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
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.