Course 424

Group Representations

Dr Timothy Murphy

Sample Paper

Attempt 6 questions. (If you attempt more, only the best 6 will be counted.) All questions carry the same number of marks. Unless otherwise stated, all groups are compact (or finite), and all representations are of finite degree over ℂ.

1. What is meant by saying that a group representation α is (a) simple, (b) semisimple?

Prove that every representation of a finite group is semisimple.

Give an example of a representation of an infinite group that is not semisimple.

Answer:

(a) 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.)

(b) 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.

(c) 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.

(d) The representation α of ℤ of degree 2 over ℂ given by

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

is not semisimple.

For the representation is not simple, since it leaves stable the 1dimensional subspace ⟨e⟩, where

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

If α were semisimple, say α = β + γ, where β, γ are of degree 1, then α(n) would be diagonalisable for all n. Since α(n) has eigenvalues 1, 1, this implies that

α(n) = I

for all n, which is not the case.

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

Determine also the representation-ring for S_{4}, ie express the product αβ of each pair of simple representations as a sum of simple representations.

Draw up the character table for the subgroup A_{4} of even permutations.

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

(c) Recall that an even class ḡ ⊂ S_{n} splits in A_{n} if and only if no odd element x ∈ S_{n} commutes with g, in which case ḡ splits into two classes of equal size. There are 3 even classes in S_{n}: 1^{4}, 2^{2}and31, containing 1, 3, 8 elements, respectively. The first two cannot split, since they contain an odd number of elements. The third class does split; for suppose x commutes with g = (abc). Then

xgx^{−1} = (x(a), x(b), x(c)) = (a, b, c).

It follows from this that

x ∈ {1, g, g^{2}}.

In particular, x is even. Thus the class 31 splits into two classes 31′ and 31′′, each containing 4 elements.

3. 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 degrees of the simple representations of S_{6}.

Answer:

(a)

(b)

(c)

(d) S_{6} has 11 classes:

1^{6}, 21^{4}, 2^{2}1^{2}, 2^{3}, 31^{3}, 321, 3^{2}, 3^{2}, 41^{2}, 51, 6.

Hence it has 11 simple representations over ℂ. It has 2 representations of degree 1: 1 and the parity representation ϵ. The natural representation ρ_{1} of degree 6 (by permutation of coordinates) splits into two simple parts:

ρ_{1} = 1 + σ_{1},

where σ_{1} is of degree 5. If α is a simple representation of odd degree, then

ϵα ≠ α.

For a transposition t has eigenvalues ±1, since t^{2} = 1. Hence

χ_{α}(t) ≠ 0.

But

χ_{ϵα}(t) = χ_{ϵ}(t)χ_{α}(t) = −χ_{α}(t).

Thus the simple representations of odd degree d divide into pairs α, ϵα. So there are an even number of representations of degree d. In particular there are at least 2 simple representations of degree 5: σ and ϵσ. We are going to draw up a partial character table for S_{6}, adding rows as we gather more material.

1^{6} 21^{4} 2^{2}1^{2} 2^{3} 31^{3} 321 3^{2} 42 41^{2} 51 6
# 1 15 45 15 40 120 40 90 90 144 120
ρ_{1} 6 4 2 0 3 1 0 2 0 1 0
σ_{1} 5 3 1 −1 2 0 −1 1 −1 0 −1
ρ_{2} 15 7 3 3 3 1 0 1 1 0 0
τ 14 6 2 2 2 0 −1 0 0 −1 −1
σ_{2} 9 3 1 3 0 0 0 −1 1 −1 0
ρ_{3} 20 8 4 0 2 2 2 0 0 0 0
θ 19 7 3 −1 1 1 1 −1 −1 −1 −1
σ_{3} 5 1 1 −3 −1 1 2 −1 −1 0 0
σ_{1}^{2} 25 9 1 1 4 0 1 1 1 0 1
ϕ 24 8 0 0 3 −1 0 0 0 −1 0

Now consider the permutation representation ρ_{2} arising from the action of S_{6} on the 15 pairs of elements. Evidently

I(ρ_{2}, 1) > 0,

since all the terms in the sum for this are ≥ 0. Let τ = ρ_{2} − 1. Then

I(τ, τ) = \frac{1}{720}(196 + 540 + 180 + 60 + 160 + 40 + 144 + 120) = 2,

while

I(τ, σ_{1}) = \frac{1}{720}(70 + 270 + 90 − 30 + 160 + 40 + 120) = 1.

Thus

σ_{2} = τ − σ_{1}

is simple. So far we have 6 simple representations:

1, ϵ, σ_{1}, ϵσ_{1}, σ_{2}, ϵσ_{2},

of degrees 1,1,5,5,9,9. Next consider the permutation representation ρ_{3} arising from the action of S_{6} on the 20 subsets of 3 elements. Evidently

I(ρ_{3}, 1) > 0,

since all the terms in the sum for this are ≥ 0. [Although not needed here, it is worth recalling that if ρ is a permutation representation arising from the action of G on the set X then I(ρ, 1) is equal to the number of orbits of the action.] Let θ = ρ_{3} − 1. Then

I(θ, θ) = \frac{1}{720}(361+735+405+15+40+120+40+90+90+144+120) = 3.

Thus θ has 3 simple parts. Now

I(θ, σ_{1}) = \frac{1}{720}(95 + 315 + 135 + 15 + 80 − 40 − 90 + 90 + 120) = 1,

while

I(θ, σ_{2}) = \frac{1}{720}(171 + 315 + 135 − 45 + 90 − 90 + 144) = 1.

It follows that

σ_{3} = θ − σ_{1} − σ_{2}

is simple. Now we have 8 simple representations:

1, ϵ, σ_{1}, ϵσ_{1}, σ_{3}, ϵσ_{3}, σ_{2}, ϵσ_{2},

of degrees 1,1,5,5,5,5,9,9. We have 3 remaining simple representations. Suppose they are of degrees a, b, c. Then

720 = 2 · 1^{2} + 4 · 5^{2} + 2 · 9^{2} + a^{2} + b^{2} + c^{2}

ie

a^{2} + b^{2} + c^{2} = 456.

Now

456 ≡ 0 mod 8.

If n is odd then n^{2} ≡ 1 mod 8. It follows that a, b, c are all even, say

a = 2d, b = 2e, c = 2f,

with

d^{2} + e^{2} + f^{2} = 114.

Since

114 ≡ 2 mod 8,

it follows that two of d, e, f are odd and one is divisible by 4. Let us suppose these are d, e, f in that order. Then

f ∈ {4, 8}.

If f = 4 then

d^{2} + e^{2} = 98 ⟹ d = e = 7,

while if f = 8 then

d^{2} + e^{2} = 50 ⟹ d = e = 5.

So the three remaining simple representations have degrees

8, 14, 14 or 10, 10, 16.

Let

ϕ = σ_{1}^{2} − 1.

Then

I(ϕ, ϕ) = \frac{1}{720}(576 + 960 + 360 + 120 + 144) = 3.

Also

I(ϕ, σ_{1}) = \frac{1}{720}(120 + 360 + 240) = 1,

while

I(ϕ, σ_{2}) = \frac{1}{720}(216 + 360 + 144) = 1.

Thus

σ_{4} = ϕ − σ_{1} − σ_{2}

is a simple representation of degree 10. We conclude that the 11 simple representations have degrees

1, 1, 5, 5, 5, 5, 9, 9, 10, 10, 16.

4. Determine the simple representations of SO(2). Suppose H is a subgroup of the compact group G of finite index. Explain how a representation β of H induces a representation β^{G} of G. Determine the simple representations of O(2). Answer:

(a) Let

R(θ) ∈ SO(2)

denote rotation through angle θ. Then the map

R(θ) ↦ e^{iθ} : SO(2) → U(1)

is an isomorphism, allowing us to identify SO(2) with U(1). This group is abelian; so every simple representation α (over ℂ) is of degree 1; and since the group is compact

im α ⊂ U(1).

ie α is a homomorphism

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.

For suppose

α : U(1) → U(1)

is a representation of U(1) distinct from all the E(n). Then

I(E_{n}, α) = 0

for all n, ie

c_{n} = \frac{1}{2π} ∫_{0}^{2π} α(e^{iθ})e^{−inθ} dθ = 0.

In other words, all the Fourier coefficients of α(e^{iθ}) vanish. But this implies (from Fourier theory) that the function itself must vanish, which is impossible since α(1) = 1.

(b) Suppose β is a representation of H in the vector space U. Express G as a union of left H-cosets:

G = g_{1}H ∪ ⋯ ∪ g_{r}H

Set

V = g_{1}U ⊕ ⋯ ⊕ g_{r}U,

ie V is the direct sum of r copies of U, labelled by g_{1}, …, g_{r}. We define the action of g ∈ G on V as follows. Suppose 1 ≤ i ≤ r. Then

gg_{i} = g_{j}h

for some j ∈ [1, r], h ∈ H. We set

g(g_{i}u) = g_{j}(hu).

That defines the action of g on the summand g_{i}U; and this is extended to V by linearity. It is readily verified that this defines a representation of G in V, and that the choice of different representatives g_{1}, …, g_{r} of the cosets would lead to an equivalent representation.

(c) Since SO(2) is a subgroup of index 2 in O(2), the representation E(n) of SO(2) = U(1) induces a representation

α_{n} = E(n)^{O(2)}

of O(2) of degree 2.

Any element of O(2) \ SO(2) is a reflection T (l) in some line l through the origin. These reflections are all conjugate, since

R(θ)T (l)R(−θ) = T (l′),

where l′ = R(θ)l. Also

T (l)R(θ)T (l) = R(−θ);

so the O(2)-conjugacy classes consist of pairs {R(±θ)}, together with the set of all reflections. Explicitly, on taking e, T e as basis for the induced representation (where T is any reflection) we see that α_{n} is given by

R(θ) ↦ (\begin{matrix}e^{iθ} & 0 \\ 0 & e^{−iθ} \end{matrix}) , T (l) ↦ (\begin{matrix}0 & 1 \\ 1 & 0. \end{matrix}) .

If n ≠ 0 this representation is simple. For

α_{n}| SO(2) = E(n) + E(−n).

It follows that the only proper subspaces stable under SO(2) are ⟨e⟩, ⟨T e⟩, and these are not stable under T. If n = 0 the representation splits into two parts:

α_{0} = 1 + ϵ,

where

ϵ(R(θ)) = 1, ϵ(T (l) = −1,

ie ϵ(S) = ±1 according as S is proper or improper. We claim that the simple representations of O(2) are precisely these representations α_{n} for n ≠ 0, together with the representations 1, ϵ of degree 1. For suppose α is a simple representation of O(2) in the vector space V. Then

α| SO(2) = E(n_{1}) + ⋯ + E(n_{r}),

ie V is the direct sum of 1-dimensional subspaces stable under SO(2). Let U = ⟨e⟩ be one such subspace. Then U carries some representation E(n), ie

R(θ)e = e^{inθ}e

for all θ. Take any reflection T. Then the subspace ⟨e, T e⟩ is stable under the full group O(2). Since α is simple,

V = ⟨e, T e⟩,

If n ≠ 0 then we see explicitly that

α = α_{n}.

If n = 0 then SO(2) acts trivially on U. If T e = e then U is 1dimensional, and α = 1. If not, then the 1-dimensional subspace ⟨e − T e⟩ carries the representation ϵ, and so α = ϵ. We conclude that these are the only simple representations of O(2).

5. Prove that SU(2) has one simple representation of each dimension 0, 1, 2, . . ..

Show that there exists a double covering Θ : SU(2) → SO(3).

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

Determine the representation-ring of SO(3), ie express the product of each pair of simple representations as a sum of simple representations.

Determine the simple representations of O(3).

Answer:

(a) 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^{m} ∈ 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_{i} ∈ 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}.

(b) We can identify SU(2) with the group

Sp(1) = {q ∈ ℍ : |q| = 1}.

[If we regard ℍ as a 2-dimensional vector space over ℂ with basis 1, j:

(z, w) ↦ z + wj,

then multiplication on the right by a quaternion defines a C-linear map, ie an element of GL(2, ℂ). Suppose q = a + bj ∈ Sp(1). Then

q^{−1} = q^{∗} = ā − bj;

and multiplication on the right by q^{−1} gives the map

z + wj ↦ (āz + b̄w) + (−bz + aw)j,

ie

(\begin{matrix}z \\ w \end{matrix}) ↦ (\begin{matrix}ā & b̄ \\ −b & a \end{matrix}) (\begin{matrix}z \\ w \end{matrix}) .

Since

|q|^{2} = |a|^{2} + |b|^{2},

this establishes an isomorphism

q ↦ (\begin{matrix}ā & b̄ \\ −b & a \end{matrix}) : Sp(1) → SU(2).]

Now let V denote the 3-dimensional real vector space of purely imaginary quaternions

v = xi + yj + zk.

Evidently

q ∈ V ⟺ q^{∗} = −q.

It follows that if q ∈ Sp(1), v ∈ V then

(qvq^{∗})^{∗} = qv^{∗}q^{∗} = −qvq^{a}st.

Hence

qvq^{∗} = qvq^{−1} ∈ V.

Thus each q ∈ Sp(1) defines a linear map

Θ(q) : v ↦ qvq^{∗} : V → V,

giving a homomorphism

Θ : Sp(1) → GL(3, ℝ).

If v ∈ V then

|v| = vv^{∗} = x^{2} + y^{2} + z^{2}.

Now

|Θ(q)v|^{2} = (qvq^{∗})(qva^{∗})^{∗}
= qvq^{a}stqv^{∗}q^{a}st
= qvv^{∗}q^{∗}
= vv^{∗}qq^{∗}
= vv^{∗}
= |v|^{2},

since vv^{∗} ∈ ℝ. Thus Θ(q) preserves the form x^{2} + y^{2} + z^{2}. Hence

Θ(q) ∈ O(3).

Since Sp(1) ≅ S^{3} is connected, so is im Θ(q). Hence

Θ(q) ∈ SO(3),

giving a homomorphism

Θ : Sp(1) → SO(3).

We have

ker Θ = {q ∈ Sp(1) : qv = vq ∀v ∈ V }.

Since any quaternion is expressible as Q = t1 + v, with t ∈ ℝ, v ∈ V. it follows that

ker Θ = {q ∈ Sp(1) : qQ = Qq ∀Q ∈ ℍ}.

It is readily verified that

Zℍ = ℝ = {t1 : t ∈ ℝ}.

Hence

ker Θ = {±1}.

To see that Θ is surjective, ie im Θ = SO(3), we note that SO(3) is generated by half-turns π(l) about an axis l. But it is readily verified that if v is a unit vector along l then

Θ(v) = π(v),

since Θ(v) leaves l fixed, and

v^{2} = −vv^{a}st = −1,

and so

Θ(v)^{2} = I.

Hence Θ defines a 2-fold covering of SO(3).

(c) Suppose

θ : G → H

is a surjective homomorphism. Then a representation

α : H → GL(V)

of H in V defines a representation

αθ : G → GL(V).

Furthermore, distinct representations of H give rise to distinct representations of G; and the representation αθ is simple if and only if α is simple, since a subspace U ⊂ V is stable under G if and only if it is stable under H. Conversely, a representation

β : G → GL(V)

arises from a representation of H in this way if and only if

ker θ ⊂ ker α;

and if it does so arise, it is from a unique representation of H. In the present case this shows that a representation of SO(3) arises from a representation α of SU(2) if and only if

α(−I) = 1.

Looking at the definition of D_{j} by the action of of SU(2) on the space of homogeneous polynomials f(z, w) of degree 2j, we see that

f(−z, −w) = (−1)^{2j}f(z, w).

Thus

D_{j}(−I) = 1 ⟺ j is a half-integer.

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

(d)

6. Define the Lie algebra ℒ G of a linear group G, showing that it is indeed a Lie algebra.

Determine the Lie algebras of SU(2) and SO(3), and show that they are isomorphic.

Are the groups isomorphic?

Answer:

(a)

7. Define the exponential e^{X} of a matrix X ∈ Mat(n, k), where k = ℝ, ℂ or ℍ.

Determine e^{X} in each of the following cases:

X = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) , X = ( \begin{matrix}0 & -1 \\ 1 & 0 \end{matrix} ) , X = ( \begin{matrix}1 & 1 \\ 0 & 1 \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 4 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 exponention of a square matrix is defined by

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

(b) i. If

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

then

X^{2} = I,

and so

e^{X} = (1 + \frac{1}{2!} + \frac{1}{4!} + ⋯)I + (\frac{1}{1!} + \frac{1}{3!} + ⋯)X

= cosh(1)I + sinh(1)X = (\begin{matrix}cosh 1 & sinh 1 \\ sinh 1 & cosh 1 \end{matrix})

ii. If

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

then

X^{2} = −I,

and so

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

iii. If

X = (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) = I + Y,

where

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

then

Y^{2} = 0 ⟹ e^{Y} = I + Y = X,

and so

e^{X} = e^{I}e^{Y}

= (\begin{matrix}e & e \\ 0 & e \end{matrix})

iv. If

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

where

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

then

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

from above, and so

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

and

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

(c) [Note that this part of the question only makes sense if k = ℝ or ℂ. One does not in general speak of the eigenvalues or eigenvectors of a matrix X over ℍ, since the solutions of Xv = qv will not in general form a subspace over ℍ.] Since e^{X} is the same whether we consider X as a real or complex matrix, we may assume that X ∈ Mat(n, ℂ).

We know that in this case X can be triangulated, ie we can find T such that

T XT^{−1} = (\begin{matrix}λ & c \\ 0 & μ \end{matrix})

But then

T X^{r}T^{−1} = (\begin{matrix}λ^{r} & c_{r} \\ 0 & μ^{r} \end{matrix})

for each r, and so

T e^{X}T^{−1} = (\begin{matrix}e^{λ} & c′ \\ 0 & e^{μ} \end{matrix})

Since Y and T Y T^{−1} have the same eigenvalues, it follows that e^{X} has eigenvalues e^{λ}, x^{μ}.

(d) i. From the last result,

det e^{X} = e^{λ} e^{μ}
= e^{λ+μ}
= e^{tr X}.

In particular,

det e^{Y} > 0

for all real Y. Since

det X = −1

in this case,

X ≠ e^{Y} .

ii. The map

x + iy ↦ (\begin{matrix}x & −y \\ y & x \end{matrix}) : ℂ → Mat(2, R)

is a homomorphism of ℝ-algebras under which

z ↦ X ⟹ e^{z} ↦ e^{X}.

The matrix

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

corresponds to the complex number i. But i = e^{z} where z = π/2i. Thus X = e^{Y} where

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

iii. We saw that

X = e^{Y}

in this case. iv. As in the second case, the matrix X corresponds to the com-

plex number

1 − i = \sqrt{2}e^{−iπ/4}.

Thus 1 − i = e^{z} where

z = log 2/2 − iπ/4.

Hence X = e^{Y}, with

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

8. Show that the connected component G_{0} of a linear group G is given by

G_{0} = {e^{X_{1}}e^{X_{2}} ⋯ e^{X_{r}}} (X_{1}, X_{2}, …, X_{r} ∈ ℒ G),

where r = 1, 2, . . ..

Explain how a representation α of a linear group G defines a representation ℒ α of ℒ G, and show that if G is connected then

ℒ α = ℒ β ⟹ α = β.

Sketch the proof that if G is simply connected then every representation of ℒ G arises from a representation of G.

Answer:

(a) It is clear that G_{0} is closed under multiplication; and it is closed under inversion, since

(e^{X_{1}} ⋯ e^{X_{r}})^{−1} = e^{−X_{r}} ⋯ e^{−X_{1}}.

Hence G_{0} is a subgroup. Also, G_{0} is connected, since

T (t) = e^{tX_{1}} ⋯ e^{tX_{r}} (0 ≤ t ≤ 1)

is a path connecting I to e^{X_{1}} ⋯ e^{X_{r}}.

Finally, G_{0} is open. For there exists an open subset U ∋ 0 in ℒ G which is mapped homeomorphically onto an open subset V = e^{U} ∋ I in G; and

e^{X_{1}} ⋯ e^{X_{r}}e^{U}

is an open neighbourhood of e^{X_{1}} ⋯ e^{X_{r}}. Since G_{0} is an open subgroup, it is also closed; so G_{0} and its complement are both open, and G_{0} is the connected component of I in G.

(b) We assume the following result:

Lemma 1 Suppose

F : G → H

is a continuous homomorphism of linear groups. Then there is a unique Lie algebra homomorphism

f : ℒ G → ℒ H

such that

F (e^{X}) = e^{f(X)}

for all X ∈ ℒ G.

Now suppose

α : G → GL(V)

is a representation of G. By the Lemma, this gives rise to a Lie algebra homomorphism

ℒ α : ℒ G → gl(V),

ie a representation of the Lie algebra ℒ G in V, such that

α(e^{X}) = e^{ℒ α(X)}

for all X ∈ ℒ G.

(c) Suppose

ℒ α = ℒ β = f,

say; and suppose T ∈ G. Then

T = e^{X_{1}} ⋯ e^{X_{r}},

since G_{0} = G. Hence

α(T) = α(e^{X_{1}}) ⋯ α(e^{X_{r}})
= e^{fX_{1}} ⋯ e^{fX_{r}}
= β(e^{X_{1}}) ⋯ β(e^{X_{r}})
= β(T).

Thus

α = β.

(d) Suppose G, H are linear groups; and suppose the Lie algebra ho-

momorphism

f : ℒ G → ℒ H

can be lifted to a homomorphism

F : G → H,

satisfying

F (e^{X}) = e^{fX}

for all X ∈ ℒ G. If

e^{X_{1}} ⋯ e^{X_{r}} = 1

is an ‘exponential relation’ in G, then

e^{fX_{1}} ⋯ e^{fX_{r}} = F (e^{X_{1}}) ⋯ F (e^{X_{r}})

= F (e^{X_{1}} ⋯ e^{X_{r}}) = 1

in H. Thus

e^{X_{1}} ⋯ e^{X_{r}} = 1 ⟹ e^{fX_{1}} ⋯ e^{fX_{r}} = 1.

Conversely if this is so, ie every exponential relation in G maps to a corresponding relation in H, then the required homomorphism F : G → H can be defined as follows: given T ∈ G, suppose

T = e^{X_{1}} ⋯ e^{X_{r}}.

Then we set

F (T) = e^{fX_{1}} ⋯ e^{fX_{r}}.

It follows at once from the hypothesis that F (T) is well-defined, ie independent of the ‘exponential product’ we choose for T, and that F is a homomorphism with ℒ F = f. i. This property always holds locally: if all the partial products

T = e^{X_{1}}, e^{X_{1}}e^{X_{2}}, e^{X_{1}}e^{X_{2}}e^{X_{3}}, …

lie in the logarithmic zone U then the corresponding relation in H holds. It is sufficient to prove this for ‘triangular relations’

e^{X}e^{Y} e^{Z} = 1.

This is established by showing that if X, Y, Z are small of size d then the ‘discrepancy’

e^{fX}e^{fY} e^{fZ} − 1

is of order d^{3}. Since a triangle of size d can be split into n^{2} triangles of size d/n, it follows that the discrepancy of a triangle in U is in fact 0.

ii. Now suppose G is simply connected, ie every loop is homotopically trivial.

9. Define the Killing form K(X, Y) of a Lie algebra ℒ.

Determine the Killing form of sl(2, R), and show that it is non-singular.

Show that if G is a compact linear group then the Killing form of G is negative definite or indefinite.

Show conversely that if the Killing form a connected linear group G is negative definite then G is compact. Is the condition of connectedness necessary here?

Answer:

(a) The Killing form is the symmetric bilinear form

K(X, Y) = tr(ad X ad Y),

where ad X is the map

Z ↦ [X, Z] : ℒ → ℒ .

(b) We have

sl(2, ℝ) = {X ∈ Mat(2, ℝ) : tr X = 0}
= ⟨H, E, F ⟩,

where

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}) .

Thus

[H, E] = HE − EH = 2E,
[H, F] = HF − F H = −2F,
[E, F] = EF − F E = H.

Now

ad H(H) = [H, H] = 0,  ad H(E) = [H, E] = 2E, ad H(F) = [H, F] = −2F,
ad E(H) = [E, H] = −2E, ad E(E) = [E, E] = 0, ad E(F) = [E, F] = H,
ad F (H) = [F, H] = 2F, ad F (E) = [F, E] = −H, ad F (F) = [F, F] = 0.

Thus ad H, ad E, ad F take matrix forms

ad H ↦ (\begin{matrix}0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & −2 \end{matrix}) , ad E ↦ (\begin{matrix}0 & 0 & 1 \\ −2 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}) , ad F ↦ (\begin{matrix}0 & −1 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{matrix})

with respect to the basis H, E, F of sl(2, ℝ). Hence

K(H, H) = tr((ad H)^{2}) = 8,
K(E, E) = tr((ad E)^{2}) = 0,
K(F, F) = tr((ad F)^{2}) = 0,
K(H, E) = tr(ad H ad E) = 0,
K(H, F) = tr(ad H ad F) = 0,
K(E, F) = tr(ad E ad F) = 4.

Thus the Killing form (with respect to the basis H, E, F) is given by the quadratic form

K(x, y, z) = K(xH + yE + zF, xH + yE + zF) = 8x^{2} + 8yz,

which is non-singular (but neither positive nor negative).

(c) Suppose G is compact. The adjoint representatiion of G in ℒ G is given by

ad g(X) = gXg^{−1}.

Since G is compact, there is an positive-definite quadratic form left invariant under this action. Choose coordinates so that this form is

x_{1}^{2} + ⋯ + x_{n}^{2}.

Then

ad g ∈ O(n).

This representation of G corresponds to the adjoint representation of ℒ G:

ad X(Z) = [X, Z].

It follows that

ad X ∈ o(n) = {S ∈ Mat(n, ℝ) : S + S′ = 0}.

But then if ad X = S

(ad X)^{2} = −S′S,

and so

tr((ad X)^{2}) = − tr(S′S) ≤ 0,

with equality only if S = 0. We conclude that K(X, Y) is negative definite or indefinite.

(d) If the Killing form K(X, Y) of a linear group G is negative-definite then it does follow that G is compact. The proof is long, and the following is more of an overview. [A complete proof can be found in Chapter 12 of Part IV of my notes.] We assume the following result:

Lemma 2 The Killing form of a compact group G is left invariant by G:

K(ad g(X), ad g(Y)) = K(X, Y).

If the Killing form is negative-definite, we can choose coordinates so that it takes the form

−(x_{1}^{2} + ⋯ + x_{n}^{2}).

Thus if we set

G_{1} = im Ad,

then

G_{1} ⊂ O(n).

[In fact, since G is connected,

G_{1} ⊂ SO(n).]

Now

ker Ad = ZG,

the centre of G. This is discrete; for

ℒ (ZG) = Z(ℒ G) = 0,

since

X ∈ ZG ⟹ ad X = 0
⟹ K(X, Y) = 0

for all Y ∈ ℒ G, in which case the Killing form will be singular. Thus

Ad : G → G_{1}

is a covering. A compactness argument shows that ZG is finitely-generated. If it is finite we are done; it is easy to see that G is compact. If not then ZG is an infinite discrete abelian subgroup; Thus

ZG = T ⊕ ℤ^{n}

where T is finite and n > 0. In particular, there is a non-trivial homomorphism

χ : ZG → ℝ.

We are going to show that this can be extended to a homomorphism

X : G → ℝ.

This will lead to a contradiction. For the kernel of the corresponding Lie algebra homomorphism will be an (n−1)-dimensional subspace of ℒ G; and it is easy to see that this subspace would in fact be an ideal, whose complement with respect to the non-singular Killing form would be a 1-dimensional ideal I ⊂ ℒ G. This would necessarily be trivial, so that K(X, Y) = 0 for all X ∈ I, Y inℒ G, contradicting the assertion that K(X, Y) is negative-definite.

To construct the extension X, we note first that a standard compactness argument shows we can find a compact subset C ⊂ G such that

Ad(C) = G_{1}.

Now let u(x) be a function on G with compact support such that

x ∈ C ⟹ u(x) > 0.

Set

w(x) = \frac{u(x)}{∑_{z∈ZG} u(zx)}

Then

w(x) > 0 and ∑_{z∈ZG} w(zx) = 1

for all x ∈ G. (We can think of w(x) as a kind of weight on G, allowing us to smooth out the given homomorphism χ.) Set

f(x) = ∑_{z∈ZG} w(zx)χ(z).

Suppose z′ ∈ ZG. As z runs over ZG, so does zz′. Hence

f(x) = ∑_{z∈ZG} w(zz′x)χ(zz′)

= ∑_{z∈ZG} w(zz′x)(χ(z) + χ(z′))

= f(z′x) + χ(z′).

In other words,

f(zx) = f(z) − χ(z).

Thus if we define the function F : G × G → ℝ by

F (x, y) = f(xy) − f(x),

then

F (zx, y) = F (x, y).

But that means we can regard F as a function on G_{1} × G, where G_{1} = G/ZG is compact. This allows us to integrate over G_{1}, and set

X(g) = ∫_{G_{1}} F (g_{1}, g).

It is a straightforward matter to verify that this function X : G → ℝ is in fact a homomorphism extending χ. [This is a very complicated argument. Hopefully nothing as horrid as this will be asked in the exam!]