Course 424

Group Representations III

Dr Timothy Murphy

EELT 3  Tuesday, 11 May 1999  14:00–16:00

Answer as many questions as you can; all carry the same number of marks. In this exam, ‘Lie algebra’ means Lie algebra over ℝ, and ‘representation’ means finite-dimensional representation over ℂ.

1. Define the exponential e^{X} of a square matrix X.

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

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

Which of these 6 matrices X are themselves expressible in the form X = e^{Y}, where Y is a real matrix? (Justify your answers in all cases.)

Answer: The exponential of a square matrix X is defined by

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

This series converges for all X ∈ Mat(n, k) by comparison with the series for e^{∥X∥}, since ∥X^{n}∥ ≤ ∥X∥^{n}.

(a) If

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

then

X^{2} = 0

and so

e^{X} = I + X = (\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) .

(b) If

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

then

X^{2} = I,

and so

e^{X} = (\begin{matrix}cosh 1 & sinh 1 \\ sinh 1 & cosh 1 \end{matrix})

(c) If

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

then

X^{2} = −I,

and so

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

(d) If

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

then

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

(e) If

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

where

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

then, since I, Y commute,

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

(f) If

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

where

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

then, since −I, Z commute,

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

(a) e^{Y} is non-singular for all Y, since e^{Y} e^{−Y} = I. Since X is singular in this case, X ≠ e^{Y}.

(b) X has eigenvalues ±1. Suppose X = e^{Y}; and suppose Y has eigenvalues λ, μ. Then X has eigenvalues e^{λ}, e^{μ}. There are two possibilities. Either λ, μ are complex conjugates, in which case the same is true of e^{λ}, e^{μ}; or else λ, μ are both real, in which case e^{λ}, e^{μ} > 0. In neither case can we get ±1. Hence X ≠ e^{Y}.

(c) By the isomorphism between the complex numbers z = x + iy and

the matrices

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

we see that

X = ℂ(i).

Since

i = e^{πi/2},

while

ℂ(e^{z}) = e^{ℂ(z)},

it follows that X = e^{Y} with

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

(d) X has eigenvalues ±1. Thus by the argument in case (b) above, X ≠ e^{Y}.

(e) We have

X = (\begin{matrix}1 & −1 \\ 1 & 1 \end{matrix}) = ℂ(1 + i).

But

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

Thus X = e^{Y} with

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

(f) X has eigenvalues −1, −1. Thus if X = e^{Y} (with Y real) then Y must have eigenvalues ±(2n + 1)πi for some integer n. In particular, Y has distinct eigenvalues, and so is semisimple (diagonalisable over ℂ). But in that case X = e^{Y} would also be semisimple. That is impossible, since a diagonalisable matrix with eigenvalues −1, −1 is necessarily −I. Hence X ≠ e^{Y}.

2. Define a linear group, and a Lie algebra; and define the Lie algebra ℒ G 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 linear group is a closed subgroup G ⊂ GL(n, ℝ) for some n.

A Lie algebra is defined by giving

(a) a vector space L;

(b) a binary operation on L, ie a map

L × L → L : (X, Y) ↦ [X, Y]

satisfying the conditions

(a) The product [X, Y] is bilinear in X, Y;

(b) The product is skew-symmetric:

[Y, X] = −[X, Y];

(c) Jacobi’s identity is satisfied:

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

for all X, Y, Z ∈ L.

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}.

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.

The dimension of a linear group G is the dimension of the real vector space ℒ G:

dim G = dim_{ℝ} ℒ G.

(a) 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}.

(b) We have

so(n) = {X ∈ Mat(n, ℝ) : X′ + X = 0, tr X = 0} = o(n),

since X′ + X = 0 ⟹ tr X = 0. Thus

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

(c) 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}.

(d) We have

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

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

dim SU(n) = dim U(n) − 1 = n^{2} − 1.

(e) We have

gl(n, ℝ) = Mat(n, ℝ).

Thus

dim GL(n, ℝ) = n^{2}.

(f) We have

sl(n, ℝ) = {X ∈ Mat(n, ℝ) : tr X = 0}.

This imposes one linear condition on X. Thus

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

(g) 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}.

(h) We have

sl(n, ℂ) = {X ∈ Mat(n, ℂ) : tr X = 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.

3. 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: 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} = I∀t}
= {X : X^{∗} = −X}
= {X : X^{∗} + X = 0}.

Also

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

Since

SU(2) = U(2) ∩ SL(2, ℂ)

it follows that

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

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} = I∀t}
= {X : X′ = −X}
= {X : X′ + X = 0}.

Also

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

Since

SO(3) = O(3) ∩ SL(3, ℝ)

it follows that

so(3) = o(3) ∩ sl(3, ℝ)
= {X : X′ + X = 0, tr X = 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⟩.

Finally, su(2) and so(3) are isomorphic under the correspondence

e ↔ −2U, f ↔ −2V, g ↔ −2W.

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

ZSU(2) = {±I} while ZSO(3) = {I}.

4. Define a representation of a Lie algebra; and show how each representation α of a linear group G gives rise to a representation ℒ α of ℒ G. Determine the Lie algebra of SL(2, ℝ); and show that this Lie algebra sl(2, ℝ) has just 1 simple representation of each dimension 1, 2, 3, . . .. Answer: 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. Suppose α is a representation of the linear group G, ie a homomorphism

α : G → GL(n, ℂ).

Under the Lie correspondence this gives rise to a Lie algebra homomorphism

A = ℒ α : L = ℒ G → gl(n, ℂ).

But now L acts on V = C^{n} by

Xv = A(X)v.

This defines a representation of L in V since

[X, Y]v = A([X, Y])v
= [AX, AY]v
= ((AX)(AY) − (AY)(AX))v
= X(Y v) − Y (Xv).

We have

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

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

Hv = λv.

Now

[H, E]v = 2Ev,

that is,

HEv − EHv = 2Ev.

In other words, since Hv = λv,

H(Ev) = (λ + 2)Ev,

ie 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 us set

e^{i} = E^{i}e_{0}.

Then the e_{i} are all eigenvectors of H. Let us set e_{i} = 0 for i outside the range [0, n − 1]. Suppose e_{0} is a μ-eigenvector. Then e_{i} is a (μ + 2i)- egenvector. Let us suppose that there are n eigenvectors in the sequence, ie

e_{n−1} ≠ 0, Ee_{n−1} = 0.

Now we show by induction that

F e_{i} = ρ_{i}e_{i−1}

for each i. 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_{n} = 0. Hence

μ = n − 1.

We conclude that the subspace

⟨e_{0}, …, e_{n−1}⟩

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_{n−1}⟩ do indeed define a representation of sl(2, ℝ); so that this Lie algebra has exactly 1 simple representation of each dimension.

5. Show that a compact connected abelian linear group of dimension n is necessarily isomorphic to the torus 𝕋^{n}.

Answer: If G is a abelian linear group then ℒ G is trivial, ie [X, Y] = 0 for all X, Y ∈ ℒ G. For e^{tX}, e^{tY} ∈ G commute for all t. If t is sufficiently small we can take logs, and deduce that tX = log(e^{tX}), tY = log(e^{tY}) commute. Hence X, Y commute.

The map

Θ : ℒ G → G

under which

X ↦ e^{X}

is a homomorphism, since

X + Y ↦ e^{X+Y} = e^{X}e^{Y} .

For any linear group G, there exist open subsets U ∋ 0 in ℒ G, V ∋ I in G such that X ↦ e^{X} defines a homeomorphism U → V.

It follows that im Θ ⊂ ℒ G is an open subgroup of G. Since G is connected, im Θ = G. Thus

G ≅ ℒ G/ ker Θ.

Moreover, ker Θ is discrete, since U ∩ ker Θ = {0}. Thus

G ≅ ℝ^{n}/K,

where K is a discrete subgroup, and n = dim G.

Lemma: A discrete subgroup K ⊂ ℝ^{n} is necessarily ≅ ℤ^{d} for some d ≤ n, ie we can find a ℤ-basis k_{1}, …, k_{d} for K such that

K = {n_{1}k_{1} + ⋯ + n_{d}k_{d} : n_{1}, …, n_{d} ∈ ℤ}.

Proof: Let k_{1} be one of the closest points to 0 in K \ {0}. Then let k_{2} be one of the closest points to the subspace ⟨k_{1}⟩ in K \ ⟨k_{1}⟩, let k_{3} be one of the closest points to the subspace ⟨k_{1}, k_{2}⟩ in K \ ⟨k_{1}, k_{2}⟩, and so on.

Then k_{1}, k_{2}, … are linearly independent. So the process must end after d ≤ n steps:

K = ⟨k_{1}, …, k_{d}⟩.

Now suppose k ∈ K, say

k = λ_{1}k_{1} + ⋯ + λ_{d}k_{d}.

We show that λ_{1}, …, λ_{d} ∈ ℤ. Let

λ_{d} = r + ϵ,

where r ∈ ℤ and |ϵ| ≤ 1/2. Then

k − rk_{d} = λ_{1}k_{1} + ⋯ + λ_{d−1}k_{d−1} + ϵk_{d}

is closer to ⟨k_{1}, …, k_{d−1}⟩ than is k_{d}. Hence ϵ = 0, ie λ_{d} ∈ ℤ. Applying the same argument to

k − λ_{d}k_{d} = λ_{1}k_{1} + ⋯ + λ_{d−1}k_{d−1},

we deduce that λ_{d−1} ∈ ℤ; and so successively λ_{d−2}, …, λ_{1} ∈ ℤ Thus k_{1}, …, k_{d} is a Z-basis for K. Extend k_{1}, …, k_{d} to a basis k_{1}, …, k_{d}, e_{1}, …, e_{n−d} of ℝ^{n}. Then

G ≅ ℝ^{n}/K ≅ ℝ/ℤ ⊕ ⋯ ⊕ ℝ/ℤ
≅ 𝕋^{d} ⊕ ℝ^{n−d}.

Since G is compact, n − d = 0, ie

G ≅ 𝕋^{n}.