Naber G.L. The Geometry of Minkowski Spacetime. An Introduction to the Mathematics of the Special Theory of Relativity
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144 3 The Theory of Spinors
Corollary 3.1.2 Let G be a matrix group that contains −G for every G ∈G
and D : G→Han irreducible representation of G.Then
D
−G
= ±D
G
. (3.1.12)
Proof: −G =(−I)G so D
−G
= D
(−I)G
= D
−I
D
G
and it will suffice to
show that D
−I
= ±I. For any G
∈G,D
−I
D
G
= D
(−I)G
= D
G
(−I)
=
D
G
D
−I
so D
−I
commutes with each D
G
,G
∈G. By Schur’s Lemma,
D
−I
= λI for some λ.ButD
−I
D
−I
= D
(−I)(−I)
= D
I
= I so (λI)(λI)=
λ
2
I = I.Thus,λ
2
=1soλ = ±1andD
−I
= ±I.
Since SL(2, C) clearly contains −G for every G ∈ SL(2, C), we find that
every irreducible representation D of SL(2, C) satisfies either D
−G
= D
G
or D
−G
= −D
G
.Aswehaveseen,thoseofthefirsttypegiverepresenta-
tions of the Lorentz group. Although those that satisfy D
−G
= −D
G
cannot
legitimately be regarded as representations of L (not being single-valued),
it has become customary to refer to such a representation of SL(2, C)asa
two-valued representation of L and we shall adhere to the custom.
The problem of determining the finite-dimensional, irreducible represen-
tations of SL(2, C) is thus seen to be a matter of considerable interest in
mathematical physics. As it happens, these representations are well-known
and rather easy to describe. Moreover, such a description is well worth the ef-
fort required to produce it since it leads inevitably to the notion of a “spinor”,
which will be our major concern in this chapter.
In order to enumerate these representations of SL(2, C)itwillbeconve-
nient to reverse our usual procedure and specify first a space of carriers and a
basis and then describe the linear transformations whose matrices relative to
this basis will constitute our representations. If m ≥ 0andn ≥ 0 are integers
we denote by P
mn
the vector space of all polynomials in z and ¯z with complex
coefficients and of degree at most m in z and at most n in ¯z, i.e.,
P
mn
= {p(z, ¯z)=p
00
+ p
10
z + p
01
¯z + p
11
z¯z + ···
+ p
mn
z
m
¯z
n
= p
rs
z
r
¯z
s
: p
rs
∈C},
with the usual coefficientwise addition and scalar multiplication, i.e., p(z, ¯z)+
q(z, ¯z)=[p
00
+ p
10
z + ···+ p
mn
z
m
¯z
n
]+[q
00
+ q
10
z + ···+ q
mn
z
m
¯z
n
]=
[p
00
+ q
00
]+[p
10
+ q
10
]z + ···+[p
mn
+ q
mn
]z
m
¯z
n
and αp(z, ¯z)=(αp
00
)+
(αp
10
)z + ···+(αp
mn
)z
m
¯z
n
. The basis implicit here is {1,z,¯z, z¯z,...,z
m
¯z
n
}
so dim P
mn
=(m +1)(n +1). Now, for each G =
*
ab
cd
+
∈ SL(2, C) we define
D
(
m
2
,
n
2
)
G
: P
mn
→ P
mn
by
D
(
m
2
,
n
2
)
G
(p(z, ¯z)) = D
(
m
2
,
n
2
)
G
(p
rs
z
r
¯z
s
)=(bz + d)
m
(
¯
b¯z +
¯
d)
n
p(w, ¯w),
3.1 Representations of the Lorentz Group 145
where
w =
az + c
bz + d
.
Then D
(
m
2
,
n
2
)
G
is clearly linear in p(z, ¯z) and maps P
mn
to P
mn
. Although
algebraically a bit messy it is straightforward to show that D
(
m
2
,
n
2
)
G
has the
properties required to determine a representation of SL(2, C). We leave the
manual labor to the reader.
Exercise 3.1.5 Show that D
(
m
2
,
n
2
)
I
is the identity transformation on P
mn
and that if G
1
and G
2
are in SL(2, C), then
D
(
m
2
,
n
2
)
G
1
G
2
= D
(
m
2
,
n
2
)
G
1
◦ D
(
m
2
,
n
2
)
G
2
.
Thus, the matrices of the linear transformations D
(
m
2
,
n
2
)
G
relative to the basis
{1,z,¯z,...,z
m
¯z
n
} for P
mn
constitute a representation of SL(2, C)whichwe
also denote
G −→ D
(
m
2
,
n
2
)
G
and call the spinor representation of type (m, n). Although it is by no means
obvious the spinor representations are all irreducible and, in fact, exhaust all
of the finite-dimensional, irreducible representations of SL(2, C)(wereferthe
interested reader to [GMS] for a proof of Theorem 3.1.3).
T
h
eore
m 3.1.3 For all m, n =0, 1, 2,..., the spinor representation D
(
m
2
,
n
2
)
of SL(2, C) is irreducible and every finite-dimensional irreducible representa-
tion of SL(2, C) is equivalent to some D
(
m
2
,
n
2
)
.
We consider a few specific examples. First suppose m =1and n =0:P
10
=
{p(z, ¯z)=p
00
+ p
10
z : p
rs
∈ C}. For G =
*
ab
cd
+
∈ SL(2, C),
D
(
1
2
,0
)
G
(p(z, ¯z)) = (bz + d)
1
(
¯
b¯z +
¯
d)
0
p(w, ¯w)
=(bz + d)
p
00
+ p
10
az + c
bz + d
=(bz + d)p
00
+(az + c)p
10
=(cp
10
+ dp
00
)+(ap
10
+ bp
00
)z
=ˆp
00
+ˆp
10
z,
where
*
ˆp
10
ˆp
00
+
=
*
ab
cd
+*
p
10
p
00
+
.
146 3 The Theory of Spinors
Thus, the representation G → D
(
1
2
,0
)
G
is given by
*
ab
cd
+
−→ D
(
1
2
,0
)
*
ab
cd
+
=
*
ab
cd
+
,
i.e., D
(
1
2
,0
)
is the identity representation of SL(2, C).
Exercise 3.1.6 Show in the same way that D
(
0,
1
2
)
is the conjugation rep-
resentation
*
ab
cd
+
−→ D
(
0,
1
2
)
*
ab
cd
+
=
*
¯a
¯
b
¯c
¯
d
+
.
Exercise 3.1.7 Show that D
(
1
2
,0
)
and D
(
0,
1
2
)
are not equivalent represen-
tations of SL(2, C), i.e., that there does not exist an invertible matrix P
such that P
−1
GP =
¯
G for all G ∈ SL(2, C). Hint:LetG =
*
i 0
0 −i
+
and
P =
*
0 i
−i 0
+
. Show that P
−1
GP =
¯
G and that P and its nonzero scalar mul-
tiples are the only matrices for which this is true. Now find a G
∈ SL(2, C)
for which P
−1
G
P = G
.
Before working out another example we include a few more observations
about D
(
1
2
,0
)
and D
(
0,
1
2
)
. First note that if G → D
G
is any representation
of SL(2, C), then the assignment G →
D
−1
G
T
=
D
T
G
−1
of the trans-
posed inverse of D
G
to each G is also a representation of SL(2, C)since
I →
D
−1
I
T
=(I
−1
)
T
= I
T
= I and G
1
G
2
→ ((D
G
1
G
2
)
−1
)
T
=
((D
G
1
D
G
2
)
−1
)
T
=
D
−1
G
2
D
−1
G
1
T
=
D
−1
G
1
T
D
−1
G
2
T
(note that inversion or
transposition alone would not accomplish this since each reverses products).
Applying this, in particular, to the identity representation D
(
1
2
,0
)
gives
G =
*
ab
cd
+
−→ (G
−1
)
T
=(G
T
)
−1
=
*
d −c
−ba
+
.
Letting =
*
0 −1
10
+
it is easily checked that
−1
= − and
(G
−1
)
T
=
−1
G.
Thus, G → (G
−1
)
T
=(G
T
)
−1
is equivalent to D
(
1
2
,0
)
and we shall denote
it
˜
D
(
1
2
,0
)
. Similarly, one can define a representation
˜
D
(
0,
1
2
)
equivalent to
conjugation by
G −→ (
¯
G
−1
)
T
=(
¯
G
T
)
−1
=
−1
¯
G.
3.1 Representations of the Lorentz Group 147
These equivalent versions of D
(
1
2
,0
)
and D
(
0,
1
2
)
as well as analogous versions
of D
(
m
2
,
n
2
)
are often convenient and we shall return to them in the next
section.
Now let m = n =1.ThenP
11
= {p(z, ¯z)=p
00
+ p
10
z + p
01
¯z + p
11
z¯z :
p
rs
∈ C} and for each G =
*
ab
cd
+
∈ SL(2, C) one has
D
(
1
2
,
1
2
)
G
(p(z, ¯z)) = (bz + d)
1
(
¯
d¯z +
¯
d)
1
(p
00
+ p
10
w + p
01
¯w + p
11
w ¯w),
where w =
az +c
bz +d
. Multiplying out and rearranging yields
D
(
1
2
,
1
2
)
G
(p(z, ¯z)) = ˆp
00
+ˆp
10
z +ˆp
01
¯z +ˆp
11
z¯z,
where
⎡
⎢
⎢
⎣
ˆp
11
ˆp
10
ˆp
01
ˆp
00
⎤
⎥
⎥
⎦
=
⎡
⎢
⎢
⎣
a¯aa
¯
b ¯ab b
¯
b
a¯ca
¯
db¯cb
¯
d
¯ac
¯
bc ¯ad
¯
bd
c¯cc
¯
d ¯cd d
¯
d
⎤
⎥
⎥
⎦
⎡
⎢
⎢
⎣
p
11
p
10
p
01
p
00
⎤
⎥
⎥
⎦
, (3.1.13)
so that
D
(
1
2
,
1
2
)
*
ab
cd
+
=
⎡
⎢
⎢
⎣
a¯aa
¯
b ¯ab b
¯
b
a¯ca
¯
db¯cb
¯
d
¯ac
¯
bc ¯ad
¯
bd
c¯cc
¯
d ¯cd d
¯
d
⎤
⎥
⎥
⎦
.
Proceeding in this manner with the notation currently at our disposal would
soon become algebraically unmanageable. For this reason we now introduce
new and powerful notational devices that will constitute the language in
which the remainder of the chapter will be written. First we rephrase the
example of D
(
1
2
,
1
2
)
in these new terms. We begin by rewriting each p(z, ¯z)as
a sum of terms of the form
φ
A
˙
X
z
A
¯z
˙
X
,
where A =1, 0and
˙
X =
˙
1,
˙
0 (the dot is used only to indicate a power of ¯z
rather than z and
˙
1,
˙
0 are treated exactly as if they were 1, 0, i.e., ¯z
˙
0
=1,
¯z
˙
1
=¯z,
˙
0+
˙
1=
˙
1, etc.). Thus,
p
00
+ p
10
z + p
01
¯z + p
11
z¯z = φ
0
˙
0
z
0
¯z
˙
0
+ φ
1
˙
0
z
1
¯z
˙
0
+ φ
0
˙
1
z
0
¯z
˙
1
+ φ
1
˙
1
z
1
¯z
˙
1
,
where φ
0
˙
0
= p
00
,φ
1
˙
0
= p
10
,φ
1
˙
1
= p
11
. With the summation convention
(over A =1, 0,
˙
X =
˙
1,
˙
0),
p(z, ¯z)=φ
A
˙
X
z
A
¯z
˙
X
.
To set up another application of the summation convention we henceforth
denote the entries in G ∈ SL(2, C)as
148 3 The Theory of Spinors
G =
G
A
B
=
#
G
1
1
G
1
0
G
0
1
G
0
0
$
and write the conjugate
¯
G of G as
¯
G =
,
¯
G
˙
X
˙
Y
-
=
#
¯
G
˙
1
˙
1
¯
G
˙
1
˙
0
¯
G
˙
0
˙
1
¯
G
˙
0
˙
0
$
.
Convention: Henceforth, conjugating a term with undotted indices dots them
all and introduces a bar, whereas conjugating a term with dotted indices
undots them and removes the bar. Whenever possible we will select undotted
index names from the beginning of the alphabet (A,B,C,...) and dotted
indices from the end (...,
˙
X,
˙
Y,
˙
Z).
Now, if we let D
(
1
2
,
1
2
)
G
(φ
A
˙
X
z
A
¯z
˙
X
)=
ˆ
φ
A
˙
X
z
A
¯z
˙
X
we find from (3.1.13)that
⎡
⎢
⎢
⎢
⎢
⎣
ˆ
φ
1
˙
1
ˆ
φ
1
˙
0
ˆ
φ
0
˙
1
ˆ
φ
0
˙
0
⎤
⎥
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎢
⎣
G
1
1
¯
G
˙
1
˙
1
G
1
1
¯
G
˙
1
˙
0
G
1
0
¯
G
˙
1
˙
1
G
1
0
¯
G
˙
1
˙
0
G
1
1
¯
G
˙
0
˙
1
G
1
1
¯
G
˙
0
˙
0
G
1
0
¯
G
˙
0
˙
1
G
1
0
¯
G
˙
0
˙
0
G
0
1
¯
G
˙
1
˙
1
G
0
1
¯
G
˙
1
˙
0
G
0
0
¯
G
˙
1
˙
1
G
0
0
¯
G
˙
1
˙
0
G
0
1
¯
G
˙
0
˙
1
G
0
1
¯
G
˙
0
˙
0
G
0
0
¯
G
˙
0
˙
1
G
0
0
¯
G
˙
0
˙
0
⎤
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎣
φ
1
˙
1
φ
1
˙
0
φ
0
˙
1
φ
0
˙
0
⎤
⎥
⎥
⎥
⎥
⎦
(3.1.14)
which all collapses quite nicely with the summation convention to
ˆ
φ
A
˙
X
= G
A
B
¯
G
˙
X
˙
Y
φ
B
˙
Y
,A=1, 0,
˙
X =
˙
1,
˙
0. (3.1.15)
For D
(
1
2
,0
)
we would write p
00
+ p
10
z = φ
0
z
0
+ φ
1
z
1
= φ
A
z
A
and
D
(
1
2
,0
)
G
(φ
A
z
A
)=
ˆ
φ
A
z
A
,where
ˆ
φ
A
= G
A
B
φ
B
,A=1, 0. (3.1.16)
Similarly, for D
(
0,
1
2
)
,p
00
+p
01
¯z = φ
˙
0
¯z
˙
0
+φ
˙
1
¯z
˙
1
= φ
˙
X
¯z
˙
X
and D
(
0,
1
2
)
G
(φ
˙
X
¯z
˙
X
)=
ˆ
φ
˙
X
¯z
˙
X
,where
ˆ
φ
˙
X
=
¯
G
˙
X
˙
Y
φ
˙
Y
,
˙
X =
˙
1,
˙
0. (3.1.17)
Notice that the 4 ×4matrixin(3.1.14) is precisely D
(
1
2
,
1
2
)
G
and that anal-
ogous statements would be true of (3.1.16)and(3.1.17) if these were written
a
s
matr
ix products. The situation changes somewhat for larger m and n so
we wish to treat one more example before describing the general case. Thus,
we let m =2andn =1.Anelementp(z, ¯z)=p
00
+ p
10
z + ···+ p
21
z
2
¯z is
to be written as a sum of terms of the form φ
A
1
A
2
˙
X
z
A
1
z
A
2
¯z
˙
X
. For example,
the constant term p
00
is written φ
00
˙
0
z
0
z
0
¯z
˙
0
so φ
00
˙
0
= p
00
and p
10
z becomes
φ
10
˙
0
z
1
z
0
¯z
˙
0
+ φ
01
˙
0
z
0
z
1
¯z
˙
0
and we take φ
10
˙
0
= φ
01
˙
0
=
1
2
p
10
,andsoon.The
result is
3.1 Representations of the Lorentz Group 149
p(z, ¯z)=φ
00
˙
0
z
0
z
0
¯z
˙
0
+ φ
10
˙
0
z
1
z
0
¯z
˙
0
+ φ
01
˙
0
z
0
z
1
¯z
˙
0
+ φ
00
˙
1
z
0
z
0
¯z
˙
1
+ φ
10
˙
1
z
1
z
0
¯z
˙
1
+ φ
01
˙
1
z
0
z
1
¯z
˙
1
+ φ
11
˙
0
z
1
z
1
¯z
˙
0
+ φ
11
˙
1
z
1
z
1
¯z
˙
1
,
wherewetake
φ
00
˙
0
= p
00
,φ
10
˙
0
= φ
01
˙
0
=
1
2
p
10
,φ
00
˙
1
= p
01
,
φ
10
˙
1
= φ
01
˙
1
=
1
2
p
11
,φ
11
˙
0
= p
20
,φ
11
˙
1
= p
21
,
so that, in particular, φ
A
1
A
2
˙
X
is symmetric in A
1
and A
2
, i.e., φ
A
2
A
1
˙
X
=
φ
A
1
A
2
˙
X
for A
1
,A
2
=1, 0. Thus, with the summation convention,
p(z, ¯z)=φ
A
1
A
2
˙
X
z
A
1
z
A
2
¯z
˙
X
= φ
A
1
A
2
˙
X
z
A
1
+A
2
¯z
˙
X
.
Now,
D
(
2
2
,
1
2
)
G
(p(z, ¯z)) =
G
1
0
z + G
0
0
2
¯
G
˙
1
˙
0
¯z +
¯
G
˙
0
˙
0
1
(φ
A
1
A
2
˙
X
w
A
1
+A
2
¯w
˙
X
),
where
w =
G
1
1
z + G
0
1
G
1
0
z + G
0
0
,
so
D
(
2
2
,
1
2
)
G
(p(z, ¯z)) = φ
A
1
A
2
˙
X
G
1
1
z + G
0
1
A
1
+A
2
G
1
0
z + G
0
0
2−A
1
−A
2
·
¯
G
˙
1
˙
1
¯z +
¯
G
˙
0
˙
1
˙
X
¯
G
˙
1
˙
0
¯z +
¯
G
˙
0
˙
0
1−
˙
X
= φ
00
˙
0
G
1
0
z + G
0
0
G
1
0
z + G
0
0
¯
G
˙
1
˙
0
¯z +
¯
G
˙
0
˙
0
+ φ
10
˙
0
G
1
1
z + G
0
1
G
1
0
z + G
0
0
¯
G
˙
1
˙
0
¯z +
¯
G
˙
0
˙
0
+ φ
01
˙
0
G
1
1
z + G
0
1
G
1
0
z + G
0
0
¯
G
˙
1
˙
0
¯z +
¯
G
˙
0
˙
0
+ φ
11
˙
0
G
1
1
z + G
0
1
G
1
1
z + G
0
1
¯
G
˙
1
˙
0
¯z +
¯
G
˙
0
˙
0
+ φ
00
˙
1
G
1
0
z + G
0
0
G
1
0
z + G
0
0
¯
G
˙
1
˙
1
¯z +
¯
G
˙
0
˙
1
+ φ
10
˙
1
G
1
1
z + G
0
1
G
1
0
z + G
0
0
¯
G
˙
1
˙
1
¯z +
¯
G
˙
0
˙
1
+ φ
01
˙
1
G
1
1
z + G
0
1
G
1
0
z + G
0
0
¯
G
˙
1
˙
1
¯z +
¯
G
˙
0
˙
1
+ φ
11
˙
1
G
1
1
z + G
0
1
G
1
1
z + G
0
1
¯
G
˙
1
˙
1
¯z +
¯
G
˙
0
˙
1
.
150 3 The Theory of Spinors
Multiplying out and collecting terms gives D
(
2
2
,
1
2
)
G
(φ
A
1
A
2
˙
X
z
A
1
+A
2
¯z
˙
X
)=
ˆ
φ
A
1
A
2
˙
X
z
A
1
+A
2
¯z
˙
X
,where
ˆ
φ
A
1
A
2
˙
X
= G
A
1
B
1
G
A
2
B
2
¯
G
˙
X
˙
Y
φ
B
1
B
2
˙
Y
,A
1
,A
2
=1, 0,
˙
X =
˙
1,
˙
0. (3.1.18)
Exercise 3.1.8 Write out all terms in the expansion of D
(
2
2
,
1
2
)
G
(p(z, ¯z)) that
contain z
2
¯z and show that they can be written in the form
G
1
B
1
G
1
B
2
¯
G
˙
1
˙
Y
φ
B
1
B
2
˙
Y
z
B
1
z
B
2
¯z
˙
Y
and so verify (3.1.18)forA
1
= A
2
=1,
˙
X =
˙
1. Similarly, find the constant
term and the terms with z,z
2
, ¯z and z¯z to verify (3.1.18) for all A
1
,A
2
and
˙
X.
Now observe that by writing the
ˆ
φ
A
1
A
2
˙
X
as a column matrix [
ˆ
φ
A
1
A
2
˙
X
]=
col[
ˆ
φ
11
˙
1
ˆ
φ
10
˙
1
ˆ
φ
01
˙
1
ˆ
φ
00
˙
1
ˆ
φ
11
˙
0
ˆ
φ
10
˙
0
ˆ
φ
01
˙
0
ˆ
φ
00
˙
0
], and similarly for the φ
B
1
B
2
˙
Y
,
(3.1.18) can be written as a matrix product
⎡
⎢
⎢
⎢
⎢
⎢
⎣
ˆ
φ
11
˙
1
ˆ
φ
10
˙
1
.
.
.
ˆ
φ
00
˙
0
⎤
⎥
⎥
⎥
⎥
⎥
⎦
=
⎡
⎢
⎢
⎢
⎢
⎢
⎣
G
1
1
G
1
1
¯
G
˙
1
˙
1
G
1
1
G
1
0
¯
G
˙
1
˙
1
G
1
0
G
1
1
¯
G
˙
1
˙
1
···
G
1
1
G
0
1
¯
G
˙
1
˙
1
G
1
1
G
0
0
¯
G
˙
1
˙
1
G
1
0
G
0
1
¯
G
˙
1
˙
1
···
.
.
.
.
.
.
.
.
.
G
0
1
G
0
1
¯
G
˙
0
˙
1
G
0
1
G
0
0
¯
G
˙
0
˙
1
G
0
0
G
0
1
¯
G
˙
0
˙
1
···
⎤
⎥
⎥
⎥
⎥
⎥
⎦
⎡
⎢
⎢
⎢
⎢
⎣
φ
11
˙
1
φ
10
˙
1
.
.
.
φ
00
˙
0
⎤
⎥
⎥
⎥
⎥
⎦
.
Unlike (3.1.14), however, the 8 ×8
coe
fficient matrix here is not D
(
2
2
,
1
2
)
G
.In-
deed, the representation G → [G
A
1
B
1
G
A
2
B
2
¯
G
˙
X
˙
Y
] which assigns this matrix
to G is not even equivalent to the spinor representation of type (2,1) since
the latter has order (2 + 1)(1 + 1) = 6, not 8. The reason is that, in writing
the elements of P
21
in the form φ
A
1
A
2
˙
X
z
A
1
z
A
2
¯z
˙
X
, we are not finding com-
ponents relative to the basis {1,z,¯z,...,z
2
¯z} since, for example, z
1
z
0
¯z
˙
1
and
z
0
z
1
¯z
˙
1
are both z ¯z. Nevertheless, it is the transformation law (3.1.18)that
is of most interest to us.
The general case proceeds in much the same way. P
mn
consists of all
p(z, ¯z)=p
00
+ p
10
z + ···+ p
mn
z
m
¯z
n
= p
rs
z
r
¯z
s
,r=0,...,m, s =0,...,n.
Each of these is written as a sum of terms of the form
φ
A
1
···A
m
˙
X
1
···
˙
X
n
z
A
1
···z
A
m
¯z
˙
X
1
···¯z
˙
X
n
,
where A
1
,...,A
m
=1, 0and
˙
X
1
,...,
˙
X
n
=
˙
1,
˙
0, and φ
A
1
...A
m
˙
X
1
...
˙
X
n
is completely symmetric in A
1
,...,A
m
(i.e., φ
A
1
···A
i
···A
j
···A
m
˙
X
1
···
˙
X
n
=
φ
A
1
···A
j
···A
i
···A
m
˙
X
1
···
˙
X
n
for all i and j) and completely symmetric in the
˙
X
1
,...,
˙
X
n
. For example,
3.1 Representations of the Lorentz Group 151
p
22
z
2
¯z
2
= φ
110···0
˙
1
˙
1
˙
0···
˙
0
z
1
z
1
z
0
···z
0
1 23 4
m
¯z
˙
1
¯z
˙
1
¯z
˙
0
···¯z
˙
0
1 23 4
n
+ φ
1010···0
˙
1
˙
0
˙
1
˙
0···
˙
0
z
1
z
0
z
1
z
0
···z
0
¯z
˙
1
¯z
˙
0
¯z
˙
1
¯z
˙
0
···¯z
˙
0
+ ···+ φ
00···011
˙
0
˙
0···
˙
0
˙
1
˙
1
z
0
z
0
···z
0
z
1
z
1
¯z
˙
0
¯z
˙
0
···¯z
˙
0
¯z
˙
1
¯z
˙
1
.
There are
m
2
n
2
terms in the sum so we may take
φ
A
1
···A
m
˙
X
1
···
˙
X
n
=
1
m
2
n
2
p
22
,
where A
1
+···+A
m
=2and
˙
X
1
+···+
˙
X
n
=
˙
2. Similarly, for each 0 ≤ r ≤ m
and 0 ≤ s ≤ n,ifA
1
,...,A
m
take the values 1 and 0 with A
1
+ ···+ A
m
= r
and
˙
X
1
,...,
˙
X
n
take the values
˙
1and
˙
0with
˙
X
1
+ ···+
˙
X
n
=˙s, we define
φ
A
1
···A
m
˙
X
1
···
˙
X
n
=
1
m
r
n
s
p
rs
.
Then
p
rs
z
r
¯z
s
=
A
1
+ ···+ A
m
= r
˙
X
1
+ ···+
˙
X
n
=˙s
A
i
=1, 0
˙
X
i
=
˙
1,
˙
0
φ
A
1
···A
m
˙
X
1
···
˙
X
n
z
A
1
···z
A
m
¯z
˙
X
1
···¯z
˙
X
n
,
where there is no sum on the left. Summing over all r =0,...,m and s =
0,...,n and using the summation convention on both sides gives
p(z, ¯z)=p
rs
z
r
¯z
s
= φ
A
1
···A
m
˙
X
1
···
˙
X
n
z
A
1
···z
A
m
¯z
˙
X
1
···¯z
˙
X
n
= φ
A
1
···A
m
˙
X
1
···
˙
X
n
z
A
1
+···+A
m
¯z
˙
X
1
+···+
˙
X
n
.
Again we observe that the φ’s are symmetric in A
1
,...,A
m
and symmetric
in
˙
X
1
,...,
˙
X
n
. Applying the transformation D
(
m
2
,
n
2
)
G
to p(z, ¯z) yields
D
(
m
2
,
n
2
)
G
φ
A
1
···A
m
˙
X
1
···
˙
X
n
z
A
1
+···+A
m
¯z
˙
X
1
+···+
˙
X
n
=
ˆ
φ
A
1
···A
m
˙
X
1
···
˙
X
n
z
A
1
+···+A
m
¯z
˙
X
1
+···+
˙
X
n
,
(3.1.19)
where
ˆ
φ
A
1
···A
m
˙
X
1
···
˙
X
n
= G
A
1
B
1
···G
A
m
B
m
¯
G
˙
X
1
˙
Y
1
···
¯
G
˙
X
n
˙
Y
n
φ
B
1
···B
m
˙
Y
1
···
˙
Y
n
,
(3.1.20)
152 3 The Theory of Spinors
and the sum is over all r =0,...,m, B
1
+···+B
m
= r with B
1
,...,B
m
=1, 0
and all s =0,...,n,
˙
Y
1
+ ···+
˙
Y
n
=˙s with
˙
Y
1
,...,
˙
Y
n
=
˙
1,
˙
0.
The transformation law (3.1.20) is typical of a certain type of “spinor”
(tho
s
e o
f “valence”
00
mn
) that we will define in the next section. For the
precise definition we wish to follow a procedure analogous to that employed
in our definition of a world tensor. The idea there was that the underlying
group being represented (L) was the group of matrices of orthogonal trans-
formations of M relative to orthonormal bases and that a world tensor could
be identified with a multilinear functional on M and its dual. By analogy
we would like to regard the elements of SL(2, C) as matrices of the structure
preserving maps of some “inner-product-like” space ß and identify “spinors”
as multilinear functionals. This is, indeed, possible, although we will have to
stretch our notion of “inner product” a bit.
Since the elements of SL(2, C)are2× 2 complex matrices, the space ß
we seek must be a 2-dimensional vector space over C.Observethatif
*
φ
1
φ
0
+
and
*
ψ
1
ψ
0
+
are two ordered pairs of complex numbers and G =
G
A
B
is in
SL(2, C) and if we define
*
ˆ
φ
1
ˆ
φ
0
+
and
*
ˆ
ψ
1
ˆ
ψ
0
+
by
#
ˆ
φ
1
ˆ
φ
0
$
=
#
G
1
1
G
1
0
G
0
1
G
0
0
$#
φ
1
φ
0
$
=
#
G
1
1
φ
1
+ G
1
0
φ
0
G
0
1
φ
1
+ G
0
0
φ
0
$
and similarly for
*
ˆ
ψ
1
ˆ
ψ
0
+
,then
ˆ
φ
1
ˆ
ψ
1
ˆ
φ
0
ˆ
ψ
0
=
#
G
1
1
G
1
0
G
0
1
G
0
0
$#
φ
1
ψ
1
φ
0
ψ
0
$
=
G
1
1
G
1
0
G
0
1
G
0
0
φ
1
ψ
1
φ
0
ψ
0
=
φ
1
ψ
1
φ
0
ψ
0
and so
ˆ
φ
1
ˆ
ψ
0
−
ˆ
φ
0
ˆ
ψ
1
= φ
1
ψ
0
− φ
0
ψ
1
. (3.1.21)
Conversely, if (3.1.21) is satisfied, then G mus
t b
e in SL(2, C). Thus, if we
define on the vector space
C
2
=
.
φ =
*
φ
1
φ
0
+
: φ
A
∈ C for A =1, 0
5
3.2 Spin Space 153
a mapping
<,>: C
2
× C
2
→ C
by
<φ,ψ>= φ
1
ψ
0
− φ
0
ψ
1
,
then the elements of SL(2, C) are precisely the matrices that preserve <,>.
Exercise 3.1.9 Verify the following properties of <,>.
1. <,> is bilinear, i.e., <φ,aψ+ bξ > = a<φ,ψ>+ b<φ,ξ>and
<aφ + bψ, ξ > = a<φ,ξ>+ b<ψ,ξ>for all a, b ∈ C and
φ, ψ, ξ ∈ C
2
.
2. <,> is skew-symmetric, i.e., <ψ,φ>= − <φ,ψ>.
3. <φ,ψ>ξ + <ξ,φ>ψ+ <ψ,ξ>φ=0forallφ, ψ, ξ ∈ C
2
.
With these observations as motivation we proceed in the next section with
an abstract definition of the underlying 2-dimensional complex vector space
ß whose multilinear functionals are “spinors”.
3.2 Spin Space
Spin space is a vector space ß over the complex numbers on which is defined
amap<,>:ß×ß → C which satisfies:
1. there exist φ and ψ in ß such that <φ,ψ>=0,
2. <ψ,φ>= − <φ,ψ>for all φ, ψ ∈ ß,
3. <aφ+ bψ, ξ > = a<φ,ξ>+ b<ψ,ξ>for all φ, ψ, ξ ∈ ßandall
a, b ∈ C,
4. <φ,ψ>ξ+ <ξ,φ>ψ+ <ψ,ξ>φ=0forallφ, ψ, ξ ∈ ß.
An element of ß is called a spin vector. The existence of a vector space of the
type described was established in Exercise 3.1.9.
Lemma 3.2.1 Each of the following holds in spin space.
(a) <φ,φ>=0 for every φ ∈ ß.
(b) <,> is bilinear, i.e., in addition to #3 in the definition we have
<φ,aψ+ bξ > = a<φ,ψ>+ b<φ,ξ>for all φ, ψ, ξ ∈ ß and
all a, b ∈ C.
(c) Any φ and ψ in ß which satisfy <φ,ψ>=0form a basis for ß. In
particular,dimß=2.
(d) There exists a basis {s
1
, s
0
} for ß which satisfies < s
1
, s
0
> =1=
− < s
0
, s
1
> (any such basis is called a spin frame for ß).
(e) If {s
1
, s
0
} is a spin frame and φ = φ
1
s
1
+ φ
0
s
0
= φ
A
s
A
, then φ
1
=
<φ,s
0
> and φ
0
= − <φ,s
1
>.
(f) If {s
1
, s
0
} is a spin frame and φ = φ
A
s
A
and ψ = ψ
A
s
A
, then
<φ,ψ>=
φ
1
ψ
1
φ
0
ψ
0
= φ
1
ψ
0
− φ
0
ψ
1
.