Naber G.L. The Geometry of Minkowski Spacetime. An Introduction to the Mathematics of the Special Theory of Relativity
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154 3 The Theory of Spinors
(g) φ and ψ in ß are linearly independent if and only if <φ,ψ>=0.
(h) If {s
1
, s
0
} and {ˆs
1
,ˆs
0
} are two spin frames with s
1
= G
1
1
ˆs
1
+ G
0
1
ˆs
0
=
G
A
1
ˆs
A
and s
0
= G
1
0
ˆs
1
+ G
0
0
ˆs
0
= G
A
0
ˆs
A
, i.e.,
s
B
= G
A
B
ˆs
A
,B=1, 0, (3.2.1)
then G =
G
A
B
=
*
G
1
1
G
1
0
G
0
1
G
0
0
+
is in SL(2, C).
(i) If {s
1
, s
0
} and {ˆs
1
,ˆs
0
} are two spin frames and φ = φ
A
s
A
=
ˆ
φ
A
ˆs
A
, then
#
ˆ
φ
1
ˆ
φ
0
$
=
#
G
1
1
G
1
0
G
0
1
G
0
0
$#
φ
1
φ
0
$
,
i.e.,
ˆ
φ
A
= G
A
B
φ
B
,A=1, 0, (3.2.2)
where the G
A
B
are given by (3.2.1).
(j) A linear transformation T :ß→ ß preserves <,> (i.e., satisfies
<Tφ,Tψ>= <φ,ψ>for all φ, ψ ∈ ß) if and only if the matrix of
T relative to any spin frame is in SL(2, C).
Proof:
Exercise 3.2.1 Prove (a) and (b).
(c) From (a) and (b) it follows that <λφ,φ>= <φ,λφ>=0 for all λ ∈ C
and all φ ∈ ß. Consequently, if <φ,ψ>= 0, neither φ nor ψ can be a
multiple of the other, i.e., they are linearly independent. Moreover, for
any ξ ∈ ß, #4 gives <φ,ψ>ξ= −<ξ,φ>ψ−<ψ,ξ>φso, since
<φ,ψ>=0,ξis a linear combination of φ and ψ so {φ, ψ} is a basis
for ß.
(d) Suppose <φ,ψ>= 0. By switching names if necessary and using #2
we may assume <φ,ψ>>0. By (c), φ and ψ form a basis for ß and
therefore so do s
1
= <φ,ψ>
−
1
2
φ and s
0
=<φ,ψ>
−
1
2
ψ.Butthen
bilinearity of <,> gives <s
1
,s
0
>= 1 and so, by #2, <s
0
,s
1
> = −1.
(e) φ = φ
1
s
1
+ φ
0
s
0
⇒<φ,s
0
> = φ
1
<s
1
,s
0
> + φ
0
<s
0
,s
0
> = φ
1
, and,
similarly, <φ,s
1
> = −φ
0
.
(f) <φ,ψ>= <φ
1
s
1
+ φ
0
s
0
,ψ
1
s
1
+ ψ
0
s
0
> = φ
1
ψ
1
<s
1
,s
1
> + φ
1
ψ
0
<s
1
,s
0
> + φ
0
ψ
1
<s
0
,s
1
> + φ
0
ψ
0
<s
0
,s
0
> = φ
1
ψ
0
−φ
0
ψ
1
.
(g) <φ,ψ>= 0 implies φ and ψ linearly independent by (c). For the converse
suppose <φ,ψ>=0.Ifφ = 0 they are obviously dependent so assume
φ = 0. Select a spin frame { s
1
,s
0
} and set φ = φ
A
s
A
and ψ = ψ
A
s
A
.
Suppose φ
1
= 0 (the proof is analogous if φ
0
= 0). By (f), <φ,ψ>=0
implies φ
1
ψ
0
− φ
0
ψ
1
=0soψ
0
=(φ
0
/φ
1
)ψ
1
and therefore
*
ψ
1
ψ
0
+
=
ψ
1
φ
1
*
φ
1
φ
0
+
,
so ψ =(ψ
1
/φ
1
)φ and φ and ψ are linearly dependent.
3.2 Spin Space 155
(h) <s
1
,s
0
> = 1 implies 1 = <G
1
1
ˆs
1
+ G
0
1
ˆs
0
,G
1
0
ˆs
1
+ G
0
0
ˆs
0
> = G
1
1
G
1
0
< ˆs
1
, ˆs
1
> + G
1
1
G
0
0
< ˆs
1
, ˆs
0
> + G
0
1
G
1
0
< ˆs
0
, ˆs
1
> + G
0
1
G
0
0
< ˆs
0
, ˆs
0
> = G
1
1
G
0
0
− G
0
1
G
1
0
=det G as required.
(i)
ˆ
φ
1
ˆs
1
+
ˆ
φ
0
ˆs
0
= φ
1
s
1
+ φ
0
s
0
= φ
1
G
1
1
ˆs
1
+ G
0
1
ˆs
0
+φ
0
G
1
0
ˆs
1
+ G
0
0
ˆs
0
=
G
1
1
φ
1
+ G
1
0
φ
0
ˆs
1
+
G
0
1
φ
1
+ G
0
0
φ
0
ˆs
0
so the result follows by equat-
ing components.
(j) Let T :ß→ ß be a linear transformation and {s
1
,s
0
} aspinframe.Let
T
A
B
be the matrix of T relative to {s
1
,s
0
}. Then, for all φ and ψ in ß,
Tφ= T
A
B
φ
B
,Tψ= T
A
B
ψ
B
and
<φ,ψ>=
φ
1
ψ
1
φ
0
ψ
0
.
Now compute
<Tφ,Tψ>=
T
1
1
φ
1
+ T
1
0
φ
0
T
1
1
ψ
1
+ T
1
0
ψ
0
T
0
1
φ
1
+ T
0
0
φ
0
T
0
1
ψ
1
+ T
0
0
ψ
0
=
#
T
1
1
T
1
0
T
0
1
T
0
0
$#
φ
1
ψ
1
φ
0
ψ
0
$
=
T
1
1
T
1
0
T
0
1
T
0
0
φ
1
ψ
1
φ
0
ψ
0
.
Thus, <Tφ, Tψ>= <φ,ψ>if and only if det
T
A
B
= 1, i.e., if and only
if
T
A
B
∈ SL(2, C).
Comparing (3.2.2)and(3.1.16) we see that spin vectors are a natural
ch
oice as
carriers for the identity representation D
(
1
2
,0
)
of SL(2, C). To find
an equally natural choice for the carrier space of the equivalent representation
˜
D
(
1
2
,0
)
we denote by ß
∗
the dual of the vector space ß and by {s
1
,s
0
} the
basis for ß
∗
dual to the spin frame {s
1
,s
0
}.Thus,
s
A
(s
B
)=δ
B
A
,A,B=1, 0. (3.2.3)
The elements of ß
∗
are called spin covectors.Foreachφ ∈ ß we define
φ
∗
∈ ß
∗
by
φ
∗
(ψ)=<φ,ψ>
for every ψ ∈ ß(φ
∗
is linear by (b) of Lemma 3.2.1).
Lemma 3.2.2 Every element of ß
∗
is φ
∗
for some φ ∈ ß.
Proof: Let f ∈ ß
∗
. Select a spin frame {s
1
,s
0
} and define φ ∈ ßby
φ = f(s
0
)s
1
− f(s
1
)s
0
. Then, for every ψ ∈ ß,φ
∗
(ψ)=<φ,ψ> = <
f(s
0
)s
1
− f(s
1
)s
0
,ψ > = f (s
0
) <s
1
,ψ > −f (s
1
) <s
0
,ψ > = −f(s
0
) <
ψ, s
1
> + f(s
1
) <ψ,s
0
> = f(s
1
)ψ
1
+f (s
0
)ψ
0
.Butf(ψ)=f(ψ
1
s
1
+ψ
0
s
0
)=
ψ
1
f(s
1
)+ψ
0
f(s
0
)=φ
∗
(ψ)sof = φ
∗
.
156 3 The Theory of Spinors
Now, for every φ ∈ ß, we may write φ = φ
A
s
A
and φ
∗
= φ
A
s
A
for some con-
stants φ
A
and φ
A
,A=1, 0. By (3.2.3), φ
∗
(s
1
)=(φ
A
s
A
)(s
1
)=φ
1
s
1
(s
1
)=
φ
1
and, similarly, φ
∗
(s
0
)=φ
0
. On the other hand, φ
∗
(s
1
)=<φ,s
1
> =
<φ
1
s
1
+ φ
0
s
0
,s
1
> = −φ
0
and, similarly, φ
∗
(s
0
)=φ
1
so we find that
)
φ
1
= −φ
0
φ
0
= φ
1
.
(3.2.4)
Now, if {ˆs
1
, ˆs
0
} is another spin frame with s
B
= G
A
B
ˆs
A
as in (3.2.1), then,
by (i) of Lemma 3.2.1, we have
#
ˆ
φ
1
ˆ
φ
0
$
=
#
G
1
1
G
1
0
G
0
1
G
0
0
$#
φ
1
φ
0
$
for every φ = φ
A
s
A
=
ˆ
φ
A
ˆs
A
in ß. Letting
#
G
1
1
G
1
0
G
0
1
G
0
0
$
=
#
G
0
0
−G
0
1
−G
1
0
G
1
1
$
=
G
A
B
−1
T
,
we find that
#
G
1
1
G
1
0
G
0
1
G
0
0
$#
φ
1
φ
0
$
=
#
G
0
0
−G
0
1
−G
1
0
G
1
1
$#
−φ
0
φ
1
$
=
#
−G
0
B
φ
B
G
1
B
φ
B
$
=
#
−
ˆ
φ
0
ˆ
φ
1
$
=
#
ˆ
φ
1
ˆ
φ
0
$
,
so
ˆ
φ
A
= G
A
B
φ
B
,A=1, 0. (3.2.5)
Consequently, spin covectors have components relative to dual spin frames
that transform under
˜
D
(
1
2
,0
)
so ß
∗
is a natural choice for a space of carriers
of this representation of SL(2, C).
Exercise 3.2.2 Verify that G
A
C
G
B
C
= G
C
A
G
C
B
= δ
A
B
and show that
ˆs
A
= G
A
B
s
B
(3.2.6)
and
ˆs
A
= G
A
B
s
B
(3.2.7)
and therefore
s
B
= G
A
B
ˆs
A
. (3.2.8)
3.2 Spin Space 157
Exercise 3.2.3 For each φ ∈ ß define φ
∗∗
:ß
∗
→ C by φ
∗∗
(f)=f (φ)for
each f ∈ ß
∗
. Show that φ
∗∗
is a linear functional on ß
∗
, i.e., φ
∗∗
∈ (ß
∗
)
∗
,and
that the map φ → φ
∗∗
is an isomorphism of ß onto (ß
∗
)
∗
.
From Exercise 3.2.3 we conclude that, just as the elements of ß
∗
are linear
functionals on ß, so we can regard the elements of ß as linear functionals on
ß
∗
. Of course, the transformation law for components relative to a double
dual basis for (ß
∗
)
∗
is the same as that for the spin frame it came from since
one takes the transposed inverse twice. The point is that we now have carrier
spaces for D
(
1
2
,0
)
and
˜
D
(
1
2
,0
)
that are both spaces of linear functionals (on
ß
∗
and ß, respectively).
Next consider a bilinear functional on, say, ß
∗
× ß
∗
: ξ :ß
∗
× ß
∗
→
C.If{s
1
,s
0
} is a spin frame and {s
1
,s
0
} its dual, then for all φ
∗
=
φ
A
s
A
and ψ
∗
= ψ
A
s
A
in ß
∗
we have ξ(φ
∗
,ψ
∗
)=ξ(φ
A
s
A
,ψ
B
s
B
)=
ξ(s
A
,s
B
)φ
A
ψ
B
. Letting ξ
AB
= ξ(s
A
,s
B
) we find that ξ(φ
∗
,ψ
∗
)=ξ
AB
φ
A
ψ
B
.
Now, if {ˆs
1
, ˆs
0
} is another spin frame with dual {ˆs
1
, ˆs
0
},then
ˆ
ξ
AB
=
ξ(ˆs
A
, ˆs
B
)=ξ
G
A
C
s
C
,G
B
D
s
D
= G
A
C
G
B
D
ξ(s
C
,s
D
)=G
A
C
G
B
D
ξ
CD
which we write as
ˆ
ξ
A
1
A
2
= G
A
1
B
1
G
A
2
B
2
ξ
B
1
B
2
,A
1
,A
2
=1, 0, (3.2.9)
and recognize as being the transformation law (3.1.20)withm =2a
ndn =0
.
Multilinear functionals on larger products ß
∗
×ß
∗
×···×ß
∗
will, in the same
way, have components which transform according to (3.1.20)forlargerm an
d
n =
0 (w
e will consider nonzero n shortly). For a bilinear ξ :ß× ß → C
we find that ξ(φ, ψ)=ξ(φ
A
s
A
,ψ
B
s
B
)=ξ(s
A
,s
B
)φ
A
ψ
B
= ξ
AB
φ
A
ψ
B
and,
in another spin frame,
ˆ
ξ
C
1
C
2
= G
C
1
D
1
G
C
2
D
2
ξ
D
1
D
2
,C
1
,C
2
=1, 0, (3.2.10)
and similarly for larger products.
Exercise 3.2.4 Verify (3.2.10). Also show that if ξ :ß
∗
×ß → C is bilinear,
{s
A
} and {ˆs
A
} are spin frames with duals {s
A
} and {ˆs
A
},ξ
A
C
= ξ(s
A
,s
C
)
and
ˆ
ξ
A
C
= ξ(ˆs
A
, ˆs
C
), then, for any φ = φ
A
s
A
=
ˆ
φ
A
ˆs
A
∈ ßandψ
∗
= ψ
A
s
A
=
ˆ
ψ
A
ˆs
A
∈ ß
∗
we have ξ(ψ
∗
,φ)=ξ
A
C
ψ
A
φ
C
=
ˆ
ξ
A
C
ˆ
ψ
A
ˆ
φ
C
and
ˆ
ξ
A
1
C
1
= G
A
1
B
1
G
C
1
D
1
ξ
B
1
D
1
,A
1
,C
1
=1, 0. (3.2.11)
All of this will be generalized shortly in our definition of a “spinor”, but
first we must construct carriers for D
(
0,
1
2
)
and
˜
D
(
0,
1
2
)
. For this we shall
require a copy
¯
ß of ß that is distinct from ß. For example, we might take
¯
ß=ß×{1} so that each element of
¯
ßisoftheform(φ, 1) for some φ ∈ ß. We
denote by
¯
φ the element (φ, 1) ∈
¯
ß. Thus,
¯
ß={
¯
φ : φ ∈ ß}.
158 3 The Theory of Spinors
We define the linear space structure on
¯
ß as follows: For
¯
φ,
¯
ψ and
¯
ξ in
¯
ßand
c ∈ C we have
¯
φ +
¯
ψ =
φ + ψ
and
c
¯
φ =
¯cφ
(this last being equivalent to
¯
λ
¯
φ =
λφ,where¯c and
¯
λ are the usual conjugates
of the complex numbers c and λ). Thus, the map φ →
¯
φ of ß to
¯
ß, which is
obviously bijective, is a conjugate (or anti-) isomorphism, i.e., satisfies
φ + ψ −→
¯
φ +
¯
ψ
and
cφ −→ ¯c
¯
φ.
The elements of
¯
ß are called conjugate spin vectors.
Let {s
1
,s
0
} be a spin frame in ß and denote by ¯s
˙
1
and ¯s
˙
0
the images of s
1
and s
0
respectively under φ →
¯
φ.Then{¯s
˙
1
, ¯s
˙
0
} is a basis for
¯
ß. Moreover,
if φ = φ
1
s
1
+ φ
0
s
0
is in ß, then
¯
φ =
¯
φ
˙
1
¯s
˙
1
+
¯
φ
˙
0
¯s
˙
0
(recall our notational
conventions from Section 3.1 concerning dotted indices, bars, etc.). Now, if
{ˆs
1
, ˆs
0
} is another spin frame, related to {s
A
} by (3.2.6), and {
¯
ˆs
˙
1
,
¯
ˆs
˙
0
} is
its image under φ →
¯
φ,thenˆs
1
= G
1
B
s
B
= G
1
1
s
1
+ G
1
0
s
0
implies
¯
ˆs
˙
1
=
¯
G
˙
1
˙
1
¯s
˙
1
+
¯
G
˙
1
˙
0
¯s
˙
0
and similarly for
¯
ˆs
˙
0
so
¯
ˆs
˙
X
=
¯
G
˙
X
˙
Y
¯s
˙
Y
,
˙
X =
˙
1,
˙
0, (3.2.12)
and so
¯s
˙
Y
=
¯
G
˙
X
˙
Y
¯
ˆs
˙
X
,
˙
Y =
˙
1,
˙
0. (3.2.13)
It follows that if
¯
φ =
¯
φ
˙
Y
¯s
˙
Y
=
¯
ˆ
φ
˙
X
¯
ˆs
˙
X
,then
¯
ˆ
φ
˙
X
=
¯
G
˙
X
˙
Y
¯
φ
˙
Y
,
˙
X =
˙
1,
˙
0, (3.2.14)
and
¯
φ
˙
Y
=
¯
G
˙
X
˙
Y
¯
ˆ
φ
˙
X
,
˙
Y =
˙
1,
˙
0. (3.2.15)
The elements of the dual
¯
ß
∗
of
¯
ß are called conjugate spin covectors and the
bases dual to {¯s
˙
X
} and {
¯
ˆs
˙
X
} are denoted {¯s
˙
X
} and {
¯
ˆs
˙
X
} respectively. Just
as before we have
¯
ˆs
˙
X
=
¯
G
˙
X
˙
Y
¯s
˙
Y
,
˙
X =
˙
1,
˙
0, (3.2.16)
and
¯s
˙
Y
=
¯
G
˙
X
˙
Y
¯
ˆs
˙
X
,
˙
Y =
˙
1,
˙
0. (3.2.17)
3.2 Spin Space 159
For each φ
∗
= φ
A
s
A
=
ˆ
φ
A
ˆs
A
∈ ß
∗
we define
¯
φ
∗
∈
¯
ß
∗
by
¯
φ
∗
=
¯
φ
˙
X
¯s
˙
X
=
¯
ˆ
φ
˙
X
¯
ˆs
˙
X
.Then
¯
ˆ
φ
˙
X
=
¯
G
˙
X
˙
Y
¯
φ
˙
Y
,
˙
X =
˙
1,
˙
0, (3.2.18)
and
¯
φ
˙
Y
=
¯
G
˙
X
˙
Y
¯
ˆ
φ
˙
X
,
˙
Y =
˙
1,
˙
0. (3.2.19)
Before giving the general definitions we once again illustrate with a specific
example. Thus, consider a multilinear functional ξ :ß×
¯
ß × ß
∗
×
¯
ß
∗
→ C.
If φ = φ
A
s
A
,
¯
ψ =
¯
ψ
˙
X
¯s
˙
X
,ζ= ζ
B
s
B
and ¯ν =¯ν
˙
Y
¯s
˙
Y
are in ß,
¯
ß, ß
∗
and
¯
ß
∗
respectively, then
ξ(φ,
¯
ψ,ζ, ¯ν)=ξ(φ
A
s
A
,
¯
ψ
˙
X
¯s
˙
X
,ζ
B
s
B
, ¯ν
˙
Y
¯s
˙
Y
)
= ξ(s
A
, ¯s
˙
X
,s
B
, ¯s
˙
Y
)φ
A
¯
ψ
˙
X
ζ
B
¯ν
˙
Y
= ξ
A
˙
X
B
˙
Y
φ
A
¯
ψ
˙
X
ζ
B
¯ν
˙
Y
,
where ξ
A
˙
X
B
˙
Y
= ξ(s
A
, ¯s
˙
X
,s
B
, ¯s
˙
Y
)arethecomponentsofξ relative to the
spin frame {s
1
,s
0
} (and the related bases for
¯
ß, ß
∗
and
¯
ß
∗
). In another spin
frame {ˆs
1
, ˆs
0
} we have ξ(φ,
¯
ψ,ζ, ¯ν)=
ˆ
ξ
A
˙
X
B
˙
Y
ˆ
φ
A
¯
ˆ
ψ
˙
X
ˆ
ζ
B
¯
ˆν
˙
Y
,where
ˆ
ξ
A
˙
X
B
˙
Y
= ξ(ˆs
A
,
¯
ˆs
˙
X
, ˆs
B
,
¯
ˆs
˙
Y
)
= ξ
G
A
A
1
s
A
1
,
¯
G
˙
X
˙
X
1
¯s
˙
X
1
,G
B
B
1
s
B
1
,
¯
G
˙
Y
˙
Y
1
¯s
˙
Y
1
= G
A
A
1
¯
G
˙
X
˙
X
1
G
B
B
1
¯
G
˙
Y
˙
Y
1
ξ
A
1
˙
X
1
B
1
˙
Y
1
which, as we shall see, is the transformation law for the components relative
to a spin frame of a “spinor of valence
11
11
”. With this we are finally
prepared to present the general definitions.
A spinor of valence
rs
mn
, also called a spinor with m undotted lower
indices, n dotted lower indices, r undotted upper indices,ands dotted upper
indices is a multilinear functional
ξ :ß×···×ß
1
23 4
r factors
×
¯
ß ×···×
¯
ß
1
23 4
s factors
×ß
∗
×···×ß
∗
1 23 4
m factors
×
¯
ß
∗
×···×
¯
ß
∗
1 23 4
n factors
−→ C.
If {s
1
,s
0
} is a spin frame (with associated bases {¯s
˙
1
, ¯s
˙
0
}, {s
1
,s
0
} and {¯s
˙
1
, ¯s
˙
0
}
for
¯
ß, ß
∗
and
¯
ß
∗
), then the components of ξ relative to {s
A
} are defined by
ξ
A
1
···A
r
˙
X
1
···
˙
X
s
B
1
···B
m
˙
Y
1
···
˙
Y
n
= ξ
s
A
1
,...,s
A
r
, ¯s
˙
X
1
,...,¯s
˙
X
s
,
s
B
1
,...,s
B
m
, ¯s
˙
Y
1
,...,¯s
˙
Y
n
,
A
1
,...,A
r
,B
1
,...,B
m
=1, 0,
˙
X
1
,...,
˙
X
s
,
˙
Y
1
,...,
˙
Y
n
=
˙
1,
˙
0.
(3.2.20)
160 3 The Theory of Spinors
Exercise 3.2.5 Show that, if {ˆs
1
, ˆs
0
} is another spin frame, then
ˆ
ξ
A
1
···A
r
˙
X
1
···
˙
X
s
B
1
···B
m
˙
Y
1
···
˙
Y
n
= G
A
1
C
1
···G
A
r
C
r
¯
G
˙
X
1
˙
U
1
···
¯
G
˙
X
s
˙
U
s
G
B
1
D
1
···
G
B
m
D
m
¯
G
˙
V
1
˙
Y
1
···
¯
G
˙
V
n
˙
Y
n
ξ
C
1
···C
r
˙
U
1
···
˙
U
s
D
1
···D
m
˙
V
1
···
˙
V
n
.
(3.2.21)
It is traditional, particularly in the physics literature, to define a “spinor
with r contravariant and m covariant undotted indices and s contravariant
and n covariant dotted indices” to be an assignment of 2
r+m+s+n
complex
numbers {ξ
A
1
···A
r
˙
X
1
···
˙
X
s
B
1
···B
m
˙
Y
1
···
˙
Y
n
} to each spin frame (or, rather, an as-
signment of two such sets of numbers {±ξ
A
1
...A
r
˙
X
1
···
˙
X
s
B
1
···B
m
˙
Y
1
···
˙
Y
n
} to each
admissible basis for M) which transform according to (3.2.21) under a change
of
b
asis.
Although our approach is more in keeping with the “coordinate-free”
fashion that is currently in vogue, most calculations are, in fact, performed
in terms of components and the transformation law (3.2.21). Observe also
tha
t
, w
hen r = s =0,(3.2.21) coincides with the transformation law (3.2.20)
fo
r t
he carriers of the representation D
(
m
2
,
n
2
)
of SL(2, C). There is a dif-
ference, however, in that the φ
A
1
···A
m
˙
X
1
···
˙
X
n
constructed in Section 3.1 are
symmetric in A
1
,...,A
m
and symmetric in
˙
X
1
,...,
˙
X
n
and no such sym-
metry assumption is made in the definition of a spinor of valence
00
mn
.
The representations of SL(2, C) corresponding to the transformation laws
(3.2.21) will, in general, be r
e
ducible,
unlike the irreducible spinor represen-
tations of Section 3.1. One final remark on the ordering of indices is apropos.
The position of an index in (3.2.20) indicates the “slot” in ξ in
to
which
the
corresponding basis element is to be inserted for evaluation. For two indices
of the same type (both upper and undotted, both lower and dotted, etc.) the
order in which the indices appear is crucial since, for example, there is no
reason to suppose that ξ(s
1
,s
0
,...)andξ(s
0
,s
1
,...) are the same. However,
since slots corresponding to different types of indices accept different sorts
of objects (e.g., spin vectors and conjugate spin covectors) there is no reason
to insist upon any relative ordering of different types of indices and we shall
not do so. Thus, for example, ξ
A
1
A
2
˙
X
1
B
1
= ξ
A
1
˙
X
1
A
2
B
1
= ξ
A
1
B
1
A
2
˙
X
1
etc., but
these need not be the same as ξ
A
2
A
1
˙
X
1
B
1
,etc.
3.3 Spinor Algebra
In this section we collect together the basic algebraic and computational tools
that will be used in the remainder of the chapter. We begin by introducing
a matrix that will figure prominantly in many of the calculations that are
before us. Thus, we let denote the 2 × 2 matrix defined by
3.3 Spinor Algebra 161
=
*
0 −1
10
+
=
*
11
10
01
00
+
=[
AB
].
Depending on the context and the requirements of the summation convention
we will also denote the entries of in any of the following ways:
=[
AB
]=[
AB
]=[¯
˙
X
˙
Y
]=[¯
˙
X
˙
Y
].
Observe that
−1
= −.Moreover,ifφ and ψ are two spin vectors and {s
1
,s
0
}
is a spin frame with φ = φ
A
s
A
and ψ = ψ
B
s
B
, then, with the summation
convention,
AB
ψ
A
φ
B
=
10
ψ
1
φ
0
+
01
ψ
0
φ
1
= φ
1
ψ
0
−φ
0
ψ
1
= <φ,ψ>.Also
let φ
∗
= φ
A
s
A
and ψ
∗
= ψ
A
s
A
be the corresponding spin covectors.
Exercise 3.3.1 Verify each of the following:
<φ,ψ>=
AB
ψ
A
φ
B
= −
AB
φ
A
ψ
B
, (3.3.1)
φ
A
=
AB
φ
B
= −φ
B
BA
, (3.3.2)
φ
A
= φ
B
BA
= −
AB
φ
B
, (3.3.3)
φ
A
ψ
A
= <φ,ψ>= −φ
A
ψ
A
, (3.3.4)
AC
BC
= δ
A
B
=
CA
CB
, (3.3.5)
(
CB
φ
B
)
CA
= φ
A
and
AC
(φ
B
BC
)=φ
A
, (3.3.6)
AB
AB
=2=
AB
AB
. (3.3.7)
Of course, each of the identities (3.3.1)–(3.3.7) has an obvious “barred and
dotted”
ve
rsion, e.g., (3.3.6) would read (¯
˙
Z
˙
Y
¯
φ
˙
Y
)¯
˙
Z
˙
X
=
¯
φ
˙
X
. In addition to
these we record several more identities that will be used repeatedly in the
sequel.
AB
CD
+
AC
DB
+
AD
BC
=0, A,B,C,D =1, 0. (3.3.8)
To prove (3.3.8) we suppose first that A = 1
. T
hus, we consider
1B
CD
+
1C
DB
+
1D
BC
.IfB = 1 this becomes
1C
D1
+
1D
1C
. C =1orD =1
gives 0 for both terms. For C =0andD =0weobtain
10
01
+
10
10
=
(−1)(1) + (−1)(−1) = 0. On the other hand, if B =0wehave
10
CD
+
1C
D0
+
1D
0C
. C = D gives 0 for each term. For C =0andD =1we
obtain
10
01
+
10
10
+
11
00
=(−1)(1) + (−1)(−1) + 0 = 0. If C =1and
D =0,
10
10
+
11
00
+
10
01
=(−1)(−1)+0+(−1)(1) = 0. Thus, (3.3.8)
is proved if A = 1 and the argument is the same if A = 0. Next we show that
if G =
G
A
B
∈ SL(2, C), then
G
A
A
1
G
B
B
1
A
1
B
1
=
AB
,A,B=1, 0. (3.3.9)
162 3 The Theory of Spinors
This follows from
G
A
A
1
G
B
B
1
A
1
B
1
= G
A
1
G
B
0
10
+ G
A
0
G
B
1
01
= G
A
0
G
B
1
− G
A
1
G
B
0
=
⎧
⎪
⎨
⎪
⎩
0 , if A = B
det G, if A =0,B=1
− det G, if A =1,B=0
=
AB
(det G)
=
AB
since det G = 1. Similarly, if G =[G
A
B
]=([G
A
B
]
−1
)
T
,
G
A
A
1
G
B
B
1
A
1
B
1
=
AB
,A,B=1, 0, (3.3.10)
and both (3.3.9)and(3.3.10) have barred and dotted versions. Observe
that
the bilinea
r form <,>:ß× ß → C is, according to our definitions in
Section 3.2, a spinor of valence
20
00
and has components in any spin
frame given by <s
A
,s
B
> = −
AB
and that (3.2.10), with ˆ
AB
=
AB
,sim-
ply confirms the appropriate transformation law. In the same way, (3.3.9)
asser
ts tha
t the
AB
can be regarded as the (constant) components of a
spinor of valence
00
20
, whereas the barred and dotted versions of these
make similar assertions about the ¯
˙
X
˙
Y
and ¯
˙
X
˙
Y
.
Exercise 3.3.2 Write out explicitly the bilinear forms (spinors) whose com-
ponents relative to every spin frame are
AB
, ¯
˙
X
˙
Y
and ¯
˙
X
˙
Y
.
The first equality in (3.3.2) asserts that, given a spin vector φ and
th
e
corresponding spin covector φ
∗
, then, relative to any spin frame, the com-
ponents of φ
∗
= φ
A
s
A
are mechanically retrievable from those of φ = φ
B
s
B
by forming the sum
AB
φ
B
. This process is called raising the index of φ
B
.
Similarly, obtaining the φ
A
from the φ
B
according to (3.3.3) by computing
φ
B
BA
is termed lowering the index of φ
B
. Due to the skew-symmetry of
the
AB
care must be exercised in arranging the order of the factors and the
placement of the indices when carrying out these processes. As an aid to the
memory, one “raises on the left and lowers on the right” with the summed in-
dices “adjacent and descending to the right”. The equalities in (3.3.6) assert
tha
t
thes
e two operations are consistent, i.e., that lowering a raised index or
vice versa returns the original component. The operations of raising and low-
ering indices extend easily to higher valence spinors. Consider, for example, a
3.3 Spinor Algebra 163
spinor ξ of valence
20
10
.Ineachspinframeξ has components ξ
AB
C
and
we now define numbers ξ
A
B
C
in this frame by
ξ
A
B
C
= ξ
A
1
B
C
A
1
A
.
In another spin frame we have
ˆ
ξ
A
B
C
=
ˆ
ξ
A
1
B
C
A
1
A
and we now show that
ˆ
ξ
A
B
C
= G
A
A
1
G
B
B
1
G
C
C
1
ξ
A
1
B
1
C
1
, (3.3.11)
so that the ξ
A
B
C
transform as the components of a spinor of valence
10
20
.
This last spinor we shall say is obtained from ξ by “lowering the first (undot-
ted) upper index”. To prove (3.3.11)weuse(3.3.9) and the transformation
law
fo
r the ξ
AB
C
as follows:
ˆ
ξ
A
B
C
=
ˆ
ξ
A
1
B
C
A
1
A
=
G
A
1
A
2
G
B
B
1
G
C
C
1
ξ
A
2
B
1
C
1
G
A1
A
3
G
A
A
4
A
3
A
4
=
G
A
1
A
2
G
A
1
A
3
G
B
B
1
G
C
C
1
G
A
A
4
ξ
A
2
B
1
C
1
A
3
A
4
= δ
A
3
A
2
G
B
B
1
G
C
C
1
G
A
A
4
ξ
A
2
B
1
C
1
A
3
A
4
=
G
B
B
1
G
C
C
1
G
A
A
4
ξ
A
3
B
1
C
1
A
3
A
4
= G
B
B
1
G
C
C
1
G
A
A
4
ξ
A
4
B
1
C
1
= G
A
A
4
G
B
B
1
G
C
C
1
ξ
A
4
B
1
C
1
= G
A
A
1
G
B
B
1
G
C
C
1
ξ
A
1
B
1
C
1
as required.
Exercise 3.3.3 With ξ as above, let ξ
ABC
=
CC
1
ξ
AB
C
1
in each spin frame.
Show that
ˆ
ξ
ABC
= G
A
A
1
G
B
B
1
G
C
C
1
ξ
A
1
B
1
C
1
and conclude that the ξ
ABC
determine a spinor of valence
30
00
.
The calculations in these last examples make it clear that a spinor of any
valence can have any one of its lower (upper) indices raised (lowered) to yield
a spinor with one more upper (lower) index. Applying this to the constant
spinors
AB
and
AB
and using (3.3.5) yields the following useful identities.
A
B
=
BC
AC
= δ
B
A
(3.3.12)
and
A
B
=
AC
CB
= −δ
A
B
. (3.3.13)