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Lie Groups I

Third Assistantship

Matrix exponential

Recall that if  K   is  R  or  C  and  Mn(K) is the set of  n × n−matrices with coefficients in  K  theexponential map

exp : Mn(K) → Mn(K),   exp(A) =∞i=0

Ai

i!

is well defined (for example by the Comparison Test).The following problems shows some of its basic properties

Problem 1.   Prove the following:1.-If  A, B  ∈ Mn(K) commute then exp(A + B) = exp(A) exp(B)2.-exp(Mn(K)) ⊆  Gln(K)3.-If  B  ∈ Gln(K) then  B exp(A)B

−1 = exp(BAB−1).

Solution   1.- For all   i   ∈   N   we have (A +  B )i = i

 j=0

i j

A jBi− j so we can use the Cauchy

Product Formula

exp(A + B) =

∞i=0

(A + B)

i

i!

=∞i=0

i j=0

i j

A jBi− j

i!

=∞i=0

i j=0

A j

 j!

Bi− j

(i − j)!

=

  ∞i=0

Ai

i!

  ∞i=0

Bi

i!

= exp(A) exp(B).

2.-From Problem 1.1 we have exp(A)exp(−A) = exp(A − A) = exp(0) = I d.3.-For all  i ∈ N we have (BAB−1)i = B AiB−1 so

B exp(A)B−1 = B

  ∞i=0

Ai

i!

B−1 =

∞i=0

BAiB−1

i!  = exp(BAB−1).

The next problem shows help to determine the Lie algebra of linear Lie groups.

Problem 2.  There is an open neighbourhood  U  of the origin such that

exp |U   : U  → exp(U ) ⊆  Gln(K)

is a diffeomorphism.

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Solution.   exp is a smooth map because it is a power series convergent in all  Mn(K). Thedifferential of exp in  A, call it  d exp(A), is the limit of the differentials of the convergent powerseries in   A   (so it exists and is smooth!). Finally   d exp(0) =   Id   so by the Inverse FunctionTheorem the result follows.Recall that the exponential map for arbitrary  K−Lie groups (G, ·) is defined in the followingway:

For  v ∈  Lie(G) there is a unique homomorphism of Lie groups

γ v   : (R, +) →  G

such that it solves the initial value problem

γ v(0) = 1 and   γ 

v(t) = γ v(t)x   for all t ∈ R   (1)

DefineexpG  : Lie(G) →  G,   exp(v) = γ v(1).

The next problem shows that in the linear case both definitions coincide.

Problem 3.   If   G   ⊆   GLn(K) is a Lie subgroup, then both definitions of exponential mapcoincide.

Solution.   Let  A ∈  Lie(G). We are going to show that if  A ∈  Lie(G) then

γ A : R → G, γ A(t) = exp(tA)

is a solution of (1). Indeed by Problem 1.1

γ 

A(t) = limh→0

γ A(t + h) − γ A(t)

h

= limh→0

exp((t + h)A) − exp(tA)

h

= limh→0

exp(tA)exp(hA) − exp(tA)

h

= exp(tA) limh→0

exp(hA) − Id

h

= γ A(t)γ 

A(0) = γ A(t)A

and  γ A(1) = exp(A) so the definitions coincide.

As we mentioned Problem 3 provides an important tool to determine the Lie algebras of G   ⊆   GLn(K). The idea is that assuming   U   and   G   are connected then Lie(G)   ⊆   Mn(K)and expG |U ∩Lie(G)  is a diffeomorphism; In particular U ∩ Lie(G) is an open set that contains theorigin so for all  v  ∈ Lie(G) there is a  λ > 0 such that  λv  ∈  U  ∩ Lie(G) so Lie(G) is determined.The following problem is a nice example of this. (In this problem we assume, as we alreadydid, that  Lie(GLn(R)) = Mn(R)).

Problem 4.  Show that

Lie(On(R)) = Lie(SOn(R)) = {A ∈ Mn(R) :   A + AT  = 0}.

Solution.   Since  SOn(R) is the connected component of  Id in  On(R) we have the first equality.Note that {A ∈ Mn(R) :   A + A

T  = 0} is a Lie subalgebra of  Mn(R) (for example if  A, B  are in

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the Lie algebra [A, B]+[A, B]T  = AB −BA+(AB)T −(BA)T  = AB −BA+(−BT A+AT B) = 0).If  U   is like in Problem 2, then

U  ∩ {A ∈ Mn(R) :   A + AT  = 0} ⊆ U  ∩ Lie(SOn(R));

Indeed if  A  ∈  U  ∩ {A  ∈  Mn(R) :   A +  AT  = 0}  then  A + AT  = 0 and  A, AT  commute so by

Problem 1.1 exp(A)T 

= exp(AT 

) = exp(A)−1

and as in Problem 2

exp |U ∩Lie(SOn(R)) : U  ∩ Lie(SOn(R)) −→ U  ∩ SOn(R)

is a diffeomorphism. In particular

{A ∈ Mn(R) :   A + AT  = 0} ⊆ Lie(SOn(R));

On the other hand

dim{A ∈ Mn(R) :   A + AT  = 0} =

n

2

 = dim SOn(R) = dim Lie(SOn(R)).

and the result follows.

Lie functor in the Linear case

As we saw last section the exponential map in the linear group case has an explicit expressionso it is easier to study the Lie functor. In this section linear group means a Lie subgroup of GLn(R).Given a linear Lie group  G ⊆  GLn(K) there is a Lie algebra associated with  G

G −→ Lie(G).

Let  G1, G2 ⊆ GLn(K) and  φ :  G1  → G2  a morphism of Lie groups then it induces a morphismbetween their Lie algebras. First we give two formulas (we do not prove them but they are easyto show; If you wish you can prove it or you can consult the proof in   Structure and Geometry of Lie groups, Hilgert and Neeb, page 59 ) that will be used in the proof: If  A, B  ∈ Mn(K)

limn→∞

exp

1

nA

exp

1

nB

n= exp(A + B) (2)

and

limn→∞

exp

1n

A

exp

1n

B

exp

−  1n

A

exp

−  1n

Bn

= exp([A, B]) (3)

Problema 5.   If   φ   :   G1   →   G2   is a morphism of linear Lie groups, then there is a funcctionLie(φ) such that the following diagram commutes

G1φ

−−−→   G2expG1expG2

Lie(G1)  Lie(φ)−−−→   Lie(G2)

Solution. First define the function  γ A(t) = expG1(tA) and note that

(φ ◦ γ A)(t) = expG2(tB) (4)

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for

B  :=  d

dt

t=0

(φ ◦ γ A)(t)

this can be seen taking   ddt

0

expG2(tB) in (4). On the other hand if we define

Lie(φ)(A) :=  d

dt

0

(φ ◦ γ A)(t)

we have by (4) that (φ ◦ γ A)(t) is a solution of the initial value problem

d

dt

t=s

γ (t) = Lie(φ)(A)γ (s), γ (0) = 1;

By definition of expG2(expG2 ◦Lie(φ))(A) = (φ ◦ expG1)(A)

Problema 6.  With the notation and definitions as above

Lie(φ) : Lie(G1) →  Lie(G2),

is a morphism of Lie algebras.

Solution.  First note that for all  λ ∈ K and  A ∈  Lie(G1)

Lie(φ)(λA) = λLie(φ)(A)

We prove Lie(φ) is additive. For any A, B  ∈ Lie(G1) take λ ∈K

such that Lie(φ)(λA), Lie(φ)(λB), Lie(φ)(λB)) ∈  U   where  U  is as in Problem 2.

expG2 Lie(φ)(λA + λB) = φ(expG1(λA + λB))

= φ

  limn→∞

expG1

1

nλA

expG1

1

nλB

n  by (2)

= limn→∞

φ

expG1

1

nλA

φ

expG1

1

nλB

n

= limn→∞

expG2

Lie(φ)

1

nλA

expG2

Lie(φ)

1

nλB

n

= expG2(Lie(φ)(λA) + Lie(λB))) by (2);

Since expG2 |U  is bijective we have

Lie(φ)(λA + λB) = Lie(φ)(λA) + Lie(φ)(λB)

soLie(φ)(A + B) = Lie(φ)(A) + Lie(φ)(B)

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Finally for any A, B  ∈ Lie(G1) take λ ∈ K such that Lie(φ)(λA), Lie(φ)(λB), Lie(φ)(λ[A+B]) ∈U   with  U  as in Problem 2

expG2(Lie(φ)(λ[A, B])) = φ(expG1(λ[A, B]))

= φ

  limn→∞

expG1

1

nλA

expG1

1

nλB

n

φ

  limn→∞

expG1

−

 1

nλA

expG1

−

 1

nλB

n  by (3)

= limn→∞

φ

expG1

1

nλA

φ

expG1

1

nλB

n

limn→∞

φ

expG1

−

 1

nλA

φ

expG1

−

 1

nλB

n

= limn→∞

expG2

Lie(φ)

1

nλA

expG2

Lie(φ)

1

nλB

n

limn→∞

expG2

Lie(φ)

− 1

n λA

expG2

Lie(φ)

− 1

n λBn

= expG2([Lie(φ)(λA), Lie(φ)(λB)]) by (3);

Since expG2 |U  is bijective we have

Lie(φ)(λ[A, B]) = [Lie(φ)(λA), Lie(φ)(λB)]

soLie(φ)([A, B]) = [Lie(φ)(A), Lie(φ)(B)].

It is not difficult to prove that Lie(φ) is the unique linear map such that the diagram of Problem5 commutes and we leave it as an exercise.Problema 7.   If  φ :  G1 → G2  and  ϕ :  G2 → G3  are homomorphisms of linear Lie groups, thenLie(ϕ ◦ φ) = Lie(ϕ) ◦ Lie(φ).

Solution. From Problem 6 we have

G1φ

−−−→   G2expG1

expG2

Lie(G1)  Lie(φ)

−−−→   Lie(G2)

andG2

ϕ−−−→   G3expG2

expG3Lie(G2)

  Lie(ϕ)−−−→   Lie(G3)

commute so the diagram

G1ϕ◦φ

−−−→   G3

expG1 expG2Lie(G1)

  Lie(ϕ)◦Lie(φ)−−−−−−−→   Lie(G3)

commutes and the uniqueness of the linear map Lie(ϕ ◦ φ) implies the result.

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Extra Exercises

Problema 8.  Show exp : Mn(K) → Mn(K) is not a polynomial.

Problema 9.  Find a Lie subalgebra  g ⊆ Mn(C) isomorphic to  Lie(U n(C)).

Problema 10.  Prove (2) and (3).

Problema 11.   Prove Lie(φ) is the unique linear map such that the diagram of Problem 5commutes.

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