3Lie algebras

III Symmetries, Fields and Particles

3.4 The exponential map

So far, we have been talking about vectors in the tangent space of the identity

e

. It turns out that the group structure means this tells us about the tangent

space of all points. To see this, we make the following definition:

Definition

(Left and right translation)

.

For each

h ∈ G

, we define the left and

right translation maps

L

h

: G → G

g 7→ hg,

R

h

: G → G

g 7→ gh.

These maps are bijections, and in fact diffeomorphisms (i.e. smooth maps with

smooth inverses), because they have smooth inverses

L

h

−1

and

R

h

−1

respectively.

In general, there is no reason to prefer left translation over right, or vice

versa. By convention, we will mostly talk about left translation, but the results

work equally well for right translation.

Then since

L

h

(

e

) =

h

, the derivative D

e

L

h

gives us a linear isomorphism

between

T

e

G

and

T

h

G

, with inverse given by D

h

L

h

−1

. Then in particular, if we

are given a single tangent vector

X ∈ L

(

G

), we obtain a tangent vector at all

points in G, i.e. a vector field.

Definition (Vector field). A vector field V of G specifies a tangent vector

V (g) ∈ T

g

G

at each point

g ∈ G

. Suppose we can pick coordinates

{x

i

}

on some subset of

G

,

and write

v(g) = v

i

(g)

∂

∂x

i

∈ T

g

G.

The vector field is smooth if

v

i

(

g

)

∈ R

are all differentiable for any coordinate

chart.

As promised, given any

X ∈ T

e

G

, we can define a vector field by using

L

∗

g

:= D

e

L

g

to move this to all places in the world. More precisely, we can define

V (g) = L

∗

g

(X).

This has the interesting property that if

X

is non-zero, then

V

is non-zero

everywhere, because L

∗

g

is a linear isomorphism. So we found that

Proposition.

Let

G

be a Lie group of dimension

>

0. Then

G

has a nowhere-

vanishing vector field.

This might seem like a useless thing to know, but it tells us that certain

manifolds cannot be made into Lie groups.

Theorem

(Poincare-Hopf theorem)

.

Let

M

be a compact manifold. If

M

has

non-zero Euler characteristic, then any vector field on M has a zero.

The Poincare-Hopf theorem actually tells us how we can count the actual

number of zeroes, but we will not go into that. We will neither prove this, nor

use it for anything useful. But in particular, it has the following immediate

corollary:

Theorem

(Hairy ball theorem)

.

Any smooth vector field on

S

2

has a zero.

More generally, any smooth vector field on S

2n

has a zero.

Thus, it follows that

S

2n

can never be a Lie group. In fact, the full statement

of the Poincare-Hopf theorem implies that if we have a compact Lie group of

dimension 2, then it must be the torus! (we have long classified the list of all

possible compact 2-dimensional manifolds, and we can check that only the torus

works)

That was all just for our own amusement. We go back to serious work.

What does this left-translation map look like when we have a matrix Lie group

G ⊆ Mat

n

(

F

)? For all

h ∈ G

and

X ∈ L

(

G

), we can represent

X

as a matrix in

Mat

n

(F). We then have a concrete representation of the left translation by

L

∗

h

X = hX ∈ T

h

G.

Indeed,

h

acts on

G

by left multiplication, which is a linear map if we view

G

as

a subset of the vector space

Mat

n

(

F

). Then we just note that the derivative of

any linear map is “itself”, and the result follows.

Now we may ask ourselves the question — given a tangent vector

X ∈ T

e

G

,

can we find a path that “always” points in the direction

X

? More concretely,

we want to find a path γ : R → T

e

G such that

˙γ(t) = L

∗

γ(t)

X.

Here we are using

L

γ(t)

to identify

T

γ(t)

G

with

T

e

G

=

L

(

G

). In the case of a

matrix Lie group, this just says that

˙γ(t) = γ(t)X.

We also specify the boundary condition γ(0) = e.

Now this is just an ODE, and by general theory of ODE’s, we know a solution

always exists and is unique. Even better, in the case of a matrix Lie group, there

is a concrete construction of this curve.

Definition

(Exponential)

.

Let

M ∈ Mat

n

(

F

) be a matrix. The exponential is

defined by

exp(M) =

∞

X

`=0

1

`!

M

`

∈ Mat

n

(F).

The convergence properties of these series are very good, just like our usual

exponential.

Theorem.

For any matrix Lie group

G

, the map

exp

restricts to a map

L

(

G

)

→

G.

Proof.

We will not prove this, but on the first example sheet, we will prove this

manually for G = SU(n).

We now let

g(t) = exp(tX).

We claim that this is the curve we were looking for.

To check that it satisfies the desired properties, we simply have to compute

g(0) = exp(0) = I,

and also

dg(t)

dt

=

∞

X

`=1

1

(` − 1)!

t

`−1

X

`

= exp(tX)X = g(t)X.

So we are done.

We now consider the set

S

X

= {exp(tX) : t ∈ R}.

This is an abelian Lie subgroup of G with multiplication given by

exp(tX) exp(sX) = exp((t + s)X)

by the usual proof. These are known as one-parameter subgroups.

Unfortunately, it is not true in general that

exp(X) exp(Y ) = exp(X + Y ),

since the usual proof assumes that

X

and

Y

commute. Instead, what we’ve got

is the Baker–Campbell–Hausdorff formula.

Theorem (Baker–Campbell–Hausdorff formula). We have

exp(X) exp(Y ) = exp

X + Y +

1

2

[X, Y ] +

1

12

([X, [X, Y ]] − [Y, [X, Y ]]) + ···

.

It is possible to find the general formula for all the terms, but it is messy.

We will not prove this.

By the inverse function theorem, we know the map

exp

is locally bijective. So

we know

L

(

G

) completely determines

G

in some neighbourhood of

e

. However,

exp

is not globally bijective. Indeed, we already know that the Lie algebra

doesn’t completely determine the Lie group, as

SO

(3) and

SU

(2) have the same

Lie algebra but are different Lie groups.

In general,

exp

can fail to be bijective in two ways. If

G

is not connected,

then

exp

cannot be surjective, since by continuity, the image of

exp

must be

connected.

Example.

Consider the groups O(

n

) and

SO

(

n

). Then the Lie algebra of O(

n

)

is

o(n) = {X ∈ Mat

n

(R) : X + X

T

= 0}.

So if X ∈ o(n), then tr X = 0. Then we have

det(exp(X)) = exp(tr X) = exp(0) = +1.

So any matrix in the image of

exp

has determinant +1, and hence can only lie

inside SO(n). It turns out that the image of exp is indeed SO(n).

More generally, we have

Proposition.

Let

G

be a Lie group, and

g

be its Lie algebra. Then the image

of g under exp is the connected component of e.

On the other hand,

exp

can also fail to be injective. This happens when

G

has a U(1) subgroup.

Example. Let G = U(1). Then

u(1) = {ix, x ∈ R}

We then have

exp(ix) = e

ix

.

This is certainly not injective. In particular, we have

exp(ix) = exp(i(x + 2π))

for any x.