Course 424 — Group Representations

Sample Exam

Dr Timothy Murphy

March 2005

Attempt 7 questions. (If you attempt more, only the best 7 will be counted.) All questions carry the same number of marks. In this paper representation means “finite-dimensional representation over ℂ”.

1. Define a group representation. What is meant by saying that 2 representations α, β are equivalent? Find all representations of S_{3} of degree 2 (up to equivalence).

What is meant by saying that a representation α is simple? Find all simple representations of D_{4} from first principles.

Answer:

(a) A representation α of a group G in a vector space V is a homo-

morphism

α : G → GL(V).

(b) The representations α, β of G in U, V respectively are said to be equivalent if U, V are isomorphic as G-spaces, that is, there exists an invertible map

t : U → V

which preserves the action of G, ie

t(gu) = g(tu)

for all g ∈ G, u ∈ U.

(c) Suppose α is a representation of

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

(taking s = (abc), t = (ab)) in the 2-dimensional vector space V.

Let e be an eigenvector of s with eigenvalue λ:

se = λe.

Since s^{3} = 1, λ ∈ {1, ω, ω^{2}} (where ω = e^{2π/3}). Let f = te. Then

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

In other words, f is also an eigenvector of s, but with eigenvalue λ^{2}. If λ^{2} ≠ λ, ie λ ≠ 1, then e, f are eigenvectors of s with different eigenvalues, and so are necessarily linearly independent. Thus in this case e, f is a basis for V, and the representation is completely determined. If λ = ω then

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

If λ = ω^{2} then f has eigenvalue (ω^{2})^{2} = ω, and we get the same representation with e, f interchanged. Thus there is just one representation in this case. We must fall into this case unless both the eigenvalues of s are 1. In the latter case, since α(g) is always semisimple (diagonalisable), we must have s → I with respect to any basis. But then we can diagonalise t without effecting s. Since t^{2} = 1,

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

Two of these representations are equivalent; so we get just 3 representations (in effect 1 + 1, 1 + ϵ, ϵ + ϵ) in this case:

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

Thus S_{3} has just 4 representations of degree 2.

(d) The representation α of G in V is said to be simple if no subspace U ⊂ V is stable under G except for U = 0, V. (The subspace U is said to be stable under G if

g ∈ G, u ∈ U ⟹ gu ∈ U.)

(e) We have

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

Let us first suppose α is a 1-dimensional representations of D_{4}. ie a homomorphism

α : D_{4} → ℂ^{∗}.

Suppose

α(s) = λ, α(t) = μ.

Then

λ^{4} = μ^{2} = 1, λμ = μλ^{3}.

The last relation gives

λ^{2} = 1.

Thus there are just 4 1-dimensional representations given by

s ↦ ±1, t ↦ ±1.

Now suppose α is a simple representation of G in the vector space V over ℂ, where dim V ≥ 2. Let e ∈ V be an eigenvector of s:

se = λe;

and let

f = te.

Then

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

ie f is a λ^{3}-eigenvector of s. It follows that the subspace

⟨e, f⟩ ⊂ V

is stable under D_{4}, since

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

Since V by definition is simple, it follows that

V = ⟨e, f⟩.

In particular, dim α = 2, and e, f form a basis for V. Since s^{4} = 1 we have λ^{4} = 1, ie λ ∈ {±1, ±i}. If λ = 1 then s would have eigenvalues 1, 1 (since 1^{3} = 1). But we know that s (ie α(s)) is diagonalisable. It follows that s = I. Similarly if λ = −1 then s has eigenvalues −1, −1 and so s = −I. In either of these cases s will always be diagonal. Since we can always diagonalise t, we can diagonalise s, t simultaneously. But in that case the representation would not be simple; for if e is a common eigenvector of s, t then the 1-dimensional space ⟨e⟩ is stable under D_{4}. Thus we are left with the cases λ = ±i. If λ = −i then on swapping e and f we would have λ = i. So we have only 1 2dimensional representation (up to equivalence):

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

In conclusion, D_{4} has just 5 simple representations: 4 of dimension 1, and 1 of dimension 2.

2. What is meant by saying that a representation α is semisimple?

State carefully, and outline the main steps in the proof of, Haar’s Theorem on the existence of an invariant measure on a compact group.

Prove that every representation of a compact group is semisimple.

Answer:

(a) The representation α of G in V is said to be semisimple if it can be expressed as a sum of simple representations:

α = σ_{1} + ⋯ + σ_{m}.

This is equivalent to the condition that each stable subspace U ⊂ V has a stable complement W:

V = U ⊕ W.

(b) Haar’s theorem states that there exists an invariant measure on a compact group G, unique up to a scalar multiple. 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).

(c) We can prove the existence of such an invariant measure in the following way. 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,

ie

F (x) = λ_{1}f(g_{1}^{−1}x) + λ_{2}f(g_{2}^{−1}x) + ⋯ + λ_{r}f(g_{r}^{−1}x).

where 0 ≤ λ_{i} ≤ 1, ∑ λ_{i} = 1 and g_{1}, g_{2}, …, g_{r} ∈ G. If F is an average of f then

i. inf f ≤ inf F ≤ sup F ≤ supf; ii. If μ is an invariant measure then μ(F) = μ(f); iii. An average of F is an average of f.

Let us set var(f) = sup f − inf f

so that

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 will follow 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) = f(g_{i}^{−1}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).

(d) Suppose α is a representation of the finite group G in the vector

space V. Let

P (u, v)

be a positive-definite hermitian form on V. Define the hermitian form Q on V by

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

Then Q is positive-definite (as a sum of positive-definite forms). Moreover Q is invariant under G, ie

Q(gu, gv) = Q(u, v)

for all g ∈ G, u, v ∈ V. For

Q(hu, hv) = \frac{1}{∥G∥} ∑_{g∈G} H(ghu, ghv)

= \frac{1}{|G|} ∑_{g∈G} H(gu, gv)

= Q(u, v),

since gh runs over G as g does. Now suppose U is a stable subspace of V. Then

U^{⊥} = {v ∈ V : Q(u, v) = 0∀u ∈ U}

is a stable complement to U. Thus every stable subspace has a stable complement, ie the representation is semisimple.

3. Define the character χ_{α} of a representation α, and show that it is a class function (ie it is constant on conjugacy classes).

Define the intertwining number I(α, β) of 2 representations α, β of a group G, and show that if G is compact then

I(α, β) = ∫_{G} \overline{χ_{α}(g)}χ_{β}(g) dg.

Prove that a representation α is simple if and only if I(α, α) = 1.

4. Draw up the character table for S_{4}.

Determine also the representation-ring for this group, ie express the product αβ of each pair of simple representation as a sum of simple representations.

Answer:

(a) S_{4} has 5 classes, corresponding to the types 1^{4}, 1^{2}2, 13, 2^{2}, 4. Thus S_{4} has 5 simple representations. Each symmetric group S_{n} (for n ≥ 2) has just 2 1-dimensional representations, the trivial representation 1 and the parity representation ϵ. Let S_{4} = Perm(X), where X = {a, b, c, d}. The action of S_{4} on X defines a 4-dimensional representation ρ of S_{4}, with character

χ(g) = |{x ∈ X : gx = x}|

In other words χ(g) is just the number of 1-cycles in g. So now we can start our character table (where the second line gives the number of elements in the class):

1^{4} 1^{2}2 13 2^{2} 4
(1) (6) (8) (3) (6)
1 1 1 1 1 1
ϵ 1 −1 1 1 −1
ρ 4 2 1 0 0

Now

I(ρ, ρ) = \frac{1}{24}(1 · 16 + 6 · 4 + 8 · 1) = 2.

It follows that ρ has just 2 simple parts. Since

I(1, ρ) = \frac{1}{24}(1 · 4 + 6 · 2 + 8 · 1) = 1,

It follows that

ρ = 1 + α,

where α is a simple 3-dimensional representation, with character given by

χ(g) = χ_{ρ}(g) − 1.

The representation ϵα is also simple, and is not equal to α since it has a different character. So now we have 4 simple characters of S_{4}, as follows:

1^{4} 1^{2}2 13 2^{2} 4
(1) (6) (8) (3) (6)
1 1 1 1 1 1
ϵ 1 −1 1 1 −1
α 3 1 0 −1 −1
ϵα 3 −1 0 −1 1

To find the 5th simple representation, we can consider α^{2}. This has character

1^{4} 1^{2}2 13 2^{2} 4
(1) (6) (8) (3) (6)
α^{2} 9 1 0 1 1

We have

I(1, α^{2}) = \frac{1}{24}(9 + 6 + 3 + 6) = 1,
I(ϵ, α^{2}) = \frac{1}{24}(9 − 6 + 3 − 6) = 0,
I(α, α^{2}) = \frac{1}{24}(27 + 6 − 3 − 6) = 1,
I(ϵα, α^{2}) = \frac{1}{24}(27 − 6 − 3 + 6) = 1.I(α^{2}, α^{2}) = \frac{1}{24}(81 + 6 + 3 + 6) = 4,

It follows that α^{2} has 4 simple parts, so that

α^{2} = 1 + α + ϵα + β,

where β is the 5th simple representation, with character given by

χ_{β}(g) = χ_{α}(g)^{2} − 1 − χ_{α}(g) − ϵ(g)χ_{α}(g).

This allows us to complete the character table:

1^{4} 1^{2}2 13 2^{2} 4
(1) (6) (8) (3) (6)
1 1 1 1 1 1
ϵ 1 −1 1 1 −1
α 3 1 0 −1 −1
ϵα 3 −1 0 −1 1
β 2 0 −1 2 0

(b) We already know how to express α^{2} in terms of the 5 simple representations. Evidently ϵβ = β since there is only 1 simple representation of dimension 2. The character of αβ is given by

\begin{aligned}& 1^{4} & 1^{2}2 & 13 & 2^{2} & 4 \\ αβ & 6 & 0 & 0 & −2 & 0 \end{aligned}

We have

I(αβ, αβ) = \frac{1}{24}(36 + 12) = 2.

Thus αβ has just 2 simple parts. These must be α and ϵα to give dimension 6:

αβ = α + ϵα.

Also we have

I(β^{2}, β^{2}) = \frac{1}{24}(16 + 8 + 48) = 3.

Thus β has 3 simple parts. So by dimension, we must have

β^{2} = 1 + ϵ + β.

Now we can give the multiplication table for the representationring:

1 ϵ  β  α  ϵα
1 1 ϵ  β  α  ϵα
ϵ ϵ 1  β  ϵα  α
β β β 1 + ϵ + β  α + ϵα  α + ϵα
α α ϵα α + ϵα 1 + β + α + ϵα ϵ + β + α + ϵα
ϵα ϵα α α + ϵα ϵ + β + α + ϵα 1 + β + α + ϵα

5. Show that the number of simple representations of a finite group G is equal to the number s of conjugacy classes in G.

Show also that if these representations are σ_{1}, …, σ_{s} then

dim^{2} σ_{1} + ⋯ + dim^{2} σ_{s} = |G|.

Determine the dimensions of the simple representations of S_{5}, stating clearly any results you assume.

Answer:

(a) Let the simple representations of G be σ_{1}, …, σ_{r}; and let χ_{i}(g) be the character of σ_{i}. The simple characters χ_{1}, …, χ_{r} are linearly independent. For if say

ρ_{1}χ_{1}(g) + ⋯ + ρ_{s}χ_{s}(g) = 0

it follows from the formula for the intertwining number that for any representation α

ρ_{1}I(α, σ_{1}) + ⋯ + ρ_{r}I(α, σ_{r}) = 0.

But on applying this with α = σ_{i} we deduce that ρ_{i} = 0 for each i. The characters are class functions:

χ(gxg^{−1}) = χ(x).

The space of class functions has dimension s, the number of classes in G. It follows that r ≤ s. To prove that r = s, it is sufficient to show that the characters span the space of class functions. Suppose g ∈ G has order e. Let [g] denote the class of g, and let C = ⟨g⟩ be the cyclic group generated by g. The group C has e 1-dimensional representations θ_{1}, …, θ_{e} given by

θ_{i} : g ↦ ω^{i}, where ω = e^{2πi/e}. Let

f(x) = θ_{0}(x) + ω^{−1}θ_{1}(x) + ω^{−2}θ_{2}(x) + ⋯ + ω^{−e+1}θ_{e−1}(x)

Then

f(g^{j}) = \begin{cases}e & if j = 1 \\ 0 & otherwise. \end{cases}

Now let us “induce up” each of the characters θ_{i} from C to G. We have

θ_{i}^{G}(x) = \frac{|G|}{|S||[x]|} ∑_{y∈[x]∩C} θ_{i}(y).

Let F (x) be the same linear combination of the induced characters that f(x) was of the θ_{i}. Then

F (x) = \frac{|G|}{|S||[x]|} ∑_{y∈[x]∩C} f(y).

Since f(y) vanishes away from g, we deduce that F (x) vanishes off the class [g], and is non-zero on that class:

F (x) \begin{cases}> 0 & if x ∈ [g], \\ = 0 & if x ∉ [g]. \end{cases}

It follows that every class function on G can be expressed as a linear combination of characters, and therefore as a linear combination of simple characters. Hence the number of simple characters is at least as great as the number of classes. We have shown therefore that the number of simple representations is equal to the number of classes.

(b) Consider the regular representation ρ of G. We have

χ_{ρ}(g) = \begin{cases}|G| & if g = e, \\ 0 & if g ≠ e. \end{cases}

Thus if α is any representation of G,

I(ρ, α) = χ_{α}(e) = dim α.

Applying this to the simple representations α = σ_{i} we deduce that

ρ = (dim σ_{1})σ_{1} + ⋯ + (dim σ_{s})σ_{s}.

Taking dimensions on each side,

|G| = (dim σ_{1})^{2} + ⋯ + (dim σ_{s})^{2}.

(c) Since S_{5} has 7 classes (corresponding to the types 1^{5}, 1^{3}2, 1^{2}3, 12^{2}, 14, 23, 5) it has 7 simple representations. We know that it has 2 representations of dimension 1, the trivial representation 1 and the parity representation ϵ. Let ρ be the natural 5-dimensional representation of S_{5} (by permutation of coordinates). Then it is easy to show directly that

ρ = 1 + σ

where σ is a simple 4-dimensional representation. Now

χ_{ρ}(1^{3}2) = 3,

since a permutation of type 1^{3}2 leaves 3 elements fixed. It follows that

χ_{σ}(1^{3}2) = 2.

Hence

ϵσ ≠ σ.

We now have 4 simple representations, of dimensions 1, 1, 4, 4. Let the other simple representations have dimensions a, b, c. Then

a^{2} + b^{2} + c^{2} = 120 − (1^{2} + 1^{2} + 4^{2} + 4^{2}) = 86.

Since m^{2} ≡ 0 or 1 mod 4, while 86 ≡ 2 mod 4, it follows that 2 of a, b, c are odd and the other even. Let t ∈ S_{5} be a transposition. Then t^{2} = 1 and so the eigenvalues of t in any representation α are ±1. It follows that if dim α is odd then χ_{α}(t) ≠ 0, and therefore

ϵα ≠ α,

since the characters of these 2 representations differ at t. In particular, representations of S_{5} of odd dimension appear in pairs σ, ϵσ. Thus 2 of a, b, c are equal odd numbers, say a = b; and

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

The only solution to this is a = 5, c = 6. We conclude that the simple representations of S_{4} have dimensions

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

6. 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}, …).

Express each product D(i)D(j) as a sum of simple representations D(k).

Answer:

(a) We know that i. if U ∈ SU(2) then U has eigenvalues

e^{±iθ} (θ ∈ ℝ).

ii. 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 shown that the conjugacy classes in SU(2) are

C(θ) (0 ≤ θ ≤ π),

where

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

(b) 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 (finite-dimensional) 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}.

(c) As we saw above, if U has eigenvalues e^{±iθ} then

χ_{j}(U) = e^{2jiθ} + e^{2(j−1}iθ + ⋯ + e^{−2jiθ}.

(d) 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}.

7. Define the exponential e^{X} of a square matrix X. Determine e^{X} in each of the following cases:

X = ( \begin{matrix}1 & 0 \\ 0 & -1 \end{matrix} ) , X = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) , X = ( \begin{matrix}0 & -1 \\ 1 & 0 \end{matrix} ) , X = ( \begin{matrix}1 & -1 \\ 1 & 1 \end{matrix} ) .

Show that if X has eigenvalues λ, μ then e^{X} has eigenvalues e^{λ}, e^{μ}. Which of the above 5 matrices X are themselves expressible in the form X = e^{Y} for some real matrix Y? (Justify your answers in all cases.) Answer:

(a) The exponential of a square matrix X is defined by

e^{X} = I + X + \frac{1}{2!}X^{2} + \frac{1}{3!}X^{3} + · · · .

[It is a straightforward matter to verify that this series always converges, using the fact that the matrix norm defined by

∥X∥^{2} = tr(X′X) = ∑_{i,j} |x_{ij}|

has the properties

i. ∥X + Y ∥ ≤ ∥X∥ + ∥Y ∥, ii. ∥λX∥ = |λ|∥X∥, iii. ∥XY ∥ ≤ ∥X∥∥Y ∥. However, I would not require this observation.] i. If

X = (\begin{matrix}1 & 0 \\ 0 & −1 \end{matrix})

then

e^{X} = (\begin{matrix}e & 0 \\ 0 & e^{−1} \end{matrix})

ii. If

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

then

X^{2} = I,

and so

e^{X} = (1 + 1/2! + 1/4! + ⋯)I + (1/1! + 1/3! + ⋯)X
= (\begin{matrix}cosh 1 & sinh 1 \\ sinh 1 & cosh 1 \end{matrix})

iii. If

X = (\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix})

then

X^{2} = −I,

and so

e^{X} = (1 − 1/2! + 1/4! − ⋯)I + (1/1! − 1/3! + ⋯)X
= (\begin{matrix}cos 1 & − sin 1 \\ sin 1 & cos 1 \end{matrix})

iv. If

X = (\begin{matrix}1 & −1 \\ 1 & 1 \end{matrix}) = I + J,

where

J = (\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix})

then (since I, J commute)

e^{X} = e^{I}e^{J}
= (\begin{matrix}e cos 1 & −e sin 1 \\ e sin 1 & e cos 1 \end{matrix})

(b) i. If

X = (\begin{matrix}1 & 0 \\ 0 & −1 \end{matrix})

then

det X = −1,

and so X ≠ e^{Y} since

det e^{Y} = e^{trY} > 0.

ii. If

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

then again

det X < 0,

and so X ≠ e^{Y}. iii. By the isomorphims between the complex numbers x + iy and

the matrices

(\begin{matrix}x & −y \\ y & x \end{matrix})

we see that

J = (\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix})

corresponds to i.

Since

i = e^{πi/2}

we see that

J = e^{Y}

with

Y = \frac{π}{2} J

= (\begin{matrix}0 & −π/2 \\ π & 0 \end{matrix}) .

iv. Similarly

X = (\begin{matrix}1 & −1 \\ 1 & 1 \end{matrix})

corresponds to the complex number 1 + i. Since

1 + i = \sqrt{2}e^{πi/4}
= e^{log 2/2+πi/4}

X = e^{Y} with

Y = (\begin{matrix}log 2/2 & −π/4 \\ π/4 & log 2/2 \end{matrix}) .

8. Define a linear group, and a Lie algebra; and define the Lie algebra LG of a linear group G, showing that it is indeed a Lie algebra. Define the dimension of a linear group; and determine the dimensions of each of the following groups:

O(n), SO(n), U(n), SU(n), GL(n, ℝ), SL(n, ℝ), GL(n, ℂ), SL(n, ℂ)?

Answer:

(a) A linear group is a closed subgroup G ⊂ GL(n, ℝ) for some n.

(b) A Lie algebra is defined by giving i. a vector space L; ii. a binary operation on L, ie a map

L × L → L : (X, Y) ↦ [X, Y]

satisfying the conditions i. The product [X, Y] is bilinear in X, Y; ii. The product is skew-symmetric:

[Y, X] = −[X, Y];

iii. Jacobi’s identity is satisfied:

[[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y] = 0

for all X, Y, Z ∈ L.

(c) Suppose G ⊂ GL(n, ℝ) is a linear group. Then its Lie algebra L = LG is defined to be

L = {X ∈ Mat(n, ℝ) : e^{tX} ∈ G ∀t ∈ ℝ}.

(d) It follows at once from this definition that

X ∈ L, λ ∈ R ⟹ λX ∈ L.

Thus to see that L is a vector subspace of Mat(n, ℝ) we must show that

X, Y ∈ L ⟹ X + Y ∈ L.

Now

(e^{X/n}e^{Y/n})^{n} ↦ e^{X+Y}

as n ↦ ∞. (This can be seen by taking the logarithms of each side.) It follows that

X, Y ∈ L ⟹ e^{X+Y} ∈ G.

On replacing X, Y by tX, tY we see that

X, Y ∈ L ⟹ e^{t(X+Y)} ∈ G
⟹ X + Y ∈ L.

Similarly

(e^{X/n}e^{Y/n}e^{−X/n}e^{−Y/n})^{n^{2}} ↦ e^{[X,Y]},

as may be seen again on taking logarithms. It follows that

X, Y ∈ L ⟹ e^{[X,Y]} ∈ G.

Taking tX in place of X, this implies that

X, Y ∈ L ⟹ e^{t[X,Y]} ∈ G
⟹ [X, Y] ∈ L.

Thus L is a Lie algebra.

(e) The dimension of a linear group G is the dimension of the real

vector space LG:

dim G = dim_{ℝ} LG.

(f) i. We have

o(n) = {X ∈ Mat(n, ℝ) : X′ + X = 0}

A skew symmetric matrix X is determined by giving the entries above the diagonal. This determines the entries below the diagonal; while those on the diagonal are 0. Thus

dim O(n) = dim o(n) = \frac{n(n − 1)}{2}.

ii. We have

so(n) = {X ∈ Mat(n, ℝ) : X′ + X = 0, trX = 0} = ø(n).

Since X′ + X = 0 ⟹ trX = 0,

dim SO(n) = dim O(n) = \frac{n(n − 1)}{2}.

iii. We have

u(n) = {X ∈ Mat(n, ℂ) : X^{∗} + X = 0}

Again, the elements above the diagonal determine those below the diagonal; while those on the diagonal are purely imaginary. Thus

dim U(n) = 2\frac{n(n − 1)}{2} + n
= n^{2}.

iv. We have

su(n) = {X ∈ Mat(n, ℂ) : X^{∗} + X = 0, trX = 0}

This gives one linear condition on the (purely imaginary) diagonal elements. Thus

dim SU(n) = dim U(n) − 1 = n^{2} − 1.

v. We have

gl(n, ℝ) = Mat(n, ℝ).

Thus

dim GL(n, ℝ) = n^{2}.

vi. We have

sl(n, ℝ) = {X ∈ Mat(n, ℝ) : trX = 0}.

This imposes one linear condition on X. Thus

dim SL(n, ℝ) = dim GL(n, ℝ) − 1 = n^{2} − 1.

vii. We have

gl(n, ℂ) = Mat(n, ℂ).

Each of the n^{2} complex entries takes 2 real values. Thus

dim GL(n, ℂ) = 2n^{2}.

viii. We have

sl(n, ℂ) = {X ∈ Mat(n, ℂ) : trX = 0}.

This imposes one complex linear condition on X, or 2 real linear conditions. Thus

dim SL(n, ℂ) = dim GL(n, ℂ) − 1 = 2n^{2} − 2.

9. Determine the Lie algebras of SU(2) and SO(3), and show that they are isomomorphic. Show that the 2 groups themselves are not isomorphic. Answer:

(a) We have

u(2) = {X ∈ Mat(2, ℂ) : e^{tX} ∈ U(2) ∀t ∈ ℝ}
= {X : (e^{tX})^{∗}e^{tX} = I ∀t}
= {X : e^{tX^{∗}} = e^{−tX} ∀t}
= {X : X^{∗} = −X}
= {X : X^{∗} + X = 0},

while

sl(2, ℂ) = {X ∈ Mat(2, ℂ) : e^{tX} ∈ SL(2, ℂ) ∀t ∈ ℝ}
= {X : det e^{tX} = 1 ∀t}
= {X : e^{ttrX} = 1 ∀t}
= {X : trX = 0}.

Since

SU(2) = U(2) ∩ SL(2, ℂ)

it follows that

su(2) = u(2) ∩ sl(2, ℂ)
= {X : X^{∗} + X = 0, trX = 0}.

The 3 matrices

e = (\begin{matrix}i & 0 \\ 0 & −i \end{matrix}) , f = (\begin{matrix}0 & −1 \\ 1 & 0 \end{matrix}) , g = (\begin{matrix}0 & i \\ i & 0 \end{matrix})

form a basis for the vector space su(2). We have

[e, f] = ef − fe = −2g,
[e, g] = eg − ge = 2f,
[f, g] = fg − gf = −2e

Thus

su(2) = ⟨e, f, g : [e, f] = −2g, [e, g] = 2f, [f, g] = −2e⟩.

(b) We have

o(3) = {X ∈ Mat(3, ℝ) : e^{tX} ∈ O(3) ∀t ∈ ℝ}
= {X : (e^{tX})′e^{tX} = I ∀t}
= {X : e^{tX′} = e^{−tX} ∀t}
= {X : X′ = −X}
= {X : X′ + X = 0},

while

sl(3, ℝ) = {X ∈ Mat(3, ℝ) : e^{tX} ∈ SL(3, ℝ) ∀t ∈ ℝ}
= {X : det e^{tX} = 1 ∀t}
= {X : e^{ttrX} = 1 ∀t}
= {X : trX = 0}.

Since

SO(3) = O(3) ∩ SL(3, ℝ)

it follows that

so(3) = o(3) ∩ sl(3, ℝ)
= {X : X′ + X = 0, trX = 0}
= {X : X′ + X = 0}

since a skew-symmetric matrix necessarily has trace 0. The 3 matrices

U = (\begin{matrix}0 & 0 & 0 \\ 0 & 0 & −1 \\ 0 & 1 & 0 \end{matrix}) , V = (\begin{matrix}0 & 0 & 1 \\ 0 & 0 & 0 \\ −1 & 0 & 0 \end{matrix}) , W = (\begin{matrix}0 & −1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix})

form a basis for the vector space so(3). We have

[V, W] = U;

and so by cyclic permutation of indices (or coordinates)

[W, U] = V, [U, V] = W.

Thus

so(3) = ⟨U, V, W : [U, V] = W, [U, W] = −V, [V, W] = U⟩.

(c) The Lie algebras su(2) and so(3) are isomorphic under the corre-

spondence

e ↔ −2U, f ↔ −2V, g ↔ −2W.

(d) However, the groups SU(2), SO(3) are not isomorphic, since

ZSU(2) = {±I} while ZSO(3) = {I}.

10. Define a representation of a Lie algebra L. What is meant by saying that such a representation is (a) simple, (b) semisimple?

Determine the Lie algebra of SL(2, ℝ), and find all the simple representations of this algebra.

Show that every representation of the group SL(2, ℝ) is semisimple, stating carefully but without proof any results you need.

Answer:

(a) Suppose L is a real Lie algebra. A representation of L in the complex vector space V is defined by giving a map

L × V → V : (X, v) ↦ Xv

which is bilinear over ℝ and which satisfies the condition

[X, Y]v = X(Y v) − Y (Xv)

for all X, Y ∈ L, v ∈ V. A representation of L in V is thus the same as a representation of the complexification L_{ℂ} of L in V.

(b) The representation is said to be simple if no proper subspace U ⊂ V is stable under L. It is said to be semisimple if V is a direct sum of simple subspaces.

(c) We have

sl(2, ℝ) = {X ∈ Mat(2, ℝ) : e^{tX} ∈ SL(2, ℝ)∀t ∈ ℝ}
= {X : det e^{tX} = 1∀t}
= {X : e^{ttrX} = 1∀t}
= {X : trX = 0}.

The 3 matrices

H = (\begin{matrix}1 & 0 \\ 0 & −1 \end{matrix}) , E = (\begin{matrix}0 & 1 \\ 0 & 0 \end{matrix}) , F = (\begin{matrix}0 & 0 \\ 1 & 0 \end{matrix})

form a basis for the vector space sl(2, ℝ). We have

[H, E] = 2E, [H, F] = −2F, [E, F] = H.

Thus

sl(2, ℝ) = ⟨H, E, F : [H, E] = 2E, [H, F] = −2F, [V, W] = H⟩.

(d) Suppose we have a simple representation of sl(2, ℝ) on V. Let v be an eigenvector of H with eigenvalue λ:

Hv = λv.

Then

[H, E]v = 2Ev,

ie

HEv − EHv = 2Ev.

In other words, since Hv = λv,

H(Ev) = (λ + 2)Ev,

that is, Ev is an eigenvector of H with eigenvalue λ + 2. By the same argument E^{2}v, E^{3}v, … are all eigenvectors of H with eigenvalues λ + 4, λ + 6, …, at least until they vanish. This must happen at some point, since V is finite-dimensional; say

E^{r+1}v = 0, E^{r}v ≠ 0.

Similarly we find that

F v, F^{2}v, …

are also eigenvectors of H (until they vanish) with eigenvalues λ − 2, λ − 4, . . .. Again we must have

F^{s+1}v = 0, F^{s}v ≠ 0

for some s. Now let us write e_{0} for F^{s}v, so that

F e_{0} = 0;

and let μ = λ − 2s, so that e_{0} is a μ-eigenvector of H:

He_{0} = μe_{0}.

Set

e_{i} = E^{i}e_{0},

so that

e_{i+1} = Ee_{i}

for i ≥ 0. Thus e_{i} is a (μ + 2i)-eigenvector of H; and if m = r + s then

e_{m} ≠ 0, e_{m+1} = Ee_{m} = 0.

By the argument above, F e_{i} is an eigenvector of H with the same eigenvalue, μ + 2(i − 1), as e_{i−1}. We shall show by induction on i that F e_{i} is a scalar multiple of e_{i−1}, say

F e_{i} = ρ_{i}e_{i−1}.

The result holds for i = 0 with ρ_{0} = 0. Suppose it holds for i = 1, 2, …, m. Then

F e_{m+1} = F Ee_{m}
= (EF − [E, F])e_{m}
= ρ_{m}Ee_{m−1} − He_{m}
= (ρ_{m} − μ − 2m)e_{m}.

This proves the result, and also shows that

ρ_{i+1} = ρ_{i} − μ − 2i for each i. It follows that

ρ_{i} = −iμ − i(i − 1).

We must have ρ_{m+1} = 0. Hence

μ = m.

We conclude that the subspace

⟨e_{0}, …, e_{m}⟩

is stable under sl(2, ℝ), and so must be the whole of V. Thus we have shown that there is at most 1 simple representation of each dimension n, and we have determined this explicitly, if it exists. In fact it is a straightforward matter to verify that the above actions of H, E, F on ⟨e_{0}, …, e_{m}⟩ do indeed define a representation of sl(2, ℝ); so that this Lie algebra has exactly 1 simple representation of each dimension.

(e) i. The representations over ℂ of a real Lie algebra L correspond

to the representations of its complexification ℂL. ii. Thus if two real Lie algebras have isomorphic complexifica-

tions then there is a one-one correspondence between their representations, with simple or semisimple representations of one corresponding to simple or semisimple representations of the other.

iii. The Lie algebras sl(n, ℝ) and su(n) have the same complexi-

fication:

ℂsl(n, ℝ) = sl(c, ℂ) = ℂsu(n).

iv. The representations of su(2) can all be lifted to representa-

tions of SU(2), since SU(2) ≅ S^{3} is simply-connected. v. The representations of SU(2) are all semisimple, since SU(2)

is compact. vi. Hence the representations of su(2 are all semisimple, and so

the representations of sl(2, ℝ) are all semisimple.