notes on lie groups

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Math 845

Notes on Lie Groups

Mark Reeder

December 22, 2010

Contents

1 Quaternions and the three-dimensional sphere 3

1.1 Hamiltion’s quaternions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 The Lie group S 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 The normalizer of Q8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 The exponential map for S 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Rotations of three-dimensional space 10

2.1 The orthogonal and special orthogonal groups . . . . . . . . . . . . . . . . . . . . . . 10

2.2 SO3 and quaternions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3 The exponential map for SO3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.4 The exponential diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Definition and basic properties of Lie groups 16

3.1 Introduction to Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.2 Manifolds defined by equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.3 Lie groups: definition and first examples . . . . . . . . . . . . . . . . . . . . . . . . . 18

4 The Lie algebra of a Lie group 20

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4.1 The tangent bundle of a manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4.2 Vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.3 The tangent bundle of a Lie group . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.4 One-parameter-subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.5 The exponential map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.6 The Adjoint representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.6.1 The product rule for paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.7 The Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

5 Abelian Lie groups 36

6 Subgroups of Lie groups 37

6.1 Closed subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

6.2 Homogeneous spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

6.3 Compact subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

7 Maximal Tori 43

7.1 The Weyl group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

7.2 The flag manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

7.3 Conjugacy of maximal tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

7.4 Fixed-points in flag manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

7.4.1 de Rham cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

8 Octonions and G2 52

8.1 Composition algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

8.2 The product on the double of a subalgebra . . . . . . . . . . . . . . . . . . . . . . . . 53

8.3 Parallelizable spheres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

8.4 Automorphisms of composition algebras . . . . . . . . . . . . . . . . . . . . . . . . . 56

8.5 The Octonions O . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

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8.6 The SU3 in Aut(O) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

8.7 The maximal torus in Aut(O) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

8.8 The SO4 in Aut(O) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

8.9 The Lie algebra of Aut(O) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

8.10 The nonsplit extension 23 · GL3(2) in Aut(O) and seven Cartan subalgebras . . . . . 63

1 Quaternions and the three-dimensional sphere

1.1 Hamiltion’s quaternions

The quaternion algebra H is a four dimensional real vector space with basis 1, i , j , k:

H = R1 ⊕ Ri ⊕ R j ⊕Rk

and multiplication rules

ij = k, jk = i, ki = j, i2 = j2 = k2 = −1,

extended to H via the associative and distributive laws. The subalgebra R = R1 is the center of H, and

every quaternion q ∈ H may be expressed as

q = t + xi + yj + zk

for unique t, x, y,z

∈R.

The conjugate of q = t + xi + yj + zk is the quaternion

q = t − xi − yj − zk.

Thus, R = {q ∈ H : q = q }. One checks that

pq = q p,

for all p, q ∈ H. The norm of q is

N (q ) = q q ∈ ROne checks that N (q ) = t2 + x2 + y2 + z 2, for q = t + xi + yj + zk . Hence N (q )

≥ 0, with equality

only for q = 0. One also checks that

N ( pq ) = N ( p)N (q ).

It follows that if q = 0 then N (q )−1 · q is a multiplicative inverse of q in H. Hence H is a division

algebra, that its set of nonzero elements

H× = H− {0}

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is a group under quaternion multiplication, and that the norm N is a homomorphism

N : H× −→ R×>0

from H× to the group R×>0 of positive real numbers under multiplication, whose kernel

ker N =

{q

∈H× : q q = 1

} =

{t + xi + yj + zk

∈H : t2 + x2 + y2 + z 2 = 1

}may be identified with the three-dimensional sphere S 3 ⊂ R4.

1.2 The Lie group S 3

From now on we write

S 3 = {q ∈ H× : q q = 1}.

Thus S 3 is a group under quaternion multiplication, fitting into the exact sequence

1 −→ S 3 −→ H× N

−→ R×

>0 −→ 1.

The group S 3 contains the quaternion group

Q8 = {±1, ±i, ± j, ±k}of order eight as a subgroup, so S 3 is nonabelian. and in fact the center of S 3 has just two elements:

Z (S 3) = {±1},

since this is already the full center of Q8. The aim for the rest of this section is to find the noncentral

conjugacy classes in S 3.

The subgroup

T = {t + xi : t2 + x2 = 1} = {eiθ : θ ∈ R}is an abelian subgroup of S 3, isomorphic to S 1, the circle group. One checks that

T = C S 3(i)

is the centralizer of i in S 3. Let N (T ) be the normalizer of T in S 3.

Lemma 1.1 We have N (T ) = T ∪ T j. Thus N (T ) consists of two circles, which are cosets of T .

Proof: The elements of order four in T are just ±i. Hence if q ∈ N (T ) we have either qiq −1 = i or

qiq −1 = −i. The former means that q ∈ T . Assume that qiq −1 = −i. We note that j ij−1 = −i as

well, so qj−1 ∈ C S 3i = T , which means q ∈ T j.

We note that Q8 < N (T ) and that

jsj−1 = s = s−1 for all s ∈ T .

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Also T j = {tj + xk : t2 + x2 = 1} lies on the equatorial two sphere

C 0 := {xi + yj + zk : x2 + y2 + z 2 = 1} ⊂ S 3.

The meaning of the subscript “0” is as follows.

As we have defined the norm of a quaternion q to be N (q ) = q q , so we define the trace of q to be

τ (q ) = 12(q + q ).

Note that τ : H → R because τ (q ) = τ (q ). In fact we have

τ (t + xi + yj + zk) = t.

Lemma 1.2 For q ∈ S 3 and all p ∈H we have τ (qpq −1) = τ ( p).

Proof: Since q

∈ S 3 we have q −1 = q . We compute

τ (qpq −1) = τ (qpq ) = 12

(qpq + qpq ) = 12

(qpq + ¯q pq ) = 12

(qpq + q pq ) = qτ ( p)q.

Since τ ( p) ∈ R it commutes with q , so we have

τ (qpq −1) = τ ( p)q q = τ ( p),

again because q ∈ S 3.

By Lemma 1.2, the restriction of τ to S 3 is a function

τ : S 3 → [−1, 1]

whose level setsC t = { p ∈ S 3 : τ ( p) = t}

are preserved under conjugation by S 3. For t = 0, the level set is the equatorial two-sphere C 0 men-

tioned above. We have

C 0 = S 3 ∩H0,

where

H0 = Ri ⊕ R j ⊕ Rk = { p ∈H : τ ( p) = 0}.

more generally, for fixed t ∈ [−1, 1], the level set

C t = t + {xi + yj + zk : x2 + y2 + z 2 = 1 − t2}is a translate of the sphere of radius √ 1 − t2 in H0. Here we are invoking the inner (dot) product on

H0 for which {i,j,k} is an orthonormal basis. We may think of C t as a sphere of constant latitude in

S 3. 1 Thus, S 3 is the disjoint union of its latitude spheres:

S 3 =

t∈[−1,1]

C t. (1)

1Of course C 1 = {1} and C −1 = {−1} are spheres of zero radius.

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Proposition 1.3 For each t ∈ [−1, 1] , the latitude sphere C t is a single conjugacy class in S 3. Hence

(1) is the partition of S 3 into conjugacy classes.

Proof: It must be shown that S 3 acts transitively on each latitude sphere C t. We first prove this for C 0.

Following Euler, we write

eiθ = cos θ + i sin θ, e jθ = cos θ + j sin θ, ekθ = cos θ + k sin θ.

Thus we have three subgroups T i, T j, T k < S 3, all isomorphic to S 1, given by

T i = {eiθ : θ ∈ R}, T j = {e jθ : θ ∈ R}, T k = {ekθ : θ ∈ R}.

Everything we have said about T i (previously called T ) holds for the other subgroups. Their normaliz-

ers are

N (T i) = T i ∪ T i j, N (T j) = T j ∪ T jk, N (T k) = T ki.

The nontrivial cosets T i j, T jk, T ki are three orthogonal great circles on the two-sphere C 0. Conjuga-

tion by j, k, i on T i T j, T k is inversion, meaning that

jeiθ j−1 = je−iθ, ke jθk−1 = e− jθ , iekθi−1 j = e−kθ.

It follows that T i conjugates the coset T i j to itself, and likewise for T j with T jk, and T k with T ki.

Explicitly, we have

eiθ · eiα j · e−iθ = ei(α+2θ) j, e jθ · e jαk · e− jθ = e j(α+2θ)k, ekθ · ekα j · e−kθ = ek(α+2θ)i.

Now take a point p ∈ C 0 and write it in spherical coordiates:

p = sin φ cos θi + sin φ sin θj + cos φk.

If we view k as the north pole of C 0 then conjugation by e jφ/2 sends k down to a point p on the same

latitude as p, and then conjugation by ekθ/2 sends p over to p. In other words, we have

ekθ/2e jφ/2 · k · e− jφ/2e−kθ/2 = p.

This proves that S 3 acts transitively on C 0 by conjugation.

Now for any t ∈ (−1, 1), define f t : C 0 → C t by

f t( p) = t + (√

1 − t2) p.

Then f t is bijective, with inverse

f −1t (q ) = q − t√

1 − t2,

and for all q ∈ S 3 we have f t(qpq −1) = f t( p). Now the transitivity on C t follows from the transitivity

on C 0, completing the proof.

We can write each t ∈ [−1, 1] as t = cos θ for θ ∈ [0, π], and we have the

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Corollary 1.4 For 0 < θ < π , the conjugacy class C cos θ meets each of T i, T j, T k in two mutually

inverse points. Namely,

C cos θ ∩ T i = {eiθ, e−iθ}, C cos θ ∩ T j = {e jθ , e− jθ}, C cos θ ∩ T k = {ekθ, e−kθ}.

Proof: The sets on the right hand side of each asserted equality consist of the points in T i, T j, T kwhose trace is cos θ.

We now understand conjugacy classes of points in S 3. The next step is conjugacy of circles. More pre-

cisely, by “circle” we mean a subgroup S < S 3 such that S S 1 via a continuous group isomorphism.

Lemma 1.5 For θ ∈ R , the subgroup eiθ of S 1 generated by eiθ is finite if θ ∈ 2πQ and is dense in

S 1 if θ /∈ 2πQ.

Proof: The group A = eiθ is finite if and only if einθ = 1 for some n ∈ Z, which is equivalent to

having θ ∈

2πQ. So if θ /∈

2πQ, the subgroup A is infinite. We prove that A is in fact dense in S 1, as

follows.

Let > 0 and subdivide S 1 into equal arcs, starting at 1, of length at most . In the infinite set A there

exist distinct points einθ and eimθ, with m = n, lying the same arc. Since A = e−imθA, it contains the

point ei(n−m)θ lying in an arc having 1 as an endpoint. The subgroup generated by ei(n−m)θ is contained

in A and meets every arc. Hence A is dense in S 1.

We revert to the notation T = T i = {eiθ : θ ∈ R}.

Proposition 1.6 Every circle in S 3 is conjugate to T .

Proof: Let S be a circle in S 3. This means S is a subgroup of S 3 and we have a continous group

isomorphism f : S 1 → S . Let s = f (eiθ), where θ ∈ R− 2πQ. Then s is dense in S , by Lemma 1.5

and the continuity of f . By Cor. 1.4, there exists q ∈ S 3 such that qsq −1 ∈ T . The conjugate element

qsq −1 also has infinite order, hence the subgroup qsq −1 is dense in T . Letting X denote the closure

of a subset X ⊂ S 3, we have

qSq −1 = q sq −1 = qsq −1 = T .

1.2.1 The normalizer of Q8

Our results in the previous section imply that i, j, k as well as T i, T j, T k are conjugate in S 3. On the

other hand, the relations in the quaternion group Q8 suggest that Q8 has an automorphism of order

three sending i → j → k → i. Is this automorphism realized by conjugation inside the normalizer

N (Q8) of Q8 in S 3? If so then, since each circle T ∗ is the centralizer of the corresponding subscript ∗,

the same element would conjugate T i → T j → T k → T i.

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Proposition 1.7 There are exactly two elements q ∈ S 3 which satisfy

qiq −1 = j, qjq −1 = k, qkq −1 = i,

namely ± 12(1 + i + j + k) , which have orders six (+) and three (−).

Proof: Letting q = t + xi + yj + zk and rewriting the equations as

qi = jq, qj = kq, qk = iq,

and equating coefficients, we find that t = x = y = z , so it suffices to determine t = τ (q ). Since we

must have t2 + x2 + y2 + z 2 = 1, it follows that t = ±12

. Since all elements of S 3 with a given t are

conjugate and

eπi/3 = 1

2 + i

√ 3, while eπi/3 = −1

2 + i

√ 3,

we see that 12(1 + i + j + k) has order six while −1

2(1 + i + j + k) has order three.

Elements q = 12(±1 ± i ± j ± k) ∈ S 3, with all possible combinations of signs, will conjugate i, j, kamongst one another, up to sign. That is, such elements q normalize Q8 and act on Q8 via outer

automorphisms. These elements, along with Q8 itself, comprise the normalizer of Q8. Thus, N (Q8)consists of the 24 quaternions

N (Q8) = {±1, ±i, ± j, ±k} ∪

1

2(±1 ± i ± j ± k)

,

and Q8 is the Sylow 2-subgroup of N (Q8). The group N (Q8) is also known as the binary tetrahedral

group for reasons that will become clear. One can show that

N (Q8) SL2(Z/3Z),

the group of 2 × 2 matrices over Z/3Z with determinant = 1.

1.3 The exponential map for S 3

The exponential map gives a canonical parametrization of compact Lie groups.

The circle group S 1 is parametrized by exponentiating the purely imaginary complex numbers iR.

Thus,

S 1 = {ez : z ∈ iR}, where ez =∞

n=0

z n

n!.

We can write each purely imaginary complex number z uniquely as z = ±iθ, where θ = |z | ≥ 0, and

we have Euler’s formula

ez = e±iθ = cos θ ± i sin θ,

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as one computes by expanding the exponential series. This parameterization of S 1 sends the open

segment

(−π, π)i = {±θi : 0 ≤ θ < π} = [0, π) · (S 1 ∩ iR)

bijectively onto S 1 − {−1} and both values ±πi are sent to −1 ∈ S 1. Thus, the map z → ez glues the

ends of the closed segment [−π, π]i together, forming a circle S 1.

Likewise, we can parametrize the 3-sphere S 3 ⊂ H, by exponentiating the pure quaternions: H0 =Ri +R j + Rk. Thus, we define

exp : H0 −→ S 3 by exp(v) =∞

n=0

vn

n!.

Let us compute this sum in closed form. As we did with S 1, we can write v = θv0, where v0 ∈ S 3∩H0,

and θ = |v|. Recall that S 3 ∩ H0 = C 0 is the conjugacy class of elements of order four; these are the

elements of H that behave like ±i, in that they square to −1. It follows that exp(v) can be computed in

the same way as e±iθ. For v2 = −θ2, so for all k ≥ 0 we have v2k = (−1)kθ2k and v2k+1 = (−1)kθ2kv.

It follows that

exp(v) =∞k=0

(−1)kθ2k

(2k)! + v

∞k=0

(−1)kθ2k

(2k + 1)! = cos θ + v

sin θ

θ = cos θ + v0 sin θ. (2)

as with Euler’s formula. This is consistent with our earlier definitions of eiθ, e jθ , ekθ; these were values

of exp on the three lines iR, jR, kR in H0.

From (2) we observe that exp maps the sphere θC 0 ⊂ H0 of radius θ to the conjugacy-class C cos θ.

Proposition 1.8 The map exp : H0 → S 3 has the following properties:

1. exp(H0) = S 3;

2. exp maps the open ball {v ∈ H0 : |v| < π} = [0, π) · C 0 bijectively onto S 3 − {−1};

3. exp(πC 0) = {−1} ⊂ S 3;

4. We have exp(qvq −1) = q exp(v)q −1 for all q ∈ S 3 and v ∈ H0.

Proof: Item 1 is implied by items 2 and 3. If q ∈ C cos θ then q = cos θ + q 0, where q 0 ∈ H0 has

squared-length

|q 0

|2 = 1

−cos2 θ = sin2 θ. The vector v0 = (sin θ)−1q 0 lies in C 0 and exp(θv0) = q .

Therefore exp(θC 0) = C cos θ.

Suppose v, v ∈ H0 have exp(v) = exp(v). Write v = θv0, v = θv0, with θ, θ ∈ [0, π) and

v0, v0 ∈ C 0. Since cos is injective on [0, π), it follows from (2) that θ = θ and v0 sin θ = v0 sin θ. If

θ = 0 then v = v = 0. Otherwise θ ∈ (0, π) and sin θ = 0, so v0 = v 0, hence v = v . If v0 is any point

in C 0, then exp(πv0) = cos π + v0 sin π = −1. This completes the proof of item 2.

Item 3 follows from the continuity of the map v → qvq −1.

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2 Rotations of three-dimensional space

2.1 The orthogonal and special orthogonal groups

Let V = Rn with the inner product

u, v =n

i=1

uivi,

where ui, vi are the coefficients of u, v with respect to the standard orthonormal basis {ei} of Rn. This

inner product is positive-definite, meaning that u, u > 0 for all nonzero vectors u ∈ Rn. The length

u is given by

|u| = u, u1/2.

The orthogonal group of V is the subgroup On ⊂ GLn(R) preserving the the lengths of vectors:

On =

{g

∈ GLn(R) :

|gu

| =

|u

|for all u

∈Rn

}.

It is useful to recognize when a matrix g belongs to On without having to check the condition |gu| = |u|for every vector u ∈ Rn.

Proposition 2.1 On For a matrix g ∈ GLn(R) , the following are equivalent.

1. g ∈ On.

2. We have gu,gv = u, v for all u, v ∈ Rn.

3. The columns of g form an orthonormal basis of Rn.

4. The product of g with its transpose is the identity matrix: g · tg = I .

Proof: The equivalence of items 1 and 2 results from the formula

u, v = 12

|u + v|2 − |u|2 − |v|2 .Applying item 2 to the orthonormal basis {ei}, we get item 3. Conversely, item 3 implies item 2 by

expanding u, v in terms of the basis {ei}. The entry in row i column j of g · tg is the inner product of

columns i and j of g, whence the equivalence of items 3 and 4.

The condition g · tg = I implies that det(g) = ±1 for all g ∈ On. The special orthogonal group is

the subgroup of determinant = 1:

SOn = {g ∈ On : det(g) = 1}.

We give On and SOn the topology inherited from the Euclidean space M n(R) = Rn2 of n × n real

matrices.

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Proposition 2.2 The subsets SOn, On ⊂ M n(R) are compact and SOn is connected, while On has two

connected components.

Proof: For 1 ≤ i ≤ j ≤ n, define functions f ij : M n(R) → R by

f ij(g) =

gi, g j

−δ ij ,

where gi, g j are the ith and j th columns of g ∈ M n(R), and δ ij = 1 or 0 according as i = j or i = j.

Then On is the set of common zeros of all the functions f ij , and SOn is the subset of On on which the

additional function det −1 is zero. All of these are polynomial, hence continous functions on M n(R),

so On and SOn are closed. Since the columns of any g ∈ On are orthonormal vectors, each entry of gbelongs to [−1, 1], hence On is a bounded subset of M n(R). Since On and SOn are closed and bounded

subsets of M n(R), it follows from the Heine-Borel theorem that On and SOn are compact.

To prove that SOn is connected, we show that every element lies in a connected subgroup. We will use

induction on n. Since SO1 = {1} and SO2 = S 1 is a circle, we may assume n ≥ 3 and that SOm is

connected for m < n.

Let g ∈ SOn, and let G = g be the closure in SOn of the subgroup generated by g. As SOn

is compact, the group G is also compact. Let λ ∈ C× be an eigenvalue of g. If λ = ±1 then a

corresponding eigenvector v lies in Rn. Scaling so that |v| = 1, and choosing an orthonormal basis of

the orthogonal complement of the line Rv, we obtain a matrix h ∈ On such that

hgh−1 ∈

1 00 SOn−1

,

which is connected, by the induction hypothesis. The conjugate by h of this subgroup is also connected,

so we have found a connected subgroup of SOn containing g.

Assume now that g has no eigenvalue equal to ±1. Let v = (v1, . . . , vn) ∈ Cn be an eigenvector of g,with eigenvalue λ ∈ C×, and let L = Cv be the complex line spanned by v. Since L is closed in Cn, it

is preserved by G, so we have a map f : G → L sending γ ∈ G to f (γ ) = γv. The map f has bounded

image, since G is compact. As f (gn) = λnv for all n ∈ Z, it follows that |λ| = 1, so λ = eiθ for some

θ ∈ R, and θ /∈ Zπ since λ = ±1. Let v = (v1, . . . , vn). Since g has real entries, we have

gv = gv = eiθv = e−iθv.

Since eiθ = e−iθ, the vectors v and v are linearly independent. Hence the vectors u = v + v and

w = i(v − v) are nonzero and linearly independent. These vectors u, w satisfy u = u and w = w , so

u, w ∈ Rn. We set c = cos θ, s = sin(θ). You can check that

gu = cu + sv, gv = −su + cv.

Since g ∈ On, we have

u, u = gu,gu = cu + sv, cu + sv = c2u, u + 2csu, v + s2v, v.

Likewise

v, v = s2u, u + 2csu, v + c2v, v,

11



and

u, v = cu + sv, −su + cv = −csu, u + (c2 − s2)u, v + csv, v.

Adding, we find u, v = 0. Hence u = |u|−1u and v = |v|−1v are an orthonormal basis of a

two-dimensional plane U ⊂ Rn.

Hence there exists h

∈ On whose first two columns are u , v and whose last n

− 2 columns are an

orthonormal basis for the orthogonal complement U ⊥

. We then have

hgh−1 ∈

SO2 00 SOn−2

,

which is connected, by the induction hypothesis. This completes the proof that SOn is connected.

Finally, On consists of two cosets of SOn, each of which is connected component.

2.2 SO3 and quaternions

Let us regard R3 as the space of the “pure” quaternions:

H0 = Ri ⊕ R j ⊕Rk = {v ∈ H : τ (v) = 0}.

The dot product may be expressed quaternionically as as

u, v = 12(uv + vu). (3)

For q ∈

S 3, let Rq : H0

→H0 be the linear map given by

Rq(v) = qvq −1.

A familiar calculation using (3) shows that

Rq(u), Rq(v) = u, v,

for all u, v ∈ H0 and q ∈ S 3. Therefore Rq ∈ O3 and we have a continuous homomorphism R : S 3 →O3, sending q → Rq. Since S 3 is connected, the image of R is connected, and therefore lies in SO3,

by Prop. 2.2. Thus, we have a homomorphism

R : S 3

−→ SO3, given by q → Rq,

where Rq(v) = qvq −1. To see this homomorphism explicitly, let q = a + bi + cj + dk ∈ S 3 and

calculate

qiq −1 = (a2 + b2 − c2 − d2)i + 2(bc + ad) j + 2(bd − ac)k

qjq −1 = 2(bc − ac)i + (a2 − b2 + c2 − d2) j + 2(cd + ab)k

qkq −1 = 2(bd + ac)i + 2(cd − ab) j + (a2 − b2 − c2 + d2)k,

12



so the matrix of Rq with respect to the basis {i,j,k} is

Rq =

a2 + b2 − c2 − d2 2(bc − ac) 2(bd + ac)

2(bc + ad) a2 − b2 + c2 − d2 2(cd − ab)2(bd − ac) 2(cd + ab) a2 − b2 − c2 + d2

. (4)

Proposition 2.3 The homomorphism R : S 3 → SO3 is surjective with ker R = {±1} equal to thecenter of S 3. In particular, every matrix in SO3 is of the form (4) for some (a,b,c,d) ∈ R4 with

a2 + b2 + c2 + d2 = 1.

Proof: We will use the following basic fact about group actions. Suppose a group G acts on a set X ,that H is another group, and that we have a homomorphism f : H → G. Assume that the subgroup

f (H ) ≤ G acts transitively on X and that there exists x ∈ X such that f (H ) contains the stabilizer

Gx = {g ∈ G : g · x = x}. Then f (H ) = G. For if g ∈ G, there is h ∈ H such that f (h) · x = g · x,

by the transitivity assumption. Then g−1f (h) ∈ Gx, so g−1f (h) = f (k) for some k ∈ H , by the

assumption that f (H ) ⊃ Gx. Thus we have g = f (hk−1) ∈ f (H ).

We apply this to the homomorphism R : S 3 → SO3, where SO3 acts on the sphere C 0 = S 2. We have

proved that R(S 3) acts transitively on C 0, and that

R(T i) =

1 00 SO2

is the stabilizer of i in SO3. It follows that R is surjective.

The kernel of R consists of those q ∈ S 3 commuting with every vector in H0. Since every quaternion

commutes with R · 1 which is the center of H, it follows that ker R is the intersection of S 3 with R · 1.

This is the unit sphere in R, that is, ker R = {±1}.

Remark 1: One can describe R more geometrically as follows. If q ∈ S 3, there is a unit vector u ∈H0

and θ ∈ [−π, π] such that

q = cos θ + u sin θ.

Since q commutes with u, it follows that Rq(u) = u so Rq is a rotation about the axis through u. To

find the angle of rotation, we note that for u, v ∈ H0 the quaternionic product uv is given by

uv = u × v − u · v ∈ H,

where × and · are the cross and dot product on R3. Note that u × v ∈ H0 and u · v ∈ R. If u · v = 0this reduces to uv = u × v, and we compute that

Rq(v) = (cos θ + u sin θ)v(cos θ − u sin θ) = cos(2θ) + sin(2θ)(u × v).

This shows that Rq is rotation about u by 2θ seen counterclockwise as u points towards you.

Remark 2: The quaternionic interpretation gives an explicit formula for the product of two rotations

in SOn. Let S, T ∈ SO3 be rotations by 2θ, 2φ about unit vectors u, v ∈ H0. Then S = R p, T = Rq,

where

p = cos θ + u sin θ, q = cos φ + v sin φ.

13



Then ST = R pq, and we compute

pq = (cos θ cos φ) − (sin θ sin φ)u · v + (sin θ cos φ)u + (cos θ sin φ)v + (sin θ sin φ)u × v.

Therefore ST is rotation by angle ψ about the axis through the vector w, where

cos ψ = (cos θ cos φ) − (sin θ sin φ)u · v,

andw = (sin θ cos φ)u + (cos θ sin φ)v + (sin θ sin φ)u × v.

Remark 3: The image under R of the binary tetrahedral group N (Q8) is the symmetry group of a

regular tetrahedron, and is isomorphic to the alternating group A4. Thus, we have an exact sequence

1 −→ {±1} −→ N (Q8) −→ A4 −→ 1.

This sequence is non-split: there is no subgroup of N (Q8) isomorphic to A4. In particular, N (Q8) and

S 4 are non-isomorphic groups of order 24. Note that the latter fits into another exact sequence (which

is now split)

1 −→

A4

−→ S 4

−→ {±1} −→

1.

Remark 4: The cosets of {±1} in S 3 are pairs of antipodal points. Each pair determines a line in R4,

so the set of antipodal pairs is the real projective space RP3. Thus, Prop. 2.3 shows that SO3 = RP3,

as topological spaces. You can also regard RP3 as the quotient of a solid ball in R3 by identifying

antipodal points on the boundary. Indeed, every element of SO3 is rotation about some axis by some

angle θ ∈ [−π, π]. The axis determines a line segment in the ball Bπ of radius π in R3 and θ determines

a point on the axis. Each θ ∈ (−π, π) gives a unique rotation about this axis, but the two values θ = ±π,

corresponding to antipodal points on the boundary of Bπ, give the same rotation. In the next section,

the exponential map will make this latter interpretation more explicit.

2.3 The exponential map for SO3

Let A be an n × n real matrix. What conditions on A ensure that the path θ → exp(θA) lies in SOn?

The condition t(exp(θA)) = (exp(θA))−1 means that

I + θ(tA) + · · · = I − θA + · · · ,

so exp(θA) ∈ On iff tA = −A. Such matrices are called skew-symmetric and their diagonal entries

are zero. In particular tr(A) = 0 so det exp(θA) = 1, so in fact exp(A) lies in SOn for any skew-

symmetric n×n matrix A. Letting son denote the set of such matrices, we therefore have an exponential

mapexp : son −→ SOn .

We now take n = 3 and calculate exp explicitly. The matrices in so3 are paremetrized by vectors

v = (x,y,z ) ∈ R3, via

v = (x,y,z ) → Av =

0 −z y

z 0 −x−y x 0

.

14



Note that v ∈ ker Av and ker A = Rv as long as v = (0, 0, 0). Let |v| =

x2 + y2 + z 2. Using the

fact that A3v = −θ2Av, we find that

exp(Av) = I +

sin |v|

|v|

Av +

1 − cos |v|

|v|2

A2v

and that exp(Av) is rotation by |v| about the axis through v, where the direction is seen counterclock-

wise as v points towards you. It follows that exp maps {Av : |v| < π} bijectively onto the complementin SO3 of the conjugacy class C of 180 degree rotations. If |v| = π then exp(Av) = exp(A−v) so expdescribes C as the sphere of radius π with antipodal points identified. That is, C is the real projective

plane.

2.4 The exponential diagram

We now have exponential maps

exp : V

−→ S 3, exp : so3

−→ SO3

and a homomorphism R : S 3 → SO3 given by Rq(v) = qvq −1. The final piece is the derivative of R,

which is a linear map

R : V −→ so3

defined as follows. For each v ∈ V , Rv ∈ so3 is the skew symmetric matrix acting on V by

R

v(u) = d

dθRexp(θv)(u)|θ=0.

Computing this explicitly using power series, we find the explicit formula

R

v(u) = vu − uv.

To see Rv as a matrix, let v = (x,y,z ) and compute Rv(i), Rv( j), Rv(k) to find that

R

v = 2

0 −z y

z 0 −x−y x 0

= A2v.

Finally, one checks that

R ◦ exp = exp ◦R.

That is, the following diagram is commutative.

V R−−−→ so3

exp expS 3

R−−−→ SO3

It follows that exp : so3 → SO3 is surjective.

Thus, the Lie groups S 3 and SO3 are parametrized by R3 via the exponential maps, just as S 1 is

parameterized by R via the usual exponential map. Moreover, the group homomorphism R : S 3 →SO3 lifts to a linear map R on the parameter spaces.

15



3 Definition and basic properties of Lie groups

Intuitively, a Lie group is a group which locally looks like Rn. We often imagine moving in a Lie

group, where small motions seem like motion in Rn. For example, we can move along circles in S 3

or we can move in SO3 by changing the axis and amount of rotation. To define all of this precisely

requires the notion of a smooth manifold.

3.1 Introduction to Manifolds

Let U be an open subset of Rn. A function f : U → Rm is differentiable on U if for each x ∈ U there

exists a linear map f x : Rn → Rm such that

limh→0

1

|h| [f (x + h) − f (x) − f x(h)] = 0.

If v

∈ Rn has length

|v

| = 1 (with respect to the standard Euclidean metric, as we have been using)

then we can take h = tv as t → 0 and the formula for f x becomes

f x(v) = limt→0

f (x + tv) − f (x)

t .

If v = ei is one of the standard basis vectors then f x(ei) = ∂f ∂xi

(x).

Now the function x → f x is a function f : U → Hom(Rn,Rm) = Rnm, called the derivative of f .We say that f : U → Rm is smooth or C ∞ if each of f, f , f , . . . is differentiable on U . The R-vector

space C ∞(U ) of all smooth functions on U is a ring containing all polynomial functions.

Now let M be a topological space. A local chart on M is a triple (ϕα, U α, M α) where U

⊂ Rn

and M α ⊂ M are open subsets and ϕα : U α → M α is a homeomorphism. Thus, ϕal parametrizesthe open subset M α of M . We call n the dimension of the chart. An atlas on M is a collection

{(ϕα, U α, M α) : α ∈ A} of local charts indexed by some set A such that M =

α∈A M α. We say the

atlas has dimension n if U α is an open subset of Rn, with the same n, for every α ∈ A.

Given two local charts (ϕα, U α, M α) and (ϕβ, U β, M β), we get a transition function

ϕ−1β ◦ ϕα : ϕ−1α (M α ∩ M β)

ϕα−→ M α ∩ M βϕ−1β−→ Rn

defined on the open set U αβ = ϕ−1α (M α ∩ M β) ⊂ Rn. An atlas {(ϕα, U α, M α) : α ∈ A} is smooth if

the transition functions ϕ−1β ◦ ϕα are smooth for each α, β ∈ A.

Example 1: Let U be an open subset of Rn. Then the identity map U → U is a smooth atlas on U ,consisting of a single chart.

Example 2: Let M = S n = {(x0, x1, . . . , xn) ∈ Rn+1 :

x2i = 1}. Take the two antipodal points

z ± = (±1, 0, . . . , 0) and let M ± = S n − {z ±}, and let U ± = Rn. For α = ±, define ϕα : Rn → S n by

ϕα(u1, . . . , un) = 1

|u|2 (α(|u|2 − 1), 2u1, . . . , 2un)

16



Then ϕ−1α : M α → Rn is given by

ϕ−1α (x0, . . . , xn) =

1

1 − αx0(x1, . . . , xn),

and the transition functions are given by

ϕ−1−α

◦ϕα(u) =

u

|u|.

This is a smooth function on ϕ−1α (M α∩M −α) = Rn−{(0, . . . , 0)}. Hence the collection {(ϕα,Rn, M α) :

α = ±} is a smooth atlas. fs

Let M and N be topological spaces with atlases {(ϕα, U α, M α) : α ∈ A} and {(ψβ, V β, N β) : β ∈ B}respectively. A function f : M → N is smooth if each composition

ψ−1β ◦ f ◦ ϕα : ϕ−1

α (M α ∩ f −1(N β)) −→ Rm

is smooth. For example, the identity M → M is smooth, and the composition of smooth functions is

smooth.

A given topological space M may admit more than one atlas. Two atlases {(ϕα, U α, M α) : α ∈ A}and {(ψβ, V α, M β) : β ∈ B} on M are considered equivalent if the identity map is smooth from one

atlas to the other in the sense just defined. That is, the atlases are equivalent if each composition

ψ−1β ◦ ϕα : ϕ−1α (M α ∩ N β) −→ Rn

is smooth.

Definition 3.1 An smooth manifold is a Hausdorff 2 topological space M with an equivalence class of

smooth atlases. We say M is n-dimensional if these atlases have dimension n.

The term “smooth manifold” is interchangeable with differentiable manifold. An equivalence class

of smooth atlases is often called a smooth structure or a differentiable structure. Having a single

smooth atlas on on a Hausdorff topological space M gives a smooth structure on M , via the equivalance

class of the given atlas.

Two manifolds M, N are diffeomorphic if there is a smooth map f : M → N having a smooth inverse

f −1 : N → M . It is possible for two manifolds to be homeomorphic but not diffeomorphic. That is,

the topological space M may admit more than one For example, the 7-sphere S 7 admits exactly 28smooth structures. 3

Lemma 3.2 If M and N are smooth manifolds with atlases {(ϕα, U α, M α) : α ∈ A} and {(ψβ, V β, N β) :

β ∈ B} , then then there is a canonical smooth structure on M ×N , given by the atlas {(ϕα×ψβ, U α×V β, M α × N β) : α ∈ A, β ∈ B}.

Proof: Left to the reader!

2Recall M is Hausdorff if given distinct points x, y ∈ M there exist disjoint open subsets U, V ⊂ M such that x ∈ U and y ∈ V .

3See Milnor, On manifolds homeomorphic to the 7-sphere Annals of Math. 1956, and Brieskorn Beispiele zur Differen-

tialtopologie von Singularit aten Invent. Math. 1966

17



3.2 Manifolds defined by equations

Let f = (f 1, . . . , f m) : Rn → Rm be a smooth function, where n ≥ m, and write n = k + m. Let

M = f −1(0) = { p ∈ Rn : f ( p) = 0.} The Implicit Function Theorem gives a sufficient condition for

M to be a smooth manifold, in terms of the derivative f of f . Recall that f is the m × n matrix of

functions

f =

∂f i∂xj

which at each point p ∈ Rn gives a linear map

f p =

∂f i∂xj

( p)

: Rn → Rm.

Theorem 3.3 (Implicit Function Theorem) Let f : Rn → Rm be a smooth function, where n ≥ m ,

and let M = f −1(0). Suppose f p(Rn) = Rm. Then there exist open sets U ⊂ Rk and V ⊂ Rm , and a

unique smooth function g : U → V such that p ∈ U × V and (U × V ) ∩ M = {(u, g(u)) : u ∈ U }.

Proof: See [Rudin’s Principles of Mathematical Analysis].

Corollary 3.4 With notation as in Thm. 3.3 assume that f p has rank m for all p ∈ M . Then M has

the structure of a smooth manifold.

Proof: Our assumption means that for each p ∈ M we can choose m columns whose determinant is

nonzero. Fix p ∈ M and order the variables so that these columns are the last m columns of f and

choose U, V,g as in Thm. 3.3. This gives a chart ϕ p : U p → M p, where U p = U , M p = (U × V ) ∩ M ,given by ϕ p(u) = (u, g(u)). Doing this for each p ∈ M gives an atlas {(ϕ, U p, M p) : p ∈ M }. One

checks that for p, q ∈ M and u ∈ ϕ p(M p ∩ M q) we have ϕ−1q ◦ ϕ p(u) = u. Hence the atlas is smooth.

3.3 Lie groups: definition and first examples

A Lie group is a smooth manifold G which is also a group whose multiplication and inverse maps

µ : G × G → G and ι : G → G

are smooth. Here G

×G has the product smooth structure, as in Lemma 24.

Let G and H be Lie groups. A Lie group homomorphism is a smooth map f : G → H which is

also a group homomorphism. And f is an isomorphism of Lie groups if f is both a diffeomorphism of

manifolds and an isomorphism of groups.

Example 1: The General Linear Group GLn(R) is the group of automorphisms of the vector space

Rn. Concretely, GLn(R) is the group of n × n invertible matrices, under matrix multiplication. To see

that GLn(R) is a Lie group, let M n(R) Rn2 be the space of all n × n matrices with entries in R.

18



The determinant det : M n(R) → R is a continuous function and GLn(R) is the open set of points in

M n(R) on which det is nonzero. Thus GLn(R) is an open subset of M n(R) and is a manifold. The

group operations are polynomial functions (or polynomial divided by det) of the matrix entries, hence

are smooth functions on GLn(R). Therefore GLn(R) is a Lie group.

The determinant map sends GLn(R) onto R×, which has two components-the positive and negative

real numbers. It follows that GLn(

R)

is disconnected. We will see that GLn(

R)

has two connected

components,

GL+n (R) = {g ∈ GLn(R) : det(g) > 0},

and its complement. Two bases of Rn have the same orientation if the matrix relating them lies in

GL+n (R). This means that one basis can be deformed continuously into the other. Thus GL+

n (R) is

orientation-preserving automorphism group of Rn.

Example 2: If U is a bounded open subset of Rn and g ∈ GLn(R) then vol(gU ) = det(g) vol(U ).

The Special Linear Group SLn(R) is the group of volume-preserving automorphisms of Rn. That is,

SLn(R) = {g ∈ GLn(R) : det(g) = 1}.

To see that SLn(R) is a Lie group we use Cor. 3.4, viewing SLn(R) = f −1(0) for the functionf : M n(R) → R defined by f (A) = det(A) − 1. If A = [aij] then

f (A) = −1 +σ∈S n

sgn(σ)a1σ(1) · · · anσ(n).

Let A j be the matrix obtained by deleting row 1 and column j from A. If det(A) = 1 then det(A j) = 0for some j. And

∂f

∂aij= ±

σ(1)= j

sgn(σ)a1σ(1) · · · anσ(n) = ±det(A j) = 0,

so the matrix f (A) has rank = 1 for all A ∈

SLn

(R) and the latter is indeed a manifold. Again, the

group operations are polynomial, so SLn(R) is a Lie group.

Example 3: The Orthogonal Group On is group of length-preserving automorphisms of the vector

space Rn. That is,

On = {g ∈ GLn(R) : |gv| = |v| ∀ v ∈ Rn} = {g ∈ GLn(R) : tgg = I }.

A matrix belongs to On exactly when its columns form an orthonormal basis of Rn. Thus, On =f −1(0), where f = (f pq) : M n(R) →Rn2 has component functions

f pq(g) = −δ pq +i gipgiq,

where δ pq = 1 if p = q and is zero otherwise. One checks that f g is surjective if g is invertible, in

particular if g ∈ On. Hence On is indeed a manifold. Again, the group operations are polynomial, so

On is a Lie group.

Since On = f −1(0) and each column of a matrix in On is a unit vector, it follows that On is a closed

and bounded subset of M n(R), hence is compact. The determinant maps On onto {±1}, so On is

disconnected. We will see that it has exactly two connected components.

19



Example 4: The Special Orthogonal Group SOn is the subgroup of GLn(R) preserving orientation,

volume and length. That is,

SOn = {g ∈ SLn(R) : |gv| = |v| ∀ v ∈ Rn} = {g ∈ SLn(R) : tgg = I }.

A matrix lies in SOn exactly when its columns form an orthonormal basis having the same orientation

as the standard orthonormal basis{

e1, . . . , e

n}. We will see that SO

n is connected; it is the component

of On containing the identity matrix.

Example 5: The group R3 is an abelian Lie group under the operation of addition. But the same

manifold R3 has another Lie group structure. The Heisenberg group

H 3 :=

1 x z

0 1 y0 0 1

: (x,y,z ) ∈ R3

is diffeomorphic to R3 but is not isomorphic to R3 as a Lie group. Indeed, H 3 is non-abelian.

4 The Lie algebra of a Lie group

Let G be a Lie group. Imagine moving along paths in G. Each smooth path γ : R → G has a position

γ (t) and velocity vector γ (t) which is “tangent” to G. The pair (γ (t), γ (t)) tells us where we are and

where we are going. Where does γ (t) live? If G were a subset of some Euclidean space Rn then we

could imagine γ (t) as a vector in Rn. But G may embed in many Euclidean spaces in many ways, and

we want a home for γ (t) that is intrinsic to G and independent of any particular realization of G in a

Euclidean space. We will see that (γ (t), γ (t)) is a path in a new manifold.

4.1 The tangent bundle of a manifold

First let U be an open subset of Rn. The tangent bundle to U is simply

T U := U ×Rn.

If we have a smooth path γ : (a, b) → U defined on an open interval (a, b) ⊂ R, then we are interested

in both the position and velocity of the path. This pair of data is a new path in the tangent bundle

T U , given by t

→ (γ (t), γ (t)). The first projection π : T U

→ U shows the position, and the fiber

π−1(u) = Rn is the vector space of all possible velocities at u. Note that T U is a union

T U = p∈U

{ p} × Rn.

of fibers of π. These fibers are called tangent spaces. Each tangent space is canonically identified with

the ambient vector space Rn containing U as an open subset.

20



Let M be a n-dimensional smooth manifold with atlas {ϕα, U α, M α) : α ∈ A}. For each α ∈ A let

ια : M α → M be the inclusion map. The tangent bundle of M is a new manifold T M equipped with

a projection map

T M π−→ M

whose fibers π−1( p) are n-dimensional vector spaces which vary smoothly. More precisely, T M is the

set of equivalence classes

T M =α∈A

(M α × Rn)/ ∼,

where (x, v) ∼ (y, w) if

ια(x) = ιβ(y) and w = (ϕ−1β ϕα)x(v).

Thus, points in T M are equivalence classes [x, v]α ∈ T M , for (x, v) ∈ M α ×Rn, with the understand-

ing that if x ∈ M α ∩ M β we have

[x, v]α = [x, (ϕ−1β ϕα)x(v)]β.

Each M α × Rn

= T M α injects into T M . Via the quotient topology, T M has an open cover

T M =α∈A

T M α

consisting of tangent bundles of the images of charts.

For each α ∈ A define

Φα : U α × Rn −→ T M α by Φα(u, v) = [ϕα(u), v)]α.

One checks that

Φ

−1

α (T M α ∩ T M β) = ϕ

−1

α (U αβ) ×Rn

,and that

Φ−1β Φα : ϕ−1

α (U αβ) × Rn −→ ϕ−1β (U αβ) × Rn

is given by ϕ−1β ϕα×(ϕ−1

β ϕα). Thus T M is a smooth 2n-dimensional manifold with atlas {(Φα, T U α, T M α) :α ∈ A}.

The projection πM : T M → M is given by πM ([x, v]α) = x. For α, β ∈ A, one checks that ϕ−1β ◦πM ◦

Φα is the composition

ϕ−1α (U αβ) × Rn proj−→ ϕ−1α (U αβ)

ϕ−1β ϕα−→ ϕ−1

β (U αβ),

which is smooth, so that πM : T M → M is smooth.

The tangent space to M at x ∈ M is the fiber

T xM := π−1M (x).

If x ∈ M α then T xM = {[x, v]α : v ∈ Rn} Rn, but this isomorphism is non-canonical, as it depends

on α.

21



In practice, the tangent bundle usually has a more concrete realization than the rather abstract general

definition just given.

Example: Consider again the n-sphere S n = {x ∈ Rn+1 : |x| = 1}. Then we may realize T S n as

T S n = {(x, v) ∈ S n × Rn+1 : x · v = 0}, (5)

where x · v is the dot product on Rn+1. To see that this realization is indeed a manifold, we use Cor.3.4 to express the right side of (5) as f −1(0), where f = (f 1, f 2) = (|x|2 − 1, x · v). We find that

f =

2x1 . . . 2xn 0 . . . 0v1 . . . vn x1 . . . xn

has rank two for all x ∈ S n, so that f −1(0) is indeed a manifold.

Proposition 4.1 If M and N are smooth manifolds and f : M → N is a smooth map, then there is a

unique smooth map f : T M → T N making the following diagram commute:

T M f

πM

T N

πN

M f N

.

If L is another manifold and g : N → L is another smooth map then

(g ◦ f ) = g ◦ f . chain rule.

Proof: Let {(ϕα, U α, M α) : α ∈ A} and {(ψβ, V β, N β) : β ∈ B} be atlases on M and N . Let

x ∈ M and choose α ∈ A such that x ∈ M α and β ∈ B such that f (x) ∈ N β . For v ∈ Rn

,define f ([x, v]α) = [f (x), (ψ−1β f ϕα)(v)]β . One checks, using the chain rule for derivatives on Rn,

that f ([x, v]α) does not depend on the choice of α such that x ∈ M α, and that the resulting map f

is smooth. It is clear that f commutes with the projections. The last assertion follows from the chain

rule on Rn.

4.2 Vector fields

Let M be a smooth n-dimensional manifold. A vector field on M is a smooth map X : M → T M such that πM

◦X is the identity on M . Intuitively, a vector field is choice of vector X ( p)

∈ T pM for

each p ∈ M , such that X ( p) varies smoothly in T M as p varies smoothly in M .

Example: Let M = S 3, viewed inside H = R4 as before, and let H0 be the subspace of H orthogonal

to R. Then for any v ∈ H0, the function X v( p) = pv is a vector field on S 3.

Let G be a Lie group. For any g ∈ G we have a map

Lg : G −→ G, given by Lg(x) = gx,

22



called left translation. This map has a derivative Lg : T G → T G, mapping each tangent space T xG

to T gxG.

A vector field X on G is called left-invariant if for all g ∈ G the following diagram commutes:

T GLg T G

G

X

Lg G

X

.

Let g = T eG denote the tangent space to G at the identity element e ∈ G. For each v ∈ g there is a

unique left-invariant vector field X v : G → T G, given by

X v(g) = L

g(v).

Conversely, if X : G → T G is any left-invariant vector field then

X (g) = L

gX (e),

so X = X v, where v = X (e). Thus, the left-invariant vector fields are in canonical bijection with

vectors in g.

4.3 The tangent bundle of a Lie group

An n-dimensional manifold M is parallelizable if there is diffeomorphism

f : M ×Rn ∼−→ T M

restricting to a linear isomorphism Rn = p × Rn → T pM for each p ∈ M . Equivalently, M is

parallelizable if there exist n vector fields X 1, . . . , X n which are linearly independent at each point in

M .

For example, S n is parallelizable exactly when n = 1, 3, 7. This is easy to see for S 1 and for S 3 the

vector fields X i, X j , X k (see example above) are linearly independent at each point. For S 7 see section

8.3.

More generally, let G be any Lie group, and choose a basis e1, . . . , en of g. Since Lg : g → T gG is an

isomorphism, the vector fields X i(g) = Lg(ei) are linearly independent in T gG for each g ∈ G. This

proves:

Proposition 4.2 Any Lie group is parallelizable.

4.4 One-parameter-subgroups

Let M be an n-dimensional manifold. A path in M is a smooth map γ : I → M defined on some open

interval in I ⊂ R. The derivative of γ is then a map on tangent bundles

γ : T I −→ T M.

23



Choose I small enough so that γ (I ) ⊂ M α for some chart (ϕα, U α, M α) on M . Then T I = I ×R and

γ (T I ) ⊂ T M α = M α × Rn U α × Rn. Taking 1 ∈ R as a basis vector, we have

γ (t, 1) = (γ (t), γ (t)),

where γ (t) is the componentwise derivative of γ (t) ⊂ U α ⊂ Rn.

Given a smooth vector field X : M → T M we seek a path γ in M such that γ (t, 1) = X (γ (t)). Thisamounts to solving a differential equation, The following result is a consequence of the local existence

and uniqueness theorems for differential equations of one variable.

Theorem 4.3 Let X : M → T M be a smooth vector field and let p ∈ M . Then there exists > 0 and

a unique smooth path γ : (−, ) → M such that

γ (t, 1) = X (γ (t)) for |t| <

and γ (0) = p.

We call γ a local integral of X As a corollary of this local uniqueness, it follows that global existence

implies global uniqueness.

Corollary 4.4 Suppose X : M → T M is a smooth vector field and I ⊂ R is an open interval

containing a closed interval [a, b]. Suppose also that we have two smooth paths γ , δ : I → M such

that

γ (t, 1) = X (γ (t)), δ (t, 1) = X (δ (t)), γ (a) = δ (a).

Then γ (t) = δ (t) for all t ∈ [a, b].

Proof: Let c = sup{t ∈ [a, b] : γ (t) = δ (t)}. By continuity, we have γ (c) = δ (c). If c < b then for all

> 0 such that (c − , c + ) ⊂ I , the paths γ and δ on are local integrals of X on (c − , c + ), hence

they agree here, by the local uniqueness of Thm. 4.3. This contradiction forces c = b.

Now let G be a Lie group. A one-parameter subgroup of G is a Lie group homomorphism

γ : R −→ G.

Note that γ (0) = e so γ (0, 1) ∈ g = T eG.

Lemma 4.5 Let γ : R → G be a one-parameter subgroup, with v := γ (0, 1) ∈ g , and let X v : G →T G be the corresponding left-invariant vector field on G given by X v(g) = L

g(v). Then for all t ∈ Rwe have

γ (t, 1) = L

γ (t)(v) = X v(γ (t)).

24



Proof: Regarding R itself as a Lie group, we can extend the vector (0, 1) ∈ T 0R to a left-invariant

vector field X (0, 1) : R → T R, given by X (0,1)(t) = Lt(0, 1), where Lt : R → R is the translation

map Lt(x) = t + x. For all x ∈ R we have Lt(x, 1) = (t + x, 1).

Since γ is a homomorphism, the diagram

RLt

γ

GLγ (t)

R

γ G

is commutative. Taking derivatives, we get the commutative diagram

T R

Lt

γ T G

Lγ (t)

T R

γ T G

.

We getγ (t, 1) = γ ◦ L

t(0, 1) = L

γ (t)(v) = X v(γ (t)),

proving the lemma.

Thus, a one-parameter subgroup γ is a global integral of the left-invariant vector field determined by

γ (0, 1) ∈ g.

Theorem 4.6 For each v ∈ g there is a unique one-parameter subgroup γ v : R → G such that

γ (0, 1) = v.

Proof: Suppose γ and δ are two one-parameter subgroups such that γ (0, 1) = v. Then γ (0) = e =δ (0) so γ (t) = δ (t) for all t ≥ 0 by Cor. 4.4. Since γ (−t) = γ (t)−1 and likewise for δ , we have

γ (t) = δ (t) for all t. This proves the uniqueness part of Thm. thm:1psg.

Now let v ∈ g. By the local existence part of Thm. 4.3, there exists > 0 and a smooth path

γ : (−, ) → G such that γ 0(0) = e and

γ 0(t, 1) = X v(γ 0(t) for |t| < .

The first step is to prove that γ 0 is a local homomorphism. That is,

γ 0(s + t) = γ 0(s)γ 0(t), for |s|, |t| < /2.

Fix s. Since both sides agree at t = 0, it is sufficient to show that both sides solve the equation

f (t, 1) = X v(f (t)).

For the left side, we have γ 0(s + t) = γ 0 ◦ Ls(t) and

[γ 0 ◦ Ls](t, 1) = γ 0 ◦ L

s(t, 1) = γ 0(t + s, 1) = X v(γ 0(t + s)).

25



For the right side, let g = γ 0(s). Then

[γ 0(s)γ 0(t)](t, 1) = [Lgγ 0](t, 1)

= L

gγ 0(t, 1)

= L

gX v(γ 0(t))

= L

g

L

γ 0(t)

(v) (by Lemma 4.5)

= L

gγ 0(t)(v)

= X v(gγ 0(t))

= X v(γ 0(s)γ 0(t)),

Thus γ 0 is indeed a local homomorphism.

The next step is to extend γ 0 to a homomorphism γ : R → G. Given t ∈ R, choose a positive integer

N such that |t/N | < /2 and set

γ (t) = γ 0(t/N )N .

If also |t/M | < /2 then γ 0(t/M )M = γ 0(t/MN )MN = γ 0(t/N )N ,

since γ 0 is a local homomorphism. Hence γ is well-defined. Similarly, γ is a homomorphism: Given

t, s ∈ R, choose N such that |t|/N and |s|/N are both < /2. Then

γ (s + t) = γ 0

s

N +

t

N

N

= γ 0(s/N )N γ 0(t/N )N = γ (s)γ (t).

Next, γ is smooth: For γ 0 is smooth on (−, ), so γ (t) = γ 0(t/N ) is smooth on (−N,N ). Finally,

γ (0, 1) = γ 0(0, 1) = X v(γ 0(0)) = X v(e) = v.

This completes the proof the theorem.

For v ∈ g, we let γ v : R → G be the unique one-parameter subgroup such that γ v(0, 1) = v.

4.5 The exponential map

Let G be a Lie group. The Exponential map for G is the map exp = expG : g → G defined by

exp(v) = γ v(1).

We wish to show that exp is a smooth map, where the vector space g is regarded as a smooth manifold.

We must examine how γ v varies with v. The simplest situation is when v is scaled.

Lemma 4.7 For all s ∈ R , we have γ sv(t) = γ v(st).

26



Proof: Let M s : R → R be the map M s(t) = st. We have M s(0, 1) = (0, s), so

(γ v ◦ M s)(0, 1) = γ v ◦ M s(0, 1) = γ v(0, s) = s · γ v(0, 1) = sv.

Since γ sv is the unique one-parameter subgroup of G such that γ sv(0, 1) = sv, we have γ v ◦ M s = γ svand the lemma follows.

Next we consider the smoothness of v → γ v. This requires an extension of the uniqueness and existencetheorem to families of vector fields.

Theorem 4.8 Let U × V ⊂ Rn × Rm be an open neighborhood of (0, 0) and let X : U × V → Rn

be a smooth map. Then there exists > 0 , a neighborhood V 0 of 0 in V and a unique smooth map

f : (−, ) × V 0 → U such that for all v ∈ V 0 the function f v(t) := f (t, v) satisfies the differential

equation

f v(t) = X (f v(t), v), for |t| < .

Proof: See references in Adams, page 8.

Proposition 4.9 Let G be a Lie group and let g = T eG. The exponential map exp : g → G , defined by

exp(v) = γ v(1) , is smooth.

Proof: Recall that G is parallelizable, via the map

X : G × g → T G given by X (g, v) = L

g(v).

The one-parameter subgroup γ v(t) solves the differential equation

γ v(t, 1) = X (γ v(t), v).

By Thm. 4.8 the map (t, v) → γ v(t) is smooth on some neighborhood (−, ) × V 0 of (0, 0) in R× g.

Then

γ N −1v(t/N )N 2 = γ v(t)

is smooth on (−N,N) × N V 0, so (t, v) → γ v(t) is smooth on R× g. Taking t = 1 we see that expis smooth.

Remark 1: For each v ∈ g, the image exp(v) = γ v(1) ∈ G is a value of the one parameter subgroup

γ v. In fact all values of γ v are obtained from exp. For Lemma 4.7 implies that

γ v(t) = γ tv(1) = exp(tv).

Remark 2: Beware that exp is not a group homomorphism if G is non-abelian.

We next show that the exponential map is functorial. Let ϕ : G → H be a Lie group homomorphism

and let g = T eG, h = T eH be the respective tangent spaces at the identity elements of G and H . Let

ϕ0 : g −→ h

be the derivative of ϕ. This is a linear map obtained by restricting ϕ : T G → T H to g.

27



Proposition 4.10 The following diagram is commutative:

g ϕ0−−−→ h

expG

expH

G ϕ

−−−→ H

Proof: In the proof we write ϕ instead of ϕ0. For all v ∈ g we have

(ϕ ◦ γ v)(0, 1) = ϕ ◦ γ v(0, 1) = ϕ(v),

so ϕ ◦ γ v = γ ϕ(v) by Thm. 4.6. Now

ϕ(expG(v)) = ϕ(γ v(1)) = γ ϕ(v)(1) = expH (ϕ(v)),

as claimed.

This is the most useful result in the theory of Lie groups. 4

Example: Let G = GLn(R). Since G is an open subset of M n(R) we canonically identify g = M n(R).

Take A ∈ g and consider the path γ : R → G given by

γ (t) = etA = I + tA + t2

2 A2 + · · · ∈ G.

Since γ (0) = A, it follows that γ = γ A and we have expG(A) = eA. Thus, the exponential map for

GLn(R) is the familiar exponential map of matrices.

Proposition 4.11 Let G be a Lie group. Then there exists a neighborhood U of 0 in g mapped diffeo-

morphically by expG onto a neighborhood of e in G.

Proof: We need another result from analysis. 5

Theorem 4.12 (Inverse Function Theorem) Let U be an open subset of Rn and let p ∈ U . Let

f : U → Rn be a smooth map whose derivative

f p =

∂f i∂x j

( p)

is invertible at p. Then there exists a neighborhood U 0 of p mapped diffeomorphically by f onto an

open neighborhood of f ( p) in Rn.

4This claim is really a challenge to find a result that is even more useful.5See Rudin Principles of Mathematical Analysis.

28



The smooth map exp : g → G has derivative expG : T g → T G. Restricting to T 0g = g, we get a

linear map exp0 : g → g. Let v ∈ g, and let h : R → g be the map h(t) = tv. Then γ v = exp ◦h, so

exp0(v) = exp0(h(0)) = γ v(0) = v,

so exp0 is the identity map, which is invertible. Prop. 4.11 now follows from the Inverse Function

Theorem.

Example: Again let G = GLn(R) with g = M n(R). A neighborhood of I in G is of the form I + U ,where U is a neighborhood of 0 in g. The series

log(I + u) = u − 12

u2 + 13

u3 − · · ·converges for u near 0 (e.g., for |u| < 1 when n = 1) and inverts the exponential map: exp(log(I +u)) = u for u near 0.

Prop. 4.11 shows that the exponential map parametrizes an open neighborhood of the identity in G. It

says nothing about how large this neighborhood is, but that does not matter, thanks to the following.

Proposition 4.13 A connected Lie group G is generated by any open neighborhood of the identity. In

particular G is generated by exp(g).

Proof: This result relies only on the continuity of the group laws, and not on the smooth structure. Let

U be an open neighborhood of e in a connected Lie group G and let H be the subgroup of G generated

by U . Since the multiplication in G is continuous, gU is an open neighborhood of g , for all g ∈ G.

Now

H =h∈H

hU and G − H =

g∈G−H

gU

are both open in G, so H is both open and closed and nonempty (we have e ∈ H ). As G is connected,it follows that H = G.

From Prop. 4.11 we have that exp(g) contains an open neighborhood of e, which generates G, so

exp(g) generates G.

Remark: We will often find that exp(g) = G, for example if G is compact. However it is not always

so, even if G is connected. For example if G = SL2(R) then the image of exp consists of matrices

exp(A) where tr(A) = 0. The eigenvalues λ, µ of A ∈ sl2(R) satisfy µ + λ = tr(A) = 0 and

λµ = det(A) ∈ R. Hence µ = −λ and λ2 ∈ R. It follows that λ is either real or purely imaginary, and

the eigenvalues e±λ of A lie on the positive real axis or the unit circle, respectively. Thus, for example

the matrix −2 00 −1/2

∈ SL2(R)

is not in the image of exp : sl2(R) → SL2(R).

Corollary 4.14 Let G and H be Lie groups with G connected. Then any Lie group homomorphism

ϕ : G → H is determined by its derivative ϕ0 : g → h.

29



Proof: If ϕ and ψ are two Lie group homomorphisms from G → H with ϕ0 = ψ0 then applying

functorality Prop. 4.10, we have

ϕ ◦ expG = expH ◦ϕ0 = expH ◦ψ

0 = ψ ◦ expG .

From Prop. 4.11 it follows that ϕ and ψ agree on a neighborhood of e in G. From Prop. 4.13 it follows

that ϕ and ψ agree everywhere.

Corollary 4.15 Let ϕ : G → H be a Lie group homomorphism and assume H is connected. If

ϕ : g → h is surjective then ϕ is surjective.

Proof: Since H is connected it is generated by expH (h), by Prop. 4.13. Since ϕ is surjective we have

expH (h) = expH (ϕ(g)) = ϕ(expG(g)). Hence the image of ϕ generates H , so ϕ is surjective.

4.6 The Adjoint representation

A representation of a group G is a homomorphism

ρ : G −→ GL(V ), (6)

where V is a vector space and GL(V ) = Aut(V ) is the group of linear automorphisms of V . If

V = Rn for some n then a choice of basis of V gives an isomorphism GL(V ) GLn(R), so a finite

dimensional representation may be regarded as a homomorphism

ρ : G −→ GLn(R).

However in most situations there is no natural choice of basis and it is better to think of a representationin the form (6).

A Lie group G has a canonical representation

Ad : G → GL(g)

called the adjoint representation, defined as follows. Take an element g ∈ G and let cg : G → G be

the conjugation map:

cg(x) = gxg−1.

Since cg(e) = e, the derivative Ad(g) := cg

maps g →

g. And since cg is a homomorphism, functoral-

ity Prop. 4.10 gives a commutative diagram

g Ad(g)−−−→ g

exp

expG

cg−−−→ G

30



Since cgh = cg ◦ ch, it follows that Ad(gh) = Ad(g) ◦ Ad(h). Thus we have a representation

AdG = Ad : G −→ GL(g), g → Ad(g).

For v ∈ g, the image Ad(g)v is the initial tangent direction of the path t → g exp(t)g−1 in G. Thus,

we have

Ad(g)v = d

dtg exp tvg−1 |t=0.

Now Ad itself is a Lie group homomorphism; its derivative at e is the linear map

ad : g → End(g), u → ad(u) = Ad(u).

For u, v ∈ g, the image ad(u)(v) is the initial tangent direction of the path t → Ad(exp(tu))v in g.

Thus, we have

ad(u)v = d

dt (Ad(exp(tu))v) |t=0.

Functorality relates Ad and ad via the commutative diagram

g ad(g)−−−→ End(g)

expG

expGL(g)G

Ad−−−→ GL(g).

Recall that expGL(g)(A) = eA = I + A + 12

A2 + · · · . Thus, the diagram expresses the equality

Ad(exp(x)) = ead(x) ∈ GL(g).

It is convenient to use the following alternative notation for ad. The Lie bracket is the mapping

[ , ] : g × g −→ g, given by [u, v] = ad(u)(v). (7)

So the adjoint representation Ad : G → GL(g) is given bv

Ad(exp(u))(v) = ead(u)(v) = v + [u, v] + 1

2[u, [u, v]] + · · · .

The adjoint representation Ad and its derivative ad are functorial.

Proposition 4.16 Let ϕ : G → H be a Lie group homomorphism. Then for every g ∈ G the following

diagram commutesg

ϕ−−−→ h

AdG(g)

AdH (ϕ(g))

g ϕ−−−→ h

and for all u, v ∈ g we have

[ϕ(u), ϕ(v)] = ϕ([u, v]).

31



Proof: For Ad we compute

ϕ ◦ AdG(g) = ϕ ◦ cg = (ϕ ◦ cg) = (cϕ(g) ◦ ϕ) = cϕ(g) ◦ ϕ = AdH (ϕ(g)) ◦ ϕ,

making the diagram commute as claimed. Next we compare two power series in t, namely

ϕ(AdG(exp(tu)v) = ϕ(v + t[u, v] +· · ·

) = ϕ(v) + tϕ([u, v]) +· · ·

,

which is equal to

AdH (ϕ(exp(tu))ϕ(v) = ϕ(v) + t[ϕ(u), ϕ(v)] + · · · .

Comparing coefficients in t we obtain [ϕ(u), ϕ(v)] = ϕ([u, v]).

Finally, the Adjoint representation is as faithful as possible.

Proposition 4.17 If G is a connected Lie group then the kernel of Ad : G → GL(g) is the center of G.

Proof: If z is in the center of G then cz is the trivial automorphism of G so its derivative Ad(z ) is theidentity map I . Conversely, for any g ∈ G and v ∈ g we have

g exp(v)g−1 = exp(Ad(g)v).

If Ad(g) = I then g centralizes exp(g), which generates G, so g is in the center of G.

4.6.1 The product rule for paths

At various points we will need to differentiate a product of two paths in a Lie group. We install the

result now for later use.

Let G be a Lie group and let γ, δ : R → G be two smooth paths in G. We then have a product path γδ ,given by (γδ )(t) = γ (t)δ (t). Our goal is to express the derivative (γδ ) in terms of γ, γ , δ , δ . We will

write γ (t) instead of γ (t, 1) etc.

For g, h ∈ G let Lg, Rh : G → G be the maps Lg(x) = gx and Rh(x) = xh. Associativity of the group

law on G means that the maps Lg and Rh commute. The product rule is as follows.

Proposition 4.18 For all t ∈ R we have

(γδ )(t) = Lγ (t)δ (t) + Rδ(t)γ (t) ∈ Gγδ(t)

Proof: Let µ : G × G → G be the product map: µ(g, h) = gh. We compute

µ : T gG ⊕ T hG → T ghG

32



as follows. A typical element of T gG is of the form Lg(v), where v ∈ g = T eG. Then

µ(L

g(v), 0) = d

dtµ(g exp(tv), h)|t=0

= d

dt [g exp(tv)h] |t=0

= d

dt [LgRh(exp(tv))]t=0

= (LgRh)(v)

= R

h ◦ L

g(v)

since Rh and Lg commute. Likewise,

µ(0, R

hu) = L

g ◦ R

h(u).

It follows that for all x ∈ T gG and y ∈ T hG we have

µ(x, y) = µ(x, 0) + µ(0, y) = R

h(x) + L

g(y).

Now γδ = µ(γ, δ ) so the chain rule gives

(γδ )(t) = µ(γ (t), δ (t)) = L

γ (t)(δ (t)) + R

δ(t)(γ (t)),

as claimed.

We use this to express the bracket [ , ] on g as the derivative of a commutator of one-parameter sub-

groups in G. Let [g, h] = ghg−1h−1 be the commutator in G.

Proposition 4.19 For all u, v ∈ g we have

[u, v] = ddt

dds [exp(tu), exp(tv)]s=0

t=0.

Proof: Using the product rule for the paths γ (s) = g(exp(sv)g−1 and δ (s) = exp(sv)−1 = exp(−sv),

we find thatd

ds [g, exp(sv)]s=0 = Ad(g)v − v,

and the result follows.

Corollary 4.20 For all u, v ∈ g we have

[u, v] = −[v, u].

Proof: A similar argument shows that

d

dt [exp(tu), h]t=0 = u = Ad(h)u.

Interchanging the order of differentiation gives the result.

33



4.7 The Lie algebra

A Lie algebra is a vector space L over a field F together with a map

[ , ] : L × L → L

called the bracket, satisfying the following three properties for all x, y, z, w ∈

L and a, b, c, d ∈

F :

(bilinearity) [ax + by,cz + dw] = ac[x, z ] + ad[x, w] + bc[y, z ] + bd[u, w];

(skew-symmetry) [x, y] = −[y, x];

(Jacobi-identity) [x, [y, z ]] = [[x, y], z ] + [y, [x, z ]].

A homomorphism of Lie algebras L, M is a linear map α : L → M preserving the bracket:

α([x, y]) = [α(x), α(y)], for all x, y ∈ L.

We let Aut(L) denote the group of automorphisms of L.

Proposition 4.21 Let G be a Lie group. Then the tangent space g = T eG , with bracket [ , ] : g×g → g

as defined in (7) , is a Lie algebra.

Proof: Since ad : g → End(g) is linear, it follows that the bracket is bilinear. Skew-symmetry was

proved in Cor. 4.20. It remains to prove the Jacobi identity. Fix t ∈ R and u ∈ g and consider the map

ϕ : G → G given by conjugation by exp(tu). We find, for any v ∈ g:

ϕ(v) = Ad(exp(tu))v = v + t[u, v] + · · · ,

so

ϕ

([u, v]) = [v, w] + t[u, [v, w]] + · · · ,but by functorality of the Lie bracket (Prop.4.16) this is equal to

[ϕ(v), ϕ(w)] = [v, w] + t ([[u, v]w] + [v, [u, w]]) + · · · .

Comparing coefficients of t gives the Jacobi identity.

From now on we call g = T eG, with bracket defined in (7), the Lie algebra of the Lie group G.

Proposition 4.22 Let G, H be Lie groups with Lie algebras g, h , and let ϕ : G → H be a Lie group

homomorphism. Then ϕ : g → h is a homomorphism of Lie algebras. In particular, Ad(G) is

contained in the automorphism group Aut(g) of g.

Proof: The first assertion was proved in Prop. 4.16 and the second is an immediate consequence.

If ϕ is injective (or more generally has discrete kernel) then by functorality Prop. 4.10 we have ϕ

injective and we can identify g with a subalgebra of h. In fact, we have

g = {x ∈ h : expH (tx) ∈ ϕ(G) for all t ∈ R}. (8)

34



The containment ⊆ follows from functorality. We will prove the other containment in Cor. 6.5 below,

under the additional assumption that ϕ(G) is closed in H . (This is no additional assumption if G is

compact. For the general case see [Warner, 3.33]). For now, let us use (8) to compute the Lie algebras

of the groups in section 3.3.

Example 1: Let G = GLn(R). We have seen that g = M n(R) and that the exponential map is the

matrix exponential exp(A) = I + A +

1

2A

2

+ · · ·. To compute the Lie bracket we can use Prop. 4.19.

This says that for A, B ∈ g, the bracket [A, B] is the coefficient of st in the group commutator

[exp(tA), exp(sB)] = (I + tA + · · · )(I + sB + · · · )(I − tA + · · · )(I − sB + · · · ),

namely,

[A, B] = AB − BA.

Example 2: Let G = SLn(R), a subgroup of H = GLn(R). B y (8) the Lie algebra g = sln(R)consists of matrices A ∈ M n(R) for which det(exp(tA)) = 1. As det(exp(tA)) = et tr(A), it follows

that

sln(R) = {

A ∈

M n(R) : tr(A) = 0}

.

Examples 3,4: Let G = On or SOn. Since exp(g) is connected, it must lie in SOn and both groups

have the same Lie algebra. g = son consisting of matrices A ∈ M n(R) such that exp(−sA) is the

transpose of exp(sA) for all s ∈ R. The transpose of an exponential matrix is the exponential of the

transpose: t exp(sA) = exp(s · tA). It follows that

son = {A ∈ M n(C) : A + tA = 0.}.

Example 5: Let G = H 3, the Heisenberg group in GL3(R). Here it is easiest to directly compute g as

the set of all tangent vectors to paths γ : R → G. Any path is of the form

γ (t) =

1 x(t) z (t)

0 1 y(t)0 0 1

and has derivative

γ (0) =

0 x(0) z (0)

0 0 y(0)0 0 0

.

Hence the Lie algebra is

h3 =

0 x z

0 0 y0 0 0

: x,y,z ∈R .

We note that A3 = 0 for all matrices A ∈ h3, so the exponential map is given by

exp(A) = I + A + 12A2,

with inverse

log(I + A) = A − 12A2.

35



5 Abelian Lie groups

If a Lie group G is abelian, then the adjoint representation is trivial and the Lie bracket on g is

zero. Nevertheless, abelian Lie groups are important for understanding the structure of non-abelian

Lie groups.

Lemma 5.1 If G is an abelian Lie group then the map expG : g → G is a group homomorphism.

Two examples of abelian Lie groups are R and S 1. The main result says that these two account for all

connected abelian Lie groups.

Proposition 5.2 Let G be a connected abelian Lie group of dimension n. Then there is 0 ≤ k ≤ nsuch that G is isomorphic to (S 1)k × Rn−k as Lie groups.

Proof: Let u, v ∈ g and define γ : R → G by

γ (t) = exp(tu)exp(tv).

Then

γ (t + s) = exp((t + s)u) exp((t + s)v) = exp(tu) exp(su)exp(tv) exp(sv).

Since G is abelian, this is equal to

exp(tu)exp(tv)exp(su)exp(sv) = γ (t)γ (s).

Therefore γ is a one-parameter subgroup. From the product formula we have γ (0) = γ u(0)+ γ v(0) =

u + v, so in fact γ (t) = exp(t(u + v)). Therefore, we have

exp(tu)exp(tv) = exp(tu + tv)

so exp is a group homomorphism, as claimed.

Let L = ker exp. Since exp is injective on a neighborhood of 0 ∈ g, it follows that L is a discrete

subgroup of g. We may write L = Ze1 ⊕ · · ·Zek for some k ≤ n, and extend to a basis {e1, . . . , en} of

g. This gives an isomorphism g ∼→ Rn sending L → Zk ⊕ 0n−k. Hence

G

g/L

Rk ⊕ Rn−k

Z

k

⊕ 0n−k

(S 1)k

×Rn−k.

A torus is a Lie group which is isomorphic to a product (S 1)n for some integer n. From Prop. 5.2 we

have the immediate

Corollary 5.3 Every compact abelian Lie group is a torus.

36



Since S 1 = R/Z, a torus T of dimension n is isomorphic to Rn/Zn. The Lie algebra t of T is Rn with

Lie bracket identically zero, and the exponential map is the projection Rn → Rn/Zn.

A key property of tori is that they are topologically cyclic. That is, a torus contains a dense cyclic

subgroup.

Proposition 5.4 Let T be a torus. Then there exists t ∈ T whose powers {tn : n ∈ Z} are dense in T .

Proof: Let T = Rn/Zn with projection map π : Rn → T and choose a countable basis {U 1, U 2, . . . }for the topology on T . A cube in Rn is a product of closed intervals [a1, b1] × · · · × [an, bn] of constant

side length |ai − bi|.Let C 0 ⊂ Rn be any cube. Inductively define a nested sequence of cubes C 0 ⊃ C 1 ⊃ · · · as follows.

If C 0 ⊃ · · · ⊃ C k−1 has been defined, let be the side length of C k−1 and choose an integer N =N (m) > 1/. Then N C k−1 is a cube of side > 1, so π(NC k−1) = T . The map πN , given by the

composition

C k−1N

−→ N C k−1π

−→ T

is continuous, so (πN )−1(U k) is an open subset of Rn contained in C k−1. Then C k is chosen to be any

cube contained in (πN )−1(U k).

Take x ∈k C k and let t = π(x) ∈ T . For all k > 1 we have

tN (k) = πN k(x) ∈ πN k(C k) ⊂ U k.

Hence the powers {t, t2, . . . } meet every open set in T .

An element of a torus T whose powers are dense in T is called a topological generator of T .

6 Subgroups of Lie groups

6.1 Closed subgroups

Let M be an n-dimensional manifold and let S be a subset of M . We say that S is a submanifold of

M 6 if there exists a subspace W ⊂ Rn and an atlas {(ϕα, U α, M α) : α ∈ A} in the smooth structure

on M such that for all α ∈ A we have either

• S ∩ M α = ∅, or

• ϕα(W ∩ U α) = S ∩ M α.

6There are various definitions of submanifold. The one used here is usually called “embedded submanifold”.

37



We say that such an atlas is good for S .

Remark 1: If S is a submanifold of M then S is a manifold, with charts (ϕα, W ∩U α, S ∩ M α) where

S ∩ M α = ∅.

Remark 2: If G is a Lie group and H ⊂ G is a submanifold and a subgroup then H is a Lie group.

However, there are subgroups of Lie groups which are Lie groups but not submanifolds. An example

is G = R2Z2 and H = {[t√ 2, t] : t ∈ R}. Here H R is dense in G.

Theorem 6.1 A closed subgroup of a Lie group is a submanifold.

Proof: Let G be a Lie group with Lie algebra g. Put a metric |x| on g, say by choosing a basis of g

and declaring it to be orthonormal.

Let H be a subgroup of G which is closed in G. The proof requires three lemmas.

We say a vector v

∈ g is “known by H ” if there exists a sequence of nonzero vectors (xn)

⊂ g such

that exp(xn) ∈ H for all n, xn → 0 and |xn|−1xn → v. Define

h := {tv : t ∈ R and v is known by H }.

Lemma 6.2 We have exp(h) ⊂ H .

Proof: Let tv ∈ h and let (xn) be a sequence for v as above. Choose integers mn such that mn =t/

|xn

| + rn, where 0

≤ rn < 1. Then mn

|xn

| → t mnxn = mn

|xn

| · |xn

|−1xn

→ tv. We have

exp(mnxn) = exp(xn)mn ∈ H for all n, so exp(mnxn) → exp(tv) ∈ H since H is closed.

Lemma 6.3 h is a vector subspace of g.

Proof: It suffices to show that h is closed under addition. Let x, y ∈ h. Choose neighborhoods

0 ∈ U ⊂ g and e ∈ V ⊂ G such that exp : U → V is a diffeomorphism, and denote the inverse by

log : V → U . Since multiplication is continuous on G we can choose a smaller neighborhood V 1 ⊂ V such that V 1 · V 1 ⊂ V .

Choose > 0 such that exp(tx) and exp(ty) lie in V 1 for all

|t| < . Then exp(tx)exp(ty)

∈ V , so we

can define a function f : (−, ) → g by

f (t) = log(exp(tx) exp(ty)).

We have f (0) = 0 and exp(f (t)) = exp(tx)exp(ty). Differentiating at t = 0, we get

f (0) = x + y.

38



Let xn = f (1/n), for n > 1/. Then exp(xn) = exp(x/n) exp(y/n) ∈ H by Lemma 6.2, xn →f (0) = 0 and

|xn|−1xn =

1/n

f (1/n)

· 1

1/nf (1/n) → 1

|x + y| · (x + y).

Hence the latter is known by H so x + y ∈ h.

Choose a vector space complement m to h, so that g = h ⊕ m. Define ϕ : g → G by ϕ(x, y) =exp(x) exp(y) where x ∈ h and y ∈ m. Since ϕ = exp on each summand h and m, it follows that

ϕ = I . Hence there is a neighborhood 0 ∈ U 0 ⊂ g such that ϕ : U 0 → ϕ(U 0) is a diffeomorphism

onto an open neighborhood ϕ(U 0) of e in G.

Lemma 6.4 There exists a neighborhood e ∈ V ⊂ ϕ(U 0) in G such that H ∩ V ⊂ ϕ(h ∩ U 0).

Proof: If no such V exists then there is a sequence hn → e in H ∩ ϕ(U 0) such that ϕ−1(hn) inhfor all n. We have ϕ−1(hn) = (xn, yn) with xn ∈ h and yn ∈ m with yn = 0 for all n. However,

hn

= exp(xn

)exp(yn

), so exp(yn

) ∈

H for all n. Since hn →

e we have (xn

, yn

) →

(0, 0). The

sequence |yn|−1yn is bounded, hence has a convergent subsequence (unk), whose limit u ∈ m satisfies

|u| = 1. But since exp(ynk) ∈ H and ynk = 0, it follows that the subsequence ynk is known by H and

therefore u ∈ h. Hence u ∈ h ∩ m = {0}, so u = 0, a contradiction.

We can now prove Thm. 6.1. Let U = ϕ−1(V ) ⊂ U 0. Then H ∩ V = ϕ(h ∩ U ). On G define an atlas

{(ϕg, U g, Gg) : g ∈ G} as follows.

For h ∈ H , set

U h = U, ϕh(u) = h · ϕ(u), Gh = hV.

Then ϕh(h ∩ U h) = h · ϕ(h ∩ U ) = H ∩ Gh.

For g /∈ H , choose a neighborhood U g of 0 in U such that gϕ(U g)∩H = ∅, using again the assumption

that H is closed. Then set Gg = g · ϕ(U g) and define ϕg : U g → Gg by ϕg(u) = g · ϕ(u). Then

Gg ∩ H = ∅. This shows that {(ϕg, U g, Gg) : g ∈ G} is a good atlas for H , so H is a submanifold of

G.

Corollary 6.5 If H is a closed subgroup of G then H is a Lie group with Lie algebra

h = {v ∈ g : exp(tv) ∈ H for all t ∈ R},

and the inclusion map H

→ G is a diffeomorphism.

Proof: The proof above constructed an atlas on H given by {(ϕh, U h, H ∩ Gh) : h ∈ H }. As this

came from an atlas on G, the inclusion H → G is a diffeomorphism. We showed in Lemma 6.2 that

h ⊂ {v ∈ g : exp(tv) ∈ H for all t ∈ R}. For the reverse containment, we may assume |v| = 1. Then

the sequence xn = (1/n)v shows that v is known by H . Since exp : h → H is a diffeomorphism near

0, it follows that h = T e(H ) is the Lie algebra of H .

39



Example: Let G = GLn(C) be the group of invertible n × n matrices with entries in C. Regarding

Cn as a real vector space R2n, we have GLn(C) ⊂ GL2n(R). Scalar multiplication by√ −1 gives

an element J ∈ GL2n(R) whose square J 2 = −I , and GLn(C) is precisely the centralizer of J in

GL2n(R). Since centralizing is a closed condition, it follows that GLn(C) is a Lie group. Likewise

SLn(C) = {g ∈ GLn(C) : det(g) = 1} is a closed subgroup of GLn(C) hence is a Lie group.

The Unitary Group Un

is the subgroup of GLn(

C)

preserving the hermitian form

u, v =i

uivi.

More explicitly,

Un = {g ∈ GLn(C) : g−1 = tg}.

These are closed conditions, so Un is a Lie group. In fact, each column of a matrix in Un lies in the

unit sphere S 2n−1 ⊂ Cn and Un is compact. Note that U1 = S 1.

The Special Unitary Group is the Lie group SUn = U n ∩ SLn(C). For n = 2 we have SU2 = S 3.

6.2 Homogeneous spaces

Let G be a Lie group and let H be a closed subgroup, and let π : G → G/H be the projection of Gonto the set G/H = {gH : g ∈ G} of left cosets of H in G. Declare a subset U ⊂ G/H to be open iff

π−1(U ) is open in G. This makes π an open mapping, and H being closed is equivalent to G/H being

a Hausdorff topological space.

Choose a neighborhoods U, V of 0 ∈ g and e ∈ G such that exp maps U diffeomorphically onto V .Recall that g = h

⊕m, where m is a vector space complement to h in g. Define

ψ : m ∩ U −→ G/H, by ϕ(u) = π exp(u).

From Lemma 6.4 it follows that ϕ is a homeomorphism onto an open neighborhood of eH in G/H .Left-translating as above, we obtain an atlas on G/H making G/H into a smooth manifold such that

the projection π : G → G/H is smooth and π maps m isomorphically onto T eH (G/H ).

Such manifolds arise as follows. A homogeneous space for G is a manifold M on which G acts

transitively, such that the action map

G × M −→ M

is smooth. Recall from the theory of group actions that for any p

∈ M with stabilizer G p in G the map

g → g · p gives a G-equivariant bijection f p : G/G p → M . Since the action is smooth, it follows that

f p is actually a diffeomorphism. Hence every homogeneous space for G is of the form G/H for some

closed subgroup H of G.

40



6.3 Compact subgroups

Let G be a Lie group and let H be a compact subgroup. Then H is closed, hence is itself a Lie group

whose Lie algebra

h = {v ∈ g : exp(tv) ∈ H for all t ∈ R}is a subalgebra of g. In the proof of Thm. 6.1, where H was only assumed to be close, we made use of

an arbitrary vector space complement to h in G. Since now H is compact, we can do better.

We first need a general property of representations of compact groups. Let V be a finite dimensional

real vector space. A positive definite inner product , on V is a bilinear map

V × V → R, (u, v) → u, vwhich is symmetric: u, v = v, u and satisfies v, v > 0 for all nonzero v ∈ V . It is a fact (which

we will not use) that for any positive definite inner product there exists a basis {vi} of V such that

vi, vi = 1 for all i and vi, v j = 0 if i = j .

Proposition 6.6 Let ρ : K → GL(V ) be a continuous representation of a compact Lie group K on a finite dimensional real vector space V . Then there exists a positive definite inner product , on V such that

ρ(k)u, ρ(k)v = u, v for all k ∈ K, u, v ∈ V.

Such an inner product is called K -invariant.

Proof: We prove this assuming the existence of Haar measure. This is a measure dk on K such that

for all continuous functions f : K → R and all h ∈ K , we have

K

f (hk) dk = K

f (k) dk. (9)

This measure dk is unique up to scalar multiple; we normalize it so that K 1 dk = 1.

Now on V we take any inner product ( , ) which is positive-definite and define a new inner-product by

averaging:

u, v =

K

(ρ(k)u, ρ(k)v) dk.

This new inner product is still positive definite, since

v, v = K

(ρ(k)v, ρ(k)v) dk > 0,

and is also invariant under K , using (9).

Remark: In the case where G is compact with finite center and V = g is the adjoint representation,

there is a canonical positive-definite inner product on g which is G-invariant, namely

u, v = −tr(ad(u) ◦ ad(v)).

This form (rather its negative) is called the Killing form on g.

41



Proposition 6.7 Let U ⊂ V be a subspace such that ρ(K )U = U . Then there exists a vector space

complement W to U in V such that ρ(K )W = W .

Proof: Define W := {v ∈ V : u, v = 0 for all u ∈ U }. From (??) it follows that ρ(k)W = W for

all k ∈ K .

A subspace U ⊂ V is K -invariant if ρ(K )U = U . A representation (ρ, V ) is irreducible if there areno proper invariant subspaces. Applying Prop. 6.7 repeatedly, we obtain:

Corollary 6.8 Every finite dimensional representation of a compact group is a direct sum of irre-

ducible representations.

Applying Prop. 6.7 to the compact subgroup H ⊂ G and its representation ρ = AdG |H obtained by

restricting AdG to H , we obtain:

Corollary 6.9 If H is a compact subgroup of a Lie group G with Lie algebra h

⊂ g , then there is a

vector space complement m ⊂ g such that AdG(H )m = m.

Thus the restriction of AdG to H gives a representation

AdG : H → GL(m).

This is called the isotropy representation of the subgroup H ⊂ G because the projection G → G/H identifies m with the tangent space to G/H at eH , where H is the isotropy group. More generally

Ad(g)m is the tangent space to T (G/H ) at gH and we may regard the entire tangent bundle of G/H as a set of pairs:

T (G/H ) = {(gH,v) : gH ∈ G/H, v ∈ Ad(g)m}.In contrast to the manifold G (where H = {1}) the manifold G/H is generally not parallelizable.

For example, if G = SO3 and H = SO2 we have G/H = S 2, on which every vector field vanishes

somewhere.

Example 1: Take G = GLn(R) and H = On(R). Then g = M n(R) decomposes as g = h⊕m, where

h = son = {A ∈ M n(R) : tA = −A}, m = {B ∈ M n(R) : tB = B},

are the spaces of skew-symmetric and symmetric matrices, respectively. Note that if h ∈ On and

B ∈ m then Ad(h)B = hBh−1 is again in m. The homogeneous space M = G/H is the set of

positive definite inner products onRn

.

Example 2: Take G = SOn+1 and let H = SOn be the subgroup fixing en+1. Then

m =

Ax =

0 x−tx 0

: x ∈ Rn

.

Identifying m = Rn, AdG |H becomes the natural representation of SOn on Rn. The homogeneous

space M = G/H is the sphere S n and the map m → S n given by Ax → exp(Ax) · en+1 is spherical

42



coordinates on S n. If we change this example slightly and take H = On fixing en+1 up to sign, then h

and m are unchanged and G/H is the real projective space Pn(R).

Example 3: The previous example generalizes as follows. Take G = SOn and let H = G ∩ (Ok × O)be the subgroup preserving a k-dimensional subspace U ⊂ Rn (and therefore also preserving the

= n − k-dimensional orthogonal complement of U ). Then

m =

0 x−tx 0

: x = k × matrix

and (h1, h2) ∈ H acts by x → h1xh−12 . The homogeneous space is the Grassmannian of k-planes in

Rn. If we take instead H = SOk × SO then h and m are unchanged but now G/H is the manifold of

oriented k-planes in R. When k = 1 an oriented line is a ray which may be identified with a point on

the sphere.

Example 4: Take G = Un+1 and H = U1 × Un. Then m = Cn with the natural action of Un and

scalar multiplication by U1, and Un+1 / U1 ×U n is the complex projective space Pn(C). Since C× is

connected and complex manifolds have a natural orientation, this example has only one version, unlike

the previous examples.

7 Maximal Tori

Let G be a compact Lie group. A maximal torus in G is a subgroup T ⊂ G which is a torus, and is

contained in no larger torus.

Lemma 7.1 A conjugate of a maximal torus is a maximal torus.

Proof: Let T be a maximal torus in G and let g ∈ G. The conjugation map cg : G → G is a

diffeomorphism of the manifold G and an automorphism of the group G, so gT g−1 is a torus. If U is

a torus containing gT g−1 then g−1U g is a torus containing T , which equals T by maximality of T , so

gT g−1 = U and so gT g−1 is a maximal torus.

Let t denote the Lie algebra of a maximal torus T in G. As G is compact, the adjoint representation

AdG decomposes as

g = t ⊕ m,

where m is the orthogonal complement of t with respect to some AdG-invariant inner product on g (for

example, the negative of the Killing form). We will study the representation of T on m, which is the

tangent space to G/T at eT .

Lemma 7.2 Every continuous finite dimensional irreducible real representation of a torus T is either

trivial (one-dimensional with trivial T -action) or is two-dimensional, of the form

ρλ(exp(x)) =

cos2πλ(x) − sin2πλ(x)sin2πλ(x) cos 2πλ(x)

for a unique linear functional λ : t →R.

43



Proof: Let ρ : T → GL(V ) be a nontrivial continuous irreducible real representation. Let t ∈ T be a

topological generator. Since T is connected and compact, the eigenvalues of ρ(t) lie on the unit circle

and at least one has infinite order. Since ρ(t) is a real matrix, the latter come in complex-conjugate

pairs e±iθ. Let V θ ⊂ V ⊗ C be the span of two eigenvectors with eigenvalues eiθ and e−iθ. Then

V θ is a four-dimensional real vector space whose fixed-points under complex conjugation are two-

dimensional. Since V is irreducible, this two-dimensional space is all of V . Again, since T is compact

and connected, ρ(T ) ⊂ SO2. The functional λ is the derivative ρ

: t → so2 = R. Since λ determinesρ, this functional is unique.

The linear functionals λ which can arise in Lemma 7.2 are precisely those which take integer values

on the lattice L = ker[exp : t → T ]. Thus, the irreducible representations of T are parametrized by the

lattice

L∗ = {λ ∈ t ∗ : λ(L) ⊂ Z}.

Lemma 7.3 A maximal torus T in G is maximal connected abelian.

Proof: Let A be a connected abelian subgroup of G containing T . Since multiplication is continuous,the closure A is also abelian and is still connected. Hence A is a torus, so T = A and therefore T = A.

Lemma 7.4 The vectors in g fixed by AdG(T ) are precisely those in t .

Proof: The vectors in t are fixed by AdG(T ) since T is abelian and AdG(T ) = AdT (T ) on t . Con-

versely, suppose u ∈ g is fixed by AdG(T ). Let H = {exp(su) : s ∈ R}. For all t ∈ T we

have

t · exp(su) · t−1 = exp(s Ad(t)u) = exp(su),

so H centralizes T . The group HT is abelian and connected, as it is the image of the map H ×T → G,

sending (h, t) → ht. From the previous lemma, HT = T , so exp(su) ∈ T for all s ∈ R. By Cor. 6.5,

we have u ∈ t .

From Lemmas 7.2 and 7.4, it follows that there are is a finite set

R ⊂ t ∗

of linear functionals α on t , closed under α → −α, such that the representation AdG : T → GL(m)

decomposes asm =

α∈R/±

mα

where mα is the two dimensional representation ρα of T on which exp(x) (for x ∈ t ) acts via the matrixcos2πα(x) − sin2πα(x)sin2πα(x) cos 2πα(x)

.

44



The elements of R are called the roots of T in G, because they determine the roots of the characteristic

polynomial det(I − AdG(t)), for t ∈ T . The roots can be seen more directly via the derivative ad of

Ad. Indeed for x ∈ t the matrix of ad(x) on mα is given by

ad(x) = 2πα(x)

0 −11 0

.

Example 1: Take G = Un, with Lie algebra

g = un = {X ∈ M n(C) : t X = −X }.

The diagonal matrices in G form a maximal torus T U n1 . The Lie algebra of T consists of diagonal

matrices with imaginary entries u =√ −1(u1, . . . , un), so t ∗ has basis {xi} where xi(u) = ui. For

1 ≤ i < j ≤ n and z ∈ C let X ij(z ) be the matrix with ij-entry equal to z , ji-entry equal to −z , and

all other entries zero. Let

mij = {X ij(z ) : z ∈ Z}.

Then mij is preserved by AdG(T ) with roots xi−x j, x j −xi and the decomposition of sun with respectto AdG(T ) is

sun = t ⊕i<j

mij

and the roots of Un are given by

R = {xi − x j : 1 ≤ i = j ≤ n}.

For SUn a maximal torus is the det = 1 subgroup of the above T , and sun is the tr = 0 subspace of

un. The subspaces m, mij and the roots are unchanged.

Example 2: Take G = SO2n (or O2n), with Lie algebra

g = so2n = {X ∈ M 2n(R) : tX = −X }.

Think of matrices in so2n as n × n matrices where each entry is a 2 × 2 matrix. For 1 ≤ i < j ≤ n and

a 2 × 2 matrix z , let X ij(z ) be the matrix with ij-entry equal to z , ji-entry equal to −tz , and all other

entries zero. For 1 ≤ i < j ≤ n let

mxi−xj = {X ij

x −yy x

: x, y ∈ R},

mxi+xj = {X ijx y

y −x

: x, y ∈ R}.

Then mxi−xj and mxi+xj is preserved by AdG(T ) with roots ±(xi − x j) and ±(xi + x j) respectively

and the decomposition of so2n with respect to AdG(T ) is

so2n = t ⊕i<j

(mxi−xj ⊕ mxi+xj)

45



and the roots of SO2n are given by

R = {xi ± x j : 1 ≤ i = j ≤ n}.

Example 3: Take G = SO2n+1 (or O2n+1), with Lie algebra

g = so2n+1 = {X ∈ M 2n+1(R) : tX = −X }.

Then G contains SO2n and a maximal torus T of SO2n is also a maximal torus of SO2n+1. We have

so2n+1 = so2n ⊕ R2n,

as in Example 2 of section 6.3. The action of T = SOn2 on R2n is a direct sum of the n irreducible

representations obtained by projecting onto each factor of T . Thus, the roots of SO2n+1 are given by

R = {xi ± x j : 1 ≤ i = j ≤ n} ∪ {xi : 1 ≤ i ≤ n}.

7.1 The Weyl group

Let G be a compact Lie group with maximal torus T . Let N (T ) = {n ∈ G : nT n−1 = T } be the

normalizer of T in G. The Weyl group is the quotient

W = N (T )/T.

Proposition 7.5 The Weyl group W is finite.

Proof: We first show that N (T ) is compact. For all t ∈ T , define maps f t, f t : G → G by f t(g) =gtg−1 and f t(g) = g−1tg. Then

N (T ) =

t∈T

f −1t (T )

∩t∈T

f −1t (T )

,

so N (T ) is closed in the compact group G hence is compact. For each n ∈ N (T ) we have a commu-

tative diagram

t

Ad(n)

−−−→ t expT

expT T

cn−−−→ T.

It follows that Ad(n)L = L, where L = ker expT . This gives a map Ad : N (T ) → Aut(L) GLn(Z),

where n = dim t . Here GLn(Z) is the discrete group of n × n matrices whose inverse is also integral.

Since N (T ) is compact, it follows that Ad(N (T )) is compact and discrete, hence is finite.

46



The kernel of Ad : N (T ) → Aut(L) is the centralizer C (T ) of T in G. Indeed, if Ad(n) is trivial on Lthen Ad(n) is trivial on t , since L spans t . For all x ∈ t we then have Ad(n)x = x so n exp(x)n−1 =exp(Ad(n)x) = exp(x) so n ∈ C (T ). The other containment is clear.

Let H = N (T )◦ be the identity component of N (T ). Since N (T )/C (T ) is the image of N (T ) in

Aut(L), it is finite. It follows that H ⊂ C (T ). Since N (T ) is compact and H is open in N (T ), the

index [N (T ) : H ]

is still finite.

I claim that H = T . Since T is abelian and connected, we have T ⊂ H . Let h be the Lie algebra of H and let x ∈ h. Let S = {exp(tx) : t ∈ R}. Since S ⊂ H ⊂ C (T ), the group generated by S and T is

abelian and connected. But T is maximal abelian connected, by Lemma 7.3. It follows that S ⊂ T , so

x ∈ t by Cor. 6.5. Hence h = t . By Cor. 4.15, the inclusion map T → H is surjective, so H = T , as

claimed. It follows that |W | = [N (T ) : T ] = [N (T ) : H ] is finite.

7.2 The flag manifold

Let G be a compact Lie group with maximal torus T . The flag manifold of G is the homogeneousspace G/T . The group G acts on G/T by left translations: Lg(xT ) = gxT . And the Weyl group acts

on G/T by right translations: Rw(xT ) = xn−1T , where w = nT . Note that Rw is well-defined and

commutes with the operators Lg for g ∈ G.

Example: Take G = S 3 and T = S 1 ⊂ C. Then N (T ) = T, j and |W | = 2 and j represents the

nontrivial element of W . The flag manifold is G/T = S 2. The left action of G on G/T factors through

the homomorphism R : S 3 → SO3 and the natural action of SO3 on S 2. In particular T acts by rotation

with axis through ±i.

The right action of W , given by R j , is the antipodal map on S 2. This action does not come from any

element of SO3. Note that R j interchanges the two fixed-points ±i of T in S 2

.

In general a normalizer preserves the fixed-point set of the normalizee. The fixed-point set of T in G/T is

[G/T ]T = {nT : n ∈ N (T )/T } = W.

Note that G/T is not a group, but W is a subset of G/T and coincides with the fixed-point set of T in

G/T .

7.3 Conjugacy of maximal tori

Always G is a compact Lie group. Our goal is to prove the following.

Theorem 7.6 Let G be a compact Lie group with maximal torus T . Then T meets every conjugacy

class in G. Equivalently,

G =

xT ∈G/T

xT x−1.

47



This is the deepest result in our course. We will prove it in the next section using cohomology. Here

we give derive some consequences of Thm. 7.6.

Corollary 7.7 Every element of G lies in a maximal torus.

Proof: Thm. 7.6 implies that every element of G lies in the maximal torus xT x−1 (see Lemma 7.1).

Corollary 7.8 All maximal tori are conjugate in G.

Proof: Let T , U be a maximal tori in G. Let u ∈ U be a topological generator. By Thm. 7.6 some

conjugate gug−1 lies in T . Then

gU g−1 = g

u

g−1 = g

u

g−1

⊂ T .

But gU g−1 is also a maximal torus, so gU g−1 = T .

Corollary 7.9 The exponential map exp : g → G is surjective.

Proof: Let g ∈ G. Then g ∈ T for some maximal torus T in G, by Cor. 7.7 Since expT : t → T is surjective there is x ∈ t such that expT (x) = g. By functorality, expG agrees with expT on T , so

g = expG(x) as well.

Corollary 7.10 A maximal torus in G is its own centralizer in G.

Proof: Let T be a maximal torus in G and let C (T ) be the centralizer of T in G. Since T is abelian, it

is clear that C (T ) ⊃ T and we must show the reverse containment.

Let g ∈ C (T ). The group H = g, T is abelian and is compact since G is compact. The identity

component H ◦ is connected abelian and contains T , so H ◦ = T . Hence T is open in the compact

group H . Therefore H/T is finite. By construction, the union of cosets

n∈ZgnT

is dense in H , hence meets every open set in H . Each coset hT is open in H , so the union meets every

coset of T in H . It follows that H is a finite union of cosets of the form gnT . That is, H/T is finite

cyclic, generated by gT . This means that gm ∈ T for some m ∈ Z.

Let t ∈ T be a topological generator, and consider the element t · g−m ∈ T . Every element in T is an

mth power, since eiθ = (eiθ/m)m. Hence there is s ∈ T such that t · g−m = sm. Since T is abelian and

g ∈ C (T ), we have t0 = (sg)m.

48



By Cor. 7.7, there exists a maximal torus U containing sg . Then t = (sg)m ∈ U so T ⊂ U , hence

T = U by maximality. We therefore have sg ∈ T . Since s ∈ T , it follows that g ∈ T and we have

proved that C (T ) ⊂ T .

Remark: Cor. 7.10 is equivalent to asserting that a maximal torus is a maximal abelian subgroup of

G. However, not all maximal abelian subgroups are tori. For example, the diagonal matrices in SOn are

isomorphic to C

n−1

2 , but the

2-torsion in a maximal torus has rank equal to the largest integer

≤ n/2.

Likewise in G2 there is a copy of C 32 , but the maximal torus has rank two. Maximal abelian subgroups

which lie in no torus of G are very interesting subgroups; they play an important role in the topology

of G and in connections between Lie groups and local Galois groups.

7.4 Fixed-points in flag manifolds

In this section we prove Thm. 7.6. Let G be a compact Lie group with maximal torus T . We wish to

show that every element g ∈ G has a conjugate in T . Equivalently, we wish to show that gxT = xT for some x

∈ G. In other words, we wish to show that the left-translation map Lg on G/T has a

fixed-point. This is not obvious. If we were considering left-translation on a circle, there would be no

fixed-point. However, on S 2, which is the flag manifold of SO3, every nontrivial element of SO3 fixes

the points in S 2 on its rotation axis.

7.4.1 de Rham cohomology

7 Let M be a compact connected oriented smooth manifold of dimension n. The cohomology of M is

a sequence of finite dimensional real vector spaces

H i

(M ), i = 0, 1, . . .

Satisfying (among others) the following properties

i) H i(M ) = 0 for i > n and H 0 H n R.

ii) H n−i(M ) H i(M ) for all i.

iii) H i(S n) = 0 for 0 < i < n.

iii) A smooth map f : M → M induces a linear map f ∗ : H i(M ) → H i(M ) for all i such that if

g : M →

M is another smooth map we have (f ◦

g)∗ = g∗

◦f ∗.

iv) If f : M × [0, 1] → M is a continuous map such that f t := f (·, t) is smooth for each t ∈ [0, 1]then f ∗0 = f ∗1 .

7For more details on cohomology of manifolds see Bott and Tu, Differential forms in algebraic topology.

49



v) If f : M → M is a smooth map with only finite many fixed-points then

ni=0

(−1)i tr(f ∗, H i(M )) = p∈M f ( p)= p

sgn det(I − f ) p,

where the right side is understood to be zero if f

has no fixed-points.

Here, if f ( p) = p then f : T M → T M restricts to a map f : T pM → T pM and det(I − f ) p is the

determinant of the linear map I − f on T pM and sgn det(I − f ) p ∈ {+1, −1, 0} is the sign of this

determinant. The formula v) is the Lefschetz fixed-point formula, and the alternating sum

L(f ) :=n

i=0

(−1)i tr(f ∗, H i(M ))

is the Lefschetz number of f . In the situation of item iv) we have L(f 1) = L(f 0). It may happen that

f 1 has infinitely many fixed-points but f 0 has only finitely many, so we can apply v) to f 0 to compute

L(f 1).

If f : M → M is the identity map then f ∗ is the identity map on M and

L(f ) =n

i=0

(−1)i dim H i(M )

is the Euler characteristic χ(M ) of M , which can be computed by triangulating M . For example, any

polyhedron drawn on the sphere S 2 with v vertices, e edges and f faces has v − e + f = 2 by Euler’s

theorem. Here χ(S 2) = 2 and we also have

dim H 0(S 2) + dim H 2(S 2) = 2.

We can also compute this using the Lefschetz fixed-point formula, even though the identity map has

infinitely many fixed-points. Let f ∈ SO3 be a nontrivial rotation by angle θ. Then f has two fixed-

points on S 2, and f acts by a rotation by θ in the tangent space at each fixed-point p, where we have

det(I − f p) = det

1 − cos θ sin θ− sin θ 1 − cos θ

= 2(1 − cos θ) > 0.

So

p∈S 2f ( p)= p

det(I

−f ) p = 1 + 1 = 2.

On the other hand, let f t be rotation about the same axis of f by tθ. Then f 1 = f and f 0 = I , so

f ∗ = f ∗0 is the identity map on H ∗(S 2) and we have

L(f ) = χ(S 2).

50



We generalize this example to prove Thm. 7.6 as follows. Let G be a compact connected Lie group

with maximal torus T . Each g ∈ G acts on G/T by Lg(xT ) = gxT . Since g is connected to the

identity e ∈ G by a path in G, we have

L(Lg) = L(Le) = χ(G/T ),

independent of g.

Let t ∈ T be a topological generator. We have txT = xT if and only if t ∈ xT x−1, which means

T = xT x−1, so x ∈ N (T ). Hence the fixed-point set of Lt in G/T is exactly

W = {nT : n ∈ N (T )}.

For each n ∈ N (T ), the derivative map Ln gives a commutative diagram

T eT (G/T ) Ln−−−→ T nT (G/T )

Ln−1tn LtT eT (G/T )

Ln−−−→ T nT (G/T ),

so that

det(I − L

t)nT = det(I − L

n−1tn)eT .

Recall the derivative of the projection π : G → G/T restricts to an isomorphism

π : m ∼−→ T eT (G/T ).

For any s ∈ T this isomorphism transforms AdG(s) on m to the map Ls on T eT (G/T ). Let s = n−1tn

as above and write s = exp(u) for u

∈ t . Then

det(I − L

n−1tn)eT =

α∈R/±

det(I − AdG(s))mα

=

α∈R/±

det

1 − cos2πα(u) sin 2πα(u)− sin2πα(u) 1 − cos2πα(u)

=

α∈R/±

2(1 − cos2πα(u)) > 0,

by Lemma 7.4. It follows that

sgn det(1

−L

t) p = +1

for every fixed-point p of Lt in G/T . For any g ∈ G we therefore have

L(Lg) = L(Lt) = |W | = 0

so Lg has a fixed-point in G/T . This completes the proof of Thm. 7.6

51



8 Octonions and G2

In this section we construct the 14-dimensional compact Lie group of type G2, using the non-associative

algebra O of octonions. The quaternions will play an important role, but unlike H, the unit sphere in

O is not a group. Instead, G2 arises as the automorphism group of O. Thus, G2 is analogous to

Aut(H) = SO3.

8.1 Composition algebras

In this section and the next we follow Conway-Smith 8 on the basic equations in composition algebras.

A quadratic form on a real vector space A is a function N : A →R such that

• N (rx) = r2N (x) for all r ∈ R and x ∈ A.

• The function (x, y) =

1

2 [N (x + y) − N (x) − N (y)] is symmetric and bilinear.

A quadratic form N is positive definite if N (x) > 0 for all nonzero x ∈ A.

Let A be an algebra with unit, not necessarily associative, containing R as a subalgebra. We say A is a

composition algebra if there is a quadratic form N : A → R which is multiplicative:

N (xy) = N (x)N (y) for all x, y ∈ A. (10)

In these notes we always assume without further mention that N is positive definite and call it the norm

on A. Since N takes positive values on A − {0}, we have N (1) = N (1 · 1) = N (1)2 so N (1) = 1 and

likewise N (−1) = 1. The associated bilinear form is given as above by

(x, y) = 12

[N (x + y) − N (x) − N (y)].

This form is symmetric by definition and satisfies (x, x) = N (x) > 0 if x = 0. This implies that if

(x, t) = 0 for all t ∈ A iff x = 0. Replacing x by x − y, we have x = y iff (x, t) = (y, t) for all t ∈ A.

We often derive equations in A using ( , ) in this way.

Define the conjugation x → x, by

x = 2(1, x) − x. (11)

We now have the following equations:

(xy,xz ) = N (x) · (y, z ) = (yx,zx). (12)

That is, left and right multiplication by x preserve the bilinear form, up to the scalar N (x). Equation

(12) is derived from (10) upon replacing y by y + z .

(xy,zw) + (xw,zy) = 2(x, y)(z, w). (13)

8On quaternions and octonions, A.K. Peters, Natick MA, 2003

52



This follows from (12) upon replacing z → w and x → x + z . We only use this when (z, w) = 0,

when we get

(xy,zw) = −(xw,zy) if (z, w) = 0. (14)

(y, xz ) = (xy,z ) = (x, z y). (15)

Thus, the adjoint of right multiplication by x is right multiplication by x, and similarly for left multi-plication. Equation (15) is proved by setting z = 1 in (13).

¯x = x. (16)

This is proved by setting y = 1 and z = t in (13).

xy = yx. (17)

Proof: use (15) repeatedly to get (yx, t) = (x,yt) = (xt, y) = (t,xy) = (t(xy), 1) = (xy,t).

N (x) = xx = xx and (x, y) = 12

(xy + yx). (18)

Proof: Using (15) we have N (x) = (x, x) = (1, xx). From (17) we have xx = xx, so (1, xx) = xxfrom (11). Now 2(x, y) = N (x + y) − N (x) − N (y) = (x + y)(x + y) − xx − yy = xy + yx.

x2 = −1 ⇔ (x, x) = 1 and (x, 1) = 0. (19)

Proof: If x2 = −1 then N (x)2 = N (x2) = N (−1) = 1 so (x, x) = N (x) = 1 and (x, 1) (12)= (x2, x) =

(−1, x) = −(x, 1), so (x, 1) = 0. Conversely, (x, 1) = 0 implies x = −x by (11) and (x, x) = 1implies 1 = xx =

−x2 by (18).

This implies that any pair of vectors orthogonal to each other and to 1 must anti-commute:

xy = −yx whenever (x, y) = (x, 1) = (1, y) = 0. (20)

Proof: Since (x, y) = 0 we have xy + yx = 0 from (18) and x = −x, y = −1 from (15).

Finally, we have the alternative law, which is a form of associativity for two elements.

x(xy) = (x2)y and (xy)y = x(y2). (21)

Proof: From (x(xy), t) c1= (xy,xt)

m2= N (x)(y, t)

norm= ((xx)y, t) we have x(xy) = (xx)y. Replacing

x by 2(x, 1)

−x and cancelling gives (21).

8.2 The product on the double of a subalgebra

Let H be any proper subalgebra of A. We let a, b, c , . . . denote typical elements of H . The restriction

of N to H makes H into a composition algebra. From (11) it follows that H is closed under the

conjugation operation a → a.

53



Choose any unit vector x ∈ A orthogonal to H . From (19) it follows that x2 = −1. The entire subspace

xH is orthogonal to H , for (xa,b) = (x, ba) = 0. Since H contains 1, we have

xa = −xa. (22)

It follows that xH = H x and

xa = ax for all a ∈

H. (23)

The double of H is the subspace H + xH ⊂ A. Since the summands are orthogonal, the dimension

of H + xH is twice that of H . In fact since (xa, xb) = (a, b), it follows that left multiplication

Lx : H → xH is an isometry.

If the double H + xH is not all of A, we may take the double of H + xH etc. This proves:

Proposition 8.1 Any composition algebra has dimension equal to a power of two.

We now come to the essential result.

Proposition 8.2 The double H + xH is closed under the product in A and we have

(a + xb)(c + xd) = (ac − db) + x(cb + ad), (24)

for all a, b, c, d ∈ H .

Proof: Equation (24) is equivalent to the following three relations:

i) a(xd) = x(ad).

ii) (xb)c = x(cb).

iii) (xb)(xd) = −db.

Let t ∈ A be arbitrary.

For i) we have

(a(xd), t) (15)= (xd, at)

(14)= − (ad, xt)

(15)= (x(ad), t) .

For ii) we have

((xb)c, t) (15)

= (xb,tc) (23)

=

bx,tc (14)

= − bc,tx (17)

= − (cb), tx (15)

=

(cb)x, t (23)

= (x(cb), t)

For iii) we have

((xb)(xd), t) (15)=

xb,t(xd)

(22)= − (xb,t(xd))

(14)= (x(xd), tb)

(15)= − (xd,x(tb))

(12)= − (d,tb)

(15)=−db, t

.

54



Equation (24) is not a definition of a product; it is a consequence of the existence of the norm N on A.

However, we can reverse the process: Starting with a composition algebra H with norm N , let x be

a symbol and let H + xH be the set of pairs a + xb with a, b ∈ H and multiplication given by (24).

Extending the norm to H + xH by

N (a + xb) = N (a) + N (b)

makes H and xH orthogonal. Now define the product on H + xH via (24):

(a + xb)(c + xd) def = (ac − db) + x(cb + ad).

Now H + xH will be a composition algebra exactly when

[N (a) + N (b)][N (c) + N (d)] = N (ac − db) + N (cb + ad).

Expanding and cancelling terms, this will hold exactly when

(cb, ad) = (ac,db),

which by (15) is equivalent to (a(cb), d) = ((ac)b, d)

for all a, b, c, d ∈ H , which is equivalent to H being associative. This proves:

Proposition 8.3 Given a composition algebra (H, N ) , the double H + xH , with product as above and

norm N (a + xb) = N (a) + N (b) , is a composition algebra exactly when H is associative.

Starting withR and the norm N (x) = x2, the doubling process produces composition algebrasC,H,O,

the complex numbers, quaternions and octonions, respectively. There the process ends, because O is

not associative. So these are the only composition algebras, a theorem first proved by Hurwitz in 1898.

8.3 Parallelizable spheres

We remarked in section 4.3 that the spheres S n are parallelizable exactly for n = 1, 3, 7. The proof that

S 3 is parallelizable works in all cases, using multiplication in composition algebras.

Let A be composition algebra of dimension 2k. Since N is positive-definite, the set S = {x ∈ A :N (x) = 1} of all unit vectors in A is a sphere of dimension 2k − 1. The tangent bundle to S is the

manifold

T S =

{(s, v)

∈ S

×A : (s, v) = 0

}.

Let V be the subspace of A orthogonal to 1. Then V = T 1S is the tangent space to S at 1. More

generally, for any x ∈ S , we have (x,vx) = (xx, v) = (1, v) = 0, so V x = T xS is the tangent space

to S at s. Hence the function

X v : S → S × A, given by X v(x) = vx

takes values in T S and is a smooth vector-field on S . Since v → X v(x) is an isomorphism V ∼→ T xS ,

it follows that S is parallelizable.

55



8.4 Automorphisms of composition algebras

An automorphism of a composition algebra A is an R- linear isomorphism σ : A → A such that

σ(xy) = σ(x)σ(y) for all x, y ∈ A. The automorphisms of A form a group Aut(A), under composi-

tion.

Lemma 8.4 If σ ∈ Aut(A) then for all x, y ∈ A we have

1. N (σ(x)) = N (x);

2. (σ(x), σ(y)) = (x, y);

3. σ(x) = σ(x).

Proof: Each item is a consequence of the one before it, so we need only prove item 1.

Let A0 be the subspace of A orthogonal to R, and let S = {x ∈ A0 : N (x) = 1} be the unit sphere inA0. From (19) we have x ∈ S if and only if x2 = −1. It follows that every σ ∈ Aut(A) preserves S .

Now let x ∈ A0 is an arbitrary nonzero element, and let n = N (x)1/2. Then y := (n−1x ∈ S so

σ(y) ∈ S so σ(x) = ny ∈ nS ⊂ A0. Hence σ preserves A0 and N (σ(x)) = n2 = N (x). And since σis an R-linear algebra automorphism we have σ(r) = r for all r ∈ R. Since R and A0 are orthogonal,

it follows that σ preserves N on all of A.

It follows that Aut(A) is a subgroup of the orthogonal group O(A, N ) = {g ∈ GL(A) : N (g(x)) =N (x) for all x ∈ A}. For each x, y ∈ A let f x,y : O(A, N ) → A be the function f x,y(g) =g(x)g(y)

− g(xy). Then G is the common zeros of the continuous functions f x,y, so G is closed in

O(A, N ). It follows that Aut(A) is a compact Lie group. All cases are tabulated below.

A Aut(A)R 1C C 2H SO3

O G2

Let H be a subalgebra of A of dimension half that of A, and let x ∈ A be orthogonal to A with

N (x) = 1. Then we have A = H + xH , an orthogonal sum, and the product in A is derived from the

product in H via the formula (24).

The following result shows how to construct automorphisms of A.

Proposition 8.5 Suppose we are given a subalgebra J ⊂ A , an algebra isomorphism τ : H → J ,and an element y ∈ A orthogonal to J . Then τ extends uniquely to an automorphism σ of A such that

σ(x) = y. Conversely, any automorphism of A arises in this way.

56



Proof: We define σ by

σ(a + xb) = τ (a) + yτ (b).

This is the only possible extension of τ to A. It is clearly R-linear and is bijective, since τ is bijective

and the sums H + xH = J + yJ are both direct. We must show that σ preserves the product in A.

The key point is that we can apply Prop. 8.2 to both (H, x) and (J, y). Thus, we get

σ ((a + xb) · (c + xd)) (24)= σ

(ac − db) + x(cb + ad)

= τ (ac − db) + yτ (cb + ad)

= τ (a)τ (c) − τ (d)τ (b) + y (τ (c)τ (b) + τ (a)τ (d))(8.4)= τ (a)τ (c) − τ (d)τ (b) + y

τ (c)τ (b) + τ (a)τ (d)

(24)= (τ (a) + yτ (b)) · (τ (c) + yτ (d))

= σ(a + xb) · σ(c + xd).

Conversely, given any automorphism σ of A, let J = σ(H ), y = σ(x). Then N (y) = N (x) = 1 and

(y, J ) = (x, H ) = 0, by Lemma autoconj. Then σ arises as in the first part of the Proposition from itsrestriction τ to H .

8.5 The Octonions O

The octonions are the double of the quaternions. Thus we have

O = H + H,

where H is the quaternion algebra with basis {1, i , j , k} and is orthogonal to H with 2 = −1. Thus,

O has R-basis 1,i,j,k,,i,j,k.

Since the quaternions are associative the octonions are a composition algebra, with norm

N (x0 + x1i + x2 j + x3k + x4 + x5i + x6 j + x7k =7

i=0

x2i .

For all q ∈ H we have q = q by (23), so from (24) the multiplication in O is given by

( p + q)(r + s) = ( pr − sq ) + (q r + sp).

In particular, we have

j(i) = k, i(k) = j, k(i) = j,and orthogonal elements u, v anticommute: uv = −vu.

Let V be the orthogonal complement of 1 in O and let S be the six-dimensional sphere in V :

S = {x ∈ V : N (x) = 1}.

We next show that any pair of orthogonal vectors in S generates a quaternion subalgebra of O. More

precisely, we have:

57



Lemma 8.6 Let u, v ∈ S be orthogonal vectors. Then the subspace J ⊂ O spanned by {1,u,v,uv}is closed under the product in O and the linear map τ : H → J sending 1 → 1, i → u, j → v, k → uvis an algebra isomorphism.

Proof: Since N (u) = N (v) = 1 and (u, 1) = (v, 1) = 0, we have u2 = v2 = −1 by (19). Likewise

N (uv) = N (u)N (v) = 1 and (1, uv) = (u, v) =

−(u, v) = 0, so (uv)2 =

−1.

From (15) we have (uv,u) = (v, uu) = (v, 1) = 0 and likewise (uv,v) = 0. Hence the vectors

1,u,v,uv are orthonormal and uv anticommutes with u and v by (20). From the alternative law (21)

we have v(uv) = −v(vu) = −(v2)u = u and (uv)u = −(vu)u = −v(u2) = v. This shows that

u,v,uv satisfy the same relations as i, j, k so the lemma is proved.

The automorphism group G := Aut(O) is a compact subgroup of O(V ) O7. Every element of Gmay be constructed as follows. Choose orthogonal vectors u, v ∈ S . By Lemma 8.6 these generate a

quaternion subalgebra J , spanned by 1,u,v,uv. Then choose x ∈ S orthogonal to u, v,uv. Then by

Prop. 8.5 there is a unique automorphism σ ∈ G such that σ(i) = u, σ( j) = v and σ() = x. Thus, we

may identify G with the manifold of triples

T = {(u,v,x) ∈ S × S × S : (u, v) = (x, u) = (x, v) = (x,uv) = 0}.

Let π : T → S be the projection π(u,v,x) = u, and let S u be the five-dimensional sphere in S orthogonal to u. Then π−1(u) may be identified with the set of pairs

T u = {(v, x) ∈ S u × S u : (x, v) = (x,uv) = 0},

and we have dim T = dim S + dim T u. Let πu : T u → S u be the projection πu(v, x) = v. Then

π−1u (v) is the three-dimensional sphere S u,v,uv in S u orthogonal to v and uv , and we have dim T u =dim S u + S u,v,uv. Therefore

dim G = dim T = dim S + dim S u + dim S u,v,uv = 6 + 5 + 3 = 14.

8.6 The SU3 in Aut(O)

9 Recall that automorphisms preserve the norm, so G = Aut(O) acts on the 6-sphere S = {x ∈ V :N (x) = 1}.

Proposition 8.7 G acts transitively on S .

Proof: Let x ∈ S . Choose u ∈ S orthogonal to x and then choose v ∈ S orthogonal to each of x,u,ux.

Then (x,uv) = −(ux,v) = 0, so x is orthogonal to each of u, v,uv hence there is an automorphism

σ ∈ G sending to x.

Let G be the stabilizer of ∈ G, and let W be the orthogonal complement of in V . Then if w ∈ W we have (w, 1) = −(w, ) = 0 and (w,) = (w, ) = (w, 1) = 0, so W = W . Since 2 = −1, we

9The definite article in the title of this and later sections indicates that the subgroup is unique up to conjugacy in Aut(O).

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can identify C with the subalgebra of O generated by {1, } and multiplication by makes W into a

C-vector space with basis {i,j,k}. Define a hermitian form on W by

v, w = (v, w) + (v,w).

For all σ ∈ G and v ∈ W we have σ(v) = σ(v), so

σ(v), σ(w) = (σ(v), σ(w))+(σ(v), σ(w)) = (σ(v), σ(w))+(σ(v), σ(w)) = (v, w)+(v,w) = v, w.

It follows that G is contained in the unitary group U(W ).

Proposition 8.8 G = SU(W ).

Since dim G = dim G − dim S = 14 − 6 = 8 = dim SU(W ), it suffices to show that G ⊂ SU(W ).

Let σ ∈ G. Since σ ∈ U(W ), there exist eigenvectors u, v,w for σ in W which are orthonormal with

respect to the hermitian form

,

. Since uv is a unit vector orthogonal to both u and v, we may take

w = uv. In fact, if σ(u) = eθu and σ(v) = eφv, then (letting c = cos θ, c = cos φ, etc.) we have

σ(uv) = σ(u)σ(v)

= (eθu)(eφv)

= (cu + su)(cv + sv)

= ccuv + ss(u)(v) + csu(v) + sc(u)v

= ccuv − ssuv − cs(uv) − sc(uv)

= [cos(θ + φ) − sin(θ + φ)]uv

= e−(θ+φ)uv.

Thus, the eigenvalues of σ are eθ, eφ and e−(θ+φ), so det σ = 1 and σ ∈ SU(W ).

Therefore the homogeneous space G2/ SU3 is equivariantly diffeomorphic to the 6-sphere S = S 6 and

we have

Corollary 8.9 Every automorphism of O fixes a point on S and acts on the orthogonal 5-sphere via

an element of SU3.

8.7 The maximal torus in Aut(O)

The above proof (with i, j, k instead of u, v, uv) shows that we have a torus T ⊂ G consisting of

automorphisms which fix and send

i → eαi, j → eβ j, k → eγk,

where α + β + γ = 0. This torus T is a maximal torus in G.

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In fact, T is a maximal torus in G as well. For suppose U is a maximal torus of G containing T . Then

U preserves the fixed-point set of T in O, which is the complex plane R⊕R. But U fixes R, hence U must preserve the orthogonal complement R of R in this plane. Since U is connected and preserves

length, it follows that U acts trivially on R, so U ⊂ G. But T is a maximal torus in G, so U = T .

The normalizer N (T ) must also preserve the fixed-point set of T in O, and hence must preserve {±}.

Now N (T )

is not connected, so it can change the sign of

. For example, the element σ2 ∈ Aut(

O)acting by +I on H and −I on H lies in N (T ) and acts on T by σ2tσ−12 = t−1. The subgroup of N (T )

fixing lies in G, and has index two in N (T ). Since the Weyl group of SU(3) is S 3, it follows that the

Weyl group W = N (T )/T is S 3 × −1 D6.

In general, the extension

1 −→ T −→ N (T ) −→ W −→ 1

is not split (eg. in S 3). In this case however, there is a copy of S 3 inside SU(3) projecting isomorphi-

cally onto the Weyl group of SU(3). This subgroup is generated by the involutions

τ =0 1 0

1 0 00 0 −1

, τ

=−1 0 0

0 0 10 1 0

.

Hence the group generated by σ2, τ , τ is isomorphic to D6 and projects isomorphically onto W .

8.8 The SO4 in Aut(O)

Let GH = {σ ∈ G : σ(H) = H} be the stabilizer of the quaternion subalgebra H spanned by

{1, i , j , k}.

Proposition 8.10 GH = SO4.

Proof: Since GH preservesH, it also preserves the orthogonal complement H. If σ ∈ GH acts trivially

on H then GH fixes so σ(u) = σ(u) = u so σ acts trivially on H as well, so σ = 1. Thus GH acts

faithfully on H (though not on H) and we have an injective homomorphism GH → O(H) O4. To

show that GH SO4 it remains to show GH is connected of dimension six.

By restriction we get another map r : GH → Aut(H) = SO3, which is seen to be surjective by taking

H = J = H in Prop. 8.5. Set K = ker r = {σ ∈ G : σ(u) = u ∀ u ∈ H}. An automorphism σ ∈ K

is completely determined by its effect on , and

σ() = uσ,

for some uσ ∈ H. Since N () = N (σ()) = 1 we have N (uσ) = 1, so uσ ∈ S 3 ⊂ H. If τ ∈ K is

another automorphism then since τ fixes uσ we have

uτσ = τ σ() = τ (uσ) = uστ () = uσ(uτ ) = (uτ uσ),

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so σ → uσ is an injective homomorphism K → S 3, which is also surjective by another application of

Prop. 8.5. Thus, GH fits into an exact sequence

1 −→ S 3 −→ GH −→ SO3 −→ 1.

In particular GH is connected and dim GH = 6. Hence GH SO4 as claimed.

Remark 1: The proof shows that the intersection

G ∩ GH = SO3 ⊂ SU3 .

is the subgroup of SU3 consisting of real matrices.

Remark 2: From Prop. 8.5, it follows that G2 acts transitively on quaternion subalgebas of O. Hence

G2/ SO4 is the eight-dimensional manifold of quaternion subalgebras of O and every automorphism

of O stabilizes a quaternion subalgebra.

8.9 The Lie algebra of Aut(O)

The automorphism group G of any finite dimensional R-algebra A is a closed subgroup of GL(A).

Hence the Lie algebra g of G is the Lie subalgebra of End(A) given by

g = {D ∈ End(A) : etD ∈ G for all t ∈ R}.

Expanding and comparing both sides of the identity (for all a, b ∈ A)

etD(ab) = [etDa][etDb]

we find that D ∈ g if and onlyD(ab) = D(a)b + aD(b).

Such an operator is called a derivation of A, and

g = Der(A)

is the Lie algebra of derivations on A with the Lie bracket [D, D] = DD − D D inherited from

End(A).

If A is a composition algebra then Aut(A) ⊂ O(A), the orthogonal group of the norm N : A → R,

so Der(A)

⊂ so(A) is represented by skew-symmetric matrices with respect to an orthonormal basis

of A. Since D(1) = D(1 · 1) = 2D(1), it follows that D(1) = 0 for any derivation. Hence we in facthave A ⊂ so(A0).

Returning to the octonions, we see that the Lie group G = Aut(O) has Lie algebra g = Der(O) ⊂ so7.

We will use this representation to find the roots R and the decomposition

g = t ⊕

α∈R/±

mα.

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As a subalgebra of g, the Lie algebra of G is

{D ∈ g : D() = 0}.

These derivations are linear with respect to the subalgebra C = R⊕R, hence they may be represented

as 3 × 3 matrices with respect to the C-basis i, j, k of the hermitian space W . Our maximal torus

T ⊂ G consists of the automorphisms t given by

t(i) = eai, t( j) = eb j, t(k) = eck, t() =

with a, b, c ∈ R such that a + b + c = 0. Its Lie algebra t consists of the derivations D ∈ g given by

D(i) = ai, D( j) = bj, D(k) = ck, D() = 0, (25)

and we have

g = t ⊕ mb−c ⊕ ma−c ⊕ ma−b,

where

mb−c =

0 0 00 0 −z 0 z 0

: z ∈ C

, ma−c =

0 0 −z 0 0 0z 0 0

: z ∈ C

, ma−b =

0 −z 0z 0 00 0 0

: z ∈ C

The remaining root spaces consist of derivations which are not C-linear, so we have to use 7 × 7 real

matrices in so7. We order the basis as (i,j,k,i,j,k,). The missing root spaces are each spread

out in two root spaces of SO7, whose maximal torus is like T but without the condition a + b + c = 0.

For example, ±(b + c) and ±a for SO7 both restrict to ±a in G. This allows us to find the derivations

comprising the remaining root spaces, as follows:

ma =

0 0 0 0 0 0 2y0 0 0 0 0 0 −2x0 0 0 0 −x −y 00 0 0 0 −y x 0

0 0 x y 0 0 00 0 y −x 0 0 0−2y 2x 0 0 0 0 0

: x, y ∈ R

mb =

0 0 0 0 −x −y 00 0 0 0 −y x 00 0 0 0 0 0 −2y0 0 0 0 0 0 2xx y 0 0 0 0 0y −x 0 0 0 0 00 0 2y −2x 0 0 0

: x, y ∈ R

mc =

0 0 −x −y 0 0 00 0 −y x 0 0 0x y 0 0 0 0 0y −x 0 0 0 0 00 0 0 0 0 0 2y0 0 0 0 0 0 2x0 0 0 0 −2y −2x 0

: x, y ∈ R

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We therefore have 12 roots

R = ±{a − b, b − c, a − c,a,b,c}.

These are linear functionals on

t = {(a,b,c) : a + b + c = 0}.

To see R more explicitly, identify t with its dual space t ∗

, using the dot product. The three linearfunctionals

(a,b,c) → a, b, c

become dot product with

13(2, −1, −1), 1

3(−1, 2, −1), 13(−1, −1, 2),

and we get

R = ±{(1, −1, 0), (0, 1, −1), (1, 0, −1), 13

(2, −1, −1), 13

(−1, 2, −1), 13

(−1, −1, 2)}.

Setting α = 13(−1, −1, 2), be = (0, 1, −1),, we have

R = ±{α, β, α + β, 2α + β, 3α + β, 3α + 2β },

the root-system of type G2. The action of the Weyl group W on t is generated by reflections about the

lines perpendicular to the roots. 10

10Picture taken from Wikipedia.

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8.10 The nonsplit extension 23·GL3(2) in Aut(O) and seven Cartan subalgebras

Let Γ be the subgroup of Aut(O) stabilizing the set

B = {±i, ± j, ±k, ±, ±i, ± j, ±k}.

Letting O7(Z) C 72 S 7 denote the stabilizer in O7 of the lattice spanned by B in O0, we have

Γ = Aut(O) ∩ O7(Z).

Since Γ permutes the seven lines Ri,R j, . . . and preserves the structure of the projective plane P2(F2)whose points are identified with these lines, it follows that this permutation action gives a homomor-

phism Γ → GL3(2), whose kernel is the subgroup of sign changes of the vectors in B . We find the

order of Γ as follows. Every element γ ∈ Γ is determined by γ (i) = u, γ ( j) = v, γ () = w . We have

14 choices for u ∈ B , then 12 choices for v, and then w can be any vector in B orthogonal to u, v,uv,

which leaves 8 choices. Hence

Γ = 14 · 12 · 8 = 23 · 168.

The count also shows there are 23 sign changes. Hence the map Γ → GL3(2) is surjective.

It is not possible to permute the seven vectors {i,j,k,i,j,k,} in 168 ways by automorphisms of

O without changing some signs. This means the extension

1 −→ C 32 −→ Γ −→ GL3(2) −→ 1

is nonsplit. However it splits over the subgroup F 21 corresponding to a Borel subgroup of PSL2(7) GL3(2). In particular, there is an element σ7 ∈ Aut(O) permuting the seven lines Ri,R j, . . . in a

cycle.

A Cartan subalgebra in g is the Lie algebra of a maximal torus in G. Each line λ ∈ {Ri,R j, . . . }generates a copy of C in O. The subalgebra of g killing λ and preserving the remaining six lines is a

Cartan subalgebra t λ. For example, if λ = , we get the Cartan subalgebra t in (25). It follows that the

Lie algebra g = Der(O) is a direct sum

g = t i ⊕ t j ⊕ t k ⊕ t i ⊕ t j ⊕ t k ⊕ t

of seven Cartan subalgebras, permuted simply-transitively by the element σ7. This is completely anal-

ogous to the decomposition

so3 = t i ⊕ t j ⊕ t k,

arising from the coordinate lines Ri, R j, Rk in the quaternion algebra H.

notes on lie groups

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