2Lie groups

III Symmetries, Fields and Particles 2.2 Matrix Lie groups
We write
Mat
n
(
F
) for the set of
n×n
matrices with entries in a field
F
(usually
R
or
C
). Matrix multiplication is certainly associative, and has an identity, namely
the identity I. However, it doesn’t always have inverses not all matrices are
invertible! So this is not a group (instead, we call it a monoid). Thus, we are
led to consider the general linear group:
Definition (General linear group). The general linear group is
GL(n, F) = {M Mat
n
(F) : det M 6= 0}.
This is closed under multiplication since the determinant is multiplicative,
and matrices with non-zero determinant are invertible.
Definition (Special linear group). The special linear group is
SL(n, F) = {M Mat
n
(F) : det M = 1} GL(n, F).
While these are obviously groups, less obviously, these are in fact Lie groups!
In the remainder of the section, we are going to casually claim that all our favorite
matrix groups are Lie groups. Proving these take some work and machinery,
and we will not bother ourselves with that too much. However, we can do it
explicitly for certain special cases:
Example. Explicitly, we can write
SL(2, R) =

a b
c d
: a, b, c, d R, ad bc = 1
.
The identity is the matrix with a = d = 1 and b = c = 0. For a 6= 0, we have
d =
1 + bc
a
.
This gives us a coordinate patch for all points where
a 6
= 0 in terms of
b, c, a
,
which, in particular, contains the identity
I
. By considering the case where
b 6
= 0, we obtain a separate coordinate chart, and these together cover all of
SL(2, R), as a matrix in SL(2, R) cannot have a = b = 0.
Thus, we see that SL(2, R) has dimension 3.
In general, by a similar counting argument, we have
dim(SL(n, R)) = n
2
1 dim(SL(n, C)) = 2n
2
2
dim(GL(n, R)) = n
2
, dim(GL(n, C)) = 2n
2
.
Most of the time, we are interested in matrix Lie groups, which will be subgroups
of GL(n, R).
Subgroups of GL(n, R)
Lemma. The general linear group:
GL(n, R) = {M Mat
n
(R) : det M 6= 0}
and orthogonal group:
O(n) = {M GL(n, R) : M
T
M = I}
are Lie groups.
Note that we write O(
n
n, R
) since orthogonal matrices make
sense only when talking about real matrices.
The orthogonal matrices are those that preserve the lengths of vectors. Indeed,
for v R
n
, we have
|Mv|
2
= v
T
M
T
Mv = v
T
v = |v|
2
.
We notice something interesting. If M O(n), we have
1 = det(I) = det(M
T
M) = det(M)
2
.
So
det
(
M
) =
±
1. Now
det
is a continuous function, and it is easy to see that
det
takes both
±
1. So O(
n
) has (at least) two connected components. Only one of
these pieces contains the identity, namely the piece
det M
= 1. We might expect
this to be a group on its own right, and indeed it is, because
det
is multiplicative.
Lemma. The special orthogonal group SO(n):
SO(n) = {M O(n) : det M = 1}
is a Lie group.
Given a frame
{v
1
, ··· , v
n
}
in
R
n
(i.e. an ordered basis), any orthogonal
matrix
M
O(
n
) acts on it to give another frame
v
a
R
n
7→ v
0
a
7→ Mv
a
R
n
.
Definition
(Volume element)
.
Given a frame
{v
1
, ··· , v
n
}
in
R
n
, the volume
element is
Ω = ε
i
1
...i
n
v
i
1
1
v
i
2
2
···v
i
n
n
.
By direct computation, we see that an orthogonal matrix preserves the sign
of the volume element iff its determinant is +1, i.e. M SO(n).
We now want to find a more explicit description of these Lie groups, at
least in low dimensions. Often, it is helpful to classify these matrices by their
eigenvalues:
Definition
(Eigenvalue)
.
A complex number
λ
is an eigenvalue of
M M
(
n
)
if there is some (possibly complex) vector v
λ
6= 0 such that
Mv
λ
= λv
λ
.
Theorem.
Let
M
be a real matrix. Then
λ
is an eigenvalue iff
λ
is an eigenvalue.
Moreover, if M is orthogonal, then |λ|
2
= 1.
Proof.
Suppose
Mv
λ
=
λv
λ
. Then applying the complex conjugate gives
Mv
λ
= λ
v
λ
.
Now suppose
M
is orthogonal. Then
Mv
λ
=
λv
λ
for some non-zero
v
λ
. We
take the norm to obtain
|Mv
λ
|
=
|λ||v
λ
|
. Using the fact that
|v
λ
|
=
|Mv
λ
|
, we
have |λ| = 1. So done.
Example.
Let
M SO
(2). Since
det M
= 1, the eigenvalues must be of the
form λ = e
, e
. In this case, we have
M = M(θ) =
cos θ sin θ
sin θ cos θ
,
where θ is the rotation angle in S
1
. Here we have
M(θ
1
)M(θ
2
) = M(θ
2
)M(θ
1
) = M(θ
1
+ θ
2
).
So we have M(SO(2)) = S
1
.
Example.
Consider
G
=
SO
(3). Suppose
M SO
(3). Since
det M
= +1, and
the eigenvalues have to come in complex conjugate pairs, we know one of them
must be 1. Then the other two must be of the form e
, e
, where θ S
1
.
We pick a normalized eigenvector
n
for
λ
= 1. Then
Mn
=
n
, and
n · n
= 1.
This is known as the axis of rotation. Similarly,
θ
is the angle of rotation. We
write M (n, θ) for this matrix, and it turns out this is
M(n, θ)
ij
= cos θδ
ij
+ (1 cos θ)n
i
n
j
sin θε
ijk
n
k
.
Note that this does not uniquely specify a matrix. We have
M(n, 2π θ) = M(n, θ).
Thus, to uniquely specify a matrix, we need to restrict the range of
θ
to 0
θ π
,
with the further identification that
(n, π) (n, π).
Also note that M(n, 0) = I for any n.
Given such a matrix, we can produce a vector
w
=
θn
. Then
w
lies in the
region
B
3
= {w R
3
: kwk π} R
3
.
This has a boundary
B
3
= {w R
3
: kwk = π}
=
S
2
.
Now we identify antipodal points on
B
3
. Then each vector in the resulting
space corresponds to exactly one element of SO(3).
Subgroups of GL(n, C)
We can similarly consider subgroups of GL(n, C). Common examples include
Definition (Unitary group). The unitary group is defined by
U(n) = {U GL(n, C) : U
U = I}.
These are important in physics, because unitary matrices are exactly those
that preserve the norms of vectors, namely kvk = kUvk for all v.
Again, if
U
U
= 1, then
|det
(
U
)
|
2
= 1. So
det U
=
e
for some
δ R
.
Unlike the real case, the determinant can now take a continuous range of values,
and this no longer disconnects the group. In fact, U(n) is indeed connected.
Definition (Special unitary group). The special unitary group is defined by
SU(n) = {U U(n) : det U = 1}.
It is an easy exercise to show that
dim[U(n)] = 2n
2
n
2
= n
2
.
For
SU
(
n
), the determinant condition imposes an additional single constraint,
so we have
dim[SU(n)] = n
2
1.
Example. Consider the group G = U(1). This is given by
U(1) = {z C : |z| = 1}.
Therefore we have
M[U(1)] = S
1
.
However, we also know another Lie group with underlying manifold
S
1
, namely
SO(2). So are they “the same”?