15Representations of compact groups
II Representation Theory
15.2 Representations of SO(3), SU(2) and U (2)
We’ve now got a complete list of representations of
S
1
and
SU
(2). We see if
we can make some deductions about some related groups. We will not give too
many details about these groups, and the proofs are just sketches. Moreover, we
will rely on the following facts which we shall not prove:
Proposition. There are isomorphisms of topological groups:
(i) SO(3)
∼
=
SU(2)/{±I} = PSU(2)
(ii) SO(4)
∼
=
SU(2) × SU(2)/{±(I, I)}
(iii) U(2)
∼
=
U(1) × SU(2)/{±(I, I)}
All maps are group isomorphisms, but in fact also homeomorphisms. To show
this, we can use the fact that a continuous bijection from a Hausdorff space to a
compact space is automatically a homeomorphism.
Assuming this is true, we obtain the following corollary;
Corollary. Every irreducible representation of SO(3) has the following form:
ρ
2m
: SO(3) → GL(V
2m
),
for some m ≥ 0, where V
n
are the irreducible representations of SU(2).
Proof.
Irreducible representations of
SO
(3) correspond to irreducible representa-
tions of
SU
(2) such that
−I
acts trivially by lifting. But
−I
acts on
V
n
as
−
1
when n is odd, and as 1 when n is even, since
ρ(−I) =
(−1)
n
(−1)
n−2
.
.
.
(−1)
−n
= (−1)
n
I.
For the sake of completeness, we provide a (sketch) proof of the isomorphism
SO(3)
∼
=
SU(2)/{±I}.
Proposition. SO(3)
∼
=
SU(2)/{±I}.
Proof sketch.
Recall that
SU
(2) can be viewed as the sphere of unit norm
quaternions H
∼
=
R
4
.
Let
H
0
= {A ∈ H : tr A = 0}.
These are the “pure” quaternions. This is a three-dimensional subspace of
H
. It
is not hard to see this is
H
0
= R
i 0
0 −i
,
0 1
−1 0
,
0 i
i 0
= R hi, j, ki,
where Rh···i is the R-span of the things.
This is equipped with the norm
kAk
2
= det A.
This gives a nice 3-dimensional Euclidean space, and
SU
(2) acts as isometries
on H
0
by conjugation, i.e.
X · A = XAX
−1
,
giving a group homomorphism
ϕ : SU(2) → O(3),
and the kernel of this map is
Z
(
SU
(2)) =
{±I}
. We also know that
SU
(2) is
compact, and O(3) is Hausdorff. Hence the continuous group isomorphism
¯ϕ : SU(2)/{±I} → im ϕ
is a homeomorphism. It remains to show that im ϕ = SO(3).
But we know
SU
(2) is connected, and
det
(
ϕ
(
X
)) is a continuous function
that can only take values 1 or
−
1. So
det
(
ϕ
(
X
)) is either always 1 or always
−
1.
But det(ϕ(I)) = 1. So we know det(ϕ(X)) = 1 for all X. Hence im ϕ ≤ SO(3).
To show that equality indeed holds, we have to show that all possible rotations
in
H
0
are possible. We first show all rotations in the
i, j
-plane are implemented
by elements of the form
a
+
bk
, and similarly for any permutation of
i, j, k
. Since
all such rotations generate SO(3), we are then done. Now consider
e
iθ
0
0 e
−iθ
ai b
−
¯
b −ai
e
iθ
0
0 e
iθ
=
ai e
2iθ
b
−
¯
be
−2iθ
−ai
.
So
e
iθ
0
0 e
−iθ
acts on
Rhi, j, ki
by a rotation in the (
j, k
)-plane through an angle 2
θ
. We can
check that
cos θ sin θ
−sin θ cos θ
,
cos θ i sin θ
i sin θ cos θ
act by rotation of 2
θ
in the (
i, k
)-plane and (
i, j
)-plane respectively. So done.
We can adapt this proof to prove the other isomorphisms. However, it is
slightly more difficult to get the irreducible representations, since it involves
taking some direct products. We need a result about products
G × H
of two
compact groups. Similar to the finite case, we get the complete list of irreducible
representations by taking the tensor products
V ⊗ W
, where
V
and
W
range
over the irreducibles of G and H independently.
We will just assert the results.
Proposition.
The complete list of irreducible representations of
SO
(4) is
ρ
m
×ρ
n
,
where m, n > 0 and m ≡ n (mod 2).
Proposition. The complete list of irreducible representations of U(2) is
det
⊗m
⊗ ρ
n
,
where
m, n ∈ Z
and
n ≥
0, and
det
is the obvious one-dimensional representation.