Download Math 845 Notes on Lie Groups

Math 845
Notes on Lie Groups
Mark Reeder
December 3, 2016
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
Binary Tetrahedral and Octahedral groups . . . . . . . . . . . . . . . . . . . .
7
The exponential map for S 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.3
2
3
4
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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
2.4
The exponential diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
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 . . . . . . . . . . . . . . . . . . . . . . . . .
19
The Lie algebra of a Lie group
21
1
4.1
The tangent bundle of a manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
4.2
Vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
4.3
The tangent bundle of a Lie group . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
4.4
One-parameter-subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
4.5
The exponential map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
4.6
The Adjoint representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
4.6.1
The product rule for paths . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
The Lie algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
4.7
5
Abelian Lie groups
36
6
Subgroups of Lie groups
38
6.1
Closed subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38
6.2
Homogeneous spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
40
6.3
Compact subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
7
8
Maximal Tori
43
7.1
The Weyl group
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
7.2
The flag manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
7.3
Conjugacy of maximal tori . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
7.4
Fixed-points in flag manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
49
7.4.1
49
de Rham cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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
2
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
1
1.1
. . . . .
64
Quaternions and the three-dimensional sphere
Hamiltion’s quaternions
The quaternion algebra H is a four dimensional real vector space with basis 1, i, j, k:
H = R1 ⊕ Ri ⊕ Rj ⊕ Rk
and multiplication rules
ij = k,
jk = i,
ki = j,
i2 = j 2 = k 2 = −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
q̄ = t − xi − yj − zk.
Thus, R = {q ∈ H : q̄ = q}. One checks that
pq = q̄ p̄,
for all p, q ∈ H. The norm of q is
N (q) = q q̄ ∈ R
One checks that N (q) = t2 + x2 + y 2 + 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 6= 0 then N (q)−1 · q̄ is a multiplicative inverse of q in H. Hence H is a division
algebra, that its set of nonzero elements
H× = H − {0}
3
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 q̄ = 1} = {t + xi + yj + zk ∈ H : t2 + x2 + y 2 + 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 q̄ = 1}.
Thus S 3 is a group under quaternion multiplication, fitting into the exact sequence
N
1 −→ S 3 −→ H× −→ 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 = CS 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 jij −1 = −i as
well, so qj −1 ∈ CS 3 i = T , which means q ∈ T j.
We note that Q8 < N (T ) and that
jsj −1 = s̄ = s−1
for all s ∈ T.
4
Also T j = {tj + xk : t2 + x2 = 1} lies on the equatorial two sphere
C0 := {xi + yj + zk : x2 + y 2 + 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 q̄, so we define the trace of q to be
τ (q) = 21 (q + q̄).
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 = q̄. We compute
τ (qpq −1 ) = τ (qpq̄) = 21 (qpq̄ + qpq̄) = 21 (qpq̄ + q̄¯p̄q̄) = 12 (qpq̄ + q p̄q̄) = qτ (p)q̄.
Since τ (p) ∈ R it commutes with q, so we have
τ (qpq −1 ) = τ (p)q q̄ = 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 sets
Ct = {p ∈ S 3 : τ (p) = t}
are preserved under conjugation by S 3 . For t = 0, the level set is the equatorial two-sphere C0 mentioned above. We have
C 0 = S 3 ∩ H0 ,
where
H0 = Ri ⊕ Rj ⊕ Rk = {p ∈ H : τ (p) = 0}.
more generally, for fixed t ∈ [−1, 1], the level set
Ct = t + {xi + yj + zk : x2 + y 2 + 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 Ct as a sphere of constant latitude in
S 3 . 1 Thus, S 3 is the disjoint union of its latitude spheres:
a
S3 =
Ct .
(1)
t∈[−1,1]
1
Of course C1 = {1} and C−1 = {−1} are spheres of zero radius.
5
Proposition 1.3 For each t ∈ [−1, 1], the latitude sphere Ct 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 Ct . We first prove this for C0 .
Following Euler, we write
eiθ = cos θ + i sin θ,
ejθ = cos θ + j sin θ,
ekθ = cos θ + k sin θ.
Thus we have three subgroups Ti , Tj , Tk < S 3 , all isomorphic to S 1 , given by
Ti = {eiθ : θ ∈ R},
Tj = {ejθ : θ ∈ R},
Tk = {ekθ : θ ∈ R}.
Everything we have said about Ti (previously called T ) holds for the other subgroups. Their normalizers are
N (Ti ) = Ti ∪ Ti j,
N (Tj ) = Tj ∪ Tj k,
N (Tk ) = Tk i.
The nontrivial cosets Ti j, Tj k, Tk i are three orthogonal great circles on the two-sphere C0 . Conjugation by j, k, i on Ti Tj , Tk is inversion, meaning that
jeiθ j −1 = je−iθ ,
kejθ k −1 = e−jθ ,
iekθ i−1 j = e−kθ .
It follows that Ti conjugates the coset Ti j to itself, and likewise for Tj with Tj k, and Tk with Tk i.
Explicitly, we have
eiθ · eiα j · e−iθ = ei(α+2θ) j,
ejθ · ejα k · e−jθ = ej(α+2θ) k,
ekθ · ekα j · e−kθ = ek(α+2θ) i.
Now take a point p ∈ C0 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 C0 then conjugation by ejφ/2 sends k down to a point p0 on the same
latitude as p, and then conjugation by ekθ/2 sends p0 over to p. In other words, we have
ekθ/2 ejφ/2 · k · e−jφ/2 e−kθ/2 = p.
This proves that S 3 acts transitively on C0 by conjugation.
Now for any t ∈ (−1, 1), define ft : C0 → Ct by
√
ft (p) = t + ( 1 − t2 )p.
Then ft is bijective, with inverse
q−t
ft−1 (q) = √
,
1 − t2
and for all q ∈ S 3 we have ft (qpq −1 ) = ft (p). Now the transitivity on Ct follows from the transitivity
on C0 , completing the proof.
We can write each t ∈ [−1, 1] as t = cos θ for θ ∈ [0, π], and we have the
6
Corollary 1.4 For 0 < θ < π, the conjugacy class Ccos θ meets each of Ti , Tj , Tk in two mutually
inverse points. Namely,
Ccos θ ∩ Ti = {eiθ , e−iθ },
Ccos θ ∩ Tj = {ejθ , e−jθ },
Ccos θ ∩ Tk = {ekθ , e−kθ }.
Proof: The sets on the right hand side of each asserted equality consist of the points in Ti , Tj , Tk
whose trace is cos θ.
We now understand conjugacy classes of points in S 3 . The next step is conjugacy of circles. More precisely, 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 heiθ i of S 1 generated by eiθ is finite if θ ∈ 2πQ and is dense in
S 1 if θ ∈
/ 2πQ.
Proof: The group A = heiθ i 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 6= n, lying the same arc. Since A = e−imθ A, 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 = Ti = {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 hsi 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 hqsq −1 i is dense in T . Letting X denote the closure
of a subset X ⊂ S 3 , we have
qSq −1 = qhsiq −1 = hqsq −1 i = T.
1.2.1
Binary Tetrahedral and Octahedral groups
2
The relations in the quaternion group show that Q8 has an automorphism of order three sending i 7→
j 7→ k 7→ i. There are also automorphisms such as i 7→ −i, j ↔ k. Can these automorphism be
realized by conjugation in S 3 ?
2
I thank Matt Sarmiento for pointing out a mistake in an earlier version of this section.
7
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 + y 2 + z 2 = 1, it follows that t = ± 12 . Since all elements of S 3 with a given t are
conjugate and cos(π/3) = 1/2 while cos(2π/3) = −1/2, we see that 12 (1 + i + j + k) has order six
while − 12 (1 + i + j + k) has order three.
The 16 quaternions q = 12 (±1 ± i ± j ± k) ∈ S 3 , with all possible combinations of signs, along with
Q8 itself, comprise a 24-element subgroup
±1 ± i ± j ± k
,
G24 = {±1, ±i, ±j, ±k} ∪
2
and Q8 is the unique (hence normal) Sylow 2-subgroup of G24 . The group G24 is usually called the
binary tetrahedral group for reasons that will become clear in the next chapter. One can show that
G24 ' SL2 (Z/3Z),
the group of 2 × 2 matrices over Z/3Z with determinant = 1.
Now G24 < N (Q8 ), the normalizer of Q8 in S 3 . To determine the size of N (Q8 ) we first note that
conjugation gives a map
N (Q8 ) → Aut(Q8 )
whose kernel is {±1}. Next, | Aut(Q8 )| = 24 because Aut(Q8 ) acts transitively on the set of orderfour subgroups of Q8 and one checks there are exactly eight automorphisms stabilizing hii. (With a
little more work
√ one can show that Aut(Q8 ) ' S4 .) It follows that |N (Q8 )| ∈ {24, 48}. But one
checks (1 + i)/ 2 ∈ N (Q8 ) − G24 , so |N (Q8 )| = 48 and we see that every automorphism of Q8 arises
via conjugation in N (Q8 ).
One can also check that N (Q8 ) = N (G24 ) = N (N (Q8 )). Hence from now on we write
G48 = N (Q8 ).
Explicitly, the elements of G48 outside G24 are given by
±1 ± i ±1 ± j ±1 ± k
±i ± j ±j ± k ±k ± i
√ , √ , √
√ , √ , √
G48 − G24 =
∪
2
2
2
2
2
2
again with all possible choices of signs. For reasons that will become clear, the group G48 is usually
called the binary octahedral group.
Note that G48 is not isomorphic to GL2 (Z/3Z): the latter has too many involutions to be a subgroup of
S 3 . However, both G48 and GL2 (3) are two-fold covers of S4 , the former via its action on Q8 and the
latter via its action on the projective line over Z/3Z.
8
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,
∞
X
zn
1
z
z
.
S = {e : z ∈ iR},
where
e =
n!
n=0
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,
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 7→ 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 + Rj + Rk . Thus, we define
exp : H0 −→ S
3
by
exp(v) =
∞
X
vn
n=0
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 = C0 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 v 2 = −θ2 , so for all k ≥ 0 we have v 2k = (−1)k θ2k and v 2k+1 = (−1)k θ2k v.
It follows that
exp(v) =
∞
X
(−1)k θ2k
k=0
(2k)!
+v
∞
X
(−1)k θ2k
k=0
(2k + 1)!
= cos θ + v
sin θ
= cos θ + v0 sin θ.
θ
(2)
as with Euler’s formula. This is consistent with our earlier definitions of eiθ , ejθ , ekθ ; these were values
of exp on the three lines iR, jR, kR in H0 .
From (2) we observe that exp maps the sphere θC0 ⊂ H0 of radius θ to the conjugacy-class Ccos 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) · C0 bijectively onto S 3 − {−1};
3. exp(πC0 ) = {−1} ⊂ S 3 ;
9
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 ∈ Ccos θ then q = cos θ + q0 , where q0 ∈ H0 has
squared-length |q0 |2 = 1 − cos2 θ = sin2 θ. The vector v0 = (sin θ)−1 q0 lies in C0 and exp(θv0 ) = q.
Therefore exp(θC0 ) = Ccos θ .
Suppose v, v 0 ∈ H0 have exp(v) = exp(v 0 ). Write v = θv0 , v 0 = θ0 v00 , with θ, θ0 ∈ [0, π) and
v0 , v00 ∈ C0 . Since cos is injective on [0, π), it follows from (2) that θ = θ0 and v0 sin θ = v00 sin θ. If
θ = 0 then v = v 0 = 0. Otherwise θ ∈ (0, π) and sin θ 6= 0, so v0 = v00 , hence v = v 0 . If v0 is any point
in C0 , 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 7→ qvq −1 .
2
Rotations of three-dimensional space
2.1
The orthogonal and special orthogonal groups
Let V = Rn with the inner product
hu, vi =
n
X
ui vi ,
i=1
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 hu, ui > 0 for all nonzero vectors u ∈ Rn . The length
u is given by
|u| = hu, ui1/2 .
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 hgu, gvi = hu, vi 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 · t g = I.
10
Proof: The equivalence of items 1 and 2 results from the formula
hu, vi =
1
2
|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 · t g is the inner product of
columns i and j of g, whence the equivalence of items 3 and 4.
The condition g · t g = 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}.
2
We give On and SOn the topology inherited from the Euclidean space Mn (R) = Rn of n × n real
matrices.
Proposition 2.2 The subsets SOn , On ⊂ Mn (R) are compact and SOn is connected, while On has two
connected components.
Proof: For 1 ≤ i ≤ j ≤ n, define functions fij : Mn (R) → R by
fij (g) = hgi , gj i − δij ,
where gi , gj are the ith and j th columns of g ∈ Mn (R), and δij = 1 or 0 according as i = j or i 6= j.
Then On is the set of common zeros of all the functions fij , 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 Mn (R),
so On and SOn are closed. Since the columns of any g ∈ On are orthonormal vectors, each entry of g
belongs to [−1, 1], hence On is a bounded subset of Mn (R). Since On and SOn are closed and bounded
subsets of Mn (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 = hgi 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
1
0
−1
hgh ∈
,
0 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
11
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 (g n ) = λn v for all n ∈ Z, it follows that |λ| = 1, so λ = eiθ for some
θ ∈ R, and θ ∈
/ Zπ since λ 6= ±1. Let v̄ = (v̄1 , . . . , v̄n ). Since g has real entries, we have
gv̄ = gv = eiθ v = e−iθ v.
Since eiθ 6= e−iθ , the vectors v and v̄ are linearly independent. Hence the vectors u = v + v̄ and
w = i(v − v̄) are nonzero and linearly independent. These vectors u, w satisfy ū = u and w̄ = w, so
u, w ∈ Rn . We set c = cos θ, s = sin(θ). You can check that
gv = −su + cv.
gu = cu + sv,
Since g ∈ On , we have
hu, ui = hgu, gui = hcu + sv, cu + svi = c2 hu, ui + 2cshu, vi + s2 hv, vi.
Likewise
hv, vi = s2 hu, ui + 2cshu, vi + c2 hv, vi,
and
hu, vi = hcu + sv, −su + cvi = −cshu, ui + (c2 − s2 )hu, vi + cshv, vi.
Adding, we find hu, vi = 0. Hence u0 = |u|−1 u and v 0 = |v|−1 v are an orthonormal basis of a
two-dimensional plane U ⊂ Rn .
Hence there exists h ∈ On whose first two columns are u0 , v 0 and whose last n − 2 columns are an
orthonormal basis for the orthogonal complement U ⊥ . We then have
SO2
0
−1
hgh ∈
,
0 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 ⊕ Rj ⊕ Rk = {v ∈ H : τ (v) = 0}.
The dot product may be expressed quaternionically as as
hu, vi = 21 (uv̄ + vū).
For q ∈ S 3 , let Rq : H0 → H0 be the linear map given by
Rq (v) = qvq −1 .
12
(3)
A familiar calculation using (3) shows that
hRq (u), Rq (v)i = hu, vi,
for all u, v ∈ H0 and q ∈ S 3 . Therefore Rq ∈ O3 and we have a continuous homomorphism R : S 3 →
O3 , sending q 7→ 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 7→ 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,
so the matrix of Rq with respect to the basis {i, j, k} is
 2

a + b2 − c 2 − d 2
2(bc − ac)
2(bd + ac)
a2 − b 2 + c 2 − d 2
2(cd − ab)  .
Rq =  2(bc + ad)
2
2(bd − ac)
2(cd + ab)
a − b2 − c2 + d2
(4)
Proposition 2.3 The homomorphism R : S 3 → SO3 is surjective with ker R = {±1} equal to the
center 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 −1 f (h) ∈ Gx , so g −1 f (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 C0 = S 2 . We have
proved that R(S 3 ) acts transitively on C0 , and that
1 0
R(Ti ) =
0 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.
13
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 = 0
this 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 = Rp , T = Rq ,
where
p = cos θ + u sin θ,
q = cos φ + v sin φ.
Then ST = Rpq , 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,
and
w = (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
S4 are non-isomorphic groups of order 24. Note that the latter fits into another exact sequence (which
is now split)
1 −→ A4 −→ S4 −→ {±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.
14
2.3
The exponential map for SO3
Let A be an n × n real matrix. What conditions on A ensure that the path θ 7→ exp(θA) lies in SOn ?
The condition t (exp(θA)) = (exp(θA))−1 means that
I + θ(t A) + · · · = I − θA + · · · ,
so exp(θA) ∈ On iff t A = −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 skewsymmetric n×n matrix A. Letting son denote the set of such matrices, we therefore have an exponential
map
exp : 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


0 −z y
0 −x .
v = (x, y, z) 7→ Av =  z
−y x
0
p
Note that v ∈ ker Av and ker A = Rv as long as v 6= (0, 0, 0). Let |v| = x2 + y 2 + z 2 . Using the
fact that A3v = −θ2 Av , we find that
1 − cos |v|
sin |v|
Av +
exp(Av ) = I +
A2v
2
|v|
|v|
and that exp(Av ) is rotation by |v| about the axis through v, where the direction is seen counterclockwise as v points towards you. It follows that exp maps {Av : |v| < π} bijectively onto the complement
in SO3 of the conjugacy class C of 180 degree rotations. If |v| = π then exp(Av ) = exp(A−v ) so exp
describes 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
R0 : V −→ so3
defined as follows. For each v ∈ V , Rv0 ∈ so3 is the skew symmetric matrix acting on V by
Rv0 (u) =
d
Rexp(θv) (u)|θ=0 .
dθ
15
Computing this explicitly using power series, we find the explicit formula
Rv0 (u) = vu − uv.
To see Rv0 as a matrix, let v = (x, y, z) and compute Rv0 (i), Rv0 (j), Rv0 (k) to find that


0 −z y
0 −x = A2v .
Rv0 = 2  z
−y x
0
Finally, one checks that
R ◦ exp = exp ◦R0 .
That is, the following diagram is commutative.
R0
V −−−→ so3



exp
expy
y
R
S 3 −−−→ 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 R0 on the parameter spaces.
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 fx0 : Rn → Rm such that
1
[f (x + h) − f (x) − fx0 (h)] = 0.
h→0 |h|
lim
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 fx0 becomes
fx0 (v) = lim
t→0
f (x + tv) − f (x)
.
t
16
If v = ei is one of the standard basis vectors then fx0 (ei ) =
∂f
(x).
∂xi
Now the function x 7→ fx0 is a function f 0 : 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 0 , f 00 , . . . 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 , Uα , Mα ) where U ⊂ Rn
and Mα ⊂ M are open subsets and ϕα : Uα → Mα is a homeomorphism. Thus, ϕa l parametrizes
the open subset Mα of M . We call n the dimension of the chart. An atlasSon M is a collection
{(ϕα , Uα , Mα ) : α ∈ A} of local charts indexed by some set A such that M = α∈A Mα . We say the
atlas has dimension n if Uα is an open subset of Rn , with the same n, for every α ∈ A.
Given two local charts (ϕα , Uα , Mα ) and (ϕβ , Uβ , Mβ ), we get a transition function
ϕ−1
β
ϕα
−1
n
ϕ−1
β ◦ ϕα : ϕα (Mα ∩ Mβ ) −→ Mα ∩ Mβ −→ R
n
defined on the open set Uαβ = ϕ−1
α (Mα ∩ Mβ ) ⊂ R . An atlas {(ϕα , Uα , Mα ) : α ∈ A} is smooth if
−1
the transition functions ϕβ ◦ 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. P 2
Example 2: Let M = S n = {(x0 , x1 , . . . , xn ) ∈ Rn+1 :
xi = 1}. Take the two antipodal points
n
z± = (±1, 0, . . . , 0) and let M± = S − {z± }, and let U± = Rn . For α = ±, define ϕα : Rn → S n by
ϕα (u1 , . . . , un ) =
1
(α(|u|2 − 1), 2u1 , . . . , 2un )
2
|u|
n
Then ϕ−1
α : Mα → R is given by
ϕ−1
α (x0 , . . . , xn ) =
1
(x1 , . . . , xn ),
1 − αx0
and the transition functions are given by
ϕ−1
−α ◦ ϕα (u) =
u
.
|u|
n
n
This is a smooth function on ϕ−1
α (Mα ∩M− α) = R −{(0, . . . , 0)}. Hence the collection {(ϕα , R , Mα ) :
α = ±} is a smooth atlas. fs
Let M and N be topological spaces with atlases {(ϕα , Uα , Mα ) : α ∈ A} and {(ψβ , Vβ , Nβ ) : β ∈ B}
respectively. A function f : M → N is smooth if each composition
−1
(Nβ )) −→ Rm
ψβ−1 ◦ f ◦ ϕα : ϕ−1
α (Mα ∩ f
is smooth. For example, the identity M → M is smooth, and the composition of smooth functions is
smooth.
17
A given topological space M may admit more than one atlas. Two atlases {(ϕα , Uα , Mα ) : α ∈ A}
and {(ψβ , Vα , Mβ ) : 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
n
ψβ−1 ◦ ϕα : ϕ−1
α (Mα ∩ Nβ ) −→ R
is smooth.
Definition 3.1 An smooth manifold is a Hausdorff 3 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 28
smooth structures. 4
Lemma 3.2 If M and N are smooth manifolds with atlases {(ϕα , Uα , Mα ) : α ∈ A} and {(ψβ , Vβ , Nβ ) :
β ∈ B}, then then there is a canonical smooth structure on M × N , given by the atlas {(ϕα × ψβ , Uα ×
Vβ , Mα × Nβ ) : α ∈ A, β ∈ B}.
Proof: Left to the reader!
3.2
Manifolds defined by equations
Let f = (f1 , . . . , fm ) : 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 0 of f . Recall that f 0 is the m × n matrix of
functions
∂f
0
f = ∂xji
which at each point p ∈ Rn gives a linear map
∂f
fp0 = ∂xji (p) : Rn → Rm .
3
Recall 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 .
4
See Milnor, On manifolds homeomorphic to the 7-sphere Annals of Math. 1956, and Brieskorn Beispiele zur Differentialtopologie von Singularitäten Invent. Math. 1966
18
Theorem 3.3 (Implicit Function Theorem) Let f : Rn → Rm be a smooth function, where n ≥ m,
and let M = f −1 (0). Suppose fp0 (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 fp0 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 0 and
choose U, V, g as in Thm. 3.3. This gives a chart ϕp : Up → Mp , where Up = U , Mp = (U × V ) ∩ M ,
given by ϕp (u) = (u, g(u)). Doing this for each p ∈ M gives an atlas {(ϕ, Up , Mp ) : p ∈ M }. One
checks that for p, q ∈ M and u ∈ ϕp (Mp ∩ Mq ) we have ϕ−1
q ◦ ϕp (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
2
that GLn (R) is a Lie group, let Mn (R) ' Rn be the space of all n × n matrices with entries in R.
The determinant det : Mn (R) → R is a continuous function and GLn (R) is the open set of points in
Mn (R) on which det is nonzero. Thus GLn (R) is an open subset of Mn (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 GLn (R) is
n
orientation-preserving automorphism group of R .
19
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 function
f : Mn (R) → R defined by f (A) = det(A) − 1. If A = [aij ] then
X
f (A) = −1 +
sgn(σ)a1σ(1) · · · anσ(n) .
σ∈Sn
Let Aj be the matrix obtained by deleting row 1 and column j from A. If det(A) = 1 then det(Aj ) 6= 0
for some j. And
X
∂f
=±
sgn(σ)a1σ(1) · · · anσ(n) = ± det(Aj ) 6= 0,
∂aij
σ(1)=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) : t gg = I}.
A matrix belongs to On exactly when its columns form an orthonormal basis of Rn . Thus, On =
2
f −1 (0), where f = (fpq ) : Mn (R) → Rn has component functions
X
fpq (g) = −δpq +
gip giq ,
i
where δpq = 1 if p = q and is zero otherwise. One checks that fg0 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 Mn (R), hence is compact. The determinant maps On onto {±1}, so On is
disconnected. We will see that it has exactly two connected components.
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) : t gg = I}.
A matrix lies in SOn exactly when its columns form an orthonormal basis having the same orientation
as the standard orthonormal basis {e1 , . . . , en }. We will see that SOn 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



 1 x z

3


0 1 y : (x, y, z) ∈ R
H3 :=


0 0 1
is diffeomorphic to R3 but is not isomorphic to R3 as a Lie group. Indeed, H3 is non-abelian.
20
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 0 (t) which is “tangent” to G. The pair (γ(t), γ 0 (t)) tells us where we are and
where we are going. Where does γ 0 (t) live? If G were a subset of some Euclidean space Rn then we
could imagine γ 0 (t) as a vector in Rn . But G may embed in many Euclidean spaces in many ways, and
we want a home for γ 0 (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 0 (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 7→ (γ(t), γ 0 (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
a
{p} × Rn .
TU =
p∈U
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.
Let M be a n-dimensional smooth manifold with atlas {ϕα , Uα , Mα ) : α ∈ A}. For each α ∈ A let
ια : Mα ,→ M be the inclusion map. The tangent bundle of M is a new manifold T M equipped with
a projection map
π
T M −→ M
whose fibers π −1 (p) are n-dimensional vector spaces which vary smoothly. More precisely, T M is the
set of equivalence classes
a
TM =
(Mα × Rn )/ ∼,
α∈A
where (x, v) ∼ (y, w) if
ια (x) = ιβ (y)
and
0
w = (ϕ−1
β ϕα )x (v).
Thus, points in T M are equivalence classes [x, v]α ∈ T M , for (x, v) ∈ Mα × Rn , with the understanding that if x ∈ Mα ∩ Mβ we have
0
[x, v]α = [x, (ϕ−1
β ϕα )x (v)]β .
Each Mα × Rn = T Mα injects into T M . Via the quotient topology, T M has an open cover
[
TM =
T Mα
α∈A
21
consisting of tangent bundles of the images of charts.
For each α ∈ A define
Φα : Uα × Rn −→ T Mα
Φα (u, v) = [ϕα (u), v)]α .
by
One checks that
−1
n
Φ−1
α (T Mα ∩ T Mβ ) = ϕα (Uαβ ) × R ,
and that
−1
−1
n
n
Φ−1
β Φα : ϕα (Uαβ ) × R −→ ϕβ (Uαβ ) × R
−1
0
is given by ϕ−1
β ϕα ×(ϕβ ϕα ) . Thus T M is a smooth 2n-dimensional manifold with atlas {(Φα , T Uα , T Mα ) :
α ∈ A}.
The projection πM : T M → M is given by πM ([x, v]α ) = x. For α, β ∈ A, one checks that ϕ−1
β ◦ πM ◦
Φα is the composition
ϕ−1
β ϕα
proj
−1
n
−1
ϕ−1
α (Uαβ ) × R −→ ϕα (Uαβ ) −→ ϕβ (Uαβ ),
which is smooth, so that πM : T M → M is smooth.
The tangent space to M at x ∈ M is the fiber
−1
Tx M := πM
(x).
If x ∈ Mα then Tx M = {[x, v]α : v ∈ Rn } ' Rn , but this isomorphism is non-canonical, as it depends
on α.
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 = (f1 , f2 ) = (|x|2 − 1, x · v). We find that
2x1 . . . 2xn 0 . . . 0
0
f =
v1 . . . 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 0 : T M → T N making the following diagram commute:
TM
πM
M
f0
f
/
TN .
/
πN
N
If L is another manifold and g : N → L is another smooth map then
(g ◦ f )0 = g 0 ◦ f 0 .
22
chain rule.
Proof: Let {(ϕα , Uα , Mα ) : α ∈ A} and {(ψβ , Vβ , Nβ ) : β ∈ B} be atlases on M and N . Let
x ∈ M and choose α ∈ A such that x ∈ Mα and β ∈ B such that f (x) ∈ Nβ . For v ∈ Rn ,
define f 0 ([x, v]α ) = [f (x), (ψβ−1 f ϕα )0 (v)]β . One checks, using the chain rule for derivatives on Rn ,
that f 0 ([x, v]α ) does not depend on the choice of α such that x ∈ Mα , and that the resulting map f 0
is smooth. It is clear that f 0 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) ∈ Tp M 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 Xv (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,
called left translation. This map has a derivative L0g : T G → T G, mapping each tangent space Tx G
to Tgx G.
A vector field X on G is called left-invariant if for all g ∈ G the following diagram commutes:
TG
O
L0g
/
TG
O .
X
G
X
Lg
/
G
Let g = Te G denote the tangent space to G at the identity element e ∈ G. For each v ∈ g there is a
unique left-invariant vector field Xv : G → T G, given by
Xv (g) = L0g (v).
Conversely, if X : G → T G is any left-invariant vector field then
X(g) = L0g X(e),
so X = Xv , 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
23
restricting to a linear isomorphism Rn = p × Rn → Tp M for each p ∈ M . Equivalently, M is
parallelizable if there exist n vector fields X1 , . . . , Xn 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 Xi , Xj , Xk (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 L0g : g → Tg G is an
isomorphism, the vector fields Xi (g) = L0g (ei ) are linearly independent in Tg G 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 0 : T I −→ T M.
Choose I small enough so that γ(I) ⊂ Mα for some chart (ϕα , Uα , Mα ) on M . Then T I = I × R and
γ 0 (T I) ⊂ T Mα = Mα × Rn ' Uα × Rn . Taking 1 ∈ R as a basis vector, we have
γ 0 (t, 1) = (γ(t), γ̇(t)),
where γ̇(t) is the componentwise derivative of γ(t) ⊂ Uα ⊂ Rn .
Given a smooth vector field X : M → T M we seek a path γ in M such that γ 0 (t, 1) = X(γ(t)). This
amounts 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 0 (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 0 (t, 1) = X(γ(t)), δ 0 (t, 1) = X(δ(t)), γ(a) = δ(a).
Then γ(t) = δ(t) for all t ∈ [a, b].
24
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 (0, 1) ∈ g = Te G.
Lemma 4.5 Let γ : R → G be a one-parameter subgroup, with v := γ 0 (0, 1) ∈ g, and let Xv : G →
T G be the corresponding left-invariant vector field on G given by Xv (g) = L0g (v). Then for all t ∈ R
we have
γ 0 (t, 1) = L0γ(t) (v) = Xv (γ(t)).
Proof: Regarding R itself as a Lie group, we can extend the vector (0, 1) ∈ T0 R to a left-invariant
vector field X( 0, 1) : R → T R, given by X(0,1) (t) = L0t (0, 1), where Lt : R → R is the translation
map Lt (x) = t + x. For all x ∈ R we have L0t (x, 1) = (t + x, 1).
Since γ is a homomorphism, the diagram
γ
/
G
R
Lt
γ
/
R
Lγ(t)
G
is commutative. Taking derivatives, we get the commutative diagram
TR
L0t
TR
γ0
γ0
/
TG
/
.
L0γ(t)
TG
We get
γ 0 (t, 1) = γ 0 ◦ L0t (0, 1) = L0γ(t) (v) = Xv (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 (0, 1) ∈ g.
Theorem 4.6 For each v ∈ g there is a unique one-parameter subgroup γv : R → G such that
γ 0 (0, 1) = v.
Proof: Suppose γ and δ are two one-parameter subgroups such that γ 0 (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.
25
Now let v ∈ g. By the local existence part of Thm. 4.3, there exists > 0 and a smooth path
γ : (−, ) → G such that γ0 (0) = e and
γ00 (t, 1) = Xv (γ0 (t)
for
|t| < .
The first step is to prove that γ0 is a local homomorphism. That is,
γ0 (s + t) = γ0 (s)γ0 (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 0 (t, 1) = Xv (f (t)).
For the left side, we have γ0 (s + t) = γ0 ◦ Ls (t) and
[γ0 ◦ Ls ]0 (t, 1) = γ00 ◦ L0s (t, 1) = γ00 (t + s, 1) = Xv (γ0 (t + s)).
For the right side, let g = γ0 (s). Then
[γ0 (s)γ0 (t)]0 (t, 1) = [Lg γ0 ]0 (t, 1)
= L0g γ00 (t, 1)
= L0g Xv (γ0 (t))
= L0g L0γ0 (t) (v)
=
(by Lemma 4.5)
L0gγ0 (t) (v)
= Xv (gγ0 (t))
= Xv (γ0 (s)γ0 (t)),
Thus γ0 is indeed a local homomorphism.
The next step is to extend γ0 to a homomorphism γ : R → G. Given t ∈ R, choose a positive integer
N such that |t/N | < /2 and set
γ(t) = γ0 (t/N )N .
If also |t/M | < /2 then
γ0 (t/M )M = γ0 (t/M N )M N = γ0 (t/N )N ,
since γ0 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
N
s
t
γ(s + t) = γ0
+
= γ0 (s/N )N γ0 (t/N )N = γ(s)γ(t).
N
N
Next, γ is smooth: For γ0 is smooth on (−, ), so γ(t) = γ0 (t/N ) is smooth on (−N, N ). Finally,
γ 0 (0, 1) = γ00 (0, 1) = Xv (γ0 (0)) = Xv (e) = v.
This completes the proof the theorem.
For v ∈ g, we let γv : R → G be the unique one-parameter subgroup such that γv0 (0, 1) = v.
26
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) = γv (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 γv varies with v. The simplest situation is when v is scaled.
Lemma 4.7 For all s ∈ R, we have γsv (t) = γv (st).
Proof: Let Ms : R → R be the map Ms (t) = st. We have Ms0 (0, 1) = (0, s), so
(γv ◦ Ms )0 (0, 1) = γv0 ◦ Ms0 (0, 1) = γv0 (0, s) = s · γv0 (0, 1) = sv.
0
Since γsv is the unique one-parameter subgroup of G such that γsv
(0, 1) = sv, we have γv ◦ Ms = γsv
and the lemma follows.
Next we consider the smoothness of v 7→ γv . This requires an extension of the uniqueness and existence
theorem 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 V0 of 0 in V and a unique smooth map
f : (−, ) × V0 → U such that for all v ∈ V0 the function fv (t) := f (t, v) satisfies the differential
equation
fv0 (t) = X(fv (t), v),
for |t| < .
Proof: See references in Adams, page 8.
Proposition 4.9 Let G be a Lie group and let g = Te G. The exponential map exp : g → G, defined by
exp(v) = γv (1), is smooth.
Proof: Recall that G is parallelizable, via the map
X : G × g → TG
given by
X(g, v) = L0g (v).
The one-parameter subgroup γv (t) solves the differential equation
γv0 (t, 1) = X(γv (t), v).
By Thm. 4.8 the map (t, v) 7→ γv (t) is smooth on some neighborhood (−, ) × V0 of (0, 0) in R × g.
Then
2
γN −1 v (t/N )N = γv (t)
27
is smooth on (−N , N ) × N V0 , so (t, v) 7→ γv (t) is smooth on R × g. Taking t = 1 we see that exp
is smooth.
Remark 1: For each v ∈ g, the image exp(v) = γv (1) ∈ G is a value of the one parameter subgroup
γv . In fact all values of γv are obtained from exp. For Lemma 4.7 implies that
γv (t) = γtv (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 = Te G, h = Te H be the respective tangent spaces at the identity elements of G and H. Let
ϕ00 : g −→ h
be the derivative of ϕ. This is a linear map obtained by restricting ϕ0 : T G → T H to g.
Proposition 4.10 The following diagram is commutative:
ϕ0
g −−−0→


expG y
h

exp
y H
ϕ
G −−−→ H
Proof: In the proof we write ϕ0 instead of ϕ00 . For all v ∈ g we have
(ϕ ◦ γv )0 (0, 1) = ϕ0 ◦ γv0 (0, 1) = ϕ0 (v),
so ϕ ◦ γv = γϕ0 (v) by Thm. 4.6. Now
ϕ(expG (v)) = ϕ(γv (1)) = γϕ0 (v) (1) = expH (ϕ0 (v)),
as claimed.
This is the most useful result in the theory of Lie groups.
5
Example: Let G = GLn (R). Since G is an open subset of Mn (R) we canonically identify g = Mn (R).
Take A ∈ g and consider the path γ : R → G given by
γ(t) = etA = I + tA +
t2 2
A
2
+ · · · ∈ G.
Since γ 0 (0) = A, it follows that γ = γA 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 diffeomorphically by expG onto a neighborhood of e in G.
5
This claim is really a challenge to find a result that is even more useful.
28
Proof: We need another result from analysis.
6
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
∂fi
0
(p)
fp =
∂xj
is invertible at p. Then there exists a neighborhood U0 of p mapped diffeomorphically by f onto an
open neighborhood of f (p) in Rn .
The smooth map exp : g → G has derivative exp0G : T g → T G. Restricting to T0 g = g, we get a
linear map exp00 : g → g. Let v ∈ g, and let h : R → g be the map h(t) = tv. Then γv = exp ◦h, so
exp00 (v) = exp00 (h0 (0)) = γv0 (0) = v,
so exp00 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 = Mn (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 − 21 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=
hU
and
G−H =
gU
g∈G−H
h∈H
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.
6
See Rudin Principles of Mathematical Analysis.
29
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
0
∈ SL2 (R)
0 −1/2
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 ϕ00 : g → h.
Proof: If ϕ and ψ are two Lie group homomorphisms from G → H with ϕ00 = ψ00 then applying
functorality Prop. 4.10, we have
ϕ ◦ expG = expH ◦ϕ00 = expH ◦ψ00 = 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
ϕ0 : g → h is surjective then ϕ is surjective.
Proof: Since H is connected it is generated by expH (h), by Prop. 4.13. Since ϕ0 is surjective we have
expH (h) = expH (ϕ0 (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 representation
in the form (6).
A Lie group G has a canonical representation
Ad : G → GL(g)
30
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) := c0g maps g → g. And since cg is a homomorphism, functorality Prop. 4.10 gives a commutative diagram
Ad(g)
g −−−→


expy
g

exp
y
cg
G −−−→ G
Since cgh = cg ◦ ch , it follows that Ad(gh) = Ad(g) ◦ Ad(h). Thus we have a representation
AdG = Ad : G −→ GL(g),
g 7→ Ad(g).
For v ∈ g, the image Ad(g)v is the initial tangent direction of the path t 7→ g exp(t)g −1 in G. Thus,
we have
d
Ad(g)v =
g exp tvg −1 |t=0 .
dt
Now Ad itself is a Lie group homomorphism; its derivative at e is the linear map
u 7→ ad(u) = Ad0 (u).
ad : g → End(g),
For u, v ∈ g, the image ad(u)(v) is the initial tangent direction of the path t 7→ Ad(exp(tu))v in g.
Thus, we have
d
(Ad(exp(tu))v) |t=0 .
ad(u)v =
dt
Functorality relates Ad and ad via the commutative diagram
ad(g)
g −−−→ End(g)



expGL(g)
expG y
y
Ad
G −−−→ 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).
So the adjoint representation Ad : G → GL(g) is given bv
1
Ad(exp(u))(v) = ead(u) (v) = v + [u, v] + [u, [u, v]] + · · · .
2
The adjoint representation Ad and its derivative ad are functorial.
31
(7)
Proposition 4.16 Let ϕ : G → H be a Lie group homomorphism. Then for every g ∈ G the following
diagram commutes
ϕ0
g −−−→


AdG (g)y
h

Ad (ϕ(g))
y H
ϕ0
g −−−→ h
and for all u, v ∈ g we have
[ϕ0 (u), ϕ0 (v)] = ϕ0 ([u, v]).
Proof: For Ad we compute
ϕ0 ◦ AdG (g) = ϕ0 ◦ c0g = (ϕ ◦ cg )0 = (cϕ(g) ◦ ϕ)0 = c0ϕ(g) ◦ ϕ0 = AdH (ϕ(g)) ◦ ϕ0 ,
making the diagram commute as claimed. Next we compare two power series in t, namely
ϕ0 (AdG (exp(tu)v) = ϕ0 (v + t[u, v] + · · · ) = ϕ0 (v) + tϕ0 ([u, v]) + · · · ,
which is equal to
AdH (ϕ(exp(tu))ϕ0 (v) = ϕ0 (v) + t[ϕ0 (u), ϕ0 (v)] + · · · .
Comparing coefficients in t we obtain [ϕ0 (u), ϕ0 (v)] = ϕ0 ([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 the
identity 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)0 in terms of γ, γ 0 , δ, δ 0 . We will
write γ 0 (t) instead of γ 0 (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.
32
Proposition 4.18 For all t ∈ R we have
0
γ 0 (t) ∈ Gγδ(t)
(γδ)0 (t) = L0γ(t) δ 0 (t) + Rδ(t)
Proof: Let µ : G × G → G be the product map: µ(g, h) = gh. We compute
µ0 : Tg G ⊕ Th G → Tgh G
as follows. A typical element of Tg G is of the form L0g (v), where v ∈ g = Te G. Then
d
µ(g exp(tv), h)|t=0
dt
d
[g exp(tv)h] |t=0
=
dt
d
=
[Lg Rh (exp(tv))]t=0
dt
= (Lg Rh )0 (v)
= Rh0 ◦ L0g (v)
µ0 (L0g (v), 0) =
since Rh and Lg commute. Likewise,
µ0 (0, Rh0 u) = L0g ◦ Rh0 (u).
It follows that for all x ∈ Tg G and y ∈ Th G we have
µ0 (x, y) = µ0 (x, 0) + µ0 (0, y) = Rh0 (x) + L0g (y).
Now γδ = µ(γ, δ) so the chain rule gives
0
(γδ)0 (t) = µ0 (γ 0 (t), δ 0 (t)) = L0γ(t) (δ 0 (t)) + Rδ(t)
(γ 0 (t)),
as claimed.
We use this to express the bracket [ , ] on g as the derivative of a commutator of one-parameter subgroups in G. Let [g, h] = ghg −1 h−1 be the commutator in G.
Proposition 4.19 For all u, v ∈ g we have
d
d
[exp(tu), exp(tv)]s=0
.
[u, v] =
dt ds
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 that
d
[g, exp(sv)]s=0 = Ad(g)v − v,
ds
and the result follows.
33
Corollary 4.20 For all u, v ∈ g we have
[u, v] = −[v, u].
Proof: A similar argument shows that
d
[exp(tu), h]t=0 = u = Ad(h)u.
dt
Interchanging the order of differentiation gives the result.
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 = Te G, 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:
ϕ0 (v) = Ad(exp(tu))v = v + t[u, v] + · · · ,
so
ϕ0 ([u, v]) = [v, w] + t[u, [v, w]] + · · · ,
but by functorality of the Lie bracket (Prop.4.16) this is equal to
[ϕ0 (v), ϕ0 (w)] = [v, w] + t ([[u, v]w] + [v, [u, w]]) + · · · .
Comparing coefficients of t gives the Jacobi identity.
From now on we call g = Te G, with bracket defined in (7), the Lie algebra of the Lie group G.
34
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 ϕ0 : 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 ϕ0
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)
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 = Mn (R) and that the exponential map is the
matrix exponential exp(A) = I + A + 21 A2 + · · · . 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). By (8) the Lie algebra g = sln (R)
consists of matrices A ∈ Mn (R) for which det(exp(tA)) = 1. As det(exp(tA)) = et tr(A) , it follows
that
sln (R) = {A ∈ Mn (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 ∈ Mn (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 · t A). It follows that
son = {A ∈ Mn (C) : A + t A = 0.}.
Example 5: Let G = H3 , 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


1 x(t) z(t)
γ(t) = 0 1 y(t)
0 0
1
and has derivative

0 x0 (0) z 0 (0)
0
y 0 (0) .
γ 0 (0) = 0
0
0
0

35
Hence the Lie algebra is



 0 x z



0 0 y : x, y, z ∈ R .
h3 =


0 0 0
We note that A3 = 0 for all matrices A ∈ h3 , so the exponential map is given by
exp(A) = I + A + 21 A2 ,
with inverse
log(I + A) = A − 21 A2 .
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 ≤ n
such 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 (0) = γu0 (0) + γv0 (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.
36
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
' (S 1 )k × Rn−k .
Zk ⊕ 0n−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.
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 {U1 , U2 , . . . }
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 C0 ⊂ Rn be any cube. Inductively define a nested sequence of cubes C0 ⊃ C1 ⊃ · · · as follows.
If C0 ⊃ · · · ⊃ Ck−1 has been defined, let be the side length of Ck−1 and choose an integer N =
N (m) > 1/. Then N Ck−1 is a cube of side > 1, so π(N Ck−1 ) = T . The map πN , given by the
composition
N
π
Ck−1 −→ N Ck−1 −→ T
is continuous, so (πN )−1 (Uk ) is an open subset of Rn contained in Ck−1 . Then Ck is chosen to be any
cube contained in (πN )−1 (Uk ).
T
Take x ∈ k Ck and let t = π(x) ∈ T . For all k > 1 we have
tN (k) = πNk (x) ∈ πNk (Ck ) ⊂ Uk .
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 .
37
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 7 if there exists a subspace W ⊂ Rn and an atlas {(ϕα , Uα , Mα ) : α ∈ A} in the smooth structure
on M such that for all α ∈ A we have either
• S ∩ Mα = ∅, or
• ϕα (W ∩ Uα ) = S ∩ Mα .
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 ∩ Uα , S ∩ Mα ) where
S ∩ Mα 6= ∅.
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
2 2
is G = R Z 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 |−1 xn → 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 mn xn = mn |xn | · |xn |−1 xn → tv. We have
exp(mn xn ) = exp(xn )mn ∈ H for all n, so exp(mn xn ) → exp(tv) ∈ H since H is closed.
Lemma 6.3 h is a vector subspace of g.
7
There are various definitions of submanifold. The one used here is usually called “embedded submanifold”.
38
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 V1 ⊂ V
such that V1 · V1 ⊂ V .
Choose > 0 such that exp(tx) and exp(ty) lie in V1 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 (0) = x + y.
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
1/n 1
1
−1
·
|xn | xn = f (1/n) →
· (x + y).
f (1/n) 1/n
|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
ϕ0 = I. Hence there is a neighborhood 0 ∈ U0 ⊂ g such that ϕ : U0 → ϕ(U0 ) is a diffeomorphism
onto an open neighborhood ϕ(U0 ) of e in G.
Lemma 6.4 There exists a neighborhood e ∈ V ⊂ ϕ(U0 ) in G such that H ∩ V ⊂ ϕ(h ∩ U0 ).
Proof: If no such V exists then there is a sequence hn → e in H ∩ ϕ(U0 ) such that ϕ−1 (hn ) 6 inh
for all n. We have ϕ−1 (hn ) = (xn , yn ) with xn ∈ h and yn ∈ m with yn 6= 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 |−1 yn is bounded, hence has a convergent subsequence (unk ), whose limit u ∈ m satisfies
|u| = 1. But since exp(ynk ) ∈ H and ynk 6= 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 ) ⊂ U0 . Then H ∩ V = ϕ(h ∩ U ). On G define an atlas
{(ϕg , Ug , Gg ) : g ∈ G} as follows.
For h ∈ H, set
Uh = U,
ϕh (u) = h · ϕ(u),
Gh = hV.
Then ϕh (h ∩ Uh ) = h · ϕ(h ∩ U ) = H ∩ Gh .
For g ∈
/ H, choose a neighborhood Ug of 0 in U such that gϕ(Ug )∩H = ∅, using again the assumption
that H is closed. Then set Gg = g · ϕ(Ug ) and define ϕg : Ug → Gg by ϕg (u) = g · ϕ(u). Then
Gg ∩ H = ∅. This shows that {(ϕg , Ug , Gg ) : g ∈ G} is a good atlas for H, so H is a submanifold of
G.
39
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 , Uh , 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 = Te (H) is the Lie algebra of H.
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
X
hu, vi =
ui v̄i .
i
More explicitly,
Un = {g ∈ GLn (C) : g −1 = t ḡ}.
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 = Un ∩ 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 G
onto 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 0 maps m isomorphically onto TeH (G/H).
40
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 Gp in G the map
g 7→ g · p gives a G-equivariant bijection fp : G/Gp → M . Since the action is smooth, it follows that
fp is actually a diffeomorphism. Hence every homogeneous space for G is of the form G/H for some
closed subgroup H of G.
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 h , i on V is a bilinear map
V × V → R,
(u, v) 7→ hu, vi
which is symmetric: hu, vi = hv, ui and satisfies hv, vi > 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
hvi , vi i = 1 for all i and hvi , vj i = 0 if i 6= 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 h , i on V
such that
hρ(k)u, ρ(k)vi = hu, vi
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
Z
Z
f (hk) dk =
f (k) dk.
(9)
K
K
This measure dk is unique up to scalar multiple; we normalize it so that
R
K
1 dk = 1.
Now on V we take any inner product ( , ) which is positive-definite and define a new inner-product by
averaging:
Z
hu, vi =
(ρ(k)u, ρ(k)v) dk.
K
41
This new inner product is still positive definite, since
Z
(ρ(k)v, ρ(k)v) dk > 0,
hv, vi =
K
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
hu, vi = − tr(ad(u) ◦ ad(v)).
This form (rather its negative) is called the Killing form on g.
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 : hu, vi = 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 are
no 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 irreducible 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.
42
Example 1: Take G = GLn (R) and H = On (R). Then g = Mn (R) decomposes as g = h ⊕ m, where
h = son = {A ∈ Mn (R) : t A = −A},
m = {B ∈ Mn (R) : t B = 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 on Rn .
Example 2: Take G = SOn+1 and let H = SOn be the subgroup fixing en+1 . Then
0 x
n
m = Ax =
: x∈R .
−t x 0
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 7→ exp(Ax ) · en+1 is spherical
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
0 x
m=
: x = k × ` matrix
−t x 0
and (h1 , h2 ) ∈ H acts by x 7→ h1 xh−1
2 . 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 ×Un 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
43
a torus containing gT g −1 then g −1 U 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
cos 2πλ(x) − sin 2πλ(x)
ρλ (exp(x)) =
sin 2πλ(x) cos 2πλ(x)
for a unique linear functional λ : t → R.
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 Vθ ⊂ V ⊗ C be the span of two eigenvectors with eigenvalues eiθ and e−iθ . Then
Vθ is a four-dimensional real vector space whose fixed-points under complex conjugation are twodimensional. 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 ρ0 : t → so2 = R. Since λ determines
ρ, 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 Ā is also abelian and is still connected. Hence Ā is a torus, so T = Ā 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. Conversely, 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),
44
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) 7→ 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 7→ −α, such that the representation AdG : T → GL(m)
decomposes as
M
m=
mα
α∈R/±
where mα is the two dimensional representation ρα of T on which exp(x) (for x ∈ t) acts via the matrix
cos 2πα(x) − sin 2πα(x)
.
sin 2πα(x) cos 2πα(x)
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
0 −1
ad(x) = 2πα(x)
.
1 0
Example 1: Take G = Un , with Lie algebra
g = un = {X ∈ Mn (C) : t X̄ = −X}.
The diagonal matrices in G form a maximal
torus T ' U1n . 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 Xij (z) be the matrix with ij-entry equal to z, ji-entry equal to −z̄, and
all other entries zero. Let
mij = {Xij (z) : z ∈ Z}.
Then mij is preserved by AdG (T ) with roots xi − xj , xj − xi and the decomposition of sun with respect
to AdG (T ) is
X
sun = t ⊕
mij
i<j
and the roots of Un are given by
R = {xi − xj : 1 ≤ i 6= 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 ∈ M2n (R) : t X = −X}.
45
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 Xij (z) be the matrix with ij-entry equal to z, ji-entry equal to −t z, and all other
entries zero. For 1 ≤ i < j ≤ n let
x −y
mxi −xj = {Xij
: x, y ∈ R},
y x
x y
mxi +xj = {Xij
: x, y ∈ R}.
y −x
Then mxi −xj and mxi +xj is preserved by AdG (T ) with roots ±(xi − xj ) and ±(xi + xj ) respectively
and the decomposition of so2n with respect to AdG (T ) is
X
so2n = t ⊕
(mxi −xj ⊕ mxi +xj )
i<j
and the roots of SO2n are given by
R = {xi ± xj : 1 ≤ i 6= j ≤ n}.
Example 3: Take G = SO2n+1 (or O2n+1 ), with Lie algebra
g = so2n+1 = {X ∈ M2n+1 (R) : t X = −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 ± xj : 1 ≤ i 6= 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.
46
Proof: We first show that N (T ) is compact. For all t ∈ T , define maps ft , fˇt : G → G by ft (g) =
gtg −1 and fˇt (g) = g −1 tg. Then
#
# "
"
\
\
fˇt−1 (T ) ,
N (T ) =
ft−1 (T ) ∩
t∈T
t∈T
so N (T ) is closed in the compact group G hence is compact. For each n ∈ N (T ) we have a commutative diagram
Ad(n)
t −−−→


expT y
t

exp
y T
c
n
→ T.
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.
The kernel of Ad : N (T ) → Aut(L) is the centralizer C(T ) of T in G. Indeed, if Ad(n) is trivial on L
then 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 homogeneous
space 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−1 T , 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 ) = hT, ji 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 Rj , is the antipodal map on S 2 . This action does not come from any
element of SO3 . Note that Rj interchanges the two fixed-points ±i of T in S 2 . 47
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 x−1 .
xT ∈G/T
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 = ghuig −1 = ghuig −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.
48
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 = hg, T i 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
[
gnT
n∈Z
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 g n T . That is, H/T is finite
cyclic, generated by gT . This means that g m ∈ 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 .
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 C2n−1 , 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 C23 , 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
8
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 ),
8
i = 0, 1, . . .
For more details on cohomology of manifolds see Bott and Tu, Differential forms in algebraic topology.
49
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 ft := f (·, t) is smooth for each t ∈ [0, 1]
then f0∗ = f1∗ .
v) If f : M → M is a smooth map with only finite many fixed-points then
n
X
(−1)i tr(f ∗ , H i (M )) =
i=0
X
sgn det(I − f 0 )p ,
p∈M
f (p)=p
where the right side is understood to be zero if f has no fixed-points.
Here, if f (p) = p then f 0 : T M → T M restricts to a map f 0 : Tp M → Tp M and det(I − f 0 )p is the
determinant of the linear map I − f 0 on Tp M and sgn det(I − f 0 )p ∈ {+1, −1, 0} is the sign of this
determinant. The formula v) is the Lefschetz fixed-point formula, and the alternating sum
n
X
L(f ) :=
(−1)i tr(f ∗ , H i (M ))
i=0
is the Lefschetz number of f . In the situation of item iv) we have L(f1 ) = L(f0 ). It may happen that
f1 has infinitely many fixed-points but f0 has only finitely many, so we can apply v) to f0 to compute
L(f1 ).
If f : M → M is the identity map then f ∗ is the identity map on M and
L(f ) =
n
X
(−1)i dim H i (M )
i=0
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 fixedpoints on S 2 , and f 0 acts by a rotation by θ in the tangent space at each fixed-point p, where we have
1 − cos θ
sin θ
0
det(I − fp ) = det
= 2(1 − cos θ) > 0.
− sin θ 1 − cos θ
50
So
X
det(I − f 0 )p = 1 + 1 = 2.
p∈S 2
f (p)=p
On the other hand, let ft be rotation about the same axis of f by tθ. Then f1 = f and f0 = I, so
f ∗ = f0∗ is the identity map on H ∗ (S 2 ) and we have
L(f ) = χ(S 2 ).
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 L0n gives a commutative diagram
L0
n
TeT (G/T ) −−−
→ TnT (G/T )



L0
L0 −1 y
y t
n
tn
L0
n
TeT (G/T ) −−−
→ TnT (G/T ),
so that
det(I − L0t )nT = det(I − L0n−1 tn )eT .
Recall the derivative of the projection π : G → G/T restricts to an isomorphism
∼
π 0 : m −→ TeT (G/T ).
For any s ∈ T this isomorphism transforms AdG (s) on m to the map L0s on TeT (G/T ). Let s = n−1 tn
as above and write s = exp(u) for u ∈ t. Then
Y
det(I − L0n−1 tn )eT =
det(I − AdG (s))mα
α∈R/±
=
Y
α∈R/±
=
Y
1 − cos 2πα(u)
sin 2πα(u)
det
− sin 2πα(u) 1 − cos 2πα(u)
2(1 − cos 2πα(u)) > 0,
α∈R/±
by Lemma 7.4. It follows that
sgn det(1 − L0t )p = +1
for every fixed-point p of Lt in G/T . For any g ∈ G we therefore have
L(Lg ) = L(Lt ) = |W | =
6 0
so Lg has a fixed-point in G/T . This completes the proof of Thm. 7.6
51
Octonions and G2
8
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 9 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) = r2 N (x) for all r ∈ R and x ∈ A.
• The function (x, y) = 12 [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) = 21 [N (x + y) − N (x) − N (y)].
This form is symmetric by definition and satisfies (x, x) = N (x) > 0 if x 6= 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 7→ x̄, by
x̄ = 2(1, x) − x.
(11)
(xy, xz) = N (x) · (y, z) = (yx, zx).
(12)
We now have the following equations:
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).
9
On quaternions and octonions, A.K. Peters, Natick MA, 2003
52
(13)
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, x̄z) = (xy, z) = (x, z ȳ).
(15)
Thus, the adjoint of right multiplication by x is right multiplication by x̄, and similarly for left multiplication. Equation (15) is proved by setting z = 1 in (13).
x̄¯ = x.
(16)
This is proved by setting y = 1 and z = t in (13).
xy = ȳx̄.
(17)
Proof: use (15) repeatedly to get (ȳx̄, t) = (x̄, yt) = (x̄t̄, y) = (t̄, xy) = (t̄(xy), 1) = (xy, t).
N (x) = xx̄ = x̄x
and
(x, y) = 12 (xȳ + yx̄).
(18)
Proof: Using (15) we have N (x) = (x, x) = (1, x̄x). From (17) we have x̄x = x̄x, so (1, x̄x) = x̄x
from (11). Now 2(x, y) = N (x + y) − N (x) − N (y) = (x + y)(x̄ + ȳ) − xx̄ − y ȳ = xȳ + yx̄.
x2 = −1 ⇔ (x, x) = 1
and (x, 1) = 0.
(19)
(12)
Proof: If x2 = −1 then N (x)2 = N (x2 ) = N (−1) = 1 so (x, x) = N (x) = 1 and (x, 1) = (x2 , x) =
(−1, x) = −(x, 1), so (x, 1) = 0. Conversely, (x, 1) = 0 implies x̄ = −x by (11) and (x, x) = 1
implies 1 = xx̄ = −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 xȳ + yx̄ = 0 from (18) and x̄ = −x, ȳ = −1 from (15).
Finally, we have the alternative law, which is a form of associativity for two elements.
x(xy) = (x2 )y
c1
m2
and
(xy)y = x(y 2 ).
(21)
norm
Proof: From (x̄(xy), t) = (xy, xt) = N (x)(y, t) = ((x̄x)y, t) we have x̄(xy) = (x̄x)y. Replacing
x̄ 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 7→ ā.
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, bā) = 0. Since H contains 1, we have
xa = −xa.
(22)
It follows that xH = Hx and
xa = āx
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 − db̄) + x(cb + ād),
(24)
for all a, b, c, d ∈ H.
Proof: Equation (24) is equivalent to the following three relations:
i) a(xd) = x(ād).
ii) (xb)c = x(cb).
iii) (xb)(xd) = −db̄.
Let t ∈ A be arbitrary.
For i) we have
(15)
(14)
(15)
(a(xd), t) = (xd, āt) = − (ād, xt) = (x(ād), t) .
For ii) we have
(14)
(17) (15)
(23)
(15)
(23)
((xb)c, t) = (xb, tc̄) = b̄x, tc̄ = − b̄c̄, tx = − (cb), tx = (cb)x, t = (x(cb), t)
For iii) we have
(15)
((xb)(xd), t) =
(22)
(14)
(15)
(12)
(15)
xb, t(xd) = − (xb, t(xd)) = (x(xd), tb) = − (xd, x(tb)) = − (d, tb) = −db̄, 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):
def
(a + xb)(c + xd) = (ac − db̄) + x(cb + ād).
Now H + xH will be a composition algebra exactly when
[N (a) + N (b)][N (c) + N (d)] = N (ac − db̄) + N (cb + ād).
Expanding and cancelling terms, this will hold exactly when
(cb, ād) = (ac, db̄),
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 with R and the norm N (x) = x2 , the doubling process produces composition algebras C, 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 = T1 S is the tangent space to S at 1. More
generally, for any x ∈ S, we have (x, vx) = (xx̄, v) = (1, v) = 0, so V x = Tx S is the tangent space
to S at s. Hence the function
Xv : S → S × A,
given by Xv (x) = vx
∼
takes values in T S and is a smooth vector-field on S. Since v 7→ Xv (x) is an isomorphism V → Tx S,
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 composition.
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(x̄).
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 in
A0 . 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−1 x ∈ S so
σ(y) ∈ S so σ(x) = ny ∈ nS ⊂ A0 . Hence σ preserves A0 and N (σ(x)) = n2 = N (x). And since σ
is 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 fx,y : O(A, N ) → A be the function fx,y (g) =
g(x)g(y) − g(xy). Then G is the common zeros of the continuous functions fx,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
1
C
C2
H
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
(24)
σ ((a + xb) · (c + xd)) = σ (ac − db̄) + x(cb + ād)
= τ (ac − db̄) + yτ (cb + ād)
= τ (a)τ (c) − τ (d)τ (b̄) + y (τ (c)τ (b) + τ (ā)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 its
restriction 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 + x1 i + x2 j + x3 k + x4 ` + x5 i` + x6 j` + x7 k` =
7
X
x2i .
i=0
For all q ∈ H we have `q = q̄` by (23), so from (24) the multiplication in O is given by
(p + q`)(r + s`) = (pr − s̄q) + (qr̄ + 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 7→ 1, i 7→ u, j 7→ v, k 7→ uv
is an algebra isomorphism.
Proof: Since N (u) = N (v) = 1 and (u, 1) = (v, 1) = 0, we have u2 = v 2 = −1 by (19). Likewise
N (uv) = N (u)N (v) = 1 and (1, uv) = (ū, v) = −(u, v) = 0, so (uv)2 = −1.
From (15) we have (uv, u) = (v, ūu) = (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) = −(v 2 )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 G
may 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 Su be the five-dimensional sphere in S
orthogonal to u. Then π −1 (u) may be identified with the set of pairs
Tu = {(v, x) ∈ Su × Su : (x, v) = (x, uv) = 0},
and we have dim T = dim S + dim Tu . Let πu : Tu → Su be the projection πu (v, x) = v. Then
πu−1 (v) is the three-dimensional sphere Su,v,uv in Su orthogonal to v and uv, and we have dim Tu =
dim Su + Su,v,uv . Therefore
dim G = dim T = dim S + dim Su + dim Su,v,uv = 6 + 5 + 3 = 14.
8.6
The SU3 in Aut(O)
10
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
¯ = (w, 1) = 0, so `W = W . Since `2 = −1, we
we have (`w, 1) = −(w, `) = 0 and (`w, `) = (w, ``)
10
The definite article in the title of this and later sections indicates that the subgroup is unique up to conjugacy in Aut(O).
58
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
hv, wi = (v, w) + `(`v, w).
For all σ ∈ G` and v ∈ W we have σ(`v) = `σ(v), so
hσ(v), σ(w)i = (σ(v), σ(w))+`(`σ(v), σ(w)) = (σ(v), σ(w))+`(σ(`v), σ(w)) = (v, w)+`(`v, w) = hv, wi.
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 h , i. Since uv is a unit vector orthogonal to both u and v, we may take
w = uv. In fact, if σ(u) = eθ` u and σ(v) = eφ` v, then (letting c = cos θ, c0 = cos φ, etc.) we have
σ(uv) = σ(u)σ(v)
= (eθ` u)(eφ` v)
= (cu + s`u)(c0 v + s0 `v)
= cc0 uv + ss0 (`u)(`v) + cs0 u(`v) + sc0 (`u)v
= cc0 uv − ss0 uv − cs0 `(uv) − sc0 `(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 7→ eα` i,
j 7→ eβ` j,
k 7→ eγ` k,
where α + β + γ = 0. This torus T` is a maximal torus in G` .
59
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 σ2 tσ2−1 = 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 S3 , it follows that the
Weyl group W = N (T )/T is S3 × h−1i ' 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 S3 inside SU(3) projecting isomorphically onto the Weyl group of SU(3). This subgroup is generated by the involutions




0 1 0
−1 0 0
τ = 1 0 0  ,
τ 0 =  0 0 1 .
0 0 −1
0 1 0
Hence the group generated by hσ2 , τ, τ 0 i 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 preserves H, 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
uτ σ ` = τ σ(`) = τ (uσ `) = uσ τ (`) = uσ (uτ `) = (uτ uσ )`,
60
so σ 7→ 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) = [etD a][etD b]
we find that D ∈ g if and only
D(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, D0 ] = DD0 − D0 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 fact
have 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
X
g=t⊕
mα .
α∈R/±
61
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) = ea` i,
t(j) = eb` j,
t(k) = ec` k,
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) = a`i,
D(j) = b`j,
D(k) = c`k,
D(`) = 0,
(25)
and we have
g` = t ⊕ mb−c ⊕ ma−c ⊕ ma−b ,
where
mb−c











 0 −z̄ 0

 0 0 −z̄
 0 0 0
= 0 0 −z̄  : z ∈ C` , ma−c = 0 0 0  : z ∈ C` , ma−b = z 0 0 : z ∈ C` .






0 0 0
z 0 0
0 z 0
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:



0
0
0
0
0
0
2y







 0


0
0
0
0
0
−2x












0 0 0 −x −y
0 
 0

 : x, y ∈ R
0
0
0
0
−y
x
0
ma = 





 0


0
x
y
0
0
0












0
0
y
−x
0
0
0






−2y 2x 0 0
0
0
0



0
0
0
0
−x
−y
0







0 0


0
0
−y
x
0












0
0
0
0 −2y 
 0 0

 : x, y ∈ R
0
0
0
0
0
0
2x
mb = 






x y

0
0
0
0
0












y
−x
0
0
0
0
0






0 0 2y −2x 0
0
0



0
0
−x
−y
0
0
0










0
0
−y
x
0
0
0











0
0
0
0
0
x y



0
0
0
0  : x, y ∈ R
mc = y −x 0



0 0


0
0
0
0
2y












0
0
0
0
0
0
2x






0 0
0
0 −2y −2x 0
62
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 linear
functionals
(a, b, c) 7→ a, b, c
become dot product with
1
(2, −1, −1),
3
1
(−1, 2, −1),
3
1
(−1, −1, 2),
3
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. 11
11
Picture taken from Wikipedia.
63
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) ' C27 o S7 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, Rj, . . . 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 homomorphism 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 −→ C23 −→ Γ −→ GL3 (2) −→ 1
is nonsplit. However it splits over the subgroup F21 corresponding to a Borel subgroup of PSL2 (7) '
GL3 (2). In particular, there is an element σ7 ∈ Aut(O) permuting the seven lines Ri, Rj, . . . in a
cycle.
A Cartan subalgebra in g is the Lie algebra of a maximal torus in G. Each line λ ∈ {Ri, Rj, . . . }
generates a copy of C in O. The subalgebra of g killing λ and preserving the remaining six lines is a
Cartan subalgebra tλ . 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 = ti ⊕ tj ⊕ tk ⊕ ti` ⊕ tj` ⊕ tk` ⊕ t`
of seven Cartan subalgebras, permuted simply-transitively by the element σ7 . This is completely analogous to the decomposition
so3 = ti ⊕ tj ⊕ tk ,
arising from the coordinate lines Ri, Rj, Rk in the quaternion algebra H.
64