Jeevanjee N. An Introduction to Tensors and Group Theory for Physicists
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194 5 Basic Representation Theory
that we can then apply our classification of complex irreps. Given any real vector
space V , we can define the complexification of V as V
C
≡ C ⊗ V , where C and
V are thought of as real vector spaces (C being a 2-dimensional real vector space
with basis {1,i}), so that if {e
i
} is a basis for V then {1 ⊗ e
i
,i ⊗ e
i
} is a (real)
basis for V
C
. Note that V
C
also carries the structure of a complex vector space, with
multiplication by i defined by
i(z ⊗v) =(iz) ⊗v, z ∈C,v∈V.
A complex basis for V
C
is then given by {1 ⊗e
i
}, and the complex dimension of V
C
is equal to the real dimension of V .
We can then define the complexification of a real representation (, V ) to be the
(complex) representation (
C
,V
C
) defined by
C
(g)
(z ⊗v) ≡z ⊗(g)v. (5.81)
We can then get a handle on the real representation (, V ) by applying our clas-
sification scheme to its complexification (
C
,V
C
). You should be aware, however,
that the irreducibility of (, V ) does not guarantee the irreducibility of (
C
,V
C
)
(see Exercise 5.40 and Example 5.39), though in many cases (
C
,V
C
) will end up
being irreducible.
Example 5.31 The complexifications of R
n
and M
n
(R)
As a warm-up to considering complexifications of representations, we consider the
complexification of a simple vector space. Consider the complexification R
n
C
of R
n
.
We can define the (obvious) complex-linear map
φ :R
n
C
→C
n
1 ⊗
a
j
e
j
+i ⊗
b
j
e
j
→
a
j
+ib
j
e
j
,a
j
,b
j
∈R
which is easily seen to be a vector space isomorphism, so we can identify R
n
C
with C
n
. One can also extend this argument in the obvious way to show that
M
n
(R)
C
=M
n
(C).
Example 5.32 The fundamental representation of so(3)
Now consider the complexification (π
C
, R
3
C
) of the fundamental representation of
so(3). As explained above, R
3
C
can be identified with C
3
, and a moment’s consider-
ation of (5.81) will show that the complexification of the fundamental representation
of so(3) is just given by the usual so(3) matrices acting on C
3
rather than R
3
.You
should check that J
z
and J
+
are given by
J
z
=
⎛
⎝
0 −i 0
i 00
000
⎞
⎠
5.9 Real Representations and Complexifications 195
J
+
=
⎛
⎝
00−1
00−i
1 i 0
⎞
⎠
and that the vector
v
0
≡e
1
+ie
2
=
⎛
⎝
1
i
0
⎞
⎠
is a highest weight vector with j =1. This, along with the fact that dim C
3
=3, al-
lows us to conclude that (π
C
, R
3
C
) (π
1
,V
1
). This is why regular three-dimensional
vectors are said to be “spin-one”. Notice that our highest weight vector v
0
=e
1
+ie
2
is just the analog of the function f
0
=x +iy ∈H
1
(R
3
), which is an equivalent rep-
resentation, and that in terms of SO(3) reps we have
C
, R
3
C
1
, H
1
R
3
.
Recall also that the adjoint representation of su(2) so(3) is equivalent to the
fundamental representation of so(3). This combined with the above results implies
ad
C
, so(3)
C
ad
C
, su(2)
C
(π
1
,V
1
)
so the adjoint representation of su(2) is “spin-one” as well.
Example 5.33 Symmetric traceless tensors
Consider the space S
2
(R
3
) of symmetric traceless second rank tensors defined in
(5.67). This is an SO(3) invariant subspace of R
3
⊗ R
3
, and so furnishes a repre-
sentation (
0
2
,S
2
(R
3
)) of SO(3). What representation is this? To find out, consider
the complexification of the associated so(3) representation, (π
2
0C
,S
2
(C
3
)), where
S
2
(C
3
) is also defined by (5.67), just with C
3
replacing R
3
. Then it is straightfor-
ward to verify that
v
0
≡(e
1
+ie
2
) ⊗(e
1
+ie
2
)
is a highest weight vector with j =2, and this along with fact that dimS
2
(C
3
) =5
(check!) implies that
π
2
0C
,S
2
C
3
(π
2
,V
2
).
This is why symmetric traceless second rank tensors on R
3
are sometimes said to
be “spin-two”.
Recall that S
2
(R
3
) arose in Example 5.23 as part of the decomposition
R
3
⊗R
3
=Rg ⊕
2
R
3
⊕S
2
R
3
which has matrix counterpart
M
3
(R) =RI ⊕A
3
(R) ⊕S
3
(R).
196 5 Basic Representation Theory
Complexifying and using the fact (cf. Example 5.22) that
2
(R
3
) is equivalent to
the adjoint representation and is hence “spin-one”, we obtain the following decom-
position:
V
1
⊗V
1
C
3
⊗C
3
M
3
(C) V
0
⊕V
1
⊕V
2
.
This is an instance of (5.68) and should be familiar from angular momentum addi-
tion in quantum mechanics.
Exercise 5.40 Consider the fundamental representation of SO(2) on R
2
, which we know
is irreducible by Exercise 5.34. The complexification of this representation is just given
by the same SO(2) matrices acting on C
2
rather than R
2
. Show that this representation is
reducible, by diagonalizing the SO(2) generator
X =
0 −1
10
.
Note that this diagonalization can now be done because both complex eigenvalues and com-
plex basis transformations are allowed, in contrast to the real case.
5.10 The Irreducible Representations of sl(2,C
C
C)
R
R
R
, SL(2,C
C
C) and
SO(3, 1)
o
In this section we will use the techniques and results of the previous section to
classify all the finite-dimensional complex irreps of sl(2, C)
R
so(3, 1), and then
use this to find the irreps of the associated groups SL(2, C) and SO(3, 1)
o
.
Let (π, V ) be a finite-dimensional complex irreducible representation of
sl(2, C)
R
. Define the operators
M
i
≡
1
2
π(S
i
) −iπ(
˜
K
i
)
,i=1, 2, 3
N
i
≡
1
2
π(S
i
) +iπ(
˜
K
i
)
,i=1, 2, 3
(5.82)
where {S
i
,
˜
K
i
}
i=1,2,3
is our usual basis for sl(2, C)
R
. One can check that the Ms and
Ns commute between each other, as well as satisfy the su(2) commutation relations
internally, i.e.
[M
i
,N
j
]=0
[M
i
,M
j
]=
3
k=1
ij k
M
k
(5.83)
[N
i
,N
j
]=
3
k=1
ij k
N
k
.
We have thus taken the complex span of the set {π(
˜
K
i
), π(S
i
)} (notice the factors
of i in (5.82)) and found a new basis for this Lie algebra of operators that makes it
5.10 The Irreducible Representations of sl(2, C)
R
, SL(2, C) and SO(3, 1)
o
197
look like two commuting copies of su(2). We can thus define the usual raising and
lowering operators
N
±
≡iN
1
∓N
2
M
±
≡iM
1
∓M
2
which then have the usual commutation relations between themselves and iM
z
, iN
z
:
[iM
z
,M
±
]=±M
±
[iN
z
,N
±
]=±N
±
[M
+
,M
−
]=2iM
z
[N
+
,N
−
]=2iN
z
.
With this machinery set up we can now use the strategy from the last section. First,
pick a vector v ∈ V that is an eigenvector of both iM
z
and iN
z
(that such a vector
exists is guaranteed by Problem 5.10). Then by applying M
+
and N
+
we can raise
the iM
z
and iN
z
eigenvalues until the raising operators give us the zero vector; let
v
0,0
denote the vector with the highest eigenvalues (which we will again refer to as
a ‘highest weight vector’), and let (j
1
,j
2
) denote those eigenvalues under iM
z
and
iN
z
, respectively, so that
M
+
(v
0,0
) =0
N
+
(v
0,0
) =0
iM
z
(v
0,0
) =j
1
v
0,0
iN
z
(v
0,0
) =j
2
v
0,0
.
(5.84)
We can then lower the eigenvalues with N
−
and M
−
to get vectors
v
k
1
,k
2
≡(M
−
)
k
1
(N
−
)
k
2
v
0,0
(5.85)
which are eigenvectors of iM
z
and iN
z
with eigenvalues j
1
− k
1
and j
2
− k
2
,re-
spectively. By finite-dimensionality of V this chain of vectors must eventually end,
though, so there exist nonnegative integers l
1
,l
2
such that v
l
1
,l
2
=0but
M
−
(v
l
1
,l
2
) =N
−
(v
l
1
,l
2
) =0. (5.86)
Calculations identical to those from the su(2) case show that l
i
=2j
i
, and that the
action of the operators M
i
, N
i
is given by
M
+
(v
0,0
) =0
N
+
(v
0,0
) =0
M
−
(v
k
1
,k
2
) =v
k
1
+1,k
2
,k
1
< 2j
1
N
−
(v
k
1
,k
2
) =v
k
1
,k
2
+1
,k
2
< 2j
2
iM
z
(v
k
1
,k
2
) =(j
1
−k
1
)v
k
1
,k
2
(5.87)
iN
z
(v
k
1
,k
2
) =(j
2
−k
2
)v
k
1
,k
2
M
−
(v
2j
1
,k
2
) =0, ∀k
2
198 5 Basic Representation Theory
N
−
(v
k
1
,2j
2
) =0, ∀k
1
M
+
(v
k
1
,k
2
) =
2j
1
k
1
−k
1
(k
1
−1)
v
k
1
−1,k
2
,k
1
=0
N
+
(v
k
1
,k
2
) =
2j
2
k
2
−k
2
(k
2
−1)
v
k
1
,k
2
−1
,k
2
=0
(this is just two copies of (5.76), one each for the M
i
and the N
i
). As in the su(2)
case, we note that the {v
k
1
,k
2
} are linearly independent and span an invariant sub-
space of V , hence must span all of V since we assumed V was irreducible. Thus,
we conclude that any complex finite-dimensional irrep of sl(2, C)
R
is of the form
(π
(j
1
,j
2
)
,V
(j
1
,j
2
)
) where V
(j
1
,j
2
)
has a basis
B ={v
k
1
,k
2
|0 ≤k
1
≤2j
1
, 0 ≤k
2
≤2j
2
}
and the operators π
(j
1
,j
2
)
(S
i
), π
(j
1
,j
2
)
(
˜
K
i
) satisfy (5.87), with 2j
1
, 2j
2
∈ Z.This
tells us that
dimV
(j
1
,j
2
)
=(2j
1
+1)(2j
2
+1).
Let us abbreviate the representations (π
(j
1
,j
2
)
,V
(j
1
,j
2
)
) as simply (j
1
, j
2
) as is
done in the physics literature. As in the su(2) case, we can show that (5.87) actually
defines a representation of sl(2, C)
R
, and using the same arguments that we did in
the su(2) case we conclude that
Proposition 5.7 The representations (j
1
, j
2
),2j
1
, 2j
2
∈ Z are, up to equivalence,
all the complex finite-dimensional irreducible representations of sl(2, C)
R
.
As in the su(2) case, we deduced (5.87) from just the sl(2, C)
R
commutation re-
lations, the finite-dimensionality of V , and the existence of a highest weight vector
v
0,0
satisfying (5.84). Thus if we are given a finite-dimensional sl(2, C)
R
represen-
tation and can find a highest weight vector v
0,0
for some (j
1
,j
2
), we can conclude
that the representation space contains an invariant subspace equivalent to (j
1
, j
2
).
Example 5.34 (π, C
2
): The fundamental (left-handed spinor) representation
Consider the left-handed spinor representation (π, C
2
) of sl(2, C)
R
, which is also
just the fundamental of sl(2, C)
R
,i.e.
π(S
1
) =S
1
=
1
2
0 −i
−i 0
π(S
2
) =S
2
=
1
2
0 −1
10
π(S
3
) =S
3
=
1
2
−i 0
0 i
π(
˜
K
1
) =
˜
K
1
=
1
2
01
10
π(
˜
K
2
) =
˜
K
2
=
1
2
0 −i
i 0
5.10 The Irreducible Representations of sl(2, C)
R
, SL(2, C) and SO(3, 1)
o
199
π(
˜
K
3
) =
˜
K
3
=
1
2
10
0 −1
.
In this case we have (check!)
M
i
=S
i
N
i
=0
and hence
M
+
=
01
00
M
−
=
00
10
(5.88)
N
+
=N
−
=0.
With this in hand you can easily check that (1, 0) ∈ C
2
is a highest weight vector
with j
1
= 1/2, j
2
= 0. Thus the fundamental representation of sl(2, C)
R
is just
(
1
2
, 0).
Example 5.35 (S
2j
π,S
2j
(C
2
)): Symmetric tensor products of left-handed spinors
As with su(2), we can build other irreps by taking symmetric tensor products. Con-
sider the 2j th symmetric tensor product representation (S
2j
π,S
2j
(C
2
)),2j ∈Z and
the vector
v
0,0
≡e
1
⊗···⊗e
1
2j times
∈S
2j
C
2
. (5.89)
Using (5.34) it is straightforward to check, as you did in Example 5.30, that this is
a highest weight vector with eigenvalues (j, 0), and so we conclude that S
2j
(C
2
)
contains an invariant subspace equivalent to (j, 0). Noting that
dimS
2j
C
2
=dim(j, 0) =2j +1
we conclude that (S
2j
π,S
2j
(C
2
)) (j, 0).
Example 5.36 ( ¯π,C
2
): The right-handed spinor representation
Consider the right-handed spinor representation (
¯
, C
2
) from Example 5.6. A quick
calculation (do it!) reveals that the induced Lie algebra representation ( ¯π,C
2
) is
given by
¯π(X)=−X
†
,X∈sl(2, C)
R
.
In particular we then have
¯π(S
i
) =S
i
since the S
i
are anti-Hermitian, as well as
200 5 Basic Representation Theory
¯π(
˜
K
i
) =¯π(iS
i
) =−i ¯π(S
i
) =−iS
i
.
We then have
M
i
=0
N
i
=S
i
and hence
N
+
=
01
00
N
−
=
00
10
(5.90)
M
+
=M
−
=0.
You can again check that (1, 0) ∈ C
2
is a highest weight vector, but this time with
j
1
=0, j
2
=1/2, and so the right-handed spinor representation of sl(2, C)
R
is just
(0,
1
2
).
Example 5.37 ( ¯π
2j
,S
2j
(C
2
)): Symmetric tensor products of right-handed spinors
As before, we can build other irreps by taking symmetric tensor products. Again,
consider the 2j th symmetric tensor product representation (S
2j
¯π,S
2j
(C
2
)),2j ∈Z
and the vector
v
0,0
≡e
1
⊗···⊗e
1
2j times
∈S
2j
C
2
. (5.91)
Again, it is straightforward to check that this is a highest weight vector with eigen-
values (0,j), and we can conclude as before that (S
2j
¯π,S
2j
(C
2
)) is equivalent to
(0, j).
So far we have used symmetric tensor products of the left-handed and right-
handed spinor representations to build the (j, 0) and (0, j) irreps. From here, getting
the general irrep (j
1
, j
2
) is easy; we just take the tensor product of (j
1
, 0) and (0, j
2
)!
To see this, let v
0,0
∈ (j, 0) and ¯v
0,0
∈ (0, k) be highest weight vectors. Then it is
straightforward to check that
v
0,0
⊗¯v
0,0
∈(j, 0) ⊗(0, k) (5.92)
is a highest weight vector with j
1
=j, j
2
=k, and so we conclude that (j, 0) ⊗(0, k)
contains an invariant subspace equivalent to (j, k). However, since
dim
(j, 0) ⊗(0, k)
=dim(j, k) =(2j +1)(2k +1)
we conclude that these representations are equivalent, and so in general (switching
notation a little),
(j
1
, j
2
) (j
1
, 0) ⊗(0, j
2
). (5.93)
5.10 The Irreducible Representations of sl(2, C)
R
, SL(2, C) and SO(3, 1)
o
201
Thus, all the irreps of sl(2, C)
R
can be built out of the left-handed spinor
(fundamental) representation, the right-handed spinor representation, and
various tensor products of the two.
As before, this classification of the complex finite-dimensional irreps of sl(2, C)
R
also yields, with minimal effort, the classification of the complex finite-dimensional
irreps of SL(2, C). Any complex finite-dimensional irrep of SL(2, C) yields a com-
plex finite-dimensional irrep of sl(2, C)
R
, which must be equivalent to (j
1
, j
2
) for
some j
1
, j
2
and since
(j
1
, j
2
)
S
2j
1
π ⊗S
2j
2
¯π,S
2j
1
C
2
⊗S
2j
2
C
2
we conclude that our original SL(2, C) irrep is equivalent to (S
2j
1
⊗ S
2j
2
¯
,
S
2j
1
(C
2
) ⊗S
2j
2
(C
2
)) for some j
1
, j
2
. Thus, the representations
S
2j
1
⊗S
2j
2
¯
, S
2j
1
C
2
⊗S
2j
2
C
2
, 2j
1
, 2j
2
∈Z
are (up to equivalence) all the complex finite-dimensional irreducible represen-
tations of SL(2, C).
How about representations of SO(3, 1)
o
? We saw that in the case of SO(3), not all
representations of the associated Lie algebra actually arise from representations of
the group, and the same is true here. Say we have a complex finite-dimensional irrep
(, V ) of SO(3, 1)
o
, and consider its induced Lie algebra representation (π, V ),
which must be equivalent to (j
1
, j
2
) for some j
1
, j
2
. Noting that
iM
z
+iN
z
=iπ(S
z
)
we have (again, be sure to distinguish π the number from π the representation!)
e
i2π ·(iM
z
+iN
z
)
v
0,0
=e
i2π(j
1
+j
2
)
v
0,0
as well as
e
i2π ·(iM
z
+iN
z
)
v
0,0
=e
−2π·π(S
z
)
v
0,0
=
e
−2πS
z
v
0,0
=(I)v
0,0
=v
0,0
so we conclude that
e
i2π(j
1
+j
2
)
=1 ⇐⇒ j
1
+j
2
∈Z, (5.94)
and thus only representations (j
1
, j
2
) satisfying this condition can arise from
SO(3, 1)
o
representations. (It is also true that for any j
1
, j
2
satisfying this condi-
tion, there exists an SO(3, 1)
o
representation with induced Lie algebra representa-
tion (j
1
, j
2
), though we will not prove that here.)
Example 5.38 R
4
: The four-vector representation of SO(3, 1)
o
The fundamental representation (, R
4
) is the most familiar SO(3, 1)
o
repre-
sentation, corresponding to four-dimensional vectors in Minkowski space. What
202 5 Basic Representation Theory
(j
1
, j
2
) does it correspond to? To find out, we first complexify the representation to
(
C
, C
4
) and then consider the induced sl(2, C)
R
representation (π
C
, C
4
). Straight-
forward calculations show that
iM
z
=
1
2
⎛
⎜
⎜
⎝
0 −i 00
i 000
0001
0010
⎞
⎟
⎟
⎠
iN
z
=
1
2
⎛
⎜
⎜
⎝
0 −i 00
i 000
00 0−1
00−10
⎞
⎟
⎟
⎠
and this, along with expressions for M
+
and N
+
that you should derive, can be used
to show that
v
0,0
=(e
1
+ie
2
) =
⎛
⎜
⎜
⎝
1
i
0
0
⎞
⎟
⎟
⎠
is a highest weight vector with (j
1
,j
2
) =(1/2, 1/2). Noting that
dimC
4
=dim
1
2
,
1
2
=4
we conclude that
π
C
, C
4
1
2
,
1
2
.
Note that j
1
+j
2
=1/2 +1/2 =1 ∈Z, in accordance with (5.94).
Before moving on to our next example, we need to discuss tensor products of
sl(2, C)
R
irreps. The nice thing here is that we can use what we know about the
tensor product of su(2) irreps to compute the decomposition of the tensor product
of sl(2, C)
R
irreps. In fact, the following is true.
Proposition 5.8 The decomposition into irreps of the tensor product of two
sl(2, C)
R
irreps (j
1
, j
2
) and (k
1
, k
2
) is given by
(j
1
, j
2
) ⊗(k
1
, k
2
) =
(l
1
, l
2
)
where |j
1
−k
1
|≤l
1
≤j
1
+k
1
, |j
2
−k
2
|≤l
2
≤j
2
+k
2
(5.95)
and each (l
1
, l
2
) consistent with the above inequalities occurs exactly once in the
direct sum decomposition.
Notice the restrictions on l
1
and l
2
, which correspond to the decomposition of
tensor products of su(2) representations. We relegate a proof of this formula to the
5.10 The Irreducible Representations of sl(2, C)
R
, SL(2, C) and SO(3, 1)
o
203
Appendix, but it should seem plausible. We have seen that one can roughly think
of an sl(2, C)
R
representations as a ‘product’ of two su(2) representations, and so
the tensor product of two sl(2, C)
R
representations (which can both be ‘factored’
into su(2) representations) should just be given by the various ‘products’ of su(2)
representations that occur when taking the tensor product of the factors. We will
apply this formula and make this concrete in the next example.
Example 5.39
2
R
4
: The antisymmetric tensor representation of SO(3, 1)
o
This is an important example since the electromagnetic field tensor F
μν
lives in this
representation. To classify this representation, we first note that
2
R
4
occurs in the
O(3, 1)-invariant decomposition
R
4
⊗R
4
=Rη
−1
⊕
2
R
4
⊕S
2
R
4
(5.96)
where
η
−1
=η
μν
e
μ
⊗e
ν
is the inverse of the Minkowski metric and
S
2
R
4
=
T ∈S
2
R
4
η
μν
T
μν
=0
(5.97)
is the set of symmetric ‘traceless’ second rank tensors, where the trace is effected
by the Minkowski metric η. Note that this is just the O(3, 1) analog of the O(n)
decomposition in Example 5.23. You should check that each of the subspaces in
(5.96) really is O(3, 1) invariant. Complexifying this yields
C
4
⊗C
4
=Cη
−1
⊕
2
C
4
⊕S
2
C
4
.
Now, we can also decompose C
4
⊗C
4
using Proposition 5.8, which yields
1
2
,
1
2
⊗
1
2
,
1
2
=(0, 0) ⊕(1, 0) ⊕(0, 1) ⊕(1, 1). (5.98)
Now, clearly Cη
−1
corresponds to (0, 0) since the former is a one-dimensional rep-
resentation and (0, 0) is the only one-dimensional irrep in the decomposition (5.98).
What about S
2
(C
4
)? Well, it is straightforward to check using the results of the
previous example that
v
0,0
=(e
1
+ie
2
) ⊗(e
1
+ie
2
) ∈S
2
C
4
is a highest weight vector with (j
1
,j
2
) = (1, 1) (it is also instructive to verify that
v
0,0
actually satisfies the condition in (5.97)). Checking dimensions then tells us
that S
2
(C
4
) (1, 1), so we conclude that
2
C
4
(1, 0) ⊕(0, 1).
This representation is decomposable, but remember that this does not imply that
2
(R
4
) is decomposable! In fact,
2
(R
4
) is irreducible. This is an unavoidable