4Representations of Lie algebras

III Symmetries, Fields and Particles

4.4 New representations from old
We are now going to look at different ways of obtaining new representations
Definition
(Conjugate representation)
.
Let
ρ
be a representation of a real Lie
algebra g on C
n
. We define the conjugate representation by
¯ρ(X) = ρ(X)
for all X g.
Note that this need not be an actual new representation.
To obtain more interesting new representations, we recall the following
definitions from linear algebra:
Definition
(Direct sum)
.
Let
V, W
be vector spaces. The direct sum
V W
is
given by
V W = {v w : v V, w W }
with operations defined by
(v
1
w
1
) + (v
2
w
2
) = (v
1
+ v
2
) (w
1
+ w
2
)
λ(v w) = (λv) (λw).
We often suggestively write v w as v + w. This has dimension
dim(V W ) = dim V + dim W.
Definition
(Sum representation)
.
Suppose
ρ
1
and
ρ
2
are representations of
g
with representation spaces
V
1
and
V
2
of dimensions
d
1
and
d
2
. Then
V
1
V
2
is
a representation space with representation ρ
1
ρ
2
given by
(ρ
1
ρ
2
)(X) · (v
1
v
2
) = (ρ
1
(X)(v
1
)) (ρ
2
(X)(v
2
)).
In coordinates, if
R
i
is the matrix for
ρ
i
, then the matrix of
ρ
1
ρ
2
is given by
(R
1
R
2
)(X) =
R
1
(X) 0
0 R
2
(X)
The dimension of this representation is d
1
+ d
2
.
Of course, sum representations are in general not irreducible! However, they
are still rather easy to understand, since they decompose into smaller, nicer bits.
In contrast, tensor products do not behave as nicely.
Definition
(Tensor product)
.
Let
V, W
be vector spaces. The tensor product
V W
is spanned by elements
v w
, where
v V
and
w W
, where we identify
v (λ
1
w
1
+ λ
2
w
2
) = λ
1
(v w
1
) + λ
2
(v w
2
)
(λ
1
v
1
+ λ
2
v
2
) w = λ
1
(v
1
w) + λ
2
(v
2
w)
This has dimension
dim(V W ) = (dim V )(dim W ).
More explicitly, if
e
1
, ··· , e
n
is a basis of
V
and
f
1
, ··· , f
m
is a basis for
W
,
then {e
i
f
j
: 1 i n, 1 j m} is a basis for V W .
Given any two maps
F
:
V V
0
and
G
:
W W
0
, we define
F G
:
V W V
0
W
0
by
(F G)(v w) = (F (v)) (G(w)),
and then extending linearly.
The operation of tensor products should be familiar in quantum mechanics,
where we combine two state spaces to get a third. From a mathematical point
of view, the tensor product is characterized by the fact that a bilinear map
V ×W U is the same as a linear map V W U.
Definition
(Tensor product representation)
.
Let
g
be a Lie algebra, and
ρ
1
, ρ
2
be representations of
g
with representation spaces
V
1
, V
2
. We define the tensor
product representation ρ
1
ρ
2
with representation space V
1
V
2
given by
(ρ
1
ρ
2
)(X) = ρ
1
(X) I
2
+ I
1
ρ
2
(X) : V
1
V
2
V
1
V
2
,
where I
1
and I
2
are the identity maps on V
1
and V
2
.
Note that the tensor product representation is not given by the “obvious”
formula ρ
1
(X) ρ
2
(X). This does not work.
Now suppose (
ρ, V
) is a reducible representation. Then it has a non-trivial
invariant subspace
U V
. Now we pick a basis of
U
and extend it to
V
. Then
ρ sends any member of U to itself, so the matrix representation of ρ looks like
R(X) =
A(X) B(X)
0 C(X)
.
However, there is no a priori reason why
B
(
X
) should vanish as well. When
this happens, we say the representation is completely reducible.
Definition
(Completely reducible representation)
.
If (
ρ, V
) is a representation
such that there is a basis of V in which ρ looks like
ρ(X) =
ρ
1
(X) 0 ··· 0
0 ρ
2
(X) ··· 0
.
.
.
.
.
.
.
.
.
.
.
.
0 0 ··· ρ
n
(X)
,
then it is called completely reducible. In this case, we have
ρ = ρ
1
ρ
2
··· ρ
n
.
Here is an important fact:
Theorem.
If
ρ
i
for
i
= 1
, ··· , m
are finite-dimensional irreps of a simple Lie
algebra
g
, then
ρ
1
··· ρ
m
is completely reducible to irreps, i.e. we can find
˜ρ
1
, ··· , ˜ρ
k
such that
ρ
1
··· ρ
m
= ˜ρ
1
˜ρ
2
··· ˜ρ
k
.
We will not prove this.