g_{j}^{−1}x)^{−1} (g_{i}^{−1}x_{0}) = (g_{i}g_{j}^{−1}x)^{−1} x_{0} ∈ U,

and so

|f(g_{i}g_{j}^{−1}x) − f(x_{0})| < ϵ.

In particular,

(g_{j}g_{i}^{−1}f)(x) < inf f + ϵ.

Let F be the average

F = \frac{1}{r^{2}} ∑_{i,j} g_{j}g_{i}^{−1}f.

Then

sup F < \frac{r^{2} − 1}{r^{2}} sup f + \frac{1}{r^{2}} (inf f + ϵ),

and so

var(F) < \frac{r^{2} − 1}{r^{2}} var(f) + \frac{1}{r^{2}} ϵ

< \frac{r^{2} − 1/2}{r^{2}} var(f),

if ϵ < var(f)/2.

Moreover this result also holds for any average of f in place of f. It follows that a succession of averages of this kind

F_{0} = f, F_{1}, …, F_{s}

will bring us to

var(F)_{s} < \frac{1}{2} var(f).

Now repeating the same argument with F_{s}, and so on, we will obtain a sequence of successive averages F_{0} = f, F_{1}, … with

var(F)_{i} ↓ 0.

It follows that

F_{i} → c

(the constant function with value c).

It remains to show that this limit value c is unique. For this we introduce right averages

H(x) = ∑_{j} μ_{j}f(xh_{j})

where 0 ≤ μ_{j} ≤ 1, ∑ μ_{j} = 1. (Note that a right average of f is in effect a left average of f̃, where f̃(x) = f(x^{−1}). In particular the results we have established for left averages will hold equally well for right averages.)

Given a left average and a right average of f, say

F (x) = ∑ λ_{i}f(g_{i}^{−1}x), H(x) = ∑ μ_{j}f(xh_{j}),

we can form the joint average

J(x) = ∑_{i,j} λ_{i}μ_{j}f(g_{i}^{−1}xh_{j}).

It is easy to see that

inf F ≤ inf J ≤ sup J ≤ sup H,
sup F ≥ sup J ≥ inf J ≥ inf H.

But if now H_{0} = f, H_{1}, … is a succession of right averages with H_{i} → d then it follows that

c = d.

In particular, any two convergent sequences of successive left averages must tend to the same limit. We can therefore set

μ(f) = c.

Thus μ(f) is well-defined; and it is invariant since f and gf have the same set of averages. Finally, if f = 1 then var(f) = 0, and f, f, f, … converges to 1, so that

μ(1) = 1.

The invariant measure on G is unique up to a scalar multiple. In other words, it is unique if we normalise the measure by specifying that

μ(1) = 1

(where 1 on the left denotes the constant function 1).

2. Prove that every simple representation of a compact abelian group is 1-dimensional and unitary.

Determine the simple representations of SO(2).

Determine also the simple representations of O(2).

Answer: Suppose α is a simple representation of the compact abelian group G in V.

Suppose g ∈ G. Let λ be an eigenvalue of g, and let E = E_{λ} be the corresponding eigenspace. We claim that E is stable under G. For suppose h ∈ G. Then

e ∈ E ⟹ g(he) = h(ge) = λhe ⟹ he ∈ E.

Since α is simple, it follows that E = V, ie gv = λv for all v, or g = λI.

Since this is true for all g ∈ G, it follows that every subspace of V is stable under G. Since α is simple, this implies that dim V = 1, ie α is of degree 1.

Thus a simple representation of G is a homomorphism α : G → ℂ^{∗}. We must show that

|α(g)| = 1

for all g ∈ G.

If |α(g)| > 1 then

|α(g^{n})| = (|g|)^{n} → ∞.

This is a contradiction, since im α ⊂ ℂ^{∗} is compact and so bounded. On the other hand, if |α(g)| < 1 then |α(g^{−1})| > 1. Hence |α(g)| = 1 for all g, ie α is unitary.

We can identify SO(2) with

U(1) = {z ∈ ℂ : |z| = 1}.

From above, a representation of U(1) is a homomorphism

α : U(1) → U(1).

For each n ∈ ℤ the map

E(n) : z → z^{n}

defines such a homomorphism. We claim that every representation of U(1) is of this form.

3. Determine the conjugacy classes in SU(2); and prove that this group has just one simple representation of each dimension.

Find the character of the representation D(j) of dimensions 2j + 1 (where j = 0, \frac{1}{2}, 1, \frac{3}{2}, …).

Determine the representation-ring of SU(2), ie express each product D(i)D(j) as a sum of simple representations D(k).

Answer: We know that

(a) if U ∈ SU(2) then U has eigenvalues

e^{±iθ} (θ ∈ ℝ).

(b) if X, Y ∈ GL(n, k) then

X ∼ Y ⟹ X, Y have the same eigenvalues.

A fortiori, if U ∼ V ∈ SU(2) then U, V have the same eigenvalues.

We shall show that the converse of the last result is also true, that is: U ∼ V in SU(2) if and only if U, V have the same eigenvalues e^{±iθ}, This is equivalent to proving that

U ∼ U(θ) = (\begin{matrix}e^{iθ} & 0 \\ 0 & e^{−iθ} \end{matrix}) ,

ie we can find V ∈ SU(2) such that

V^{−1}UV = U(θ).

To see this, let v be an e^{iθ}-eigenvalue of U. Normalise v, so that v^{∗}v = 1; and let w be a unit vector orthogonal to v, ie w^{∗}w = 1, v^{∗}w = 0. Then the matrix

V = (vw) ∈ Mat(2, ℂ)

is unitary; and

V^{−1}UV = (\begin{matrix}e^{iθ} & x \\ 0 & e^{−iθ} \end{matrix})

But in a unitary matrix, the squares of the absolute values of each row and column sum to 1. It follows that

|e^{iθ}|^{2} + |x|^{2} = 1 ⟹ x = 0,

ie

V^{−1}UV = U(θ).

We only know that V ∈ U(2), not that V ∈ SU(2). However

V ∈ U(2) ⟹ | det V | = 1 ⟹ det V = e^{iϕ}.

Thus

V′ = e^{−iϕ/2}V ∈ SU(2)

and still

(V′)^{−1}UV = U(θ).

To summarise: Since U(−θ) ∼ U(θ) (by interchange of coordinates), we have show that if

C(θ) = {U ∈ SU(2) : U has eigenvalues e^{±iθ}}

then the conjugacy classes in SU(2) are

C(θ) (0 ≤ θ ≤ π).

Now suppose m ∈ ℕ, Let V (m) denote the space of homogeneous polynomials P (z, w) in z, w. Thus V (m) is a vector space over ℂ of dimension m + 1, with basis z^{m}, z^{m−1}w, …, w^{m}.

Suppose U ∈ SU(2). Then U acts on z, w by

(\begin{matrix}z \\ w \end{matrix}) ↦ (\begin{matrix}z′ \\ w′ \end{matrix}) = U (\begin{matrix}z \\ w \end{matrix}) .

This action in turn defines an action of SU(2) on V (m):

P (z, w) ↦ P (z′, w′).

We claim that the corresponding representation of SU(2) — which we denote by D_{m/2} — is simple, and that these are the only simple (finitedimensional) representations of SU(2) over ℂ.

To prove this, let

U(1) ⊂ SU(2)

be the subgroup formed by the diagonal matrices U(θ). The action of SU(2) on z, w restricts to the action

(z, w) ↦ (e^{iθ}z, e^{−iθ}w)

of U(1). Thus in the action of U(1) on V (m),

z^{m−r}w^{r} ↦ e^{(m−2r)iθ}z^{m−r}w^{r},

It follows that the restriction of D_{m/1} to U(1) is the representation

D_{m/2}|U(1) = E(m) + E(m − 2) + ⋯ + E(−m)

where E(m) is the representation

e^{iθ} ↦ e^{miθ}

of U(1).

In particular, the character of D_{m/2} is given by

χ_{m/2}(U) = e^{miθ} + e^{(m−2}iθ + ⋯ + e^{−miθ}

if U has eigenvalues e^{±iθ}.

Now suppose D_{m/2} is not simple, say

D_{m/2} = α + β.

(We know that D_{m/2} is semisimple, since SU(2) is compact.) Let a corresponding split of the representation space be

V (m) = W_{1} ⊕ W_{2}.

Since the simple parts of D_{m/2}|U(1) are distinct, the expression of V (m) as a direct sum of U(1)-spaces,

V (m) = ⟨z^{m}⟩ ⊕ ⟨z^{m−1}w⟩ ⊕ ⋯ ⊕ ⟨w^{m}⟩

is unique. It follows that W_{1} must be the direct sum of some of these spaces, and W_{2} the direct sum of the others. In particular z^{m} ∈ W_{1} or z^{n} ∈ W_{2}, say z^{m} ∈ W_{1}. Let

U = \frac{1}{\sqrt{2}} (\begin{matrix}1 & −1 \\ 1 & 1 \end{matrix}) ∈ SU(2).

Then

(\begin{matrix}z \\ w \end{matrix}) ↦ \frac{1}{\sqrt{2}} (\begin{matrix}z + w \\ −z + w \end{matrix})

under U. Hence

z^{m} ↦ 2^{−m/2}(z + w)^{m}.

Since this contains non-zero components in each subspace ⟨z^{m−r}w^{r}⟩, it follows that

W_{1} = V (m),

ie the representation D_{m/2} of SU(2) in V (m) is simple.

To see that every simple (finite-dimensional) representation of SU(2) is of this form, suppose α is such a representation. Consider its restriction to U(1). Suppose

α|U(1) = e_{r}E(r)+e_{r−1}E(r−1)+⋯+e_{−r}E(−r) (e_{r}, e_{r−1}, …, e_{−r} ∈ ℕ).

Then α has character

χ(U) = χ(θ) = e_{r}e^{riθ} + e_{r−1}e^{(r−1)iθ} + ⋯ + e_{−r}e^{−riθ}

if U has eigenvalues e^{±iθ}.

Since U(−θ) ∼ U(θ) it follows that

χ(−θ) = χ(θ),

and so

e_{−i} = e_{i},

ie

χ(θ) = e_{r}(e^{riθ} + e^{−riθ}) + e_{r−1}(e^{(r−1)iθ} + e^{−(r−1)iθ}) + · · · .

It is easy to see that this is expressible as a sum of the χ_{j}(θ) with integer (possibly negative) coefficients:

χ(θ) = a_{0}χ_{0}(θ) + a_{1/2}χ_{1/2}(θ) + ⋯ + a_{s}χ_{s}(θ) (a_{0}, a_{1/2}, …, a_{s} ∈ ℤ).

Using the intertwining number,

I(α, α) = a_{0}^{2} + a_{1/2}^{2} + ⋯ + a_{s}^{2}

(since I(D_{j}, D_{k}) = 0). Since α is simple,

I(α, α) = 1.

It follows that one of the coefficients a_{j} is ±1 and the rest are 0, ie

χ(θ) = ±χ_{j}(θ)

for some half-integer j. But

χ(θ) = −χ_{j}(θ) ⟹ I(α, D_{j}) = −I(D_{j}, D_{j}) = −1,

which is impossible. Hence

χ(θ) = χ_{j}(θ),

and so (since a representation is determined up to equivalence by its character)

α = D_{j}.

Finally, we show that

D_{j}D_{k} = D_{j+k} + D_{j+k−1} + ⋯ + D_{|j−k|}.

It is sufficient to prove the corresponding result for the characters

χ_{j}(θ)χ_{k}(θ) = χ_{j+k}(θ) + χ_{j+k−1}(θ) + ⋯ + χ_{|j−k|}(θ).

We may suppose that j ≥ k. We prove the result by induction on k. If k = 0 the result is trivial, since χ_{0}(θ) = 1. If k = 1/2 then

χ_{j}(θ)χ_{1/2}(θ) = (e^{2jiθ} + e^{2(j−1)iθ} + e^{−2jiθ}) (e^{iθ} + e^{−iθ})
= (e^{(2j+1)iθ} + e^{−(2j−1)iθ}) + (e^{(2j−1)iθ} + e^{−(2j+1)iθ})
= χ_{j+1/2}(θ) + χ_{j−1/2}(θ),

as required. Suppose k ≥ 1. Then

χ_{k}(θ) = χ_{k−1}(θ) + (e^{kiθ} + e^{−kiθ}).

Thus applying our inductive hypothesis,

χ_{j}(θ)χ_{k}(θ) = χ_{j+k−1}(θ) + ⋯ + χ_{j−k+1} + χ_{j}(θ)(e^{kiθ} + e^{−kiθ}).

But

χ_{j}(θ)(e^{kiθ} + e^{−kiθ}) = (e^{2jiθ} + e^{2(j−1)iθ} + e^{−2jiθ}) (e^{kiθ} + e^{−kiθ})
= χ_{j+k}(θ) + χj − k(θ),

giving the required result

χ_{j}(θ)χ_{k}(θ) = χ_{j+k−1}(θ) + ⋯ + χ_{j−k+1} + χ_{j+k}(θ) + χj − k(θ)
= χ_{j+k}(θ) + ⋯ + χ_{j−k}.

4. Show that there exists a surjective homomorphism

Θ : SU(2) → SO(3)

with finite kernel.

Hence or otherwise determine all simple representations of SO(3).

Determine also all simple representations of O(3).

Answer: The set of skew-hermitian 2 × 2 matrices

S = (\begin{matrix}ia & −b + ic \\ b + ic & id \end{matrix}) (a, b, c, d ∈ ℝ)

forms a 4-dimensional real vector space U. The group SU(2) acts on this space by

(U, S) ↦ U^{−1}SU = U^{∗}SU,

since

(U^{∗}SU)^{∗} = U^{∗}S^{∗}U = −U^{∗}SU.

The 3-dimensional subspace W ⊂ U formed by trace-free skew-hermitian matrices

T = (\begin{matrix}ix & −y + iz \\ y + iz & −ix \end{matrix}) (x, y, z ∈ ℝ)

is stable under SU(2) since

tr(U^{∗}T U) = tr(U^{−1}T U) = tr T = 0.

Thus W carries a representation of SU(2) of degree 3, corresponding to a homomorphism

Θ : SU(2) → GL(3, ℝ).

Moreover, this homomorphism preserves the positive-definite quadratic form

det T = x^{2} + y^{2} + z^{2}

on W since

det(U^{∗}T U) = det(U^{−1}T U) = det T.

Hence

im Θ ⊂ O(3).

Finally, SU(2) ≅ S^{3} is connected; and so therefore is its image. But SO(3) is an open subgroup of O(3). Hence

im Θ ⊂ SO(3).

Thus our homomorphism takes the form

Θ : SU(2) → SO(3).

It remains to show that Θ has a finite kernel, and is surjective. If

U ∈ ker Θ

then

U^{−1}T U = T

for all T ∈ W. Each S ∈ U can be expressed in the form

S = T + ρI,

where T ∈ W and ρ = tr S/2. It follows that

U^{−1}SU = S

for all skew-hermitian S ∈ U. Hence

U^{−1}HU = H

for all hermitian H, since H is hermitian if and only if S = iH is skew-hermitian. It follows from this that

U^{−1}XU = X

for all X ∈ Mat(2, ℂ), since every X is expressible in the form

X = H + S,

with H = (X + X^{∗})/2 hermitian and S = (X − X^{∗})/2 skew-hermitian. But it is a simple matter to see that the only such U are U = ±I. Thus

ker Θ = {±I}.

To see that Θ is surjective, we note that if

U(θ) = (\begin{matrix}e^{iθ} & 0 \\ 0 & e^{−iθ} \end{matrix})

then

U(θ)^{−1} (\begin{matrix}ix & −y + iz \\ y + iz & −ix \end{matrix}) U(θ) = (\begin{matrix}ix & e^{−2iθ}(−y + iz) \\ e^{2iθ}(y + iz) & −ix \end{matrix}) ,

ie

ΘU(θ) = (\begin{matrix}1 & 0 & 0 \\ 0 & cos 2θ & − sin 2θ \\ 0 & sin 2θ & cos 2θ \end{matrix}) = R(Ox, 2θ),

rotation about Ox through angle 2θ. In particular, im Θ contains all rotations about Ox.

Now let

U = \frac{1}{\sqrt{2}} (\begin{matrix}1 & −1 \\ 1 & 1 \end{matrix}) .

Then U ∈ SU(2) and

U^{−1} (\begin{matrix}ix & −y + iz \\ y + iz & −ix \end{matrix}) U = (\begin{matrix}iz & −y − ix \\ y − ix & −iz \end{matrix}) .

Thus

α(U) = (\begin{matrix}0 & 0 & 1 \\ 0 & 1 & 0 \\ −1 & 0 & 0 \end{matrix}) = R(Oy, π/2).

Writing P = R(Oy, π/2),

P^{−1}R(Ox, θ)P = R(Oz, θ).

Thus im Θ contains all rotations about Oz as well as Ox. But is is easy to see that every rotation R ∈ SO(3) is expressible as a product of rotations about Ox and Oz. Hence

im Θ = SO(3),

ie Θ is surjective.

Thus

SO(3) = SU(2)/{±i}.

It follows that the representations of SO(3) are just the representations α of SO(2) such that

α(−I) = I.

In particular, the simple representations of SO(3) are those simple representations D_{j} of SU(2) such that D_{j}(−I) = I. But D_{j} is defined by the action of SU(2) on the polynomials

P (z, w) = c_{0}z^{2j} + c_{1}z^{2j−1}w + ⋯ + c_{2j}w^{2j}.

It is clear that

P (−z, −w) = P (z, w)

for all P of degree 2j if and only if 2j is even, ie j is an integer.

Thus the simple representations of SO(3) are D_{0}, D_{1}, D_{2}, … of degrees 1, 3, 5, . . ..

Since

O(3) = SO(3) × C_{2},

where C_{2} = {±I}, the simple representations of O(3) are of the form α × β, where α is a simple representation of SO(3), and β is a simple representation of C_{2}. Thus the simple representations of O(3) are D_{j} × 1 and D_{j} ×ϵ, where j ∈ ℕ and ϵ is the representation −I → −1 of C_{2}.

5. Explain the division of simple representations of a finite or compact group G over ℂ into real, essentially complex and quaternionic. Give an example of each (justifying your answers).

Show that if α is a simple representation with character χ then the value of

∫_{G} χ(g^{2}) dg

determines which of these three types α falls into.

Answer: Suppose α is a simple representation of G over ℂ. Then α is said to be real if

α = β_{ℂ}

for some representation of G over ℝ. If this is so then the character

χ_{α}(g) = χ_{β}(g)

is real. We say that α is quaternionic if its character is real, but it is not real. Finally, we say that α is essentially complex if its character is not real.

The trivial character 1 of any group is real, since it is the complexification of the trivial character over ℝ.

The 1-dimensional character θ of the cyclic group C_{3} = ⟨g⟩ given by

θ : g ↦ ω = e^{2π/3}

is essentially complex, since its character θ is not real.

Consider the quaternion group

Q_{8} = {±1, ±i, ±j, ±k}.

We can regard the quaternions ℍ as a 2-dimensional vector space over ℂ = ⟨1, i⟩. The action of Q_{8} on ℍ by multiplication on the left defines a 2-dimensional representation α of D_{8}. We assert that this is a simple quaternionic representation.

It is certainly simple, since otherwise ℍ would have a 1-dimensional subspace ⟨q⟩ stable under D_{8}, and therefore under ℍ, since D_{8} spans ℍ. But that is impossible since

x = (xq^{−1})q

for any x ∈ ℍ. The simple representations of D_{4} must have dimensions 1, 1, 1, 1, 2 (since ∑ dim_{i}^{2} = 8). It follows that

α^{∗} = α

since there is only 1 2-dimensional simple representation. Hence χ_{α} is real.

It remains to show that α is not real. Consider the 4-dimensional representation β of D_{8} over ℝ, defined by the same action of D_{4} on ℍ. This is easily seen to be simple, by the argument above. It follows that β_{ℂ} is either simple, or splits into 2 simple representations over ℂ of dimension 2. The only possibility is that

β_{ℂ} = 2α.

Now if α were real, say

α = γ_{ℂ}

we would deduce that β = 2γ which is impossible, since β is simple.

Now suppose α is a simple representation of G in V. Then (α^{∗})^{2} is the representation arising from the action of G on the space of bilinear forms on V.

But

α^{∗} = α ⟺ χ_{α} is real.

Thus

I(1, (α^{∗})^{2}) = I(α, α^{∗}) = \begin{cases}1 & if α is real or quaternionic \\ 0 & if α is essentially complex \end{cases} .

In other words, there is just 1 invariant bilinear form (up to a scalar multiple) if α is real or quaternionic, and no such form if α is essentially complex.

Now the space of bilinear forms splits into the direct sum of symmetric (or quadratic) and skew-symmetric forms, since each bilinear form B(u, v) can be expressed as

B(u, v) = \frac{1}{2} (B(u, v) + B(v, u)) + \frac{1}{2} (B(u, v) − B(v, u)) ,

where the first form is symmetric and the second skew-symmetric.

It follows that

(α^{∗})^{2} = ϕ + ψ,

where ϕ is the representation of G in the space of symmetric forms, and ψ the representation in the space of skew-symmetric forms.

If α is essentially complex, there is no invariant symmetric or skewsymmetric form. But if α is real or quaternionic, there must be just 1 invariant form, either symmetric or skew-symmetric. We shall see that in fact there is an invariant symmetric form if and only if α is real.

Certainly if α is real, say α = β_{ℂ}, where β is a representation in the real vector space U, then we know that there is an invariant positivedefinite form on U, and this will give an invariant quadratic form on V = U_{ℂ}.

Conversely, suppose α is a quaternionic simple representation on V. Then β = α_{ℝ} is simple. For

(α_{ℝ})_{ℂ} = α + α^{∗}

for any representation α over ℂ. Thus if β = γ + γ′ then (with α quaternionic)

2α = γ_{ℂ} + γ′_{ℂ},

and it will follow that

α = γ_{ℂ} = γ′_{ℂ},

so that α is real.

Since β is simple, there is a unique invariant quadratic form P on V_{ℝ}, and this form is positive-definite. But if there were an invariant quadratic form Q on V this would give an invariant quadratic form on V_{ℝ}, which would not be positive-definite, since we would have

Q(iu, iu) = −Q(u).

Thus if α is quaternionic, then there is no invariant quadratic form on V, and therefore there is an invariant skew-symmetric form.

It follows that we can determine which class α falls into by computing

I(1, ϕ) and I(1, ψ).

To this end we compute the characters of ϕ and ψ.

Suppose g ∈ G. Then we can diagonalise g, ie we can find a basis e_{1}, …, e_{n} of V consisting of eivenvectors, say

ge_{i} = λ_{i}e_{i}.

The space of quadratic forms is spanned by the n(n + 1)/2 forms

x_{i}x_{j} (i ≤ j),

where x_{1}, …, x_{n} are the coordinates with respect to the basis e_{1}, …, e_{n}. It follows that

χ_{ϕ}(g) = ∑ λ_{i}^{2} + ∑_{i<j} λ_{i}λ_{j}.

Now

χ_{α}(g) = ∑ λ_{i}, χ_{α}(g^{2}) = ∑ λ_{i}^{2}.

It follows that

χ_{p}hi(g) = \frac{1}{2} (χ_{α}(g)^{2} + χ_{a}lpha(g^{2})) .

We deduce from this that

I(1, ϕ) = \frac{1}{2∥G∥} ∑_{g∈G} (χ_{α}(g)^{2} + χ_{α}(g^{2})) .

Since

I(1, ϕ) + I(1, ψ) = I(1, (α^{∗})^{2})

= \frac{1}{∥G∥} ∑_{g} χ_{α}(g^{−1})^{2}

= \frac{1}{∥G∥} ∑_{g} χ_{α}(g)^{2},

it follows that

I(1, ψ) = \frac{1}{2∥G∥} ∑_{g∈G} (χ_{α}(g)^{2} − χ_{α}(g^{2})) .

Putting all this together, we conclude that

\frac{1}{∥G∥} ∑_{g∈G} χ_{α}(g^{2}) = I(1, ϕ) − I(1, ψ)

= \begin{cases}1 & if α is real, \\ −1 & if α is quaternionic, \\ 0 & if α is essentially complex. \end{cases}