and Cartan. It is also usually expressed slightly di¬erently. If x1 , x2 , . . . , xn is a basis for

g with structure constants ci , then ω can be written in the form

jk

ω = x1 ω 1 + · · · + xn ω n

where the ω i are R-valued left-invariant 1-forms and Proposition 9 can then be expanded

to give

dω i = ’ 1 ci ω j § ω k ,

2 jk

which is the most common form in which the structure equations are given. Note that the

identity d(d(ω i )) = 0 is equivalent to the Jacobi identity.

An Extended Example: 2- and 3-dimensional Lie Algebras. It is clear that

up to isomorphism, there is only one (real) Lie algebra of dimension 1, namely g = R with

the zero bracket. This is the Lie algebra of the connected Lie groups R and S 1. (You are

asked to prove in an exercise that these are, in fact, the only connected one-dimensional

Lie groups.)

The ¬rst interesting case, therefore, is dimension 2. If g is a 2-dimensional Lie alge-

bra with basis x1 , x2 , then the entire Lie algebra structure is determined by the bracket

[x1 , x2 ] = a1 x1 +a2 x2 . If a1 = a2 = 0, then all brackets are zero, and the algebra is abelian.

If one of a1 or a2 is non-zero, then, by switching x1 and x2 if necessary, we may assume

that a1 = 0. Then, considering the new basis y1 = a1 x1 + a2 x2 and y2 = (1/a1 )x2 , we

get [y1 , y2 ] = y1 . Since the Jacobi identity is easily veri¬ed for this Lie bracket, this does

de¬ne a Lie algebra. Thus, up to isomorphism, there are only two distinct 2-dimensional

Lie algebras.

The abelian example is, of course, the Lie algebra of the vector space R2 (as well as

the Lie algebra of S 1 — R, and the Lie algebra of S 1 — S 1 ).

An example of a Lie group of dimension 2 with a non-abelian Lie algebra is the matrix

Lie group

ab

a ∈ R+ , b ∈ R .

G=

01

In fact, it is not hard to show that, up to isomorphism, this is the only connected non-

abelian Lie group (see the Exercises).

Now, let us pass on to the classi¬cation of the three dimensional Lie algebras. Here,

the story becomes much more interesting. Let g be a 3-dimensional Lie algebra, and let

x1 , x2 , x3 be a basis of g. Then, we may write the bracket relations in matrix form as

( [x2 , x3 ] [x3 , x1 ] [x1 , x2 ] ) = ( x1 x2 x3 ) C

where C is the 3-by-3 matrix of structure constants. How is this matrix a¬ected by a

change of basis? Well, let

( y1 y2 y3 ) = ( x1 x2 x3 ) A

L.2.18 29

where A ∈ GL(3, R). Then it is easy to compute that

( [y2 , y3 ] [y3 , y1 ] [y1 , y2 ] ) = ( [x2 , x3 ] [x3 , x1 ] [x1 , x2 ] ) Adj(A)

where Adj(A) is the classical adjoint matrix of A, i.e., the matrix of 2-by-2 minors. Thus,

A’1 = (det(A))’1 tAdj(A).

(Do not confuse this with the adjoint mapping de¬ned earlier!) It then follows that

( [y2 , y3 ] [y3 , y1 ] [y1 , y2 ] ) = ( y1 y2 y3 ) C ,

where

C = A’1 C Adj(A) = det(A) A’1 C tA’1 .

It follows without too much di¬culty that, if we write C = S + a, where S is a symmetric

ˆ

3-by-3 matrix and

« « 1

’a3 a2

0 a

ˆ = a3 ’a1 a = a2 ,

a 0 where

’a2 a1 a3

0

then C = S + a , where

S = det(A) A’1 S tA’1 a = tAa.

and

Now, I claim that the condition that the Jacobi identity hold for the bracket de¬ned

by the matrix C is equivalent to the condition Sa = 0. To see this, note ¬rst that

[[x2 , x3 ], x1 ] + [[x3 , x1 ], x2 ] + [[x1 , x2 ], x3 ]

1 2 3 1 2 3 1 2 3

= [C1 x1 + C1 x2 + C1 x3 , x1 ] + [C2 x1 + C2 x2 + C2 x3 , x2 ] + [C3 x1 + C3 x2 + C3 x3 , x3 ]

= (C3 ’ C2 )[x2 , x3 ] + (C1 ’ C3 )[x3 , x1 ] + (C2 ’ C1 )[x1 , x2 ]

2 3 3 1 1 2

= 2a1 [x2 , x3 ] + 2a2 [x3 , x1 ] + 2a3 [x1 , x2 ]

= 2 ( [x2 , x3 ] [x3 , x1 ] [x1 , x2 ] ) a

= 2 ( x1 x2 x3 ) Ca,

and Ca = (S + a)a = Sa since ˆ a = 0. Thus, the Jacobi identity applied to the basis

ˆ a

x1 , x2 , x3 implies that Sa = 0. However, if y1 , y2 , y3 is any other triple of elements of g,

then for some 3-by-3 matrix B, we have

( y1 y2 y3 ) = ( x1 x2 x3 ) B,

and I leave it to the reader to check that

[[y2 , y3 ], y1 ]+[[y3 , y1 ], y2 ]+[[y1 , y2 ], y3 ] = det(B) ([[x2 , x3 ], x1 ] + [[x3 , x1 ], x2 ] + [[x1 , x2 ], x3 ])

L.2.19 30

in this case. Thus, Sa = 0 implies the full Jacobi identity.

There are now two essentially di¬erent cases to treat. In the ¬rst case, if a = 0,

then the Jacobi identity is automatically satis¬ed, and S can be any symmetric matrix.

However, two such choices S and S will clearly give rise to isomorphic Lie algebras if and

only if there is an A ∈ GL(3, R) for which S = det(A) A’1 S tA’1 . I leave as an exercise

for the reader to show that every choice of S yields an algebra (with a = 0) which is

equivalent to exactly one of the algebras made by one of the following six choices:

« « «

000 010 010

0 0 0 1 0 0 1 0 0

000 000 001

.

« « «

100 100 100

0 0 0 0 1 0 0 1 0

000 000 001

On the other hand, if a = 0, then by a suitable change of basis A, we see that we can

assume that a1 = a2 = 0 and that a3 = 1. Any change of basis A which preserves this

normalization is seen to be of the form

«1

A1 A1 A1

2 3

A2 A2 A2 .

A= 1 2 3

0 0 1

Since Sa = 0 and since S is symmetric, it follows that S must be of the form

«

s11 s12 0

S = s12 s22 0.

0 0 0

Moreover, a simple calculation shows that the result of applying a change of basis of the

above form is to change the matrix S into the matrix

«

s11 s12 0

S = s12 s22 0

0 00

where

’A1 ’A2

A2 A2

1

s11 s12 s11 s12

2 2 2 1

= 12 .

’A2 ’A1

A1 A2 ’ A1 A2 1 1

s12 s22 A1 s12 s22 A1

1 2

21

It follows that s11 s22 ’ (s12 )2 = s11 s22 ’ (s12 )2 , so there is an “invariant” to be dealt

with. We leave it to the reader to show that the upper left-hand 2-by-2 block of S can be

brought by a change of basis of the above form into exactly one of the four forms

0 0 1 0 σ 0 σ 0

’σ

0 0 0 0 0 σ 0

where σ > 0 is a real positive number.

L.2.20 31

To summarize, every 3-dimensional Lie algebra is isomorphic to exactly one of the

following Lie algebras: Either

[x2 , x3 ] = x1 [x2 , x3 ] = x2

sl(2, R) : [x3 , x1 ] = x1

so(3) : [x3 , x1 ] = x2 or

[x1 , x2 ] = x3 [x1 , x2 ] = x3

or an algebra of the form

[x2 , x3 ] = b11 x1 + b12 x2

[x3 , x1 ] = b21 x1 + b22 x2

[x1 , x2 ] = 0

where the 2-by-2 matrix B is one of the following

0 0 1 0 1 0 0 1

0 0 0 0 0 1 1 0

01 1 1 σ 1 σ 1

’1 0 ’1 ’1 ’1 ’σ

0 σ

and, in the latter two cases, σ is a positive real number. Each of these eight latter types

can be represented as a subalgebra of gl(3, R) in the form

±«

’b11 z

(1 + b21 )z

x

g = b22 z y x, y, z ∈ R

(1 ’ b12 )z

0 0 z

I leave as an exercise for the reader to show that the corresponding subgroup of GL(3, R)

is a closed, embedded, simply connected matrix Lie group whose underlying manifold is

di¬eomorphic to R3 .

Actually, it is clear that, because of the skew-symmetry of the bracket, only n n of2

these constants are independent. In fact, using the dual basis x1 , . . . , xn of g— , we can

write the expression for the Lie bracket as an element β ∈ g — Λ2 (g— ), in the form

β = 1 ci xi — xj § xk .

2 jk

The Jacobi identity is then equivalent to the condition J (β) = 0, where

J : g — Λ2 (g— ) ’ g — Λ3 (g— )

is the quadratic polynomial map given in coordinates by

cij cm + cjk cm + cki cm xm — xi § xj § xk .

1

J (β) = k i j

6

L.2.21 32

Exercise Set 2:

Lie Groups

1. Show that for any real vector space of dimension n, the Lie group GL(V ) is isomorphic

to GL(n, R). (Hint: Choose a basis b of V , use b to construct a mapping φb : GL(V ) ’

GL(n, R), and then show that φb is a smooth isomorphism.)

2. Let G be a Lie group and let H be an abstract subgroup. Show that if there is an open

neighborhood U of e in G so that H © U is a smooth embedded submanifold of G, then H

is a Lie subgroup of G.

3. Show that SL(n, R) is an embedded Lie subgroup of GL(n, R). (Hint: SL(n, R) =

det’1 (1).)

4. Show that O(n) is an compact Lie subgroup of GL(n, R). (Hint: O(n) = F ’1 (In ),

where F is the map from GL(n, R) to the vector space of n-by-n symmetric matrices given

by F (A) = t A A. Taking note of Exercise 2, show that the Implicit Function Theorem

applies. To show compactness, apply the Heine-Borel theorem.) Show also that SO(n) is

an open-and-closed, index 2 subgroup of O(n).

5. Carry out the analysis in Exercise 3 for the complex matrix Lie group SL(n, C) and the

analysis in Exercise 4 for the complex matrix Lie groups U(n) and SU(n). What are the

(real) dimensions of all of these groups?

6. Show that the map µ: O(n) — An — Nn ’ GL(n, R) de¬ned by matrix multiplication

is a di¬eomorphism although it is not a group homomorphism. (Hint: The map is clearly

smooth, you must only compute an inverse. To get the ¬rst factor ν1 : GL(n, R) ’ O(n) of

the inverse map, think of an element b ∈ GL(n, R) as a row of column vectors in Rn and

let ν1 (b) be the row of column vectors which results from b by apply the Gram-Schmidt

orthogonalization process. Why does this work and why is the resulting map ν1 smooth?)

Show, similarly that the map

µ: SO(n) — An © SL(n, R) — Nn ’ SL(n, R)

is a di¬eomorphism. Are there similar factorizations for the groups GL(n, C) and SL(n, C)?

(Hint: Consider unitary bases rather than orthogonal ones.)

7. Show that

a ’b

SU(2) = aa + bb = 1 .

ba

Conclude that SU(2) is di¬eomorphic to the 3-sphere and, using the previous exercise,

that, in particular, SL(2, C) is simply connected, while π1 SL(2, R) Z.