ma4e0 lie groups - university of warwick...ma4e0 lie groups lecture notes autumn 2012 0 topological...

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 MatthewEgginton.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 [email protected]

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→ Gdefined by m(g1, g2) = g1g2 and inversion becomes i : G→ G defined by i(g) = g−1. If we suppose that there isa 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. S1 = (x, y) ∈ R2 : x2 + y2 = 1 = z ∈ C : |z| = 1 with m(z1, z2) = z1z2 and i(z) = z. Note that this iscompact, connected and commutative

4. If G1, G2 are topological groups then G1 ×G2 with the product topology is a topological group.

5. T k = S1 × ...× S1

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 elementgenerates 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.

ProofLg1 Lg2

(h) = g1g2h = Lg1g2(h)

and alsoLe(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 = AkA = 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⋃i∈ZA

k 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 openset 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 ∪Uk is an open subsetof G and i fG is connected then ∪Uk = G.

Proof Uk is open since it can be written as ⋃g∈Uk−1

Lg(U)

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 ⊂ Uk for all k and so ∪V k ⊂ ∪Uk. 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 = ∪gH 6=HgH is open in G. Since G is connected we

have G \H = ∅. Since G ⊂ ∪Uk ⊂ G we have G = ∪Uk. Q.E.D.

Examples include R? ⊃ R+ and the latter is an open subgroup and R? = R+ ∪ (−1)R+. and so R+ is closedas well. Any closed subgroup of R+, eg 1 or 2kk ∈ Z.

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1 Manifolds

Definition 1.1 A chart (U, φ) of dimension n on an open set U of a topological space M is a homeomorphismφ of U onto an open set in Rn.

We get functions x1, ..., xn on U such that, for p ∈ U φ(p) =

x1(p)...

xn(p)

. These are called the coordinate

functions of the chart.Examples includeM = 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 S1 with the angle, or with stereographicprojection 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 classCk 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 Ck as maps fromopen sets of Rn.

Ck means, for k > 0 that all partial derivatives of orders up to k are continuous, for k = 0 that the functionsare 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

antipodal points and set the centre of the circle to be the origin. Then we have that y1+x = φ(x,y)

2 by considering

similar triangles. Thus we get that x = 4−t24+t2 and y = 4t

4+t2 , and thus we have found φ−1(t). Also we have thatψ(x,y)

2 = y1−x and so ψ φ−1(t) = 4

t by direct calculation. This is a map from φ(U ∩ V ) which is R \ 0 since

U = S1 \ (−1, 0)) and V = S1 \ (1, 0)) and so the change of coordinates map is C∞.

Exercise: Check other examples of charts are compatible for suitable k.

Definition 1.3 An atlas A of dimension n and class Ck on a topological space M is a collection of charts whichare dimension n and pairwise compatible of class Ck 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 Ck compatible with all ofits charts.

Remark Every atlas is contained in a unique maximal atlas of the same dimension and class.

Example 1.2 On S1 we have two charts given by projections from two distinct points covering S1.

Definition 1.4 A manifold structure of dimension n and class Ck on a topological space M is a choice ofmaximal atlas of class Ck and dimension n.

A manifold structure can be specified by a non maximal atlas, e.g. the atlas of two charts on S1. We can thenuse 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∞ (Cω even).

2. Any finite dimensional real vector space V by picking a basis b = (v1, ..., vn). We get a map φb : V → Rnby φb(v) = (a1, ..., an) if v =

∑aivi. 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 = b1g where g is an n × n real matrix that is invertible. Then we have

v = b1

a1...an

= b2

c1...cn

= b1g

a1...an

and so

a1...an

= g

c1...cn

and so φb1 = g φb2 and g is Ck for all k

and so the charts are compatible and hence in the same maximal atlas.

3. Sn the the n sphere and one can generalise stereographic projection. φ(x) = 2(x1,...,xn)1−x0

is projection ontothe plane tangent at (1, 0, ..., 0) from U1 = x ∈ Sn|x0 6= 1.

We can consider charts from the implicit function theorem. Let F : Rn+k → Rk be a C∞ map theny ∈ F (Rn+k is called a regular value if for all x ∈ Rn+k with F (x) = y the derivative DxF is onto as a linearmap Rn+k → Rk.

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Theorem 1.5 F−1(y) = x ∈ Rn+k|F (x) = y when y is a regular value is an n dimensional manifold. Chartscan be obtained by taking n of the domain coordinates on a suitable open set and the remaining k domaincoordinates are C∞ functions of these.

Example 1.4 S1 = F−1(1) where F (x, y) = x2 + y2. D(x,y)F (h, k) = 2xh + 2yk and since x2 + y2 = 1 x andy cannot simultaneously be zero and so the derivative is onto so 1 is a regular value. Then we solve x2 + y2 = 1for either x and y. This gives x =

√1− y2 and we can do this likewise for Sn.

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 aring 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 onM ×N which determines a product manifold structure of dimension m+ n.

For example Tn = S1 × ...× S1 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 m(g, h) = g−1h 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. S1 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. Tn 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) sinceit is the preimage of R? under the determinant map. It has a global chart using the n2 matrix entries ascoordinates, 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 = j2 = k2 = −1 and k = ij = −ji and so is askew field. H = R4 as a real vector space. An element in H is written as q = q01 + q1i + q2j + q3k andq = q01− q1i− q2j − q3k and qq = ||q||21 and qq′ = qq′. H? = H \ 0 under quaternion multiplication isa group with quadratic multiplication map and invers i(q) = q/||q||2 and so both are smooth and so it is aLie group of dimension 4.

Quaternions of unit length for a subgroup Sp(1) = S3 as a manifold. Sp(1) is a manifold and the Euclideancoordinates are smooth functions when restricted to Sp(1) and so it is a Lie group of dimension 3.

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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 gT gy ⇐⇒ x · (gT gy−y) = 0 ⇐⇒ gT gy−y = 0 ⇐⇒ gT g = In.Thus

O(n) = g ∈ GL(n,R)|gT g = In = g ∈Mn×n(R)|gT g = InSet F (g) = gT g for g defined on Mn×n(R). What is the target. Note gT 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 ofF as a map Mn×n(R)→ Sn(R). Take g ∈ O(n) and consider F (g + th) for h ∈Mn×n(R). Then

DgF (h) =d

dtF (g + th)|t=0 = gTh+ hT g

For g fixed, is every element of Sn(R) of this form for a suitable h. Take A ∈ Sn(R) and try to solvegTh + hT g = A for some h or solve gTh = 1

2A and hT g = 12A, but these are the same as A = AT . Thus

h = 12gA. Thus DgF 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 amanifold structure of dimension n2− dim(Sn(R)) = n2− 1

2n(n− 1). Also O(n) has a group structure sinceg1, g2 ∈ O(n) then (g1g2)T g1g2 = In. The matrix entries are functions on O(n) and are smooth. Hencemultiplication and inverses are smooth in manifold structure. Thus O(n) is a Lie group. O(1) = ±1.

Now consider O(2). If g =

(a bc d

)then a2 + c2 = 1 = b2 + d2 and ab + cd = 0. so

(bd

)⊥(ac

)and

so is a multiple of

(−ca

), e.g.

(bd

)= λ

(−ca

)and this gives λ2 = 1. Hence O(2) is the matrices of the

form

(a −λcc λa

)with λ = ±1 and a2 + c2 = 1. Note that det(g) = λ and so λ depends continuously on g.

Hence O(2) has two connected components, with g ∈ O(2)|det(g) = 1 as a subgroup and as a manifold isa 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 haveSL(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

dtdet(g + th)|t=0 =

d

dtdet(g(In + tg−1h))|t=0 =

d

dtdet(In + tg−1h)|t=0 = tr(g−1h) = n 6= 0

and so 1 is a regular value of det so SL(n,R) is a Lie group of dimension n2 − 1.

13. SO(n) = O(n)∩SL(n,R) = g ∈Mn×n(R)|gT g = In,det(g) = 1. For example SO(1) = 1 and SO(2) =(a −cc a

)∈M2×2(R)|a2 + c2 = 1

which is the same as O(2). Note that g ∈ SO(n) =⇒ det(g) = ±1

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). detis 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 thatO(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 holdsfor 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)

so xi · xj = δij, and so xn ∈ Sn−1 so rotate xn to the vector

0...01

this will rotate the other vectors

simultaneously and end up with a matrix of the form

... . . . . . . 0...

......

...0 . . . 0 1

. Repeating we eventually get to

SO(2) which is connected. Hence SO(n) is connected for all n.

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We have maps O(n) → Sn−1 by picking any column. Putting these maps together we get a map O(n) →×nSn−1 which is continuous and injective and gives O(n) its usual topology. The image consists of thesubset of this product of pairwise orthogonal vectors. The image is closed hence compact,so O(n) is acompact Lie group. So is SO(n).

14. Pseudo -orthogonal groups O(p, q) and SO(p, q).

On Rp+q let Bp,q(x, y) =∑p

1 xiyi −∑p+qp+1 xiyi = xTAp,qy. 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) ⇐⇒ gTAp,qg = Ap,q and we can define F : M(p+q)×(p+q)(R) → Sp+q(R) byF (g) = gTAp+qg so O(p, q) = F−1(Ap,q) and is a regular value. This makes O(p, q) into a Lie group ofdimension 1

2 (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).

We consider O(1, 1). Take an element here and write it as g =

(a bc d

)and then we get that a2 − c2 = 1,

b2 − d2 = −1 and ab − cd = 0 and so

(b−d

)= −λ

(−ca

)and with λ2 = 1. Now det(g) = −λ and so for

SO(1, 1) we have the form

(a cc a

)with a2−c2 = 1. This is a hyperbolic function, and so SO(1, 1) has two

connected components and O(1, 1) has four connected components. Now O(1, 1)0 corresponds to det(g) = 1and a > 0 and then

t 7→(

cosh t sinh tsinh t cosh t

)is a homomorphism of groups from R → O(1, 1)0 which is a diffeomorphism. This is an isomorphism ofLie 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 homomorphismof groups and a smooth map of manifolds. An isomorphism of Lie groups is a bijection which is ahomomorphism of Lie groups, as well as its inverse.

15. An antisymmetric bilinear form can only be non singular when its dimension is even. So take R2n and set

Jn =

(0 −InIn 0

)and Ω(x, y) = xTJny. If x =

(x′

x′′

)and y =

(y′

y′′

)then Ω(x, y) =

(x′T y′′ − x′′T y′x′y′′ − x′′y′

)Remark If V is a real 2n dimensional vector space and w is an antisymmetric bilinear form on V whichis non singular We define

Sp(2n,R) = g ∈ GL(2n,R)|Ω(gx, gy) = Ω(x, y)

Then g ∈ Sp(2n,R) ⇐⇒ gTJng = Jn and set F (g) = gTJng and then (gTJng)T = −gTJng and soF maps M2n×2n(R) into A2n(R) = A ∈ M2n×2n(R)|AT = −A which is a vector space of dimension122n(2n − 1) = n(2n − 1). Then one can show Jn is a regular value of F and so Sp(2n,R) is a Lie groupof 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 haveall positive signs as can multiply by i.

17. Unitary group On Cn one can form the Hermitian inner product 〈z, v〉 = zT v. A complex linear mapg : Cn → Cn is unitary if 〈gz, gv〉 = 〈z, v〉 for all z, v ∈ Cn. The set of unitary matrices is denoted byU(n). We can write the inner product as 〈z, v〉 = zT (gT g)v and we denote g? = gT . 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 areal vector space of dimension n2. If A = X + iY then A? == XT − iY T = X + iY , and so X = XT andY = −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 basisof Cn so we get a map U(n) → ×nS2n−1 which is a homomorphism onto its image, which is n-tuplesz1, ..., zn ∈ Cn with 〈zi, zj〉 = 0 for i 6= j. The image is a closed set in ×nS2n−1 hence is compact, so U(n)is a compact Lie group.

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U(1) = z ∈ C||z| = 1 = S1

If

(a bc d

)is in U(2) then |a|2 + |c|2 = 1 = |b|2 + |d|2 and ab + cd = 0. Thus b = −λc and d = λa with

|λ| = 1 and so

(a bc d

)is determined by λ ∈ S1 and

(ac

)∈ S3 with λ = det

(a bc d

)and so as a manifold

U(2) is diffeomorphic to S1 × S3.

Note that g? = gT and so det g? = det g = det g and so |det g| = 1. If we set

SU(n) = U(n) ∩ SL(n,C) =−1det(1)

by viewing det : U(n)→ S1. Then one can check that 1 is a regular value of det and one gets SU(n) as amanifold of dimension n2 − 1 with matrix entries defining smooth functions, and so is a Lie group.

SU(1) = 1

We now consider SU(2). An element of U(2) is written

(a bc d

)with |a|2 + |c|2 = 1 = |b|2 + |d|2 and

ab+ cd = 0. Then we get, using similar arguments to above, that

U(2) =

(a −λcc λa

)∣∣∣∣ |a|2 + |c|2 = 1, |λ| = 1

' S3 × S1

and note that det g = λ for g ∈ U(2) and so SU(2) is diffeomorphic to S3 and S3 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

〈z, v〉p,q =

p∑1

zkvk −q∑1

zp+kvp+k = z?Ap,qv

and setU(p, q) = g ∈M(p+q)×(p+q)(C)|g?Ap,qg = Ap,q

and observe that Ap,q is a regular value of F (g) = g?Ap,qg 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)→ S1 and so SU(p, q)is a Lie group of dimension (p+ q)2 − 1.

SU(1, 1) is the isometry group of the Poincare 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)|gT 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) isn+ 1

2n(n− 1) and the dimension of Mn×n(H) is 4n2 and so the dimension of Sp(n) = n(2n+ 1) which isthe 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 S4n−1 and Sp(n) maps into ×nS4n−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 determinantfunction 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,qg = Ap,qwith 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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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 alsocalled 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 exampleA = EndV the set of linear maps V → V . This is called the commutator bracket. It measures how nonzero the commutativity of the algebra is. For instance if V = Rn and then End(V ) = Mn×n(R) and so thisis 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)). IfM = 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 constantc ∈ 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 ′(a) + (x− a)2ga(x)

where ga ∈ C∞(R). We can rearrange this and then check that it is smooth

D(f)(x) = 0 +D(x− a)f ′(a) +D((x− a)2ga(x))

= D(x− a)f ′(a) +D(x− a)(x− a)ga(x) + (x− a)D(x− a)ga(x) + (x− a)2D(ga(x))

and if we set x = a we getD(f)(a) = D(x− a)(a)f ′(a) + 0

and so D(f)(a) = h(a)f ′(a) for some h ∈ C∞(R) for all a ∈ R and so D = h ddx . If Di = hi

ddx for i = 1, 2 then

[D1, D2] = (h1h′2 − h2h′1)

d

dx

and this is not the trivial Lie algebra. X (M) is the Lie algebra of vector fields onM and [x, y] is the Lie bracketof 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= Lg1g2

and Le = IdG and so Lg−1 = L−1g 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 Liesubalgebra 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 takingdγ(t)dt or γ(t). How though can we define γ(t) for γ : (a, b)→M a smooth map?

We can take a chart on M around a point on γ and take the tangent vector in Rn of φ γ, say ddt (φ γ(t)) =

v ∈ Rn. If we take another chart (V, ψ) around the same point then in general ddt (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 = Dφγ(t)(ψ φ−1)v. Fix a point x ∈M and considerobjects 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 = Dφγ(t)(ψ φ−1)v.

Definition 4.1 The tangent space TxM is the set of equivalence classes of ∼x. We denote by [U, φ, v]x theequivalence 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

anda[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 γ(t0) = dγ(t)dt |t=t0 by taking a chart

(U, φ) around x and setting

γ(t0) =

[U, φ,

d(φ γ(t))

dt|t=t0

]x

Theorem 4.1 If M is a manifold of dimension n then for all x ∈M TxM is a vector space of dimension n. IfF : M → N is smooth then there is a natural linear map dxF : TxM → 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 dxF

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 → TxM by v 7→ [U, φ, v]x which is linear andalso 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 ' TxMif we pick a chart. Then

dxF [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:

(F γ)·(t) = [V, ψ,d

dtψ F γ(t)]Fγ(t)

= [V, ψ,d

dtψ F φ−1 φ γ(t)]Fγ(t)

= [V, ψ,Dφγ(t)ψ F φ−1d

dtφ γ(t)]Fγ(t)

= dγ(t)F [U, φ,d

dtφ γ(t)]γ(t)

= dγ(t)F (γ(t))

Q.E.D.

For example consider Rn with identity chart. This allows a canonical identification TxRn ' Rn and then dγdt

is the same ordinary derivative.

Definition 4.3 Let f ∈ C∞(M) and let x ∈M. If X ∈ TxM 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 : TxM→ Tf(x)R with Tf(x)R = R, and write the X to the left, X(f) = df(X).

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Proposition 4.4 Fixing x ∈ TxM the map f 7→ X(f) (C∞(M) → R) is linear and satisfies the product rule,namely

X(fg) = 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 ∈ TxM such thatD(f) = X(f) for all f ∈ C∞(M). To see this let (U, φ) be a chart about x such that the coordinate functions

xi. Then x1, ..., xn are the restrictions of globally defined f1, ..., fn ∈ C∞(M). Then v =

D(f1)...

D(fn)

∈ Rn

and X = [U, φ, v]x.

3. This means that we could alternatively define TxM as the linear maps C∞(M) → R that satisfy theproduct 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 picka basis b = e1, ..., en for V we get a chart (M, φb) as follows. Take x ∈ M and expand x in the basis

so that x =∑ni=1 xi(x)ei and put φb =

(x1(x)...xn(x)

). The x1, ..., xn are coordinate functinos of this chart.

M then becomes a manifold using the atlas (M, φb). If v ∈ V then v =∑viei and v =

(v1...vn

)and

[M, φb, v]x ∈ TxM. The map v 7→ [M, φb, v]x is a linear isomorphism of vector spaces. This is independentof the basis. The corresponding derivation at x we write as vx and for f ∈ C∞(M), vx(f) = d

dtf(x+tv)|t=0.

2. Suppose thatM = F−1(c) for F : Rn+k → Rk smooth and c a regular value of f , andM given the manifoldstructure by the implicit function theorem. Take X ∈ TxM and any curve γ through x, say with γ(t0) = xsuch that X = γ(t0). To do this pick a chart (U, φ) about x and then X = [U, φ, v]x for v ∈ Rn and putγ(t) = φ−1(φ(x) + (t− t0)v) and so d

dtφ(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′ ∈ Rn+k. But γ(t) ∈ Mfor all t and so F (γ(t)) = c and hence (Dγ(t0)F )(v′) = 0 i.e. v′ ∈ KerDxF , and DxF : Rn+k → Rk

and DxF is surjective so dim KerDxF = n. We have a map TxM → Rn+k defined by X 7→ ddtγ(t)|t=t0

which is linear and injective and maps TxM to Ker(DxF ) and so we can use Ker(DxF ) as a model forTxM. For example, Sn ⊂ Rn+1 with F (x) = x · x. For x ∈ F−1(1) we have DxF (v) = 2x · v and soKerDxF = 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 sothe space is A ∈Mn×n(R) : trA = 0

7. SL(n,C) has TInSL(n,C) = A ∈Mn×n(C) : trA = 0

8. O(n) = g ∈ Mn×n(R) : gT g = In. Define F (g) = gT g : Mn×n(R) → Sn(R) and now note thatDInF (A) = AT +A and kerDInF = A ∈Mn×n(R) : AT +A = 0 = An(R)

9. TInU(n) = A ∈Mn×n(R) : A? +A = 0

Recall we have diffeomorphisms Lg : G→ G such that Lg(h) = gh with the properties that Lg1Lg2

= Lg1g2

and Le = Id. Similarly for on the right with Rg(h) = hg but we have Rg1Rg2

= Rg2g1and Re = Id.

Lemma 4.5 TgG = deLg(TeG) 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 eachx ∈M we pick an Xx ∈ TxM.

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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 definea function X(f) by X(f)(x) = Xx(f) and it satisfies the Leibniz rule, since

X(f1f2)(x) = Xx(f1f2) = Xx(f1)f2(x) + f1Xx(f2(x)) = X(f1)f2(x) + f1(x)X(f2) = (X(f1)f2 + f1X(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 ∈ TxM 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 linkbetween 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 aresaid to be F -related if dxF : TxM→ TF (x)N is such that

dxF (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 areF -related for i = 1, 2 then [X1, X2] is F -related to [Y1, Y2].

Proof Take f ∈ C∞(N ) ThendxF (Xi)x(f) = (Xi)x(f F )

anddxF (Xi)x = (Yi)F (x)

and so(Yi)F (x)(f) = (Xi)x(f F )

and is the same as Yi(f)(F (x)) orYi(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 dxF [X1, X2]x = [Y1, Y2]F (x) as required Q.E.D.

5 The Lie Algebra of a Lie Group

Definition 5.1 A vector field X on a Lie group G is called left invariant if

dhLg(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) = (deLg(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 TeG. Define ε : X (G)G → TeG by ε(X) = Xe. This is linear.

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Theorem 5.3 ε : X (G)G → TeG defined by ε(X) = Xe is a linear isomorphism.

Proof Let X ∈ ker ε and so Xe = 0. But Xg = deLgXe = 0 for all g and so X = 0 and so ε is injective.

Now take ξ ∈ TeG and set ξh = deLh(ξ) for all h ∈ G. Then

dhLg(ξh) = dhLg(deLh(ξ)) = de(Lg Lh)(ξ) = deLgh(ξ) = ξgh

Hence ξ is left invariant and so smooth by the above lemma. Thus ξ ∈ X (G)G and ε(ξe) = ξe = deLe(ξ) = ξ andso ε is surjective. Q.E.D.

Definition 5.4 The left invariant vector field ξ determined by ξ ∈ TeG is called the left invariant extensionof ξ, with deLg(ξ) = ξ.

If ξ, η ∈ TeG then ξ, η are in X (G)G which is a Lie algebra so we can form [ξ, η] which is left invariant and so ofthe form h for some h ∈ TeG with h = ε([ξ, η]). This makes ε : X (G)G → TeG. We call TeG with this bracketthe Lie algebra of G.Remark We could equally use right invariant vector fields and get a second bracket on TeG denoted by [·, ·]rightwith ξh = deRh(ξ) and [ξ, η]right = −[ξ, η]e

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.

[ξ, η]e(f) = ξe(η(f))− ηe(ξ(f))

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,∞), S1, Tn, C? = S1 × (0,∞), Cn.

We need a technique for computing with vector fields. If (U, φ) is a chart with coordinates x1, ..., xn soxi = ti φ where ti are Cartesian coordinates on Rn.

If f is a smooth function on the manifold we can form(∂fφ−1

∂ti

) φ ∈ C∞(U) is a vector field on U denoted

by ∂∂xi

Lemma 5.6 If X ∈ X (M) then X|U =∑ni=1 a

i ∂∂xi

with ai ∈ C∞(U).

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

in a vector space, TInGL(n,R) = Mn×n(R) as vector spaces. A ∈Mn×n(R) corresponds to ddt (In + tA)|t=0. A is

the left invariant extension and then

Ag = (dInLg)A = (dInLg)d

dt(In + tA)|t=0 =

d

dtLg(In + tA)|t=0 =

d

dt(g + tgA)|t=0

and applying this to coordinate functions gives

Ag(xij) =d

dt(g + tgA)ij |t=0 =

d

dtgij + t

∑k

gikAkj |t=0 =∑k

gikAkj =∑k

xik(g)Akj

and so A(xij) =∑k xikAkj Then

[A, B](xij) : = A(B(xij))− B(A(xij))

= A(∑k

xikBkj)− B(∑k

xikAkj)

=∑k

A(xik)Bkj −∑k

B(xik)Akj

=∑k,l

xilAlkBkj −∑k,l

xilBlkAkj

=∑l

xil∑k

(AlkBkj −BlkAkj)

=∑l

xil(AB −BA)lj

= ˜AB −BA(xij)

Using the lemma below we get [A, B] = ˜AB −BA and so the Lie algebra of GL(n,R) is Mn×n(R) with theLie 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 TInGis identified with kerDInF for F mapping Mn×n(R) to a suitable set, and then if A ∈ kerDInF then there is acurve In + tA + O(t2) with A associated to d

dtIn + tA + O(t2)|t=0. The higher order terms do not change theargument, 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

degF : TeGG→ TeHH 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 deGF : g → h is a Lie algebra homomorphism.

Proof Take ξ ∈ g and consider ξ its left invariant extension:

dgF ξg = dgF (deGLg(ξ)) = deG(F Lg)(ξ)

and thenF Lg(h) = F (Lg(h)) = F (gh) = F (g)F (h) = LF (g)(F (h))

and then we get that

dgF (ξ) = deG(LF (g) F )(ξ) = deHLF (g)deGF (ξ) = ˜deGF (ξ)F (g)

and so ξ and ˜deGF (ξ) are F related. Take ξ, η ∈ g and then [ξ, η] is F related to [ ˜deGF (ξ), ˜deGF (η)] and so

dgF ([ξ, η]g) = [ ˜deGF (ξ), ˜deGF (η)]F (g)

and setting g = eG we getdeGF ([ξ, η]g) = [deGF (ξ), deGF (η)]h

Q.E.D.

Definition 5.9 We denote deGF 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 S1 are abelian and one dimen-sional. Also SU(2) and SO(3) have isomorphic Lie algebras but different centres. U(1) × SU(2) and U(2) arediffeomorphic 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 alsoF 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 → his an isomorphism.

By the inverse function theorem there exists a U ⊂ G and V ⊂ H both open and eG ∈ U , eH ∈ V and Fa diffeomorphism of U into V . Hence F is an open map so F (G) is an open subgroup of H. H is connected soF (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 isone to one on U and eG ∈ U and so K ∩ U = eg so K is discrete in the relative topology. The sets kk∈Kthen 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 aconnected 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 isone, and call it F . Then F? : SO(3) → SU(2) is a homomorphism of Lie algebras. F? is non zero otherwise Fwould be constant. Do some algebra and see that F? has to be surjective and so a linear isomorphism. ThereforeF 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 somet0 ∈ (a, b) is said to be an integral curve through x ∈M if γ(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 withγ(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).γ(t) = [U, φ,Γ(t)]γ(t). with Γ(t) a curve in Rn. which is d

dtφ(γ(t)) and so we have γ(t)Xγ(t). Thus

d

dtφ(γ(t)) = v(φ(γ(t))) (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 uniquenesstheorem 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 thatφ(γ(t+ a)) is a solution and so we can always find an integral curve through a given point at a convenient valueof t. so if γ1(t1) = x = γ2(t2) then we can shift one so that γ2(t) = γ2(t + t2 − t1) is an integral curve withγ2(t1) = x and so γ1(t) = γ2(t) for common values of t near t1. Thus you can get a large solution form combiningγ1 and γ2

Definition 6.3 An integral curve γ of a vector field X is called a maximal integral curve if any integral curvepassing through a point of γ is just γ restricted to a subinterval (after shifting parameters to agree at a commonpoint). 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

(x(t), y(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 thent = 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 haveγ(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 throughg at t = t0 and h ∈ G then γ(t) = hγ(t) is an integral curve on (a, b) passing through hg at t = t0.

Proof

γ(t) =d

dthγ(t) =

d

dtLhγ(t) = dγ(t)Lh(γ(t)) = XLh(γ(t))=Xγ(t)

and so γ is an integral curve andγ(t0) = hγ(t0) = hg

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 Xwith 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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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)overlap and we have a common solution on the overlap γ−1X (s)γX(s+ t) and γX(t). The new solution is definedon (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 eat t = 0. Then lemmas 6.5, 6.6 imply that gγX(t− t0) is the integral curve through g at time t0. The propertyγX(s)γX(t) = γX(t + s) says γX : R → G is a homomorphism of groups and γX a smooth curve implies thatγX : 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 denoted1-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) forall 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 inductionσ(n) = nσ(1). We can apply this argument to σ(nx) to get σ(nx) = nσ(x). Then nσ(mn x) = σ(mnn x) = mσ(x)and so σ(mn x) = m

n σ(x) and since σ is continuous and rationals are dense in R we get that σ(x) = xσ(1) for allx ∈ R. This gives that σ is linear and hence smooth. We can also show that σ(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→ σ(0) gives us a map. To map 3 to 2 noteξ 7→ γξ 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 = σ(0) and form γξ It is enough to show that σ is the integral curve of ξ through e.It should be clear that σ(0) = e. Then

σ(t) =d

dsσ(t+ s)|s=0 =

d

dsσ(t)σ(s)|s=0 = deLσ(t)

d

dsσ(s)|s=0 = deLσ(t)ξ = ξσ(t)

and so σ is an integral curve of ξ so σ = γξ by uniqueness. Q.E.D.

According to the theorem, γξ(1) ∈ G is determined by ξ ∈ g.

Definition 6.10 We define expG : g→ G by expG ξ = γξ(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 dependingsmoothly on the parameters. Hence the ODE for ξ depends linearly on ξ. Picking a basis ξ1, ..., ξk for g we haveξ = a1ξ1 + ...+ akξk and so we are looking for solutions of

γa(t) = ξγ(t) = (a1ξ1 + ...+ ak ξk)γa(t)

where a = (a1, ..., ak) Q.E.D.

Proposition 6.12γξ(s) = exp(sξ)

Proof Consider t 7→ γξ(st) and note this is smooth and γξ(st1)γξ(st2) = γξ(s(t1 + t2)) and so γξ(st) = γη(t)for some η and

η = γξ(0) =d

dtγξ(st)|t=0 =

d

dt(st)|t=0γξ(0) = sξ

and therefore γξ(xt) = γsξ(t) and then set t = 1 to conclude Q.E.D.

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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 standardidentity T0Rn = Rn

2. G = R? with g = R and then T1R? = R with the identification ddt (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, andwe then have ξx = f(x) d

dx and we want to know when this is left invariant.

ξx = d1Lxξ = d1Lxd

dt(1 + tξ)|t=0

=d

dtLx(1 + tξ)|t=0

=d

dt(x+ txξ)|t=0

= (xξ)x

and so f(x) = xξ.

The integral curve σ(t) through 1 at time t = 0 will satisfy σ(t) = ξσ(t) and be given by a function x(t) and

so x(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 vectorddt (In + tξ)|t=0 at In ∈ GL(n,R). We have

ξg = dInLgd

dt(In + tξ)|t=0 =

d

dt(g + tgξ)|t=0 = gξg

where the over-line means directional derivative.

If g(t) is the integral curve through In at t = 0 then it satisfies

g(t) = ξg(t) = g(t)ξg(t)

and g(t) as a tangent vector is g(t)g(t) and the derivative is that of a function. Thus g(t) = g(t)ξ. given a

matrix ξ we form ξ2, ξ3, ... and hence we form

eξ = In +∞∑k=1

ξk

k!

which is a smooth function in ξ with ddte

tξ = 0 +∑∞k=1 k

tk−1ξk

k! = etξξ and so

d

dt(g(t)e−tξ) =

d

dt(g(t))e−tξ + g(t)

d

dt(e−tξ) = g(t)ξe−tξ + g(t)(−e−tξξ) = 0

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 homomor-phism 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

dtF (expG(tξ))|t=0 = F?ξ

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. HenceexpO(n) ξ = expGL(n,R) ξ for ξ ∈ o(n).

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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 : T0g→ TeG

is a linear map. However T0g = g (using directional derivatives) and also TeG = g and so we can view d0 exp :g→ g as a linear map. Then we have the following.

Theorem 6.17d0 exp = Idg

Proof Given ξ ∈ g there is a curve in g, say γ(t) with γ(0) = 0 and γ(0) = ξ, for instance γ(t) = tξ. Then

d0 exp γ(0) =d

dtexp(γ(t))|t=0 =

d

dtexp(tξ)|t=0 = ξ ∈ TeG

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 andexpU → 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 = expU where exp : U → V is adiffeomorphism. 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 groupssuch 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 diffeomor-phism from a neighbourhood e 3 V ⊂ G into U ⊂ g. If we pick a basis ξ1, ..., ξN for g then for g ∈ V we haveexp−1 g =

∑Ni=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 withthe 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 aroundg. 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. Whenthen 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 andonly 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 isgenerated 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 thereforeexp(tξ) exp(tη) = exp(tζ) for some ζ ∈ g. We have

ζ =d

dtexp(tξ) exp(tη)|t=0 = ξ + η

using the lemma below. Thus exp is a homomorphism of Lie groups. Q.E.D.

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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

0 ∈ (a, b) then f(0) =∑Ni=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 ) Q.E.D.

Examples of connected abelian Lie groups are Rn, Tn,Rk × Tn−k 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 around0 with Γ ∩ B = 0. Then either Γ = 0 or there are k linearly independent vectors v1, ..., vk ∈ Rn such that

Γ = ∑ki=1mivi|mi ∈ Z

Proof We proceed by induction. Take n = 1. Then if Γ 6= 0 there exists a non zero element v ∈ Γ of minimallength. Since Γ is closed and ||v|| ≥ R (radius of B), we claim that Γ = Zv. We know that 0 6= w ∈ Γ thenw = λv for λ ∈ R and assume that λ > 0, and then we can write λ = p + r for p ∈ N and 0 ≤ r < 1. Thenv ∈ 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 sequenceγm ∈ 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−lmv1 →0 in Rn. Let lm = hm + rm for some hm ∈ Z, and rm ∈ [0, 1]. Then γm − lmv1 = (γm − hmv1) − rmv1. Thensequence rm ∈ [0, 1) ⊂ [0, 1] and the latter is compact and so rm has a convergent subsequence. Hence we cansuppose we have a sequence γm and coset representation γm such that there is a sequence rm ∈ [0, 1] convergentwith γm− rmv1 → 0 in Rn, but γm ∈ Rv1. Let r = lim rm. γm is convergent in Rn to rv1. Γ is a closed subgroupof 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′ in Rn/Rv1 with B′∩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 giveΓ/Zv1. Q.E.D.

Proof (of theorem 6.24) G connected abelian implies that exp : g → G is a homomorphism and exp is asurjective map. Let Γ = ker exp which is a subgroup of g which is closed and since exp is a diffeomorphism near0 ∈ g then there exists an open U ⊂ g with 0 ∈ U and Γ ∩ U = 0. It is enough to study such subgroups γ ofRn.

We thus know from the above lemma that there exist v1, ..., vk linearly independent such that Γ = Zv1 + ...+Zvk. Put g1 = spanRv1, ..., vk which has dimension k. Pick g2 ⊂ g such that g = g1 ⊕ g2 and then Γ ⊂ g1 andexp : g2 → G will be an isomorphism onto its image G2 ⊂ G. Put G1 = exp(g1) which is a subgroup of G andG = G1 ×G2. G2

∼= Rn−k and G1∼= g/Γ ∼= Rk/Zk ∼= (R/Z)k = T k and so G ∼= T k × Rn−k. Q.E.D.

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

dxf : TxM→ 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

dxf : TxM→ 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 andboth f and f−1 are smooth.

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A diffeomorphism is both an immersion and a submersion. Something which is both an immersion and asubmersion 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 manifoldM 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 TxN ⊂ TxM.

Definition 7.3 A submanifold f : N → M is called embedded if f is a homeomorphism to f(N ) with therelative 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 commonperiod for eiat and eibt. The periods of these are t1 = 2πn

a and t2 = 2πmb and a common period would be

when t1 = t2. For this we would need ba ∈ 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 S1 as a subgroup of Tn we send z 7→ (zm1 , ..., zmn). So long as there are no common divisors ofm1, ...,mn then this will be injective and an immersion. Thus S1 is a submanifold of Tn 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 whereH 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. dhf : ThH → Tf(h)G is injective.

Given ξ ∈ h, deHLhξ ∈ ThH and all arise in this way. Then observe that

(f LH)(h′) = f(hh′) = f(h)f(h′) = Lf(h)f(h′) = (Lf(h) f)(h′)

and so we get that

dhf(deHLhξ) = deH (f Lh)(ξ) = deH (Lf(h) f)(ξ) = deGLf(h)deHf(ξ) = deGLf(h)(f?(ξ))

Both of deGLf(h) and deHLh are invertible and so dhf is injective if and only if f? is injective. Thu f is animmersion 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 andf? are injective.

This then characterises Lie subgroups as homomorphisms f with both f and f? injective. Then f(H) is asubgroup 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 existsh(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(h′) = expG ξ′

for ξ′ 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(t0ξ)f(h) expG(ζ)

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and note Ψt0 is smooth.

d(eH ,0)Ψt0(η, ρ) =d

dsΨt0(expH sη, sρ)|s=0

=d

dsexpG(t0ξ)f(expH(sη))expG(sρ))|s=0

= deGLexpG(t0ξ)(f?(η) + ρ))

therefore d(eH ,0)Ψt0 is surjective onto TexpG(t0ξG and the domain is h ⊕ m which has the same dimension sod(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 suchthat if V = Ψt0(U1 × U2) then V is open in G, eH ∈ U1, s ∈ U2 and expG(tξ) ∈ V and Ψt0 is a diffeomorphismof U1 × U2 onto V .

Ψ−1t0 : V → U1×U2 will be smooth and we put Φt0 = π1Ψ−1t0 and then Φt0 is a smooth map G ⊃ V → U1 ⊂ H.There is an interval around t0 ∈ (a′, b′) such that (a′, b′) ⊂ (a, b) and expG(tξ) ∈ V . Then t ∈ (a′, b′) we haveh(t) = Φt0(expG(tξ)) and so h(t) is smooth on (a′, b′). t0 is arbitrary and so h(t) is smooth on (a, b).

Pick c ∈ (a, b) such that h(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)(h(c))

but

df(h(c))f(h(c)) = f(h(c))· =d

dt(expG(tξ))|t=c = deGLexpG(cξ)(ξ)

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 isdetermined 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 ofs ∈ g and V of eG ∈ G with expG : U → V a diffeomorphism such that expG(U ∩ h) = V ∩H. Then H with therelative topology has an atlas built using exp−1G on V ∩H and left translating making H into a Lie subgroup ofG.

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 thatexp(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 givesξ + η = ξ + η + ζ(t)

and so ζ(0) = 0. ζ is smooth in t around t = 0 and so we have a Taylor expansion with remainder

ζ(t) = ζ(0) + tζ(0) +1

2t2R(t)

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) = limN→∞

(exp

(ξ

N

)exp

( ηN

))N

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Proof When N ≥ 1δ we have

exp

(ξ

N

)exp

( ηN

)= exp

(ξ

N+

η

N+ ζ

(1

N

))and so (

exp

(ξ

N

)exp

( ηN

))N= exp

(ξ + η +Nζ

(1

N

))and Nζ

(1N

)→ 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 uniquemanifold 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 (

expG

(ξ

N

)expG

( ηN

))N∈ H for all 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 bytη 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 andV = 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 fora U ⊂ g.

Fix W ⊂ g where expG : W → expG(W ) is a diffeomorphism, and fix a norm on g. Consider open balls B 1k

(0)

for large enough k ∈ N to ensure B 1k

(0) ⊂W . We have

expG(B 1k

(0) ∩ h) 6= expG(B 1k

(0)) ∩H

If ξ ∈ B 1k

(0)∩h then expG ξ ∈ expG(B 1k

(0))∩H and so expG(B 1k

(0))∩H is strictly larger than expG(B 1k

(0)∩h)

Let hk ∈ H ∩ expG(B 1k

(0)), but hk 6∈ expG(B 1k

(0) ∩ h). Then hk = expG ξk for ξk ∈ B 1k

(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 1k

(0) ⊂W . Due to bijectivity of expG on W we have ηk = ξk and so ηk ∈ B 1k

(0)∩h and so hk ∈ expG(B 1k

(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 buthk 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 diffeomorphismof 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 andrelabel the subsequence ηk/tk.

Let η be the limit of this sequence. Let t > 0, nk(t) = bt/tkc and so

t/tk − 1 ≤ nk(t) ≤ t/tk

and thus we have thatt− tk ≤ tknk(t) ≤ t

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but since tk → 0 as k →∞ we have tknk(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

limk→∞

expG(nk(t)ηk) = expG(tη)

and solimk→∞

(expG(ηk))nk (t) = expG(tη)

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 acontradiction. 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(ξ′k) expG(η′k)

with ξ′k ∈Wh and η′k ∈Wm, but hk 6∈ expGW1 and so η′k 6= 0 for all k. Also expG η′k ∈ H since hk, expG ξ

′k ∈ H

and then we have expG η′k ∈ expGWm ∩H = e and so η′k = 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 isa 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 isclosed, 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 Liesubgroup 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 gσ? = ξ ∈Mn×n(R)| − ξT = 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.

Jn =

(0 −InIn 0

)and so J2

n = −In and define σ1(g) = JngJ−1n , which is an inner automorphism. Then

Gσ1 = (U −VV U

)|U + iV ∈ GL(n,C) and so GL(n,C) is a Lie subgroup of GL(2n,R). If instead

σ2(g) = Jn(g−1)TJ−1n then Gσ2 = Sp(2n,R).

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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 H is a subgroup.

Proof Take g, h ∈ H and suppose gk → g and hk → h for sequences hk, gk ∈ H. Then

gh = m(g, h) = limk→∞

m(gk, hk)

and m(gk, hk) ∈ H and so gh ∈ H. Continuity of the inverse gives i(H) ⊂ H so H 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 enoughto 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 × Tn−k for some k and n = dimG.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 = S1.Continuous homomorphisms R → R are smooth since they are linear. for R → S1 let σ : R → S1 ⊂ C be

continuous and σ(s) +σ(t) = σ(s+ t) as a complex valued function. Use the result from complex function theoryso 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 sog(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 imageof (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 theorem8.4 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′ ∈ gV then g−1g′ ∈ V and

F (g′) = F (gg−1g′) = F (g)F (g−1g′) = LF (g) F Lg−1(g′)

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(t1ξ1)... expG(tnξn). Let

h = (h1, ..., hn) ∈ Rn and then

d

dtα(0 + th)|t=0 =

d

dtexpG(th1ξ1)... expG(thnξn)|t=0 = h1ξ1 + ...+ hnξn

and so d0α is the identification of Rn with g given by a basis, therefore d0α is a linear isomorphism and hence αis n onto a neighbourhood V of eG in G. For (t1, ..., tn) ∈ U ,

F (α(t1, ..., tn)) = F (expG(t1ξ1)... expG(tnξn)) = F (expG(t1ξ1))...F (expG(tnξn))

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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9 Distributions and Frobenius Theorem

Definition 9.1 1. A k-dimensional tangent distribution D on a manifold M is a k-dimensional subspaceDx ⊂ TxM for each x ∈M.

2. We say D is smooth if each point inM 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 ∈ TxRn|X(tk+1) = ... = X(tn) = 0 = X ∈ TxRn|X =∑ki=1 ai

∂∂ti|x. Then Γ(D) = C∞(Rn) ∂

∂t1+ ...+ C∞(Rn) ∂

∂tkand so D is smooth and involutive.

2. G a Lie group, h ⊂ g a k-dimensional linear subspace. Put Dg = deLg(h) ⊂ TgG. Then D is a k-

dimensional distribution and if ξ1, ..., ξk is a basis for h then Dg = span(ξ1,g, ..., ξk,g) and so D is smooth.

We have ξi ∈ Γ(D) and so D is involutive so [ξi, ξj ] ∈ Γ(D) but [ξi, ξj ] = [ξi, ξj ] and so [ξi, ξj ] ∈ h so h is aLie 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 submanifoldfor D is a k-dimensional submanifold (N , φ) with φ : N →M with N a k-dimensional connected manifold andφ an immersion of N into M such that dyφ(TyN ) = Dφ(y) for all y ∈ N .

Note that we have a notion of maximal integral submanifold, analogous to maximal integral curves of a vectorfield.

Proposition 9.3 If D is a smooth distribution on M such that for each x ∈M there is an integral submanifoldφ : 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 X, Y on N with dy′φ(Xy′) = Xφ(y′) and dy′φ(Yy′) = Yφ(y′) and so near y, X and (Y ) are φ-related

to X and Y respectively. This implies [X, Y ] and [X,Y ] are φ-related and therefore [X,Y ]x = [X,Y ]φ(Y ) =

dyφ([X, Y ]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 Mand 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 manifoldM 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 withdeHφ(TeHH) = h and deHφ : TeHH → 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 involutiveso 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 waywhich 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 translatesof integral manifolds are integral manifolds. Consider (H,Lg−1 φ) which is a submanifold of G and an integralsubmanifold. 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(h1h2) = 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

dtΦ(γ(t), 0 + tη)|t=0 =

d

dtφ(γ(t)) exp(tη)|t=0 = dhφ(x) + deLφ(h)η

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and both maps have zero kernel so d(h,0)Φ has zero kernel as a linear map ThH ⊕m→ g and hence d(h,0)Φ is anisomorphism.

Hence by inverse function theorem there is a neighbourhood Uh of h in H, Wh of 0 in m and V h of φ(h) inG with V h = Φ(Uh ×Wh) and Φ : Uh ×Wh → V h a diffeomorphism.

Let Φ−1 on V h be given by maps ψh1 : V h → Uh and ψh2 : V h → Wh. These are smooth maps andψh1 (φ(h′) exp 0) is the first component of Φ−1(φ(h′) exp 0) = Φ−1(Φ(h′, 0)) = (h′, 0) and so

ψh1 (φ(h′)) = h′

for h′ ∈ Uh. Let h1, h2 ∈ H then for h′1 near h1 and h′2 near h2 then

h′1h′2 = ψ

h′1h′2

1 (φ(h′1h′2)) = ψ

h′1h′2

1 (φ(h′1)φ(h′2))

and this is a composition of smooth functions. Then

mH(h′1, h′2) = ψ

h′1h′2

1 (mH(φ(h′1), φ(h′2)))

on (Uh × Uh) ∩ (mG φ × φ)−1(Uh1h2). This is an open set and contains h1h2 so mH is smooth on H × H.Similarly iH : H → H is smooth. Let φ(eH) = e and so deH (TeHH) = 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, upto 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 thedimension 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 kerσ = 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 groupwhose Lie algebra is g. Lie subgroups of GL(n,R) are called linear Lie groups. For example all classical matrixgroups 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 smoothmap. G is a symmetry group of M. For example rotations are the symmetry group of sphere. O(n + 1)acting on Sn is the symmetry group of spherical geometry. This is an example of a homogeneous geometry.Sn = O(n+ 1)/O(n).

• Integration on Lie groups

• Representation theory

• Internal symmetry groups.

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