MA4E0 Lie Groups
December 5, 2012
Contents
0 Topological Groups
1
1 Manifolds
2
2 Lie Groups
3
3 The Lie Algebra
6
4 The Tangent space
8
5 The Lie Algebra of a Lie Group
10
6 Integrating Vector fields and the exponential map
13
7 Lie Subgroups
17
8 Continuous homomorphisms are smooth
22
9 Distributions and Frobenius Theorem
23
10 Lie Group topics
24
These notes are based on the 2012 MA4E0 Lie Groups course, taught by John Rawnsley, typeset by Matthew
Egginton.
No guarantee is given that they are accurate or applicable, but hopefully they will assist your study.
Please report any errors, factual or typographical, to m.egginton@warwick.ac.uk
i
MA4E0 Lie Groups Lecture Notes Autumn 2012
0
Topological Groups
Let G be a group and then write the group structure in terms of maps; multiplication becomes m : G × G → G
defined by m(g1 , g2 ) = g1 g2 and inversion becomes i : G → G defined by i(g) = g −1 . If we suppose that there is
a topology on G as a set given by a subset T ⊂ P (G) with the usual rules. We give G × G the product topology.
Then we require that m and i are continuous maps. Then G with this topology is a topological group
Examples include
1. G any group with the discrete topology
2. Rn with the Euclidean topology and m(x, y) = x + y and i(x) = −x
3. S 1 = {(x, y) ∈ R2 : x2 + y 2 = 1} = {z ∈ C : |z| = 1} with m(z1 , z2 ) = z1 z2 and i(z) = z̄. Note that this is
compact, connected and commutative
4. If G1 , G2 are topological groups then G1 × G2 with the product topology is a topological group.
5. T k = S 1 × ... × S 1
We now state a first result and prove it using the following few lemmas.
Proposition 0.1 If G is a topological group which is connected then any open set containing the identity element
generates G as a group, i.e. every element of G is a finite product of elements of the open set.
Lemma 0.2 If G is a topological group and g ∈ G then Lg (h) = gh = m(g, h) defines a continuous map G → G.
In fact Lg is a homeomorphism.
Proof
Lg1 ◦ Lg2 (h) = g1 g2 h = Lg1 g2 (h)
and also
Le (h) = eh = h = IdG (h)
and so Lg ◦ Lg−1 = Le = IdG = Lg−1 ◦ Lg and so Lg is bijective and has a continuous inverse.
Q.E.D.
Definition 0.3 If G is a group and A, B ⊂ G then we define
1. AB = {ab|a ∈ A, b ∈ B}
2. A−1 = {a−1 |a ∈ A}
3. we inductively define Ak+1 = Ak A = AAk for k ∈ N and A−k = (A−1 )k
Lemma 0.4 Suppose that A ⊂ G and then
• if e ∈ A then Ak ⊂ Ak+1
• A−k = (Ak )−1
• If e ∈ A and A−1 = A then
S
i∈Z
Ak is a subgroup of G, called the subgroup of G generated by A.
Lemma 0.5 If G is a topological group and U is an open subset of G with e ∈ U then V = U ∩ U −1 is an open
set with V = V −1 and e ∈ V .
Theorem 0.6 If G is a topological group and U is an open subset such that e ∈ U then ∪U k is an open subset
of G and i fG is connected then ∪U k = G.
Proof U k is open since it can be written as
[
Lg (U )
g∈U k−1
and since Lg is a homeomorphism so Lg (U ) is open hence the above is a union of open sets, and so is open.
If we set V = U ∩ U −1 and note this is a subset of U and so V k ⊂ U k for all k and so ∪V k ⊂ ∪U k . V is open
and H := ∪V k is a subgroup of G.
H has cosets Lg (H) which are open sets and so G \ H = ∪gH6=H gH is open in G. Since G is connected we
have G \ H = ∅. Since G ⊂ ∪U k ⊂ G we have G = ∪U k .
Q.E.D.
Examples include R? ⊃ R+ and the latter is an open subgroup and R? = R+ ∪ (−1)R+ . and so R+ is closed
as well. Any closed subgroup of R+ , eg {1} or {2k k ∈ Z}.
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MA4E0 Lie Groups Lecture Notes Autumn 2012
1
Manifolds
Definition 1.1 A chart (U, φ) of dimension n on an open set U of a topological space M is a homeomorphism
φ of U onto an open set in Rn .


x1 (p)
We get functions x1 , ..., xn on U such that, for p ∈ U φ(p) =  ... . These are called the coordinate
xn (p)
functions of the chart.
Examples include M = Rn = U the identity map. This is called the canonical chart. One could instead have
U to be an open set, and φ to be the inclusion map. Another example is S 1 with the angle, or with stereographic
projection as the chart.
If (U, φ) and (V, ψ) are two charts then ψ ◦ φ−1 |φ(U ∩V ) is the change of coordinates map between the two.
This is a homeomorphism φ(U ∩ V ) to ψ(U ∩ V ).
Definition 1.2 Two charts (U, φ) and (V, ψ) of dimension n on a topological space M are compatible of class
C k if either U ∩ V = ∅ or if U ∩ V 6= ∅ then the maps ψ ◦ φ−1 |φ(U ∩V ) and φ ◦ ψ −1 |ψ(U ∩V ) are C k as maps from
open sets of Rn .
C k means, for k > 0 that all partial derivatives of orders up to k are continuous, for k = 0 that the functions
are continuous, and for k = ω that the function is real analytic.
Example 1.1 INSERT FIGURE Two stereographic projections on the circle. Suppose we project from two
y
= φ(x,y)
by considering
antipodal points and set the centre of the circle to be the origin. Then we have that 1+x
2
similar triangles. Thus we get that x =
ψ(x,y)
2
y
1−x
−1
4−t2
4+t2
and y =
4t
4+t2 ,
and thus we have found φ−1 (t). Also we have that
4
t
=
and so ψ ◦ φ (t) = by direct calculation. This is a map from φ(U ∩ V ) which is R \ {0} since
U = S \ {(−1, 0))} and V = S 1 \ {(1, 0))} and so the change of coordinates map is C ∞ .
1
Exercise: Check other examples of charts are compatible for suitable k.
Definition 1.3 An atlas A of dimension n and class C k on a topological space M is a collection of charts which
are dimension n and pairwise compatible of class C k such that the domains of the charts cover M .
An atlas is said to be maximal if it contains all charts of dimension n which are C k compatible with all of
its charts.
Remark Every atlas is contained in a unique maximal atlas of the same dimension and class.
Example 1.2 On S 1 we have two charts given by projections from two distinct points covering S 1 .
Definition 1.4 A manifold structure of dimension n and class C k on a topological space M is a choice of
maximal atlas of class C k and dimension n.
A manifold structure can be specified by a non maximal atlas, e.g. the atlas of two charts on S 1 . We can then
use charts from the maximal atlas when needed.
Example 1.3
1. Rn with atlas of one chart {(Rn , IdRn )} of dimension n and class C ∞ (C ω even).
n
2. Any finite dimensional real vector
P space V by picking a basis b = (v1 , ..., vn ). We get a map φb : V → R
by φb (v) = (a1 , ..., an ) if v =
ai vi . V gets a topology by making φb a homeomorphism. Then (V, φb ) is
a chart of dimension n on V . This construction is independent of the choice of bases, b2 = (w1 , ..., wn )
and b1 = b. Thenwe 
have b2 
= b1
g where gis an
 n ×n real
 matrix that is invertible. Then we have
a1
c1
a1
a1
c1
 
 
 
 
 
v = b1  ...  = b2  ...  = b1 g  ...  and so  ...  = g  ...  and so φb1 = g ◦ φb2 and g is C k for all k
an
cn
an
an
cn
and so the charts are compatible and hence in the same maximal atlas.
3. S n the the n sphere and one can generalise stereographic projection. φ(x) =
the plane tangent at (1, 0, ..., 0) from U1 = {x ∈ S n |x0 6= 1}.
2(x1 ,...,xn )
1−x0
is projection onto
We can consider charts from the implicit function theorem. Let F : Rn+k → Rk be a C ∞ map then
y ∈ F (Rn+k is called a regular value if for all x ∈ Rn+k with F (x) = y the derivative Dx F is onto as a linear
map Rn+k → Rk .
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MA4E0 Lie Groups Lecture Notes Autumn 2012
Theorem 1.5 F −1 (y) = {x ∈ Rn+k |F (x) = y} when y is a regular value is an n dimensional manifold. Charts
can be obtained by taking n of the domain coordinates on a suitable open set and the remaining k domain
coordinates are C ∞ functions of these.
Example 1.4 S 1 = F −1 (1) where F (x, y) = x2 + y 2 . D(x,y) F (h, k) = 2xh + 2yk and since x2 + y 2 = 1 x and
y cannot simultaneously be zero andpso the derivative is onto so 1 is a regular value. Then we solve x2 + y 2 = 1
for either x and y. This gives x = 1 − y 2 and we can do this likewise for S n .
Definition 1.6 A continuous map F : M → N between two manifolds is smooth if for any pair of charts (U, φ)
and (V, ψ) of M and N respectively with f −1 (V ) ∩ U 6= ∅ the map of open sets
ψ ◦ f ◦ φ−1 : φ(f −1 (V ) ∩ U ) → ψ(V )
is C ∞ as a map between open sets of Euclidean space.
f is called a diffeomorphism if it is a homeomorphism and both f and f −1 are smooth. If such a map exists,
then we say M and N are diffeomorphic.
When N = R we refer to smooth maps as smooth functions.
The set of all smooth functions on a manifold M is denoted by C ∞ (M ). This is a real vector space and a
ring with pointwise properties
Definition 1.7 If M and N are manifolds of dimension m and n respectively then we can form a product chart
(U × V, φ × ψ) from charts on M and N by
φ × ψ(x, y) = (φ(x), ψ(y)) ∈ Rm × Rn = Rm+n
and then this is a homeomorphism of U × V onto an open set of Rm+n . This construction gives an atlas on
M × N which determines a product manifold structure of dimension m + n.
For example T n = S 1 × ... × S 1 is a product manifold of dimension n and is compact.
We generally assume that our manifolds are Hausdorff topological spaces and, for Lie groups, we assume that
the topology is second countable, i.e. there is a countable basis for the topology.
2
Lie Groups
Definition 2.1 A Lie group G is a set on which there are two structures:
1. G is a group
2. G is a C ∞ manifold with Hausdorff, second countable topology such that m : G × G → G given by m(g, h) =
gh and i : G → G given by i(g) = g −1 are both smooth maps
Remark Sometimes one map m̃(g, h) = g −1 h is used in the definition. This is equivalent to the one given above.
Example 2.1
1. G a finite group is a Lie group with the discrete topology
2. Rn is a Lie group under addition with the standard (canonical) smooth structure
3. S 1 is a group under addition of angle or complex multiplication, and is a Lie group
4. If G and H are Lie groups then G × H is a Lie group
5. T n is a Lie group and is abelian and connected.
6. R? is a group under multiplication so is a 1 dimensional Lie group that is not connected
7. GL(n, R) is a Lie group. It is a group under matrix multiplication, and is an open set in Mn×n (R) since
it is the preimage of R? under the determinant map. It has a global chart using the n2 matrix entries as
coordinates, and the group operations are polynomials in these coordinates, and so are smooth.
8. GL(n, C) is a Lie group of dimension 2n2
9. The quaternions H. H = R1 + Ri + Rj + Rk where i2 = j 2 = k 2 = −1 and k = ij = −ji and so is a
skew field. H = R4 as a real vector space. An element in H is written as q = q0 1 + q1 i + q2 j + q3 k and
q̄ = q0 1 − q1 i − q2 j − q3 k and q q̄ = ||q||2 1 and qq 0 = qq 0 . H? = H \ {0} under quaternion multiplication is
a group with quadratic multiplication map and invers i(q) = q̃/||q||2 and so both are smooth and so it is a
Lie group of dimension 4.
Quaternions of unit length for a subgroup Sp (1) = S 3 as a manifold. Sp (1) is a manifold and the Euclidean
coordinates are smooth functions when restricted to Sp (1) and so it is a Lie group of dimension 3.
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MA4E0 Lie Groups Lecture Notes Autumn 2012
10. GL(n, H) is a Lie group of dimension 4n2
11. Orthogonal Group O(n). This is the group of operations that preserves the inner product on Rn .
O(n) = {g ∈ GL(n, R)|(gx) · (gy) = x.y∀x, y ∈ Rn }
Rewrite conditions so we can use the inverse function theorem to get charts. View x, y as column vectors,
and so x · y = xT y. Then (gx) · (gy) = xT g T gy ⇐⇒ x · (g T gy − y) = 0 ⇐⇒ g T gy − y = 0 ⇐⇒ g T g = In .
Thus
O(n) = {g ∈ GL(n, R)|g T g = In } = {g ∈ Mn×n (R)|g T g = In }
Set F (g) = g T g for g defined on Mn×n (R). What is the target. Note g T g is not an arbitrary n × n matrix.
It is symmetric. Let Sn (R) = {A ∈ Mn×n (R)|AT = A} and note that In ∈ Sn (R). Is In a regular value of
F as a map Mn×n (R) → Sn (R). Take g ∈ O(n) and consider F (g + th) for h ∈ Mn×n (R). Then
Dg F (h) =
d
F (g + th)|t=0 = g T h + hT g
dt
For g fixed, is every element of Sn (R) of this form for a suitable h. Take A ∈ Sn (R) and try to solve
g T h + hT g = A for some h or solve g T h = 21 A and hT g = 12 A, but these are the same as A = AT . Thus
h = 12 gA. Thus Dg F is surjective for every g ∈ O(n).
Since this holds forall g ∈ O(n), O(n) = F −1 (In ) and In is a regular value of F and hence O(n) gets a
manifold structure of dimension n2 − dim(Sn (R)) = n2 − 12 n(n − 1). Also O(n) has a group structure since
g1 , g2 ∈ O(n) then (g1 g2 )T g1 g2 = In . The matrix entries are functions on O(n) and are smooth. Hence
multiplication and inverses are
O(1) = {±1}.
smooth
in manifold structure. Thus O(n) is a Lie group.
a b
b
a
Now consider O(2). If g =
then a2 + c2 = 1 = b2 + d2 and ab + cd = 0. so
⊥
and
c d
d
c
−c
b
−c
so is a multiple of
, e.g.
=λ
and this gives λ2 = 1. Hence O(2) is the matrices of the
a
d
a
a −λc
form
with λ = ±1 and a2 + c2 = 1. Note that det(g) = λ and so λ depends continuously on g.
c λa
Hence O(2) has two connected components, with {g ∈ O(2)| det(g) = 1} as a subgroup and as a manifold is
a circle. So O(2) is a disjoint union of two circles.
12. Special Linear Group SL(n, R) = {g ∈ GL(n, R)| det(g) = 1}. We look at det : Mn×n (R) → R. We have
SL(n, R) = det−1 (1) but is it a regular value of det? If g ∈ SL(n, R) and h ∈ Mn×n (R) we need to compute
Dg det(h) =
d
d
d
det(g + th)|t=0 =
det(g(In + tg −1 h))|t=0 =
det(In + tg −1 h)|t=0 = tr(g −1 h) = n 6= 0
dt
dt
dt
and so 1 is a regular value of det so SL(n, R) is a Lie group of dimension n2 − 1.
T
13. SO(n)
= O(n)
∩ SL(n, R) = {g ∈ Mn×n (R)|g g = In , det(g) = 1}. For example SO(1) = {1} and SO(2) =
a −c
∈ M2×2 (R)|a2 + c2 = 1 which is the same as O(2). Note that g ∈ SO(n) =⇒ det(g) = ±1
c a
and hence SO(n) is an open set in O(n) since SO(n) = det−1 (0, 2) viewing det as a function on O(n). det
is a polynomial in the matrix entries so is smooth hence continuous and so det−1 (0, 2) is open
Lemma 2.2 If G is a Lie group and H ⊂ G is a subgroup and an open set then H is a Lie group
Proof Take charts from G which intersect H and take (U ∩ H, φ|U ∩H ) as a chart on H.
Q.E.D.
If G is a Lie group this makes G0 , the component containing the identity, into a Lie group. Also note that
O(n)0 ⊂ O(n) and O(n) ⊃ SO(n) ⊃ SO(n)0 and also SO(2) = O(2)0 = SO(2)0 . We can show this holds
for all n by showing SO(n) is connected.
SO(n − 1) sits inside SO(n). Given g ∈ SO(n) with g viewedasn column vectors g = (x1 , ..., xn )
0
 .. 
 
so xi · xj = δij , and so xn ∈ S n−1 so rotate xn to the vector  .  this will rotate the other vectors
0
1


..
 . . . . . . . 0
.
..
..
.. . Repeating we eventually get to
simultaneously and end up with a matrix of the form 
 ..
.
.
.
0 ... 0 1
SO(2) which is connected. Hence SO(n) is connected for all n.
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MA4E0 Lie Groups Lecture Notes Autumn 2012
We have maps O(n) → S n−1 by picking any column. Putting these maps together we get a map O(n) →
×n S n−1 which is continuous and injective and gives O(n) its usual topology. The image consists of the
subset of this product of pairwise orthogonal vectors. The image is closed hence compact,so O(n) is a
compact Lie group. So is SO(n).
14. Pseudo -orthogonal groups O(p, q) and SO(p, q).
Pp
Pp+q
On Rp+q let Bp,q (x, y) = 1 xi yi − p+1 xi yi = xT Ap,q y. We then define
O(p, q) = {g ∈ M(p+q)×(p+q) (R)|Bp,q (gx, gy) = Bp,q (x, y)}
It is clear that g ∈ O(p, q) ⇐⇒ g T Ap,q g = Ap,q and we can define F : M(p+q)×(p+q) (R) → Sp+q (R) by
F (g) = g T Ap+q g so O(p, q) = F −1 (Ap,q ) and is a regular value. This makes O(p, q) into a Lie group of
dimension 21 (p + q)(p + q − 1).
Observe that O(p, 0) = O(p) and also define SO(p, q) = O(p, q) ∩ SL(p + q, R).
a b
We consider O(1, 1). Take an element here and write it as g =
and then we get that a2 − c2 = 1,
c
d
b
−c
b2 − d2 = −1 and ab − cd = 0 and so
= −λ
and with λ2 = 1. Now det(g) = −λ and so for
−d
a
a c
SO(1, 1) we have the form
with a2 − c2 = 1. This is a hyperbolic function, and so SO(1, 1) has two
c a
connected components and O(1, 1) has four connected components. Now O(1, 1)0 corresponds to det(g) = 1
and a > 0 and then
cosh t sinh t
t 7→
sinh t cosh t
is a homomorphism of groups from R → O(1, 1)0 which is a diffeomorphism. This is an isomorphism of
Lie groups.
O(1, 3) is the Lorentz group, occurring in special relativity.
Definition 2.3 A homomorphism of Lie groups φ : H → G is a map which is a homomorphism
of groups and a smooth map of manifolds. An isomorphism of Lie groups is a bijection which is a
homomorphism of Lie groups, as well as its inverse.
15. An antisymmetric
when its dimension
is even.So take R2n and
bilinear form can only be nonsingular
set
0
0
0 −In
x
y
x0T y 00 − x00T y 0
T
Jn =
and Ω(x, y) = x Jn y. If x =
and y =
then Ω(x, y) =
In
0
x00
y 00
x0 y 00 − x00 y 0
Remark If V is a real 2n dimensional vector space and w is an antisymmetric bilinear form on V which
is non singular We define
Sp(2n, R) = {g ∈ GL(2n, R)|Ω(gx, gy) = Ω(x, y)}
Then g ∈ Sp(2n, R) ⇐⇒ g T Jn g = Jn and set F (g) = g T Jn g and then (g T Jn g)T = −g T Jn g and so
F maps M2n×2n (R) into A2n (R) = {A ∈ M2n×2n (R)|AT = −A} which is a vector space of dimension
1
2 2n(2n − 1) = n(2n − 1). Then one can show Jn is a regular value of F and so Sp(2n, R) is a Lie group
of dimension (2n)2 − n(2n − 1) = n(2n + 1). This group is called the real symplectic group
16. Complex versions GL(n, C) , etc. O(p, q, C) is the same as O(p + q, C) as one C bilinear form would have
all positive signs as can multiply by i.
17. Unitary group On Cn one can form the Hermitian inner product hz, vi = z̄ T v. A complex linear map
g : Cn → Cn is unitary if hgz, gvi = hz, vi for all z, v ∈ Cn . The set of unitary matrices is denoted by
U (n). We can write the inner product as hz, vi = z̄ T (ḡ T g)v and we denote g ? = ḡ T . Then g ∈ U (n) ⇐⇒
g ? g = In . This is a group under matrix multiplication.
Define F (g) = g ? g and then note that (g ? g)? = g ? g and let Hn (C) = {A ∈ Mn×n (C)|A? = A} which is a
real vector space of dimension n2 . If A = X + iY then A? == X T − iY T = X + iY , and so X = X T and
Y = −Y T and so corresponds to symmetric and antisymmetric and so the dimension is the sum.
In is a regular value of F : Mn×n (C) → Hn (C) and so U (n) is a Lie group of dimension 2n2 − n2 = n2 .
We can write g ? g in terms of the columns of g. This is that the columns of g form an orthogonal basis
of Cn so we get a map U (n) → ×n S 2n−1 which is a homomorphism onto its image, which is n-tuples
z 1 , ..., z n ∈ Cn with hz i , z j i = 0 for i 6= j. The image is a closed set in ×n S 2n−1 hence is compact, so U (n)
is a compact Lie group.
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MA4E0 Lie Groups Lecture Notes Autumn 2012
U (1) = {z ∈ C||z| = 1} = S 1
a
c
b
is in U (2) then |a|2 + |c|2 = 1 = |b|2 + |d|2 and āb + c̄d = 0. Thus b = −λc̄ and d = λā with
d
a b
a
a b
1
3
|λ| = 1 and so
is determined by λ ∈ S and
∈ S with λ = det
and so as a manifold
c d
c
c d
1
3
U (2) is diffeomorphic to S × S .
If
Note that g ? = ḡ T and so det g ? = det ḡ = det g and so | det g| = 1. If we set
−1
SU (n) = U (n) ∩ SL(n, C) = det(1)
by viewing det : U (n) → S 1 . Then one can check that 1 is a regular value of det and one gets SU (n) as a
manifold of dimension n2 − 1 with matrix entries defining smooth functions, and so is a Lie group.
SU (1) = {1}
a b
We now consider SU (2). An element of U (2) is written
with |a|2 + |c|2 = 1 = |b|2 + |d|2 and
c d
āb + c̄d = 0. Then we get, using similar arguments to above, that
a −λc̄ 2
2
|a|
+
|c|
=
1,
|λ|
=
1
' S3 × S1
U (2) =
c λā and note that det g = λ for g ∈ U (2) and so SU (2) is diffeomorphic to S 3 and S 3 is diffeomorphic to Sp(1)
and so SU (2) and Sp(1) are diffeomorphic.
18. We here consider U (p, q) and SU (p, q). We define
hz, vip,q =
p
X
1
z¯k vk −
q
X
z̄p+k vp+k = z ? Ap,q v
1
and set
U (p, q) = {g ∈ M(p+q)×(p+q) (C)|g ? Ap,q g = Ap,q }
and observe that Ap,q is a regular value of F (g) = g ? Ap,q g and so U (p, q) is a Lie group of dimension
(p + q)2 . For g ∈ U (p, q) we have | det g| = 1 and 1 is a regular value of det : U (p, q) → S 1 and so SU (p, q)
is a Lie group of dimension (p + q)2 − 1.
SU (1, 1) is the isometry group of the Poincaré disc. It is three dimensional and related to SL(2, R). SU (2, 2)
is the conformal group and is important in Penrose’s twister theory.
19. Sp(n) are called the quaternary unitary groups. Defined like U (n) but using quaternions Hn .
Sp(n) = {g ∈ Mn×n (H)|ḡ T g = In }
Now In is a regular value of F : Mn×n (H) → Hn (H) and Sp(n) = F −1 (In ). The dimension of Hn (H) is
n + 21 n(n − 1) and the dimension of Mn×n (H) is 4n2 and so the dimension of Sp(n) = n(2n + 1) which is
the same dimension as that of Sp(2n, R).
If g ∈ Sp(n) then the columns of g are unit vectors in Hn so are in S 4n−1 and Sp(n) maps into ×n S 4n−1
with image a closed set of pairwise orthogonal vectors, and so is compact. Now note O(n) ⊂ U (n) ⊂ Sp(n)
since R ⊂ C ⊂ H.
These are the compact exceptional groups, together with their det = 1 versions. There is no determinant
function on H as it makes no sense on non commutative rings.
There are five more compact simple groups G2 , F4 , E6 , E7 , E8 . Sp(p, q) can also be defined by g ? Ap,q g = Ap,q
with g ∈ M(p+q)×(p+q) (H)
3
The Lie Algebra
Definition 3.1 A real Lie algebra is a real vector space V with binary operation [·, ·] : V × V → V satisfying
1. it is bilinear
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MA4E0 Lie Groups Lecture Notes Autumn 2012
2. [x, y] = −[y, x]
3. [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 called the Jacobi identity
Observe that these relations are homogeneous.
Example 3.1
1. We can take any real vector space and set [x, y] = 0. This is the trivial Lie algebra, or also
called the abelian Lie algebra
2. A an associate Lie algebra over R and set [a, b] = ab − ba. We make A into a Lie algebra, for example
A = EndV the set of linear maps V → V . This is called the commutator bracket. It measures how non
zero the commutativity of the algebra is. For instance if V = Rn and then End(V ) = Mn×n (R) and so this
is the Lie algebra with the commutator.
3. If (V, [, ]) is a Lie algebra then a subspace U of V becomes a Lie algebra so long as [x, y] ∈ U for x, y ∈ U .
This is called a Lie subalgebra. For instance if A a commutator algebra, then End(A) is a Lie algebra.
Denote by Der(A) the space of linear maps D : A → A which satisfy the Leibniz rule,
D(ab) = D(a)b + aD(b)
This is obviously a subspace of End(A) and we claim that it is a Lie subalgebra. Let D1 , D2 ∈ Der(A).
Then
[D1 , D2 ](a, b) = D1 (D2 (ab)) − D2 (D1 (ab))
= D1 (D2 (a)b + aD2 (b)) − D2 (D1 (a)b + aD1 (b))
= ... =
[D1 , D2 ](a)b + a[D1 , D2 ](b)
and so is in Der(A).
Our main use of this last example is with A = C ∞ (M) for M a manifold. Let X (M) = Der(C ∞ (M )). If
M = R and D ∈ X (M) then D(1) = D(1 × 1) = D(1) × 1 + 1 × D(1) = 2D(1) and so D(c) = 0 for any constant
c ∈ R.
If we take f ∈ C ∞ (R and a ∈ R then we can use Taylor expansion to get
f (x) = f (a) + (x − a)f 0 (a) + (x − a)2 ga (x)
where ga ∈ C ∞ (R). We can rearrange this and then check that it is smooth
D(f )(x) = 0 + D(x − a)f 0 (a) + D((x − a)2 ga (x))
= D(x − a)f 0 (a) + D(x − a)(x − a)ga (x) + (x − a)D(x − a)ga (x) + (x − a)2 D(ga (x))
and if we set x = a we get
D(f )(a) = D(x − a)(a)f 0 (a) + 0
d
d
. If Di = hi dx
for i = 1, 2 then
and so D(f )(a) = h(a)f 0 (a) for some h ∈ C ∞ (R) for all a ∈ R and so D = h dx
[D1 , D2 ] = (h1 h02 − h2 h01 )
d
dx
and this is not the trivial Lie algebra. X (M) is the Lie algebra of vector fields on M and [x, y] is the Lie bracket
of vector fields x, y ∈ X (M).
If σ : M → M is a diffeomorphism and X ∈ X (M ) then we can form an action by
(σ · X)(f ) = (X(f ◦ σ)) ◦ σ −1
Then if X ∈ X (M) then σ · X ∈ X (M).
If G is a Lie group define Lg : G → G for g ∈ G by Lg (h) = gh. This is a C ∞ map. Note Lg1 ◦ Lg2 = Lg1 g2
and Le = IdG and so Lg−1 = L−1
g and so is invertible. We say X ∈ X (M) is left invariant if Lg · X = X for all
g ∈ G. Denote by X (G)G the set of left invariant vector fields. Since X 7→ Lg · X is linear we have that X (G)G
is a subspace of X (G). Note that Lg · ([X, Y ]) = [Lg · X, Lg · Y ] for all X, Y ∈ X (G) and so X (G)G is a Lie
subalgebra of X (G).
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4
The Tangent space
In Rn if we have a smooth curve γ : (a, b) → Rn we can associate a tangent vector to each point on γ by taking
dγ(t)
dt or γ̇(t). How though can we define γ̇(t) for γ : (a, b) → M a smooth map?
d
(φ ◦ γ(t)) =
We can take a chart on M around a point on γ and take the tangent vector in Rn of φ ◦ γ, say dt
d
n
v ∈ R . If we take another chart (V, ψ) around the same point then in general dt (ψ ◦ γ(t)) = w is different but
ψ ◦ γ(t) = ψ ◦ φ−1 ◦ φ(γ(t)) = ψ ◦ φ−1 (φ ◦ γ(t)) and so w = Dφ◦γ(t) (ψ ◦ φ−1 )v. Fix a point x ∈ M and consider
objects of the form (U, φ, v) where (U, φ) is a chart about x and v ∈ Rn . Define an equivalence relation ∼x by
(U, φ, v) ∼x (V, ψ, w) if w = Dφ◦γ(t) (ψ ◦ φ−1 )v.
Definition 4.1 The tangent space Tx M is the set of equivalence classes of ∼x . We denote by [U, φ, v]x the
equivalence class of (U, φ, v) at x.
This becomes a vector space by picking a chart (U, φ) at x and defining
[U, φ, v]x + [U, φ, w]x = [U, φ, v + w]x
and
a[U, φ, v]x = [U, φ, av]x
This is well defined as the derivative of a linear map.
Definition 4.2 If γ : (a, b) → M is a smooth curve with x = γ(t0 ) we define γ̇(t0 ) =
(U, φ) around x and setting
d(φ ◦ γ(t))
γ̇(t0 ) = U, φ,
|t=t0
dt
x
dγ(t)
dt |t=t0
by taking a chart
Theorem 4.1 If M is a manifold of dimension n then for all x ∈ M Tx M is a vector space of dimension n. If
F : M → N is smooth then there is a natural linear map dx F : Tx M → TF (x) N which satisfies the chain rule,
namely if F : M → N and E : N → P and x ∈ M then
dx (E ◦ F ) = dF (x) E ◦ dx F
Furthermore if γ : (a, b) → M is a smooth curve then
dγ(t) F (γ(t)) = (F ◦ γ)· (t)
Proof Fix x ∈ M and (U, φ) a chart around x. Define a map Rn → Tx M by v 7→ [U, φ, v]x which is linear and
also define [V, ψ, w]x 7→ Dψ(x) φ ◦ ψ −1 (w) and this is also linear. Note that these are inverses and so Rn ' Tx M
if we pick a chart. Then
dx F [U, φ, v]x = [V, ψ, Dφ(x) (ψ ◦ F ◦ φ−1 )(v)]F (x)
which is independent of charts on M and N . This formula also works to show that the chain rule holds:
d
ψ ◦ F ◦ γ(t)]F ◦γ(t)
dt
d
= [V, ψ, ψ ◦ F ◦ φ−1 ◦ φ ◦ γ(t)]F ◦γ(t)
dt
d
= [V, ψ, Dφ◦γ(t) ψ ◦ F ◦ φ−1 φ ◦ γ(t)]F ◦γ(t)
dt
d
= dγ(t) F [U, φ, φ ◦ γ(t)]γ(t)
dt
= dγ(t) F (γ̇(t))
(F ◦ γ)· (t) = [V, ψ,
Q.E.D.
n
n
n
For example consider R with identity chart. This allows a canonical identification Tx R ' R and then
is the same ordinary derivative.
Definition 4.3 Let f ∈ C ∞ (M) and let x ∈ M. If X ∈ Tx M then we set X(f ) ∈ R to be equal to
(Dφ(x) f ◦ φ−1 )(v)
if (U, φ) is a chart about x and X = [U, φ, v]x
This is the identification of df : Tx M → Tf (x) R with Tf (x) R = R, and write the X to the left, X(f ) = df (X).
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MA4E0 Lie Groups Lecture Notes Autumn 2012
Proposition 4.4 Fixing x ∈ Tx M the map f 7→ X(f ) (C ∞ (M) → R) is linear and satisfies the product rule,
namely
X(f g) = X(f )g(x) + f (x)X(g)
Note that this should be clear since it is the directional derivative for a fixed chart.
Remark
1. This says X is a derivation at x from C ∞ (M) → R.
2. Suppose D : C ∞ (M) → R is linear and satisfies the product rule, then there is an x ∈ Tx M such that
D(f ) = X(f ) for all f ∈ C ∞ (M). To see this let (U, φ) be a chart about x such that the coordinate
functions


D(f1 )


xi . Then x1 , ..., xn are the restrictions of globally defined f1 , ..., fn ∈ C ∞ (M). Then v =  ...  ∈ Rn
D(fn )
and X = [U, φ, v]x .
3. This means that we could alternatively define Tx M as the linear maps C ∞ (M) → R that satisfy the
product rule.
Example 4.1
1. M an open set in a vector space V (for example GL(n, R) in Mn×n (R)). When we pick
a basis b = {e1 , ..., en } for V we get a chart (M,!φb ) as follows. Take x ∈ M and expand x in the basis
x1 (x)
Pn
so that x = i=1 xi (x)ei and put φb = .
. The x1 , ..., xn are coordinate functinos of this chart.
..x (x)
n
!
v1
P
and
M then becomes a manifold using the atlas {(M, φb )}. If v ∈ V then v =
vi ei and v = .
..v
n
[M, φb , v]x ∈ Tx M. The map v 7→ [M, φb , v]x is a linear isomorphism of vector spaces. This is independent
d
of the basis. The corresponding derivation at x we write as vx and for f ∈ C ∞ (M), vx (f ) = dt
f (x+tv)|t=0 .
2. Suppose that M = F −1 (c) for F : Rn+k → Rk smooth and c a regular value of f , and M given the manifold
structure by the implicit function theorem. Take X ∈ Tx M and any curve γ through x, say with γ(t0 ) = x
such that X = γ̇(t0 ). To do this pick a chart (U, φ) about x and then X = [U, φ, v]x for v ∈ Rn and put
d
γ(t) = φ−1 (φ(x) + (t − t0 )v) and so dt
φ(γ(t))|t=t0 = v.
Then we can view this as a curve in Rn+k and take its derivative at t0 to get a v 0 ∈ Rn+k . But γ(t) ∈ M
for all t and so F (γ(t)) = c and hence (Dγ(t0 ) F )(v 0 ) = 0 i.e. v 0 ∈ KerDx F , and Dx F : Rn+k → Rk
d
γ(t)|t=t0
and Dx F is surjective so dim KerDx F = n. We have a map Tx M → Rn+k defined by X 7→ dt
which is linear and injective and maps Tx M to Ker(Dx F ) and so we can use Ker(Dx F ) as a model for
Tx M. For example, S n ⊂ Rn+1 with F (x) = x · x. For x ∈ F −1 (1) we have Dx F (v) = 2x · v and so
KerDx F = {v ∈ Rn+1 |x · v = 0}, i.e. all vectors perpendicular to the radius vector x.
3. GL(n, R) has Mn×n (R) as the tangent space at In .
4. GL(n, C) has Mn×n (C) as the tangent space at In .
5. GL(n, H) has Mn×n (H) as the tangent space at In .
6. SL(n, R) = {g ∈ Mn×n (R) : det g = 1}. Now the tangent space at In is the kernel of DIn det = tr and so
the space is {A ∈ Mn×n (R) : trA = 0}
7. SL(n, C) has TIn SL(n, C) = {A ∈ Mn×n (C) : trA = 0}
8. O(n) = {g ∈ Mn×n (R) : g T g = In }. Define F (g) = g T g : Mn×n (R) → Sn (R) and now note that
DIn F (A) = AT + A and ker DIn F = {A ∈ Mn×n (R) : AT + A = 0} = An (R)
9. TIn U (n) = {A ∈ Mn×n (R) : A? + A = 0}
Recall we have diffeomorphisms Lg : G → G such that Lg (h) = gh with the properties that Lg1 ◦ Lg2 = Lg1 g2
and Le = Id. Similarly for on the right with Rg (h) = hg but we have Rg1 ◦ Rg2 = Rg2 g1 and Re = Id.
Lemma 4.5 Tg G = de Lg (Te G) since Lg (e) = g.
Definition 4.6 A vector field on a manifold M is a choice of tangent vector at each point of M, i.e. for each
x ∈ M we pick an Xx ∈ Tx M.
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Given a vector field X on M we can differentiate f ∈ C ∞ (M) to give a function x 7→ Xx (f ), i.e. we can define
a function X(f ) by X(f )(x) = Xx (f ) and it satisfies the Leibniz rule, since
X(f1 f2 )(x) = Xx (f1 f2 ) = Xx (f1 )f2 (x) + f1 Xx (f2 (x)) = X(f1 )f2 (x) + f1 (x)X(f2 ) = (X(f1 )f2 + f1 X(f2 ))(x)
X is linear but X may not be a derivation since we do not know X(f ) ∈ C ∞ (M).
Definition 4.7 A vector field X on M is said to be smooth if X(f ) ∈ C ∞ (M) for all f ∈ C ∞ (M).
If X is smooth then X is a derivation.
Conversely if D : C ∞ (M) → C ∞ (M) is a derivation then consider, for a given x ∈ M, the map f 7→ D(f )(x)
from C ∞ (M) → R. Then it defines a derivation at x and so a tangent vector Xx ∈ Tx M with D(f )(x) = Xx (f ).
Then D determines a vector field X which is smooth. Thus derivations of C ∞ (M) are smooth vector fields.
Thus X (M) is now the vector space of smooth vector fields with the bracket of derivations. This is the link
between the algebraic point of view (derivations) and the geometric point of view (tangent space).
Definition 4.8 Let F : M → N be a smooth map of smooth manifolds. Vector fields X on M and Y on N are
said to be F -related if dx F : Tx M → TF (x) N is such that
dx F (Xx ) = YF (x)
for all x ∈ M.
Theorem 4.9 If Xi ∈ X (M) and Yi ∈ X (N ) for i = 1, 2 and F : M → N is a smooth map and Xi and Yi are
F -related for i = 1, 2 then [X1 , X2 ] is F -related to [Y1 , Y2 ].
Proof Take f ∈ C ∞ (N ) Then
dx F (Xi )x (f ) = (Xi )x (f ◦ F )
and
dx F (Xi )x = (Yi )F (x)
and so
(Yi )F (x) (f ) = (Xi )x (f ◦ F )
and is the same as Yi (f )(F (x)) or
Yi (f ) ◦ F = Xi (f ◦ F )
(4.1)
This is the derivation version of being F -related. Now we have
[X1 , X2 ](f ◦ F ) = X1 (X2 (f ◦ F )) − X2 (X1 (f ◦ F ))
= X1 (Y2 (f ) ◦ F ) − X2 (Y1 (f ) ◦ F )
= Y1 (Y2 (f )) ◦ F − Y2 (Y1 (f )) ◦ F
= ([Y1 , Y2 ](f )) ◦ F
Evaluating at x leads to dx F [X1 , X2 ]x = [Y1 , Y2 ]F (x) as required
5
Q.E.D.
The Lie Algebra of a Lie Group
Definition 5.1 A vector field X on a Lie group G is called left invariant if
dh Lg (Xh ) = Xgh
for all g, h ∈ G.
Note that this means X is Lg related to X for all g ∈ G.
Lemma 5.2 If X is a vector field on a Lie group which is left invariant then X is smooth
Proof Take f ∈ C ∞ (G) and then
X(f )(g) = Xg (f ) = Xge (f ) = (de Lg (Xe ))(f ) = Xe (f ◦ Lg )
and also note that
(f ◦ Lg )(h) = f (Lg (h)) = f (gh) = f (m(g, h)) = (f ◦ m)(g, h)
and so differentiating in h in the Xe direction leaves something smooth in g and so Xf ∈ C ∞ (G).
Q.E.D.
To find out how big X (G)G is we compare it to Te G. Define ε : X (G)G → Te G by ε(X) = Xe . This is linear.
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Theorem 5.3 ε : X (G)G → Te G defined by ε(X) = Xe is a linear isomorphism.
Proof Let X ∈ ker ε and so Xe = 0. But Xg = de Lg Xe = 0 for all g and so X = 0 and so ε is injective.
Now take ξ ∈ Te G and set ξ˜h = de Lh (ξ) for all h ∈ G. Then
dh Lg (ξ˜h ) = dh Lg (de Lh (ξ)) = de (Lg ◦ Lh )(ξ) = de Lgh (ξ) = ξ˜gh
Hence ξ˜ is left invariant and so smooth by the above lemma. Thus ξ ∈ X (G)G and ε(ξ˜e ) = ξ˜e = de Le (ξ) = ξ and
so ε is surjective.
Q.E.D.
Definition 5.4 The left invariant vector field ξ˜ determined by ξ ∈ Te G is called the left invariant extension
˜
of ξ, with de Lg (ξ) = ξ.
˜ η̃ are in X (G)G which is a Lie algebra so we can form [ξ,
˜ η̃] which is left invariant and so of
If ξ, η ∈ Te G then ξ,
˜ η̃]). This makes ε : X (G)G → Te G. We call Te G with this bracket
the form h for some h ∈ Te G with h = ε([ξ,
the Lie algebra of G.
Remark We could equally use right invariant vector fields and get a second bracket on Te G denoted by [·, ·]right
ˆ η̂]e
with ξˆh = de Rh (ξ) and [ξ, η]right = −[ξ,
Corollary 5.5 If G is an abelian Lie group then [ξ, η] = 0
Note the bracket can be zero for non abelian Lie groups.
˜ η̃]e (f ) = ξ˜e (η̃(f )) − η̃e (ξ(f
˜ ))
[ξ,
means [·, ·] only contains information about G near the identity. It gives the best information when G is connected.
Example 5.1 Examples of abelian Lie groups: Rn , R? , (0, ∞), S 1 , T n , C? = S 1 × (0, ∞), Cn .
We need a technique for computing with vector fields. If (U, φ) is a chart with coordinates x1 , ..., xn so
xi = ti ◦ φ where ti are Cartesian coordinates on Rn .
◦φ−1
◦ φ ∈ C ∞ (U ) is a vector field on U denoted
If f is a smooth function on the manifold we can form ∂f∂t
i
by
∂
∂xi
Lemma 5.6 If X ∈ X (M) then X|U =
Pn
i=1
∂
with ai ∈ C ∞ (U ).
ai ∂x
i
Example 5.2 GL(n, R) ⊂ Mn×n (R) is an open set. We have coordinates xij (g) = gij . since GL(n, R) is open
d
in a vector space, TIn GL(n, R) = Mn×n (R) as vector spaces. A ∈ Mn×n (R) corresponds to dt
(In + tA)|t=0 . Ã is
the left invariant extension and then
Ãg = (dIn Lg )A = (dIn Lg )
d
d
d
(In + tA)|t=0 = Lg (In + tA)|t=0 = (g + tgA)|t=0
dt
dt
dt
and applying this to coordinate functions gives
Ãg (xij ) =
X
X
X
d
d
(g + tgA)ij |t=0 = gij + t
gik Akj |t=0 =
gik Akj =
xik (g)Akj
dt
dt
k
and so Ã(xij ) =
P
k
k
k
xik Akj Then
[Ã, B̃](xij ) : = Ã(B̃(xij )) − B̃(Ã(xij ))
X
X
= Ã(
xik Bkj ) − B̃(
xik Akj )
k
=
X
k
A(xik )Bkj −
k
=
X
X
=
X
l
B̃(xik )Akj
k
xil Alk Bkj −
k,l
=
X
X
xil Blk Akj
k,l
xil
X
(Alk Bkj − Blk Akj )
k
xil (AB − BA)lj
l
^
= AB
− BA(xij )
^
Using the lemma below we get [Ã, B̃] = AB
− BA and so the Lie algebra of GL(n, R) is Mn×n (R) with the
Lie bracket as the commutator.
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Lemma 5.7 If X, Y ∈ X (M) and for an atlas of charts (U, φ) we have X(xi ) = Y (xi ) for the coordinates
(x1 , ..., xn ) on U then X = Y .
Example 5.3 For all classical groups SL(n, R), O(n) etc, the same calculation works in the sense that TIn G
is identified with ker DIn F for F mapping Mn×n (R) to a suitable set, and then if A ∈ ker DIn F then there is a
d
In + tA + O(t2 )|t=0 . The higher order terms do not change the
curve In + tA + O(t2 ) with A associated to dt
argument, so all brackets are the commutators.
We denote Mn×n (R) with the commutator bracket by gl(n, R) and likewise for the others.
If G and H are Lie groups , F : G → H is a homomorphism of Lie groups then F (eG ) = eH and so
deg F : TeG G → TeH H is there for a linear map from the Lie algebra of g to h.
Theorem 5.8 If F : G → H is a Lie group homomorphism then deG F : g → h is a Lie algebra homomorphism.
Proof Take ξ ∈ g and consider ξ˜ its left invariant extension:
dg F ξ˜g = dg F (deG Lg (ξ)) = deG (F ◦ Lg )(ξ)
and then
F ◦ Lg (h) = F (Lg (h)) = F (gh) = F (g)F (h) = LF (g) (F (h))
and then we get that
˜ = de (LF (g) ◦ F )(ξ) = de LF (g) de F (ξ) = de^
dg F (ξ)
F (ξ)F (g)
G
H
G
G
˜ η̃] is F related to [de^
and so ξ˜ and de^
F (ξ) are F related. Take ξ, η ∈ g and then [ξ,
F (ξ), de^
F (η)] and so
G
G
G
˜ η̃]g ) = [de^
dg F ([ξ,
F (ξ), de^
F (η)]F (g)
G
G
and setting g = eG we get
deG F ([ξ, η]g ) = [deG F (ξ), deG F (η)]h
Q.E.D.
Definition 5.9 We denote deG F by F? : g → h.
Corollary 5.10 Homomorphic Lie groups have isomorphic Lie algebras,
However, the converse is not true. There are a few examples of this. R and S 1 are abelian and one dimensional. Also SU (2) and SO(3) have isomorphic Lie algebras but different centres. U (1) × SU (2) and U (2) are
diffeomorphic as manifolds but not isomorphic as groups, and have the same Lie algebras.
If F : G → H and E : K → G are homomorphisms of Lie groups we get F? : g → h and E? : K → G and also
F ◦ E : K → H is a homomorphism of Lie groups so we get (F ◦ E)? : k → h is a Lie algebra homomorphism.
By the chain rule (F ◦ E)? = F? ◦ E? . Then IdG : G → G has derivative (IdG )? = Idg .
Suppose G and H are compact connected Lie groups and F : G → H i as homomorphism such that F? : g → h
is an isomorphism.
By the inverse function theorem there exists a U ⊂ G and V ⊂ H both open and eG ∈ U , eH ∈ V and F
a diffeomorphism of U into V . Hence F is an open map so F (G) is an open subgroup of H. H is connected so
F (G) = H and so F is surjective.
Let K = ker(F ). This is a normal subgroup of G which is closed in G. G is compact so K is compact. F is
one to one on U and eG ∈ U and so K ∩ U = {eg } so K is discrete in the relative topology. The sets {{k}}k∈K
then form an open covering of K, and so K is a finite group.
Fix k ∈ K and consider the map G → K given by g 7→ gkg −1 ∈ K which is continuous, and so it has a
connected image. Hence gkg −1 = k for all g ∈ G and so K ⊂ Z(G).
For example there are no trivial homomorphisms from SO(3) → SU (2). To show this, suppose that there is
one, and call it F . Then F? : SO(3) → SU (2) is a homomorphism of Lie algebras. F? is non zero otherwise F
would be constant. Do some algebra and see that F? has to be surjective and so a linear isomorphism. Therefore
F is onto with a finite kernel in the centre of SO(3) but the centre of SO(3) is {I3 } so F is an isomorphism.
This is not possible since Z(SU (2)){±I2 }.
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6
Integrating Vector fields and the exponential map
Definition 6.1 Let X ∈ X (M ) and x ∈ M then a smooth curve γ : (a, b) → M in M with γ(to ) = x for some
t0 ∈ (a, b) is said to be an integral curve through x ∈ M if γ̇(t) = Xγ(t) for all t ∈ (a, b)
Theorem 6.2 If X ∈ X (M) and x ∈ M then there exists an ε > 0 and γ : (−ε, ε) → M a smooth curve with
γ(0) = x and γ an integral curve of X.
If γi : (a, b) → M are two integral curves of X with γ1 (t0 ) = x = γ2 (t0 ) then γ1 (t) = γ2 (t) for all t ∈ (a, b).
Proof Take a chart (U, φ) with x ∈ U . Then
Xφ−1 (y) = [U, φ, v(y)]φ−1 (y)
for y ∈ φ(U ) a vector valued function v on φ(U ) ⊂ Rn . X is smooth implies taht v is smooth on φ(U ).
d
φ(γ(t)) and so we have γ̇(t)Xγ(t) . Thus
γ̇(t) = [U, φ, Γ(t)]γ(t) . with Γ(t) a curve in Rn . which is dt
d
φ(γ(t)) = v(φ(γ(t)))
dt
(6.1)
and for γ(t) to pass through x at t0 we need
φ(γ(t0 )) = φ(x)
and so φ(γ(t)) is a solution of an ODE with initial value conditions at t0 and so the existence and uniqueness
theorem for ODES gives existence of integral curves for at least a short time.
Q.E.D.
Note there is no explicit t in the RHS of (6.1) so it is autonomous. Hence φ(γ(t)) is a solution implies that
φ(γ(t + a)) is a solution and so we can always find an integral curve through a given point at a convenient value
of t. so if γ1 (t1 ) = x = γ2 (t2 ) then we can shift one so that γ˜2 (t) = γ2 (t + t2 − t1 ) is an integral curve with
γ˜2 (t1 ) = x and so γ1 (t) = γ̃2 (t) for common values of t near t1 . Thus you can get a large solution form combining
γ1 and γ2
Definition 6.3 An integral curve γ of a vector field X is called a maximal integral curve if any integral curve
passing through a point of γ is just γ restricted to a subinterval (after shifting parameters to agree at a common
point). X is said to be complete if the interval of definition of any maximal integral curve is (−∞, ∞).
∂
Example 6.1 Consider R2 with X(x,y) = ∂x
. Then v = (x, y) = (1, 0). If (x(t), y(t)) is an integral curve then
(ẋ(t), ẏ(t)) = (1, 0) and we want the integral cuve through (x0 , y0 ) at t0 .
This gives the integral cuve as (t − t0 + x0 , y0 ) for t ∈ (−∞, ∞) so X is a complete vector field.
If we were on R \ {0} with the same X then the integral curves have the same formula but if y0 = 0 then
t = t0 − x0 gives (0, 0) which is not allowed. We have two integral curves for either t > t0 − x0 or t < t0 − x0 .
On Rn \ {0} there are many complete vector fields.
Theorem 6.4 If G is a Lie group then all left invariant vector fields are complete
Lemma 6.5 If M is a manifold and X ∈ X (M) and γ : (a, b) → M is an integral curve then for c ∈ R we have
γ̃(t) = γ(t + c) is an integral curve on (a − c, b − c).
Proof Done before
Q.E.D.
Lemma 6.6 Let X be a left invariant vector field on a Lie group G. If γ : (a, b) → G is an integral curve through
g at t = t0 and h ∈ G then γ̄(t) = hγ(t) is an integral curve on (a, b) passing through hg at t = t0 .
Proof
d
d
hγ(t) = Lh γ(t) = dγ(t) Lh (γ̇(t)) = XLh (γ(t))=Xγ̄(t)
dt
dt
and so γ̄ is an integral curve and
γ̄(t0 ) = hγ(t0 ) = hg
γ̄(t) =
Q.E.D.
Lemma 6.7 Let X be a left invariant vector field on a Lie group G and γX be the maximal integral curve of X
with γX (0) = e. Then γX is defined on (−∞, ∞) and satisfies γX (s)γX (t) = γX (s + t) for all s, t.
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MA4E0 Lie Groups Lecture Notes Autumn 2012
Proof We prove that γX (s)γX (t) = γX (s + t) for those s, t where both sides are defined. i.e. if γX is defined
(a, b) then we take s, t, s+t ∈ (a, b). Fix s ∈ (a, b) and consider γ1 (t) = γX (s)γX (t) on (a, b) and γ2 (t) = γX (s+t)
on (a − s, b − s) for t such that t, s + t ∈ (a, b).
By the previous lemmas, γ1 and γ2 are integral curves and at t = 0 both pass through γX (s) so γ1 (t) = γ2 (t)
for all t ∈ (a, b) ∪ (a − s, b − s), i.e. t, s + t ∈ (a, b).
Now suppose b < ∞, and we shift by an amount s ∈ (a, 0) then b − s > b and intervals (a, b) and (0, b − a)
−1
overlap and we have a common solution on the overlap γX
(s)γX (s + t) and γX (t). The new solution is defined
on (a, b − a) which is a contradiction, so b = ∞.
The case a = −∞ is similar.
Q.E.D.
Proof (of theorem 6.4) Let X be a left invariant vector field and γX (t) the maximal integral curve through e
at t = 0. Then lemmas 6.5, 6.6 imply that gγX (t − t0 ) is the integral curve through g at time t0 . The property
γX (s)γX (t) = γX (t + s) says γX : R → G is a homomorphism of groups and γX a smooth curve implies that
γX : R → G is a homomorphism of Lie groups.
Q.E.D.
Definition 6.8 A Lie group homomorphism σ : R → G is called a 1-parameter subgroup of G, and is denoted
1-PSG.
If G is a Lie group and σ is a continuous homomorphism then we say σ is a continuous 1-PSG.
Example 6.2 If G = R then we are looking for continuous maps σ : R → R with σ(s) + σ(t) = σ(s + t) for
all s, t ∈ R. We have that σ(0) = 0. Now, for n ∈ Z we have σ(n + 1) = σ(n) + σ(1) and hence by induction
mn
σ(n) = nσ(1). We can apply this argument to σ(nx) to get σ(nx) = nσ(x). Then nσ( m
n x) = σ( n x) = mσ(x)
m
m
and so σ( n x) = n σ(x) and since σ is continuous and rationals are dense in R we get that σ(x) = xσ(1) for all
x ∈ R. This gives that σ is linear and hence smooth. We can also show that σ̇(0) = σ(1).
Theorem 6.9 The following three sets are in bijection for a Lie group G.
1. The set of 1-PSG of a Lie group G.
2. The maximal integral curves of left invariant vector fields
3. The Lie algebra g.
Proof To map 2 to 1 note γX is a 1-PSG. To map 1 to 3 note σ 7→ σ̇(0) gives us a map. To map 3 to 2 note
ξ 7→ γξ̃ is one. Then we show the action of these in turn gives the identity, so they are all bijections.
Given σ a 1-PSG set ξ = σ̇(0) and form γξ̃ It is enough to show that σ is the integral curve of ξ˜ through e.
It should be clear that σ(0) = e. Then
σ̇(t) =
d
d
d
σ(t + s)|s=0 =
σ(t)σ(s)|s=0 = de Lσ(t) σ(s)|s=0 = de Lσ(t) ξ = ξ˜σ(t)
ds
ds
ds
and so σ is an integral curve of ξ˜ so σ = γξ̃ by uniqueness.
Q.E.D.
According to the theorem, γξ̃ (1) ∈ G is determined by ξ ∈ g.
Definition 6.10 We define expG : g → G by expG ξ = γξ̃ (1). This is called the exponential map of G.
Proposition 6.11 expG : g → G is a smooth map
Proof A smooth family of ODEs (jointly smooth in parameters and manifold variables) has solutions depending
smoothly on the parameters. Hence the ODE for ξ˜ depends linearly on ξ. Picking a basis ξ1 , ..., ξk for g we have
ξ = a1 ξ1 + ... + ak ξk and so we are looking for solutions of
γ̇a (t) = ξ˜γ(t) = (a1 ξ˜1 + ... + ak ξ˜k )γa (t)
where a = (a1 , ..., ak )
Q.E.D.
Proposition 6.12
γξ̃ (s) = exp(sξ)
Proof Consider t 7→ γξ̃ (st) and note this is smooth and γξ̃ (st1 )γξ̃ (st2 ) = γξ̃ (s(t1 + t2 )) and so γξ̃ (st) = γη̃ (t)
for some η and
d
d
η = γ̇ξ̃ (0) = γξ̃ (st)|t=0 = (st)|t=0 γ̇ξ̃ (0) = sξ
dt
dt
and therefore γξ̃ (xt) = γsξ
Q.E.D.
e (t) and then set t = 1 to conclude
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Corollary 6.13 exp(sξ) exp(tξ) = exp((t + s)ξ)
It is NOT true that exp ξ exp η = exp(ξ + η) in general.
Corollary 6.14 ξ˜ has integral curve through g at time t0 given by g exp((t − t0 )ξ)
Example 6.3
1. For Rn , and 1-PSG σ has the form σ(v) = vσ(1) and so exp v = v under the standard
identity T0 Rn = Rn
2. G = R? with g = R and then T1 R? = R with the identification
d
dt (1
+ tv)|t=0 ↔ v
?
For ξ ∈ R we want a left invariant extension to R . We use the coordinate functions x from R? ⊂ R, and
d
and we want to know when this is left invariant.
we then have ξ˜x = f (x) dx
d
ξ˜x = d1 Lx ξ = d1 Lx (1 + tξ)|t=0
dt
d
= Lx (1 + tξ)|t=0
dt
d
= (x + txξ)|t=0
dt
= (xξ)x
and so f (x) = xξ.
The integral curve σ(t) through 1 at time t = 0 will satisfy σ̇(t) = ξ˜σ(t) and be given by a function x(t) and
so ẋ(t) = x(t)ξ and so x(t) = Aeξt with x(0) = A = 1 and so exp ξ = x(1) = eξ .
3. Suppose G = GL(n, R) and g = Mn×n (R). Then ξ ∈ Mn×n (R) corresponds with the tangent vector
d
dt (In + tξ)|t=0 at In ∈ GL(n, R). We have
d
d
ξ˜g = dIn Lg (In + tξ)|t=0 = (g + tgξ)|t=0 = gξ g
dt
dt
where the over-line means directional derivative.
If g(t) is the integral curve through In at t = 0 then it satisfies
ġ(t) = ξ˜g(t) = g(t)ξ g(t)
and ġ(t) as a tangent vector is ġ(t)g(t) and the derivative is that of a function. Thus ġ(t) = g(t)ξ. given a
matrix ξ we form ξ 2 , ξ 3 , ... and hence we form
eξ = In +
∞
X
ξk
k=1
which is a smooth function in ξ with
d tξ
dt e
=0+
P∞
k=1
kt
k!
k−1 k
ξ
k!
= etξ ξ and so
d
d
d
(g(t)e−tξ ) = (g(t))e−tξ + g(t) (e−tξ ) = g(t)ξe−tξ + g(t)(−e−tξ ξ) = 0
dt
dt
dt
and so g(t)e−tξ is a constant A and so g(t) = Aetξ and for t = 0 we have In = A and so expGL(n,R) ξ = eξ .
4. We can show that, for example, if ξ ∈ o(n) then eξ ∈ O(n) and so expO(n) ξ = eξ .
Theorem 6.15 If F : G → H is a homomorphism of Lie groups and F? : g → h is the corresponding homomorphism of Lie algebras then
F (expG ξ) = expH (F? ξ)
Proof Consider t 7→ F (expG (tξ)) which is a 1-PSG of H, hence of the form expH (tη) for some η ∈ h given by
η=
d
F (expG (tξ))|t=0 = F? ξ
dt
Q.E.D.
Corollary 6.16 Example 4 above. Observe that if we have a matrix group, the inclusion map O(n) ,→ GL(n, R)
is a homomorphism of Lie groups whose derivative is the inclusion o(n) ,→ gl(n, R) of Lie algebras. Hence
expO(n) ξ = expGL(n,R) ξ for ξ ∈ o(n).
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MA4E0 Lie Groups Lecture Notes Autumn 2012
Example 6.4 det : GL(n, R) → R? is a homomorphism of Lie groups and det? = tr hence det(eξ ) = edet? ξ =
etrξ
exp : g → G is a smooth map with 0 ∈ g and e ∈ G with exp(0) = e. Then
d0 exp : T0 g → Te G
is a linear map. However T0 g = g (using directional derivatives) and also Te G = g and so we can view d0 exp :
g → g as a linear map. Then we have the following.
Theorem 6.17
d0 exp = Idg
Proof Given ξ ∈ g there is a curve in g, say γ(t) with γ(0) = 0 and γ̇(0) = ξ, for instance γ(t) = tξ. Then
d0 exp γ̇(0) =
d
d
exp(γ(t))|t=0 =
exp(tξ)|t=0 = ξ ∈ Te G
dt
dt
i.e. d0 exp ξ = ξ.
Q.E.D.
Corollary 6.18 There are hence neighbourhoods U of 0 ∈ g and V of e ∈ G such that V = expG U and
exp U → V is a diffeomorphism.
Proof Inverse function theorem since Idg is an isomorphism
Q.E.D.
Corollary 6.19 If G is connected then every element is a finite product of exponentials, i.e. g ∈ G =⇒ g =
exp ξ1 ... exp ξN for ξi ∈ g.
Proof Any open set around e generates G. Take one of the form V = exp U where exp : U → V is a
diffeomorphism.
Q.E.D.
Corollary 6.20 If G is a connected Lie group and Fi : G → H for i = 1, 2 are homomorphisms of Lie groups
such that Fi? : g → h are equal, i.e. F1? = F2? then F1 = F2
Proof Take g ∈ G then by the previous corollary g = exp ξ1 ... exp ξN and so
F1 (g) = F1 (exp(ξ1 )...F1 (exp(ξN ))
= exp(F1? (ξ1 ))... exp(F1? (ξN ))
= exp(F2? (ξ1 ))... exp(F2? (ξN ))
= F2 (exp(ξ1 )...F2 (exp(ξN ))
= F2 (g)
Q.E.D.
Let exp : U → V be a diffeomorphism and exp−1 : V → U be the inverse function. Then this is a diffeomorphism fromPa neighbourhood e 3 V ⊂ G into U ⊂ g. If we pick a basis ξ1 , ..., ξN for g then for g ∈ V we have
N
exp−1 g = i=1 xi (g)ξi and this will define a smooth function on V which are the components of a smooth map
into Rn , ψ : V → Rn which is a diffeomorphism onto an open set in Rn . (V, ψ) is a chart on G compatible with
the atlas and these charts depend on the choice of open set U in g and the basis of g. These are all compatible.
If g ∈ G and then g ∈ gV so we can left translate to any element. Then (gV, ψ ◦ Lg−1 ) will be a chart around
g. These are all compatible, so we get an atlas for G built from the exponential map (problem sheet 6).
Any real vector space is a group under addition so the Lie algebra of a Lie group is itself a Lie group. When
then is exp : g → G a homomorphism of Lie groups, i.e. exp(ξ + η) = exp(ξ) exp(η).
Proposition 6.21 Let G be connected Lie group and then exp : g → G is a homomorphism of Lie groups if and
only if G is abelian.
Proof Suppose exp : g → G is a homomorphism. Then exp(ξ) exp(η) = exp(ξ + η). G is connected and so G is
generated by exponentials and so G is abelian.
Now suppose that G is abelian and connected. Then t 7→ exp(tξ) exp(tη) is a 1-PSG of G therefore
exp(tξ) exp(tη) = exp(tζ) for some ζ ∈ g. We have
ζ=
d
exp(tξ) exp(tη)|t=0 = ξ + η
dt
using the lemma below. Thus exp is a homomorphism of Lie groups.
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Q.E.D.
MA4E0 Lie Groups Lecture Notes Autumn 2012
Lemma 6.22 If f : (a, b) → M is given by a function F : (a, b) × ... × (a, b) → M where f (t) = F (t, t, ..., t) and
PN
0 ∈ (a, b) then f (0) = i=1 fi (0) where fi (t) = F (0, ..., t, ..., 0) with the t in the i-th spot.
Proof The chain rule for f = F ◦ ∆ where ∆ is the diagonal map, ∆(t) = (t, t, ..., t).
Q.E.D.
Corollary 6.23 IF g ∈ G is a connected and abelian Lie group then exp : g → G is surjective.
Proof IF g ∈ G then g = exp(ξ1 )... exp(ξN ) = exp(ξ1 + ... + ξN )
n
n
k
Examples of connected abelian Lie groups are R , T , R × T
n−k
Q.E.D.
of dimension n.
Theorem 6.24 Up to isomorphism, the above list are the only connected abelian Lie groups.
Lemma 6.25 Let Γ ⊂ Rn be an additive subgroup which is closed and for which there is an open ball B around
0 with Γ ∩ B = {0}. Then either Γ = {0} or there are k linearly independent vectors v1 , ..., vk ∈ Rn such that
Pk
Γ = { i=1 mi vi |mi ∈ Z}
Proof We proceed by induction. Take n = 1. Then if Γ 6= {0} there exists a non zero element v ∈ Γ of minimal
length. Since Γ is closed and ||v|| ≥ R (radius of B), we claim that Γ = Zv. We know that 0 6= w ∈ Γ then
w = λv for λ ∈ R and assume that λ > 0, and then we can write λ = p + r for p ∈ N and 0 ≤ r < 1. Then
v ∈ Γ =⇒ pv ∈ Γ and therefore w − pv ∈ Γ and now
||w − pv|| = r||v|| < ||v||
and so, since v was minimal, we have that r = 0 and so w = pv ∈ Zv.
The general case.
Suppose we know the result for n − 1 and Γ 6= {0}. Then there is some non-zero v1 ∈ Γ of minimal length.
Then Rv1 ⊃ Rv1 ∩ Γ is the case of n = 1 and so Rv1 ∩ Γ = Zv1 .
Consider Rn /Rv1 ⊃ Γ/Zv1 . We need to show that Γ/Zv1 satisfies the same assumptions as Γ. If it doesn’t
then there is a non zero element sequence in Γ/Zv1 tending to 0. Using coset representation there is a sequence
γm ∈ Γ such that γm 6∈ Zv1 and γm + Zv1 → 0 in Rn /Rv1 , i.e. γm + Rv1 → 0 in Rn /Rv1 .
To tend to zero in the quotient Euclidean space means we can find a sequence in lm ∈ Z such that γm −lm v1 →
0 in Rn . Let lm = hm + rm for some hm ∈ Z, and rm ∈ [0, 1]. Then γm − lm v1 = (γm − hm v1 ) − rm v1 . Then
sequence rm ∈ [0, 1) ⊂ [0, 1] and the latter is compact and so rm has a convergent subsequence. Hence we can
suppose we have a sequence γm and coset representation γm such that there is a sequence rm ∈ [0, 1] convergent
with γm − rm v1 → 0 in Rn , but γm ∈ Rv1 . Let r = lim rm . γm is convergent in Rn to rv1 . Γ is a closed subgroup
of Rn and so rv1 ∈ Γ. Therefore rv1 ∈ Zv1 ∩ [0, 1]v1 and so r = 0 or r = 1.
If r = 0 then γm → 0 in Rn .
If r = 1 then γm − v1 → 0 in Rn and both are sequence in Γ. Since Γ ∩ B = {0} there is an M such that
m ≥ M implies that γm = 0 or γm − v1 = 0. But γm is a representative of a non-zero coset.
Hence there exists a B 0 in Rn /Rv1 with B 0 ∩ Γ/Zv1 = {0}. Then by induction there exist v2 , ..., vk in Rn /Rv1
such that v2 + Rv1 , ..., vk + Rv1 are linearly independent and integer combinations of v2 + Zv1 , ..., vk + Zvk give
Γ/Zv1 .
Q.E.D.
Proof (of theorem 6.24) G connected abelian implies that exp : g → G is a homomorphism and exp is a
surjective map. Let Γ = ker exp which is a subgroup of g which is closed and since exp is a diffeomorphism near
0 ∈ g then there exists an open U ⊂ g with 0 ∈ U and Γ ∩ U = {0}. It is enough to study such subgroups γ of
Rn .
We thus know from the above lemma that there exist v1 , ..., vk linearly independent such that Γ = Zv1 + ... +
Zvk . Put g1 = spanR {v1 , ..., vk } which has dimension k. Pick g2 ⊂ g such that g = g1 ⊕ g2 and then Γ ⊂ g1 and
exp : g2 → G will be an isomorphism onto its image G2 ⊂ G. Put G1 = exp(g1 ) which is a subgroup of G and
G = G1 × G2 . G2 ∼
Q.E.D.
= Rn−k and G1 ∼
= g/Γ ∼
= Rk /Zk ∼
= (R/Z)k = T k and so G ∼
= T k × Rn−k .
Corollary 6.26 Any connected compact Abelian Lie group is isomorphic to a torus.
7
Lie Subgroups
Definition 7.1 A smooth map of manifolds f : M → N is an immersion if
dx f : Tx M → Tf (x) N
has zero kernel, i.e. it is injective, for all x ∈ M.
A smooth map of manifolds f : M → N is a submersion if
dx f : Tx M → Tf (x) N
is surjective for all x ∈ M. A smooth map of manifolds f : M → N is a diffeomorphism if it is a bijection and
both f and f −1 are smooth.
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MA4E0 Lie Groups Lecture Notes Autumn 2012
A diffeomorphism is both an immersion and a submersion. Something which is both an immersion and a
submersion is locally a diffeomorphism. Locally an immersion looks like an axis in a product, e.g. M ,→ M × N .
A submersion is locally a projection, e.g. M × N → M.
Definition 7.2 A submanifold of a manifold M is a smooth map f : M → N which is an injective immersion,
i.e. both f and df are injective.
Thus crossings are not allowed in a submanifold. We often view N ⊂ M and Tx N ⊂ Tx M.
Definition 7.3 A submanifold f : N → M is called embedded if f is a homeomorphism to f (N ) with the
relative topology.
Example 7.1
1. Linear subspaces of a finite dimensional vector space are embedded submanifolds
2. R → T 2 by t 7→ (eiat , eibt ) is an immersion so long as (a, b) 6= (0, 0) and injective when there is no common
2πm
and a common period would be
period for eiat and eibt . The periods of these are t1 = 2πn
a and t2 =
b
b
when t1 = t2 . For this we would need a ∈ Q. When it isn’t, then t 7→ (eiat , eibt ) is injective. This is a
submanifold and sits R as a subgroup of T 2 , a line of irrational slope. This is dense.
3. To see S 1 as a subgroup of T n we send z 7→ (z m1 , ..., z mn ). So long as there are no common divisors of
m1 , ..., mn then this will be injective and an immersion. Thus S 1 is a submanifold of T n and is embedded.
Definition 7.4 If G is a Lie group then a Lie submanifold f : H → G is a smooth map of manifolds where
H is a Lie group and f is a homomorphism.
The above uses the properties, in Lie group terms of
1. H and G are Lie groups.
2. f : H → G is a homomorphism of Lie groups.
3. f is injective.
4. dh f : Th H → Tf (h) G is injective.
Given ξ ∈ h, deH Lh ξ ∈ Th H and all arise in this way. Then observe that
(f ◦ LH )(h0 ) = f (hh0 ) = f (h)f (h0 ) = Lf (h) f (h0 ) = (Lf (h) ◦ f )(h0 )
and so we get that
dh f (deH Lh ξ) = deH (f ◦ Lh )(ξ) = deH (Lf (h) ◦ f )(ξ) = deG Lf (h) deH f (ξ) = deG Lf (h) (f? (ξ))
Both of deG Lf (h) and deH Lh are invertible and so dh f is injective if and only if f? is injective. Thu f is an
immersion if and only if f? : g → h is injective.
Proposition 7.5 A homomorphism of Lie groups f : H → G is an immersion of manifolds if and only if f and
f? are injective.
This then characterises Lie subgroups as homomorphisms f with both f and f? injective. Then f (H) is a
subgroup of G and f? (h) is a Lie subalgebra of g.
Theorem 7.6 If f : H → G is a Lie subgroup then
f? (h) = {ξ ∈ g| expG (tξ) ∈ f (H) for t in some open interval }
Proof Firstly f? (h) = {ξ ∈ g|...} since if η ∈ h put ξ = f? (η) and then expG (tξ) = expG (tf? (η)) = f (expH (tη)) ∈
f (H) for all t.
We prove the converse, suppose ξ ∈ g has expG (tξ) ∈ f (H) for all t ∈ (a, b). For t ∈ (a, b) there exists
h(t) ∈ H with f (h(t)) = expG (tξ).
We show that h(t) is a smooth curve in H. We do this by (locally ) finding a solution map for f (h0 ) = expG ξ 0
for ξ 0 near ξ in f? (h).
f? (h) is a linear subspace of g so we can find a subspace m ⊂ g such that g = f? (h) ⊕ m and then H × m is a
manifold of the same dimension as G.
Pick t0 ∈ (a, b) and consider Ψt0 : H × m → G given by
Ψt0 (h, ζ) = expG (t0 ξ)f (h) expG (ζ)
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MA4E0 Lie Groups Lecture Notes Autumn 2012
and note Ψt0 is smooth.
d
Ψt (expH sη, sρ)|s=0
ds 0
d
=
expG (t0 ξ)f (expH (sη))expG (sρ))|s=0
ds
= deG LexpG (t0 ξ) (f? (η) + ρ))
d(eH ,0) Ψt0 (η, ρ) =
therefore d(eH ,0) Ψt0 is surjective onto TexpG (t0 ξ G and the domain is h ⊕ m which has the same dimension so
d(eH ,0) Ψt0 is a linear isomorphism.
Therefore Ψt0 is a diffeomorphism near (eH , 0) and therefore there exist open sets U1 ⊂ H and U2 ⊂ m such
that if V = Ψt0 (U1 × U2 ) then V is open in G, eH ∈ U1 , s ∈ U2 and expG (tξ) ∈ V and Ψt0 is a diffeomorphism
of U1 × U2 onto V .
−1
Ψ−1
t0 : V → U1 ×U2 will be smooth and we put Φt0 = π1 ◦Ψt0 and then Φt0 is a smooth map G ⊃ V → U1 ⊂ H.
0 0
0 0
There is an interval around t0 ∈ (a , b ) such that (a , b ) ⊂ (a, b) and expG (tξ) ∈ V . Then t ∈ (a0 , b0 ) we have
h(t) = Φt0 (expG (tξ)) and so h(t) is smooth on (a0 , b0 ). t0 is arbitrary and so h(t) is smooth on (a, b).
Pick c ∈ (a, b) such that ḣ(c) ∈ Th(c) H and so η = dh(c) Lh(c)−1 (h(c)) ∈ h and then
f? (η) = f? (dh(c) Lh(c)−1 (h(c))) = (df (h(c)) Lf (h(c)−1 ) ) ◦ (ff (h(c)) f )(ḣ(c))
but
d
(expG (tξ))|t=c = deG LexpG (cξ) (ξ)
dt
df (h(c)) f (ḣ(c)) = f (h(c))· =
and f (h(c)−1 ) = f (h(c))−1 = expG (−cξ) and so
f? (η) = ξ
Q.E.D.
This means that f? (h) = {ξ ∈ g| expG (tξ) ∈ f (H)∀t} and f (expH (tη)) = exp tf? (η). This means expH is
determined by f? and expG hence an atlas for H is determined by f? and the topology of H.
Proposition 7.7 Let H be a subgroup of a Lie group G and h ⊂ g a subspace such that we have open sets U of
s ∈ g and V of eG ∈ G with expG : U → V a diffeomorphism such that expG (U ∩ h) = V ∩ H. Then H with the
relative topology has an atlas built using exp−1
G on V ∩ H and left translating making H into a Lie subgroup of
G.
Lemma 7.8 Let G be a Lie group, ξ, η ∈ g, then there exists a ζ(t) ∈ g for |t| < δ for some δ > 0 such that
exp(tξ) exp(tη) = exp(tξ + tη + ζ(t))
for |t| < δ, and ζ(t)/t2 → 0 as t → 0.
Proof Take U ⊂ g, 0 ∈ U such that exp : U → V = exp(U ) is a diffeomorphism then choose δ such that
exp(tξ) exp(tη) ∈ V for |t| < δ. Then set
ζ(t) = exp−1 (exp(tξ) exp(tη) − tξ − tη)
for |t| < δ. Then ζ is a smooth function of t and ζ(0) = 0. Also
exp(tξ) exp(tη) = exp(tξ + tη + ζ(t))
and differentiating at t = 0 gives
ξ + η = ξ + η + ζ̇(t)
and so ζ̇(0) = 0. ζ is smooth in t around t = 0 and so we have a Taylor expansion with remainder
1
ζ(t) = ζ(0) + tζ̇(0) + t2 R(t)
2
where R(t) is bounded as t → 0. clearly R(t) =
ζ(t)
t2 .
Q.E.D.
Corollary 7.9 If ξ, η ∈ g then
exp(ξ + η) = lim
N →∞
exp
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ξ
N
exp
η N
N
MA4E0 Lie Groups Lecture Notes Autumn 2012
Proof When N ≥
1
δ
we have
exp
ξ
N
ξ
N
exp
and so
exp
and N ζ
1
N
exp
η
ξ
η
1
= exp
+
+ζ
N
N
N
N
η N
1
= exp ξ + η + N ζ
N
N
→ 0 as N → ∞.
Q.E.D.
Theorem 7.10 Let H be a closed subgroup of a Lie group G then H with the relative topology has a unique
manifold structure making the inclusion map i : H →,→ G a Lie subgroup.
Proof Take a subset of g which would be a Lie algebra of H if the result were true, i.e. set
h = {ξ ∈ g| expG (tξ) ∈ H∀t ∈ R}
We show h is a linear subspace:
Take ξ ∈ h adn c ∈ R, and consider t 7→ expG (t(cξ)) = expG ((ct)ξ) ∈ H for all t and so cξ ∈ h.
Take ξ, η ∈ h and consider
η N
ξ
∈ H for all N
expG
expG
N
N
and this converges in G as N → ∞. Since H is closed the limit expG (ξ + η) is in H. Replacing ξ by tξ and η by
tη results in expG (tξ + tη) ∈ H for all t and so ξ + η ∈ h.
We now prove that conditions for proposition 7.7 hold for this subspace, i.e. there exists open sets U ⊂ g and
V = expG (U ) such that expG : U → V is a diffeomorphism and expG (U ∩ h) = expG (U ) ∩ H.
We proceed by contradiction. Suppose no such U exists, i.e. we never have expG (U ∩ h) = expG (U ) ∩ H for
a U ⊂ g.
Fix W ⊂ g where expG : W → expG (W ) is a diffeomorphism, and fix a norm on g. Consider open balls B k1 (0)
for large enough k ∈ N to ensure B k1 (0) ⊂ W . We have
expG (B k1 (0) ∩ h) 6= expG (B k1 (0)) ∩ H
If ξ ∈ B k1 (0) ∩ h then expG ξ ∈ expG (B k1 (0)) ∩ H and so expG (B k1 (0)) ∩ H is strictly larger than expG (B k1 (0) ∩ h)
Let hk ∈ H ∩ expG (B k1 (0)), but hk 6∈ expG (B k1 (0) ∩ h). Then hk = expG ξk for ξk ∈ B k1 (0) and so (ξk ) → 0
in g as k → ∞ and therefore (hk ) → e in G.
We claim that hk 6∈ expG (W ∩ h). Otherwise hk = expG ηk with ηk ∈ W ∩ h and hk = expG ξk with ξk ∈
B k1 (0) ⊂ W . Due to bijectivity of expG on W we have ηk = ξk and so ηk ∈ B k1 (0)∩h and so hk ∈ expG (B k1 (0)∩h)
which contradicts our choice of hk , and so the claim holds.
Thus we have a sequence hk in H with (hk ) → e in G and an open set W1 = W ∩ h containing 0 in h but
hk 6∈ expG (W1 ) for all k large.
Pick a subspace m ⊂ g supplementary to h in g making g = h ⊕ m. Define α : h × m → G by
α(ξ, η) = expG (ξ) expG (η)
Then α is smooth and has derivative at (0, 0) given by
d(0,0) α(ξ, η) = ξ + η
which is the identification of h ⊕ m with g and so is a linear isomorphism. Hence α is a diffeomorphism near
(0, 0) so we have open sets Wh and Wm around the origin in h and m respectively so that α is a diffeomorphism
of Wh × Wm with α(Wh × Wm ) which is an open set in G containing e.
We claim(to prove the claim above!!) that we can shrink Wm until H and expG (Wm ) meet only at e.
Suppose not, then we can find a sequence ηk in Wm so that expG ηk ∈ H \ {e} and ηk → 0. Put tk = ||ηk || > 0
and then ηk /tk is a vector of norm 1 for each k and hence we have a sequence ηk /tk in the unit sphere in Wm
which is compact, and so this sequence has a convergent subsequence. We throw away the other terms and
relabel the subsequence ηk /tk .
Let η be the limit of this sequence. Let t > 0, nk (t) = bt/tk c and so
t/tk − 1 ≤ nk (t) ≤ t/tk
and thus we have that
t − tk ≤ tk nk (t) ≤ t
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MA4E0 Lie Groups Lecture Notes Autumn 2012
but since tk → 0 as k → ∞ we have tk nk (t) → t as k → ∞.
Multiplying the left side by ηk /tk and the right side by η we obtain
nk (t)ηk → tη
as k → ∞. Applying expG we have that
lim expG (nk (t)ηk ) = expG (tη)
k→∞
and so
lim (expG (ηk ))nk (t) = expG (tη)
k→∞
but expG (ηk ) ∈ H =⇒ (expG (ηk ))nk (t) ∈ H =⇒ expG (tη) ∈ H and also from group property of H,
expG (−tη) ∈ H for t > 0. Thus expG (tη) ∈ H for t ∈ R and thus ηinH but since η 6= 0 and η ∈ m we have a
contradiction. Thus the claim holds, i.e. we can shrink Wm so that expG (Wm ) ∩ H = {e}.
Now to conclude the proof of the first claim (using the second claim!!). Since hk → e in G and expG (Wh ×Wm )
is open and contains e then hk ∈ expG (Wh × Wm ) for all k ≥ k0 , and hence
hk = expG (ξk0 ) expG (ηk0 )
with ξk0 ∈ Wh and ηk0 ∈ Wm , but hk 6∈ expG W1 and so ηk0 6= 0 for all k. Also expG ηk0 ∈ H since hk , expG ξk0 ∈ H
and then we have expG ηk0 ∈ expG Wm ∩ H = {e} and so ηk0 = 0 This is a contradiction.
Thus the first claim is established, i.e. we can find U open in g and V = expG U such that expG : U → V is
a diffeomorphism and expG (U ∩ h) = V ∩ H.
By proposition 7.7 we then have a manifold structure on H making it into a Lie group with Lie algebra h.
Then i : H ,→ G is a Lie subgroup.
Q.E.D.
Corollary 7.11 H has Lie algebra h = {ξ ∈ g| expG (tξ) ∈ H∀t}
The following is an example of how this is used
Theorem 7.12 If F : G → H is a homomorphism of Lie groups then
ker(F ) = {g ∈ G|F (g) = eH }
is a Lie subgroup of G with Lie algebra given by ker(F? )
Proof ker(F ) = F −1 ({eH }) and by continuity of F and Hausdorff property of H and G we have this set is
closed, and it is also a subgroup of G. Thus it is a Lie subgroup of G with Lie algebra
{ξ ∈ g| expG (tξ) ∈ ker(F )∀t} = {ξ ∈ g|F (expG (tξ)) = eH ∀t}
but since F (expG (tξ)) = expH (tF? (ξ)) we have F? (ξ) = 0 as required.
Q.E.D.
Example 7.2 det : GL(n, R) → R? and then ker det = SL(n, R) and det? = tr and so sl(n, R) = {ξ ∈
gl(n, R)|trξ = 0}
Theorem 7.13 If σ : G → G is an automorphism, then the fixed point set Gσ = {g ∈ G|σ(g) = g} is a Lie
subgroup with Lie algebra gσ?
Example 7.3
1. G = GL(n, R) with σ(g) = (g −1 )T is an automorphism and Gσ = O(n), and g σ? = {ξ ∈
Mn×n (R)| − ξ T = ξ} = An (R).
2. G = GL(n, C) with σ(g) = (g −1 )T then Gσ = U (n).
3. R2n = Rn ⊕ Rn and G = GL(2n, R) and write 2n × 2n matrices as 2 × 2 matrices of n × n blocks.
0 −In
Jn =
and so Jn2 = −In and define σ1 (g) = Jn gJn−1 , which is an inner automorphism. Then
In
0
U −V
Gσ1 = {
|U + iV ∈ GL(n, C)} and so GL(n, C) is a Lie subgroup of GL(2n, R). If instead
V
U
σ2 (g) = Jn (g −1 )T Jn−1 then Gσ2 = Sp(2n, R).
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MA4E0 Lie Groups Lecture Notes Autumn 2012
8
Continuous homomorphisms are smooth
The basic idea is to look at continuous 1-PSG of Rn . These are linear σ(t) = tv with v = σ(1), and hence smooth.
Lemma 8.1 If G is a Lie group and H is a subgroup then the topological closure H̄ is a subgroup.
Proof Take g, h ∈ H̄ and suppose gk → g and hk → h for sequences hk , gk ∈ H. Then
gh = m(g, h) = lim m(gk , hk )
k→∞
and m(gk , hk ) ∈ H and so gh ∈ H̄. Continuity of the inverse gives i(H̄) ⊂ H̄ so H̄ is a subgroup.
Q.E.D.
Lemma 8.2 If σ : R → G is a continuous 1-PSG then σ(R) is a connected closed abelian Lie subgroup of G.
Proof By the previous lemma, it is a closed subgroup, and hence a Lie subgroup. σ(R) is connected henve σ(R)
is connected. σ(R) is abelian and so σ(R) is abelian.
Q.E.D.
Corollary 8.3 If σ : R → G is a continuous 1-PSG then σ is a continuous 1-PSG of σ(R). Thus it is enough
to prove that σ is smooth when G is a connected abelian Lie group.
Theorem 8.4 Any continuous 1-PSG of a Lie subgroup is smooth
Proof If G is connected and abelian then G is isomorphic to Rk × T n−k for some k and n = dim G.
Since maps into products are continuous or smooth or homomorphism if and only if the components are
continuous or smooth or homomorphisms we need only consider the cases where G = R and G = S 1 .
Continuous homomorphisms R → R are smooth since they are linear. for R → S 1 let σ : R → S 1 ⊂ C be
continuous and σ(s) + σ(t) = σ(s + t) as a complex valued function. Use the result from complex function theory
so there exists a unique g : R → R continuous with σ(t) = eig(t) with g(0) = 0. We then have
ei(g(s)+g(t)) = eig(s+t)
and so
g(s + t) − g(s) − g(t) ∈ 2πZ
so we have a continuous map R × R → 2πZ and so the image must be connected which must contain the image
of (0, 0) which is 0. Hence
g(s + t) − g(s) − g(t) = 0
for all s, t. This implies g(t) = tg(1) so σ is smooth.
Q.E.D.
Theorem 8.5 If G and H are Lie groups and F : G → H is a continuous homomorphism then F is smooth.
Proof Let expG (tξ) be a smooth 1-PSG of G. Then F (expG (tξ)) is a continuous 1-PSG of H and by theorem
8.4 is smooth.
We claim that F is smooth on G if and only if it is smooth on an open set V around eH .
If F is smooth on V and g ∈ G then gV is an open set around g. If g 0 ∈ gV then g −1 g 0 ∈ V and
F (g 0 ) = F (gg −1 g 0 ) = F (g)F (g −1 g 0 ) = LF (g) ◦ F ◦ Lg−1 (g 0 )
which is a composition of smooth maps hence is smooth. Thus the claim is established.
Take a basis ξ1 , ..., xın for g and consider α : Rn → G defined by α(t1 , ..., tn ) = expG (t1 ξ1 )... expG (tn ξn ). Let
h = (h1 , ..., hn ) ∈ Rn and then
d
d
α(0 + th)|t=0 =
expG (th1 ξ1 )... expG (thn ξn )|t=0 = h1 ξ1 + ... + hn ξn
dt
dt
and so d0 α is the identification of Rn with g given by a basis, therefore d0 α is a linear isomorphism and hence α
is n onto a neighbourhood V of eG in G. For (t1 , ..., tn ) ∈ U ,
F (α(t1 , ..., tn )) = F (expG (t1 ξ1 )... expG (tn ξn )) = F (expG (t1 ξ1 ))...F (expG (tn ξn ))
which is smooth in (t1 , ..., tn ) by theorem 8.4. Therefore F is smooth on V , and hence smooth everywhere.
Q.E.D.
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MA4E0 Lie Groups Lecture Notes Autumn 2012
9
Distributions and Frobenius Theorem
Definition 9.1
1. A k-dimensional tangent distribution D on a manifold M is a k-dimensional subspace
Dx ⊂ Tx M for each x ∈ M.
2. We say D is smooth if each point in M has a neighbourhood U with k-smooth vector fields X1 , ..., Xk ∈ X (U )
with (X1 )x , ..., (Xk )x forming a basis at each x ∈ U for Dx .
3. A vector field X on M is said to belong to D if Xx ∈ Dx for all x ∈ M.
4. D is said to be involutive if the set of vector fields belonging to D is a Lie subalgebra of X (M).
5. The set of smooth vector fields belonging to D is denoted by Γ(D).
We can rewrite involutive to be [Γ(D), Γ(D)] ⊂ Γ(D).
Example 9.1
1. M = Rn and put Dx = {X ∈ Tx Rn |X(tk+1 ) = ... = X(tn ) = 0} = {X ∈ Tx Rn |X =
Pk
∂
∞
n ∂
∞
n ∂
i=1 ai ∂ti |x }. Then Γ(D) = C (R ) ∂t1 + ... + C (R ) ∂tk and so D is smooth and involutive.
2. G a Lie group, h ⊂ g a k-dimensional linear subspace. Put Dg = de Lg (h) ⊂ Tg G. Then D is a kdimensional distribution and if ξ1 , ..., ξk is a basis for h then Dg = span(ξ˜1,g , ..., ξ˜k,g ) and so D is smooth.
^
˜ ˜
We have ξ˜i ∈ Γ(D) and so D is involutive so [ξ˜i , ξ˜j ] ∈ Γ(D) but [ξ
i , ξj ] = [ξi , ξj ] and so [ξi , ξj ] ∈ h so h is a
Lie subalgebra of g. The converse is also true.
Hence left invariant involutive smooth distributions are in bijection with Lie subalgebras.
Definition 9.2 Let D be a smooth k-dimensional tangent distribution on M, then an integral submanifold
for D is a k-dimensional submanifold (N , φ) with φ : N → M with N a k-dimensional connected manifold and
φ an immersion of N into M such that dy φ(Ty N ) = Dφ(y) for all y ∈ N .
Note that we have a notion of maximal integral submanifold, analogous to maximal integral curves of a vector
field.
Proposition 9.3 If D is a smooth distribution on M such that for each x ∈ M there is an integral submanifold
φ : N → M with x ∈ φ(N ) then D is involutive.
Proof Let X, Y ∈ Γ(D), x ∈ M, φ : N → M with y ∈ N and φ(y) = x and (, φ) an integral submanifold. Then
there exists X̃, Ỹ on N with dy0 φ(X̃y0 ) = Xφ(y0 ) and dy0 φ(Ỹy0 ) = Yφ(y0 ) and so near y, X̃ and ˜(Y ) are φ-related
to X and Y respectively. This implies [X̃, Ỹ ] and [X, Y ] are φ-related and therefore [X, Y ]x = [X, Y ]φ(Y ) =
dy φ([X̃, Ỹ ]y ) ∈ Dx . Since x is arbitrary, [X, Y ] ∈ Γ(D)
Q.E.D.
Theorem 9.4 (Frobenius) Let D be a smooth involutive k-dimensional tangent distribution on a manifold M
and then each x ∈ M lies on a locally unique integral submanifold.
Definition 9.5 A maximal integral submanifold (or leaf ) (N , φ) of an involutive distribution D on a manifold
M is a connected integral submanifold whose image is not a proper subset of any other integral submanifold.
Theorem 9.6 If G is a Lie group, h ⊂ g a Lie subalgebra then there is a Lie subgroup (H, φ) of G with
deH φ(TeH H) = h and deH φ : TeH H → h an isomorphism of Lie algebras.
Proof If G is a Lie group, h ⊂ g a Lie subalgebra, and D the left invariant extension of h then D is involutive
so we let (H, φ) be the corresponding leaf of D with eG ∈ φ(H).
We know φ : H → G is smooth, injective with injective differential. We need to show H is a group in a way
which is smooth and making φ a Lie group homomorphism.
First step is to show φ(H) is a subgroup of G. Take g ∈ φ(H). The left invariance of D means left translates
of integral manifolds are integral manifolds. Consider (H, Lg−1 ◦ φ) which is a submanifold of G and an integral
submanifold. Also e = Lg−1 (g) ∈ Lg−1 (φ(H)) so Lg−1 (φ(H)) = φ(H) hence φ(H) is a subgroup of G.
We make H into a group by defining φ(h1 h2 ) = φ(h1 )φ(h2 ). We claim this makes H into a Lie subgroup.
Pick a supplement m for h in g, so g = h ⊕ m and define Φ : H × m → G by
Φ(h, ξ) = φ(h) exp ξ
Then we have
d(h,0) Φ(X, η) =
d
d
Φ(γ(t), 0 + tη)|t=0 = φ(γ(t)) exp(tη)|t=0 = dh φ(x) + de Lφ(h) η
dt
dt
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MA4E0 Lie Groups Lecture Notes Autumn 2012
and both maps have zero kernel so d(h,0) Φ has zero kernel as a linear map Th H ⊕ m → g and hence d(h,0) Φ is an
isomorphism.
Hence by inverse function theorem there is a neighbourhood U h of h in H, W h of 0 in m and V h of φ(h) in
G with V h = Φ(U h × W h ) and Φ : U h × W h → V h a diffeomorphism.
Let Φ−1 on V h be given by maps ψ1h : V h → U h and ψ2h : V h → W h . These are smooth maps and
ψ1h (φ(h0 ) exp 0) is the first component of Φ−1 (φ(h0 ) exp 0) = Φ−1 (Φ(h0 , 0)) = (h0 , 0) and so
ψ1h (φ(h0 )) = h0
for h0 ∈ U h . Let h1 , h2 ∈ H then for h01 near h1 and h02 near h2 then
h0 h0
h0 h0
h01 h02 = ψ1 1 2 (φ(h01 h02 )) = ψ1 1 2 (φ(h01 )φ(h02 ))
and this is a composition of smooth functions. Then
h0 h0
mH (h01 , h02 ) = ψ1 1 2 (mH (φ(h01 ), φ(h02 )))
on (U h × U h ) ∩ (mG ◦ φ × φ)−1 (U h1 h2 ). This is an open set and contains h1 h2 so mH is smooth on H × H.
Similarly iH : H → H is smooth. Let φ(eH ) = e and so deH (TeH H) = h = De .
Q.E.D.
An application of this is that every Lie algebra (real, finite dimensional) is the Lie algebra of a Lie group, up
to isomorphism. The method to show this uses representation theory.
Definition 9.7 A representation of a Lie algebra g is a homomorphism σ : g → gl(n, R) where n is called the
dimension of the representation.
σ([ξ, η]) = σ(ξ)σ(η) − σ(η)σ(ξ)
Note gl(n, R) = Mn×n (R) ⊂ Mn×n (C).
Definition 9.8 A representation is faithful if ker σ = {0}.
Thus for a faithful representation g is isomorphic to a Lie subalgebra of gl(n, R).
Theorem 9.9 (Ado) Every finite dimensional real Lie algebra has a faithful representation.
A consequence of this is every finite dimensional Lie algebra g is a Lie subalgebra of gl(n, R) for some n.
Hence there is a Lie subgroup φ : H → GL(n, R) whose Lie algebra h is isomorphic to g. Thus H is a Lie group
whose Lie algebra is g. Lie subgroups of GL(n, R) are called linear Lie groups. For example all classical matrix
groups are linear. Also note SL(2n, R) has the same Lie algebra as many non-linear Lie groups.
10
Lie Group topics
• Actions of Lie groups. G acts on M becomes a map σ : G × M → M by σ(g, x) = g · x is a smooth
map. G is a symmetry group of M. For example rotations are the symmetry group of sphere. O(n + 1)
acting on S n is the symmetry group of spherical geometry. This is an example of a homogeneous geometry.
S n = O(n + 1)/O(n).
• Integration on Lie groups
• Representation theory
• Internal symmetry groups.
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