4Representations of Lie algebras

III Symmetries, Fields and Particles 4.3 Representations of su(2)
We are now going to study the complex representations of
su
(2), which are
equivalently the representations of the complexification
su
C
(2). In this section,
we will write
su
(2) when we actually mean
su
C
(2) for brevity. There are a
(i)
We’ve already done this before, in the guise of “spin”/“angular momentum”
in quantum mechanics.
(ii)
The representations are pretty easy to classify, and form a motivation for
our later general classification of representations of complex simple Lie
algebras.
(iii)
Our later work on the study of simple Lie algebras will be based on our
knowledge of representations of su(2).
Recall that the uncomplexified su(2) has a basis
su(2) = span
R
T
a
=
1
2
a
: a = 1, 2, 3
,
where
σ
a
are the Pauli matrices. Since the Pauli matrices are independent over
C, they give us a basis of the complexification of su(2):
su
C
(2) = span
C
{σ
a
: a = 1, 2, 3}.
However, it is more convenient to use the following basis:
H = σ
3
=
1 0
0 1
, E
±
=
1
2
(σ
1
±
2
) =
0 1
0 0
,
0 0
1 0
.
We then have
[H, E
±
] = ±2E
±
, [E
+
, E
] = H.
We can rewrite the first relation as saying
H
(E
±
) = ±2E
±
.
Together with the trivial result that
H
(
H
) = 0, we know
H
has eigenvalues
±2 and 0, with eigenvectors E
±
, H. These are known as roots of su(2).
Now let’s look at irreducible representations of
su
(2). Suppose (
V, ρ
) is a
representation. By linear algebra, we know ρ(H) has an eigenvector v
λ
, say
ρ(H)v
λ
= λv
λ
.
The eigenvalues of
ρ
(
H
) are known as the weights of the representation
ρ
. The
operators E
±
are known as step operators. We have
ρ(H)ρ(E
±
)v
λ
= (ρ(E
±
)ρ(H) + [ρ(H), ρ(E
±
)])v
λ
= (ρ(E
±
)ρ(H) + ρ([H, E
±
])v
λ
= (λ ± 2)ρ(E
±
)v
λ
.
So
ρ
(
E
±
)
v
λ
are also eigenvectors of
ρ
(
H
) with eigenvalues
λ ±
2, provided
ρ
(
E
±
)
v
λ
are non-zero. This constrains what can happen in a finite-dimensional
representation. By finite-dimensionality, we cannot have infinitely many eigen-
values. So at some point, we have to stop. In other words, a finite-dimensional
representation must have a highest weight Λ C, with
ρ(H)v
Λ
= Λv
Λ
, ρ(E
+
)v
Λ
= 0.
Now if we start off with
v
Λ
we can keep applying
E
to get the lower weights.
Indeed, we get
v
Λ2n
= (ρ(E
))
n
v
Λ
for each
n
. Again, this sequence must terminate somewhere, as we only have
finitely many eigenvectors. We might think that the irrep would consist of the
basis vectors
{v
Λ
, v
Λ2
, v
Λ4
, ··· , v
Λ2n
}.
However, we need to make sure we don’t create something new when we act on
these by
ρ
(
E
+
) again. We would certainly get that
ρ
(
E
+
)
v
Λ2n
is an eigenvector
of eigenvalue Λ
2
n
+ 2, but we might get some other eigenvector that is not
v
Λ2n+2
. So we have to check.
We can compute
ρ(E
+
)v
Λ2n
= ρ(E
+
)ρ(E
)v
Λ2n+2
= (ρ(E
)ρ(E
+
) + [ρ(E
+
), ρ(E
)])v
Λ2n+2
= ρ(E
)ρ(E
+
)v
Λ2n+2
+ 2n + 2)v
Λ2n+2
.
using the fact that [
E
+
, E
] =
H
. We now get a recursive relation. We can
analyze this by looking at small cases. When n = 1, the first term is
ρ(E
)ρ(E
+
)v
Λ2n+2
= ρ(E
)ρ(E
+
)v
Λ
= 0,
by definition of v
Λ
. So we have
ρ(E
+
)v
Λ2
= Λv
Λ
.
When n = 2, we have
ρ(E
+
)v
Λ4
= ρ(E
)ρ(E
+
)v
Λ2
+ 2)v
Λ2
= (2Λ 2)v
Λ2
.
In general, ρ(E
+
)v
Λ2n
is some multiple of v
Λ2n+2
, say
ρ(E
+
)v
Λ2n
= r
n
V
Λ2n+2
.
Plugging this into our equation gives
r
n
= r
n1
+ Λ 2n + 2,
with the boundary condition
ρ
(
E
+
)
v
Λ
= 0, i.e.
r
1
= Λ. If we solve this recurrence
relation, we determine an explicit relation for r
n
given by
r
n
= (Λ + 1 n)n.
Now returning to the problem that the sequence must terminate, we figure that
we also need to have a lowest weight of the form Λ
2
n
. Hence we must have
some non-zero vector v
Λ2n
6= 0 such that
ρ(E
)v
Λ2n
= 0.
For this to be true, we need
r
N+1
= 0 for some non-negative integer
N
. So we
have
+ 1 (N + 1))(N + 1) = 0.
This is equivalent to saying
N)(N + 1) = 0.
Since
N
+ 1 is a non-negative integer, we must have Λ
N
= 0, i.e. Λ =
N
. So
in fact the highest weight is an integer!
In summary, we’ve got
Proposition.
The finite-dimensional irreducible representations of
su
(2) are
labelled by Λ Z
0
, which we call ρ
Λ
, with weights given by
{−Λ, Λ + 2, ··· , Λ 2, Λ}.
The weights are all non-degenerate, i.e. each only has one eigenvector. We have
dim(ρ
Λ
) = Λ + 1.
Proof.
We’ve done most of the work. Given any irrep, we can pick any eigenvector
of
ρ
(
H
) and keep applying
E
+
to get a highest weight vector
v
Λ
, then the above
computations show that
{v
Λ
, v
Λ2
, ··· , v
Λ
}
is a subspace of the irrep closed under the action of
ρ
. By irreducibility, this
must be the whole of the representation space.
The representation
ρ
0
is the trivial representation;
ρ
1
is the fundamental one,
and ρ
2
Now what do these tell us about representations of the Lie group
SU
(2)? We
know
SU
(2) is simply connected, so by our discussion previously, we know the
complex representations of
SU
(2) are exactly the representations of
su
(2). This
is not too exciting.
We know there are other Lie groups with Lie algebra
su
(2), namely
SO
(3).
Now that
SO
(3) is not simply connected, when does a representation of
su
(2)
give us a representation of SO(3)? Recall that we had
SO(3)
=
SU(2)
Z/2Z
.
So an element in
SO
(3) is a pair
{A, A}
of elements in
SU
(2). Given a
representation
ρ
Λ
of
su
(2), we obtain a corresponding representation
D
Λ
of
SU
(2). Now we get a representation of
SO
(3) if and only if
D
Λ
respects the
identification between A and A. In particular, we need
D
Λ
(I) = D
Λ
(I). ()
On the other hand, if this were true, then multiplying both sides by
D
Λ
(
A
) we
get
D
Λ
(A) = D
Λ
(A).
So (
) is a necessary and sufficient condition for
D
Λ
to descend to a representation
of SO(3).
We know that
I = exp(H).
So we have
D
Λ
(I) = exp(ρ
Λ
(H)).
We know that ρ
Λ
has eigenvalues Λ S
Λ
. So D
Λ
(I) has eigenvalues
exp(λ) = (1)
λ
= (1)
Λ
,
since
λ
and Λ differ by an even integer. So we know
D
Λ
(
I
) =
D
Λ
(
I
) if and
only if Λ is even. In other words, we get a representation of
SO
(3) iff Λ is even.
In physics, we have already seen this as integer and half-integer spin. We
have integer spin exactly when we have a representation of
SO
(3). The fact that
half-integer spin particles exist means that “spin” is really about
SU
(2), rather
than SO(3).