3Lie algebras
III Symmetries, Fields and Particles
3.3 Lie algebras from Lie groups
We now try to get a Lie algebra from a Lie group G, by considering T
e
(G).
Theorem.
The tangent space of a Lie group
G
at the identity naturally admits
a Lie bracket
[ ·, ·] : T
e
G × T
e
G → T
e
G
such that
L(G) = (T
e
G, [ ·, ·])
is a Lie algebra.
Definition
(Lie algebra of a Lie group)
.
Let
G
be a Lie group. The Lie algebra
of
G
, written
L
(
G
) or
g
, is the tangent space
T
e
G
under the natural Lie bracket.
The general convention is that if the name of a Lie group is in upper case
letters, then the corresponding Lie algebra is the same name with lower case
letters in fraktur font. For example, the Lie group of SO(n) is so(n).
Proof.
We will only prove it for the case of a matrix Lie group
G ⊆ Mat
n
(
F
).
Then
T
I
G
can be naturally identified as a subspace of
Mat
n
(
F
). There is then
an obvious candidate for the Lie bracket — the actual commutator:
[X, Y ] = XY − Y X.
The basic axioms of a Lie algebra can be easily (but painfully) checked.
However, we are not yet done. We have to check that if we take the bracket
of two elements in
T
I
(
G
), then it still stays within
T
I
(
G
). This will be done by
producing a curve in G whose derivative at 0 is the commutator [X, Y ].
In general, let
γ
be a smooth curve in
G
with
γ
(0) =
I
. Then we can Taylor
expand
γ(t) = I + ˙γ(0)t + ¨γ(0)t
2
+ O(t
3
),
Now given
X, Y ∈ T
e
G
, we take curves
γ
1
, γ
2
such that
˙γ
1
(0) =
X
and
˙γ
2
(0) =
Y
.
Consider the curve given by
γ(t) = γ
−1
1
(t)γ
−1
2
(t)γ
1
(t)γ
2
(t) ∈ G.
We can Taylor expand this to find that
γ(t) = I + [X, Y ]t
2
+ O(t
3
).
This isn’t too helpful, as [
X, Y
] is not the coefficient of
t
. We now do the slightly
dodgy step, where we consider the curve
˜γ(t) = γ(
√
t) = I + [X, Y ]t + O(t
3/2
).
Now this is only defined for
t ≥
0, but it is good enough, and we see that its
derivative at
t
= 0 is [
X, Y
]. So the commutator is in
T
I
(
G
). So we know that
L(G) is a Lie algebra.
Example.
Let
G
=
GL
(
n, F
), where
F
=
R
or
C
. Then
L
(
GL
(
n, F
)) =
gl
(
n, F
) =
Mat
n
(
F
) because we know it must be an
n × n
-dimensional sub-
space of Mat
n
(F).
More generally, for a vector space
V
, say of dimension
n
, we can consider
the group of invertible linear maps
V → V
, written
GL
(
V
). By picking a basis
of
V
, we can construct an isomorphism
GL
(
V
)
∼
=
GL
(
n, F
), and this gives us a
smooth structure on
GL
(
V
) (this does not depend on the basis chosen). Then
the Lie algebra gl(V ) is the collection of all linear maps V → V .
Example. If G = SO(2), then the curves are of the form
g(t) = M(θ(t)) =
cos θ(t) −sin θ(t)
sin θ(t) cos θ(t)
∈ SO(2).
So we have
˙g(0) =
0 −1
1 0
˙
θ(0).
Since the Lie algebra has dimension 1, these are all the matrices in the Lie
algebra. So the Lie algebra is given by
so(2) =
0 −c
c 0
, c ∈ R
.
Example.
More generally, suppose
G
=
SO
(
n
), and we have a path
R
(
t
)
∈
SO(n).
By definition, we have
R
T
(t)R(t) = I.
Differentiating gives
˙
R
T
(t)R(t) + R
T
(t)
˙
R(t) = 0.
for all t ∈ R. Evaluating at t = 0, and noting that R(0) = I, we have
X
T
+ X = 0,
where
X
=
˙
R
(0) is a tangent vector. There are no further constraints from
demanding that
det R
= +1, since this is obeyed anyway for any matrix in O(
n
)
near I.
By dimension counting, we know the antisymmetric matrices are exactly the
matrices in L(O(n)) or L(SO(n)). So we have
o(n) = so(n) = {X ∈ Mat
n
(R) : X
T
= −X}.
Example.
Consider
G
=
SU
(
n
). Suppose we have a path
U
(
t
)
∈ SU
(
n
), with
U(0) = I. Then we have
U
†
(t)U(t) = I.
Then again by differentiation, we obtain
Z
†
+ Z = 0,
where Z =
˙
U(0) ∈ su(n). So we must have
su(n) ⊂ {Z ∈ Mat
n
(C) : Z
†
= −Z}.
What does the condition
det U
(
t
) = 1 give us? We can do a Taylor expansion by
det U(t) = 1 + tr Z · t + O(t
2
).
So requiring that det U(t) = 1 gives the condition
tr Z = 0.
By dimension counting, we know traceless anti-Hermitian matrices are all the
elements in the Lie algebra. So we have
su(n) = {Z ∈ Mat
n
(C), Z
†
= −Z, tr Z = 0}.
Example.
We look at
SU
(2) in detail. We know that
su
(2) is the 2
×
2 traceless
anti-Hermitian matrices.
These are given by multiples of the Pauli matrices
σ
j
, for
j
= 1
,
2
,
3, satisfying
σ
i
σ
j
= δ
ij
I + iε
ijk
σ
k
.
They can be written explicitly as
σ
1
=
0 1
1 0
, σ
2
=
0 −i
i 0
, σ
3
=
1 0
0 −1
One can check manually that the generators for the Lie algebra are given by
T
a
= −
1
2
iσ
a
.
Indeed, each
T
a
is in
su
(2), and they are independent. Since we know
dim su
(2) =
3, these must generate everything.
We have
[T
a
, T
b
] = −
1
4
[σ
a
, σ
b
] = −
1
2
iε
abc
σ
c
= f
ab
c
T
c
,
where the structure constants are
f
ab
c
= ε
abc
.
Example.
Take
G
=
SO
(3). Then
so
(3) is the space of 3
×
3 real anti-symmetric
matrices, which one can manually check are generated by
˜
T
1
=
0 0 0
0 0 −1
0 1 0
,
˜
T
2
=
0 0 1
0 0 0
−1 0 0
,
˜
T
3
=
0 −1 0
1 0 0
0 0 0
We then have
(
˜
T
a
)
bc
= −ε
abc
.
Then the structure constants are
[
˜
T
a
,
˜
T
b
] = f
ab
c
˜
T
c
,
where
f
ab
c
= ε
abc
.
Note that the structure constants are the same! Since the structure constants
completely determine the brackets of the Lie algebra, if the structure constants
are the same, then the Lie algebras are isomorphic. Of course, the structure
constants depend on which basis we choose. So the real statement is that if
there are some bases in which the structure constants are equal, then the Lie
algebras are isomorphic.
So we get that
so
(3)
∼
=
su
(2), but
SO
(3) is not isomorphic to
SU
(2). Indeed,
the underlying manifold is
SU
(2) is the 3-sphere, but the underlying manifold of
SO
(3) has a fancy construction. They are not even topologically homeomorphic,
since SU(2) is simply connected, but SO(3) is not. More precisely, we have
π
1
(SO(3)) = Z/2Z
π
1
(SU(2)) = {0}.
So we see that we don’t have a perfect correspondence between Lie algebras and
Lie groups. However, usually, two Lie groups with the same Lie algebra have
some covering relation. For example, in this case we have
SO(3) =
SU(2)
Z/2Z
,
where
Z/2Z = {I, −I} ⊆ SU(2)
is the center of SU(2).
We can explicitly construct this bijection as follows. We define the map
d : SU(2) → SO(3) by
d(A)
ij
=
1
2
tr(σ
i
Aσ
j
A
†
) ∈ SO(3).
This is globally a 2-to-1 map. It is easy to see that
d(A) = d(−A),
and conversely if
d
(
A
) =
d
(
B
), then
A
=
−B
. By the first isomorphism theorem,
this gives an isomorphism
SO(3) =
SU(2)
Z/2Z
,
where Z/2Z = {I, −I} is the center of SU(2).
Geometrically, we know
M
(
SU
(2))
∼
=
S
3
. Then the manifold of
SO
(3) is
obtained by identifying antipodal points of the manifold.