Naber G.L. The Geometry of Minkowski Spacetime. An Introduction to the Mathematics of the Special Theory of Relativity
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174 3 The Theory of Spinors
= δ
c
a
δ
˙
X
˙
Z
δ
A
C
σ
c
C
˙
Z
= σ
a
A
˙
X
,
wherewehaveused(3.3.5) and its barred and dotted equivalent.
Exer
ci
se 3.4.4 Prove (3.4.13), (3.4.14)and(3.4.15).
Similar
exer
cises in index gymnastics yield the analogues of (3.4.8)and
(3.4.10):
Λ
a
b
G
A
B
¯
G
˙
X
˙
Y
σ
b
B
˙
Y
= σ
a
A
˙
X
(3.4.16)
and
G
A
B
¯
G
˙
X
˙
Y
σ
a
B
˙
Y
=Λ
α
a
σ
α
A
˙
X
. (3.4.17)
Exercise 3.4.5 Prove (3.4.16)and(3.4.17) and use (3.4.16) to show that
G
A
B
¯
G
˙
X
˙
Y
σ
a
A
˙
X
=Λ
a
b
σ
b
B
˙
Y
. (3.4.18)
Given v ∈M, {e
a
} and {s
A
} we have constructed a spinor V
A
˙
X
=
σ
a
A
˙
X
v
a
.Theσ
a
A
˙
X
allow us to retrieve the v
a
from the V
A
˙
X
. Indeed, mul-
tiplying on both sides of V
A
˙
X
= σ
b
A
˙
X
v
b
by σ
a
A
˙
X
and summing as indi-
cated gives
V
A
˙
X
σ
a
A
˙
X
= σ
a
A
˙
X
σ
b
A
˙
X
v
b
= −δ
a
b
v
b
= −v
a
,
so
v
a
= −V
A
˙
X
σ
a
A
˙
X
,a=1, 2, 3, 4. (3.4.19)
Note that if the V
A
˙
X
were the components of an arbitrary spinor of valence
11
00
, then the numbers −V
A
˙
X
σ
a
A
˙
X
would, in general, be complex and
so would not be the components of any world vector. However, we show
next that if V
A
˙
X
is Hermitian, then the −V
A
˙
X
σ
a
A
˙
X
are real and, moreover,
determine a world vector. Indeed,
V
A
˙
X
σ
a
A
˙
X
= V
1
˙
1
σ
a
1
˙
1
+ V
1
˙
0
σ
a
1
˙
0
+ V
0
˙
1
σ
a
0
˙
1
+ V
0
˙
0
σ
a
0
˙
0
= V
1
˙
1
σ
a
1
˙
1
+ V
1
˙
0
σ
a
1
˙
0
+ V
0
˙
1
σ
a
0
˙
1
+ V
0
˙
0
σ
a
0
˙
0
= V
˙
1
1
σ
a
1
˙
1
+ V
˙
0
1
σ
a
˙
0
1
+ V
˙
1
0
σ
a
˙
1
0
+ V
˙
0
0
σ
a
0
˙
0
=
¯
V
1
˙
1
¯σ
a
1
˙
1
+
¯
V
0
˙
1
¯σ
a
0
˙
1
+
¯
V
1
˙
0
¯σ
a
1
˙
0
+
¯
V
0
˙
0
¯σ
a
0
˙
0
= V
1
˙
1
σ
a
1
˙
1
+ V
0
˙
1
σ
a
0
˙
1
+ V
1
˙
0
σ
a
1
˙
0
+ V
0
˙
0
σ
a
0
˙
0
= V
A
˙
X
σ
a
A
˙
X
and so V
A
˙
X
σ
a
A
˙
X
is real.
3.4 Spinors and World Vectors 175
Exercise 3.4.6 Show that if V
A
˙
X
is a spinor that satisfies
¯
V
A
˙
X
= −V
A
˙
X
,
then V
A
˙
X
σ
a
A
˙
X
is pure imaginary so iV
A
˙
X
is Hermitian.
Now, given a Hermitian spinor V of valence
11
00
,aspinframe{s
A
} and an
admissible basis {e
a
}, we define a vector v ∈Mby specifying its components
in every admissible basis in the following way: Write V = V
A
˙
X
s
A
⊗ ¯s
˙
X
and
define the components v
a
of v relative to {e
a
} by
v
a
= −V
A
˙
X
σ
a
A
˙
X
,a=1, 2, 3, 4.
Next suppose {ˆe
a
} is another admissible basis for M, related to {e
a
} by
Λ ∈L.LetΛ=Λ
±G
= Spin(±G )andlet{ˆs
A
} be the spin frame related to
{s
A
} by G (or −G). Then V =
ˆ
V
A
˙
X
ˆs
A
⊗
¯
ˆs
˙
X
,where
ˆ
V
A
˙
X
= G
A
B
¯
G
˙
X
˙
Y
V
B
˙
Y
(−G gives the same components). We define the components of v relative to
{ˆe
a
} by
ˆv
a
= −
ˆ
V
A
˙
X
σ
a
A
˙
X
,a=1, 2, 3, 4.
To justify the definition we must, as usual, verify that the v
a
transform
correctly, i.e., that Λ
a
b
v
b
= −
ˆ
V
A
˙
X
σ
a
A
˙
X
.But
−
ˆ
V
A
˙
X
σ
a
A
˙
X
= −G
A
B
¯
G
˙
X
˙
Y
V
B
˙
Y
σ
a
A
˙
X
= −
G
A
B
¯
G
˙
X
˙
Y
σ
a
A
˙
X
V
B
˙
Y
= −
Λ
a
b
σ
b
B
˙
Y
V
B
˙
Y
by (3.4.18)
=Λ
a
b
−V
B
˙
Y
σ
b
B
˙
Y
=Λ
a
b
v
b
as required. We summarize:
Theorem 3.4.1 Let {e
a
} be an admissible basis for M and {s
A
} aspin
frame for ß. The map which assigns to each vector v ∈M(v = v
a
e
a
) its
spinor equivalent (V = V
A
˙
X
s
A
⊗ ¯s
˙
X
,whereV
A
˙
X
= σ
a
A
˙
X
v
a
) is one-to-one
and onto the set of all Hermitian spinors of valence
11
00
.
Recall (Section 3.1) that every v ∈Mgives rise to a v
∗
∈M
∗
(the
dual of M) defined by v
∗
(u)=v · u and that every element of M
∗
, i.e.,
every covector, arises in this way from some v ∈M.Moreover,if{e
a
} is an
admissible basis for M and {e
a
} is its dual basis for M
∗
and if v = v
a
e
a
,
then v
∗
= v
a
e
a
,wherev
a
= η
aα
v
α
.Now,forA =1, 0and
˙
X =
˙
1,
˙
0, define
V
A
˙
X
= σ
a
A
˙
X
v
a
. (3.4.20)
176 3 The Theory of Spinors
Exercise 3.4.7 Show that
V
A
˙
X
= V
B
˙
Y
BA
¯
˙
Y
˙
X
, (3.4.21)
where V
B
˙
Y
= σ
b
B
˙
Y
v
b
.
Since V
B
˙
Y
are the components, relative to a spin frame {s
A
}, of a spinor of
valence
11
00
and (3.4.21) exhibits the V
A
˙
X
as the result of two successive
contracted outer products of this spinor (with and ¯), we conclude that
the V
A
˙
X
are the components, relative to {s
A
}, of a spinor of valence
00
11
which we call the spinor equivalent of the covector v
∗
.
Exercise 3.4.8 Show that, in another spin frame {ˆs
A
} related to {s
A
} by
(3.2.1)and(3.2.6),
ˆ
V
A
˙
X
= σ
a
A
˙
X
ˆv
a
, (3.4.22)
where ˆv
a
=Λ
a
b
v
b
, ΛbeingΛ
±G
.
Theorem 3.4.2 Let {e
a
} be an admissible basis for M and {s
A
} aspin
frame for ß. The map which assigns to each covector v
∗
∈M
∗
(v
∗
= v
a
e
a
)
its spinor equivalent (V
A
˙
X
s
A
⊗¯s
˙
X
,whereV
A
˙
X
= σ
a
A
˙
X
v
a
) is one-to-one and
onto the set of all Hermitian spinors of valence
00
11
.
Exercise 3.4.9 Complete the proof of Theorem 3.4.2.
Now, let us fix an admissible basis {e
a
} and a spin frame {s
A
}.Letv =
v
a
e
a
and u = u
a
e
a
be in M and V = V
A
˙
X
s
A
⊗ ¯s
˙
X
and U = U
A
˙
X
s
A
⊗ ¯s
˙
X
the
spinor equivalents of v and u. We compute U
A
˙
X
V
A
˙
X
=(σ
a
A
˙
X
u
a
)(σ
b
A
˙
X
v
b
)=
(u
a
v
b
)(σ
a
A
˙
X
σ
b
A
˙
X
)=u
a
v
b
(−δ
a
b
)=−u
a
v
a
= −η
ab
u
b
v
a
= −u · v so
U
A
˙
X
V
A
˙
X
= −u · v. (3.4.23)
Observe that if we let
[V
A
˙
X
]=
#
V
1
˙
1
V
1
˙
0
V
0
˙
1
V
0
˙
0
$
=
1
√
2
*
v
3
+ v
4
v
1
+ iv
2
v
1
− iv
2
−v
3
+ v
4
+
,
then det[V
A
˙
X
]=−
1
2
v · v so
V
A
˙
X
V
A
˙
X
=2det[V
A
˙
X
]=−v · v. (3.4.24)
Consequently, if v is null, det[V
A
˙
X
] = 0 so, assuming v =0, [V
A
˙
X
]has
rank 1.
3.4 Spinors and World Vectors 177
Exercise 3.4.10 Show that if
*
ab
cd
+
is a 2 × 2 complex matrix of rank 1,
then there exist pairs (φ
1
,φ
0
)and(ψ
1
,ψ
0
) of complex numbers such that
*
ab
cd
+
=
*
φ
1
φ
0
+
,
¯
ψ
˙
1
¯
ψ
˙
0
-
=
#
φ
1
¯
ψ
˙
1
φ
1
¯
ψ
˙
0
φ
0
¯
ψ
˙
1
φ
0
¯
ψ
˙
0
$
.
Consequently, if v ∈Mis null and nonzero we may write V
A
˙
X
= φ
A
¯
ψ
˙
X
for
A =1, 0and
˙
X =
˙
1,
˙
0. Observe that, in another spin frame,
ˆ
V
A
˙
X
= G
A
B
¯
G
˙
X
˙
Y
V
B
˙
Y
= G
A
B
¯
G
˙
X
˙
Y
φ
B
¯
ψ
˙
Y
=
G
A
B
φ
B
¯
G
˙
X
˙
Y
¯
ψ
˙
Y
.
Thus, if we define
ˆ
φ
A
= G
A
B
φ
B
and
¯
ˆ
ψ
˙
X
=
¯
G
˙
X
˙
Y
¯
ψ
˙
Y
,then
ˆ
V
A
˙
X
=
ˆ
φ
A
¯
ˆ
ψ
˙
X
.
Consequently, if we let φ be the spin vector whose components in {s
A
} are φ
A
and
¯
ψ be the conjugate spin vector whose components in {¯s
A
} are
¯
ψ
˙
X
,then
V is the outer product φ ⊗
¯
ψ of φ and
¯
ψ. Even more can be said, however.
Exercise 3.4.11 Suppose z
1
and z
2
are two complex numbers for which
z
1
¯z
2
is real. Show that one of z
1
or z
2
is a real multiple of the other.
Now, V
1
˙
1
and V
0
˙
0
are both real (±v
3
+ v
4
)soφ
1
¯
ψ
˙
1
and φ
0
¯
ψ
˙
0
are real
and, since v is null, but not zero, not both can be zero. Exercise 3.4.11 gives
an r
1
∈ R such that either ψ
1
= r
1
φ
1
or φ
1
= r
1
ψ
1
and also an r
0
∈ R such
that either ψ
0
= r
0
φ
0
or φ
0
= r
0
ψ
0
. Since at least one of r
1
or r
0
is nonzero
we may assume without loss of generality that
*
ψ
1
ψ
0
+
=
*
r
1
φ
1
r
0
φ
0
+
.
We claim that, in fact, there exists a single real number r such that
*
ψ
1
ψ
0
+
= r
*
φ
1
φ
0
+
. (3.4.25)
To prove this we first suppose φ
1
=0.Thenψ
1
=0so
*
ψ
1
ψ
0
+
=
*
0
ψ
0
+
=
r
0
*
0
φ
0
+
= r
0
*
φ
1
φ
0
+
. Similarly, if φ
0
=0,then
*
ψ
1
ψ
0
+
= r
1
*
φ
1
φ
0
+
. Now, suppose
neither φ
1
nor φ
0
is zero. Then, since V
1
˙
0
= V
0
˙
1
,
178 3 The Theory of Spinors
φ
1
¯
ψ
˙
0
= φ
0
¯
ψ
˙
1
,
¯
φ
˙
1
ψ
0
= φ
0
¯
ψ
˙
1
, (3.4.26)
¯
φ
˙
1
φ
0
ψ
0
=
¯
ψ
˙
1
.
Thus,
¯
ψ
˙
1
=0wouldgiveψ
0
=0andψ
1
=0so[V
A
˙
X
]=
*
00
00
+
and this gives
v
1
= v
2
= v
3
= v
4
= 0, contrary to our assumption that v = 0. Similarly,
ψ
0
= 0 implies v = 0, again a contradiction. Thus, ψ
1
and ψ
0
are nonzero so
(3.4.26)gives
¯
φ
˙
1
φ
0
=
¯
ψ
˙
1
ψ
0
=
r
1
¯
φ
˙
1
r
0
φ
0
(since r
1
∈ R). Consequently, r
1
= r
0
so
*
ψ
1
ψ
0
+
=
*
r
1
φ
1
r
1
φ
0
+
= r
1
*
φ
1
φ
0
+
and
(3.4.25)isprovedwithr = r
1
. From this it follows that
V
A
˙
X
= φ
A
¯
ψ
˙
X
= φ
A
(r
¯
φ
˙
X
)
= ±
|r|
1
2
φ
A
|r|
1
2
¯
φ
˙
X
(+ if r>0and− if r<0). Now we define a spin vector ξ by ξ
A
= |r|
1
2
φ
A
(relative to {s
A
}). Then
¯
ξ
˙
X
= |r|
1
2
¯
φ
˙
X
since |r|
1
2
is real. Thus,
V
A
˙
X
= ±ξ
A
¯
ξ
˙
X
.
Finally, observe that v
3
+ v
4
= V
1
˙
1
= rφ
1
¯
φ
˙
1
and −v
3
+ v
4
= V
0
˙
0
= rφ
0
¯
φ
˙
0
so
v
4
=
1
2
r
|φ
1
|
1
2
+ |φ
0
|
1
2
and, in particular, r>0 if and only if v
4
> 0. We have therefore proved
Theorem 3.4.3 Let {e
a
} be an admissible basis for M and {s
A
} aspin
frame for ß.Letv ∈Mbe a nonzero null vector, v = v
a
e
a
,andVitsspinor
equivalent, V = V
A
˙
X
s
A
⊗ ¯s
˙
X
=(σ
a
A
˙
X
v
a
)s
A
⊗ ¯s
˙
X
. Then there exists a spin
vector ξ such that:
(a) If v is future-directed, then
V
A
˙
X
= ξ
A
¯
ξ
˙
X
,
and,
(b) If v is past-directed, then
V
A
˙
X
= −ξ
A
¯
ξ
˙
X
.
3.5 Bivectors and Null Flags 179
Notice that ξ in the theorem is certainly not unique since if ν
A
= e
iθ
ξ
A
(θ ∈ R), then ¯ν
˙
X
= e
−iθ
¯
ξ
˙
X
so ν
A
¯ν
˙
X
= ξ
A
¯
ξ
˙
X
= V
A
˙
X
.
Observe that the process we have just described can be reversed as well.
That is, given a nonzero spin vector ξ
A
we define the spinor V
A
˙
X
= ξ
A
¯
ξ
˙
X
.
Then det[V
A
˙
X
]=ξ
1
¯
ξ
˙
1
ξ
0
¯
ξ
˙
0
− ξ
1
¯
ξ
˙
0
ξ
0
¯
ξ
˙
1
= 0 so the vector equivalent v
a
=
−σ
a
A
˙
X
V
A
˙
X
gives a null vector v ∈Mand, moreover, v
4
= −V
A
˙
X
σ
4
A
˙
X
=
−
−
1
√
2
V
1
˙
1
σ
4
1
˙
1
+ V
0
˙
0
σ
4
0
˙
0
=
1
√
2
(V
1
˙
1
+ V
0
˙
0
)=
1
√
2
(ξ
1
¯
ξ
˙
1
+ ξ
0
¯
ξ
˙
0
)=
1
√
2
(|ξ
1
|
2
+ |ξ
0
|
2
) > 0sov is future-directed. Thus, every nonzero spin vector
ξ gives rise in a natural way to a future-directed null vector v which we will
call the flagpole of ξ and which will play a prominant role in the geometrical
representation of ξ that we construct in the next section.
3.5 Bivectors and Null Flags
We recall (Section 2.7) that a bivector on M is a real-valued bilinear form
˜
F : M×M→ R that is skew-symmetric (
˜
F (u, v)=−
˜
F (v, u) for all u
and v in M). Thus,
˜
F is a skew-symmetric world tensor of covariant rank
2 and contravariant rank 0. We have already seen that bivectors are useful
for the description of electromagnetic fields and will return to their role in
electromagnetic theory in the next section. For the present our objective
is to find a “spinor equivalent” for an arbitrary bivector, show how a spin
vector gives rise, in a natural way, to a bivector and construct from it a
geometrical representation (“up to sign”) for an arbitrary nonzero spin vector.
This geometrical picture of a spin vector, called a “null flag”, emphasizes
what is perhaps its most fundamental characteristic, that is, an essential
“two-valuedness”.
Now fix an admissible basis {e
a
} for M and a spin frame {s
A
} for ß.
The components of
˜
F relative to {e
a
} are given by F
ab
=
˜
F (e
a
,e
b
) and, by
skew-symmetry, satisfy
F
ab
=
1
2
(F
ab
− F
ba
)=F
[ab]
. (3.5.1)
For A, B =1, 0and
˙
X,
˙
Y =
˙
1,
˙
0 we define
F
A
˙
XB
˙
Y
= F
AB
˙
X
˙
Y
= σ
a
A
˙
X
σ
b
B
˙
Y
F
ab
and take these to be the components of the spinor equivalent of
˜
F relative
to { s
A
}. Thus, in another spin frame {ˆs
A
}, related to {s
A
} by (3.2.1)and
(3.2.6),
ˆ
F
A
˙
XB
˙
Y
= G
A
A
1
¯
G
˙
X
1
˙
X
G
B
B
1
¯
G
˙
Y
1
˙
Y
F
A
1
˙
X
1
B
1
˙
Y
1
= G
A
A
1
¯
G
˙
X
1
˙
X
G
B
B
1
¯
G
˙
Y
1
˙
Y
σ
α
A
1
˙
X
1
σ
β
B
1
˙
Y
1
F
αβ
180 3 The Theory of Spinors
=
G
A
A
1
¯
G
˙
X
1
˙
X
σ
α
A
1
˙
X
1
G
B
B
1
¯
G
˙
Y
1
˙
Y
σ
β
B
1
˙
Y
1
F
αβ
=(Λ
a
α
σ
a
A
˙
X
)
Λ
b
β
σ
b
B
˙
Y
F
αβ
by (3.4.17)
= σ
a
A
˙
X
σ
b
B
˙
Y
Λ
a
α
Λ
b
β
F
αβ
,
where Λ = Λ
G
.Thus,
ˆ
F
A
˙
XB
˙
Y
= σ
a
A
˙
X
σ
b
B
˙
Y
ˆ
F
ab
. (3.5.2)
We list some useful properties of the spinor equivalent of a bivector.
F
ab
= σ
a
A
˙
X
σ
b
B
˙
Y
F
A
˙
XB
˙
Y
,a,b=1, 2, 3, 4, (3.5.3)
¯
F
A
˙
XB
˙
Y
= F
A
˙
XB
˙
Y
, i.e., F
A
˙
XB
˙
Y
is Hermitian, (3.5.4)
F
B
˙
YA
˙
X
= −F
A
˙
XB
˙
Y
. (3.5.5)
The proof of (3.5.5) proceeds as follows: F
B
˙
YA
˙
X
= σ
a
B
˙
Y
σ
b
A
˙
X
F
ab
=
σ
a
B
˙
Y
σ
b
A
˙
X
(−F
ba
)=−σ
b
A
˙
X
σ
a
B
˙
Y
F
ba
= −σ
a
A
˙
X
σ
b
B
˙
Y
F
ab
= −F
A
˙
XB
˙
Y
.
Exercise 3.5.1 Prove (3.5.3)and(3.5.4).
Now
we
use (3.5.5)towrite
F
A
˙
XB
˙
Y
=
1
2
[F
A
˙
XB
˙
Y
− F
B
˙
YA
˙
X
]
=
1
2
[F
A
˙
XB
˙
Y
− F
B
˙
XA
˙
Y
+ F
B
˙
XA
˙
Y
− F
B
˙
YA
˙
X
]
=
1
2
[F
A
˙
XB
˙
Y
− F
B
˙
XA
˙
Y
]+
1
2
[F
B
˙
XA
˙
Y
− F
B
˙
YA
˙
X
].
Observe that by (3.3.14),
AB
CD
F
C
˙
XD
˙
Y
=
δ
C
A
δ
D
B
− δ
D
A
δ
C
B
F
C
˙
XD
˙
Y
=
F
A
˙
XB
˙
Y
− F
B
˙
XA
˙
Y
and, similarly, ¯
˙
X
˙
Y
¯
˙
U
˙
V
F
B
˙
UA
˙
V
= F
B
˙
XA
˙
Y
− F
B
˙
YA
˙
X
so
F
A
˙
XB
˙
Y
=
1
2
AB
CD
F
C
˙
XD
˙
Y
+
1
2
¯
˙
X
˙
Y
¯
˙
U
˙
V
F
B
˙
UA
˙
V
=
AB
1
2
CD
F
C
˙
XD
˙
Y
+¯
˙
X
˙
Y
1
2
˙
U
˙
V
F
B
˙
UA
˙
V
=
AB
1
2
F
C
˙
X
C
˙
Y
+¯
˙
X
˙
Y
1
2
F
B
˙
UA
˙
U
F
A
˙
XB
˙
Y
=
AB
1
2
F
C
˙
X
C
˙
Y
+¯
˙
X
˙
Y
1
2
F
˙
UB
˙
U
A
(3.5.6)
Now define φ
AB
by
φ
AB
=
1
2
F
˙
UA
˙
U
B
,A,B=1, 0.
3.5 Bivectors and Null Flags 181
Then we claim that
φ
BA
= φ
AB
(3.5.7)
and
¯
φ
˙
X
˙
Y
=
1
2
F
C
˙
X
C
˙
Y
. (3.5.8)
To prove (3.5.7) we compute
φ
BA
=
1
2
F
˙
U
B
˙
U
A
= −
1
2
F
˙
U
A
˙
UB
by (3.5.5)
= −
1
2
,
¯
˙
U
˙
V
F
˙
VA
˙
W
B
¯
˙
W
˙
U
-
= −
1
2
,
F
˙
VA
˙
W
B
¯
˙
U
˙
V
¯
˙
W
˙
U
-
= −
1
2
,
F
˙
V
A
˙
W
B
−¯
˙
U
˙
V
¯
˙
U
˙
W
-
= −
1
2
,
F
˙
VA
˙
W
B
−δ
˙
V
˙
W
-
=
1
2
F
˙
VA
˙
V
B
= φ
AB
.
Exercise 3.5.2 Prove (3.5.8).
With t
his w
e may write (3.5.6)as
F
A
˙
XB
˙
Y
=
AB
¯
φ
˙
X
˙
Y
+ φ
AB
¯
˙
X
˙
Y
. (3.5.9)
We observe next that the process which just led us from
˜
F to F
A
˙
XB
˙
Y
to
φ
AB
can be reversed in the following sense: Given a symmetric spinor φ
AB
of valence
00
20
we can define F
A
˙
XB
˙
Y
=
AB
¯
φ
˙
X
˙
Y
+ φ
AB
¯
˙
X
˙
Y
and obtain a
spinor of valence
00
22
which satisfies (3.5.4)since
¯
F
A
˙
XB
˙
Y
= (F
˙
AX
˙
BY
)=
(F
X
˙
AY
˙
B
)=(
XY
¯
φ
˙
A
˙
B
+ φ
XY
¯
˙
A
˙
B
)=¯
˙
X
˙
Y
φ
AB
+
¯
φ
˙
X
˙
Y
AB
=
AB
¯
φ
˙
X
˙
Y
+
φ
AB
¯
˙
X
˙
Y
= F
A
˙
XB
˙
Y
,and(3.5.5)sinceF
B
˙
YA
˙
X
=
BA
¯
φ
˙
Y
˙
X
+ φ
BA
¯
˙
Y
˙
X
=
(−
AB
)
¯
φ
˙
X
˙
Y
+ φ
AB
(−¯
˙
X
˙
Y
)=−F
A
˙
XB
˙
Y
. Now define F
ab
by (3.5.3), i.e.,
F
ab
= σ
a
A
˙
X
σ
b
B
˙
Y
F
A
˙
XB
˙
Y
.
Relative to another spin frame {ˆs
A
}, related to {s
A
} by (3.2.1)and(3.2.6),
ˆ
F
A
˙
XB
˙
Y
= G
A
A
1
¯
G
˙
X
˙
X
1
G
B
B
1
¯
G
˙
Y
˙
Y
1
F
A
1
˙
X
1
B
1
˙
Y
1
.
Exercise 3.5.3 Show that σ
a
A
˙
X
σ
b
B
˙
Y
ˆ
F
A
˙
XB
˙
Y
=Λ
a
α
Λ
b
β
F
αβ
,whereΛ=Λ
G
.
Thus, defining
ˆ
F
ab
= σ
a
A
˙
X
σ
b
B
˙
Y
ˆ
F
A
˙
XB
˙
Y
, we find that the F
ab
transform as
the components of a bivector and we may define
˜
F : M×M→R by
182 3 The Theory of Spinors
˜
F (u, v)=
˜
F (u
a
e
a
,v
b
e
b
)
=
˜
F (e
a
,e
b
)u
a
v
b
= F
ab
u
a
v
b
relative to any admissible basis. Thus, every symmetric spinor φ of valence
00
20
gives rise, in a natural way, to a bivector
˜
F .
Next we use the information accumulated thus far to construct a geomet-
rical representation (“up to sign”) of an arbitrary nonzero spin vector ξ.We
begin, as at the end of Section 3.4, by constructing the flagpole v of ξ (the
future-directed null vector equivalent of V
A
˙
X
= ξ
A
¯
ξ
˙
X
). Observe that every
spin vector in the family {e
iθ
ξ : θ ∈ R} has the same flagpole as ξ since, if ξ
A
is replaced by e
iθ
ξ
A
,then
¯
ξ
˙
X
becomes e
−iθ
¯
ξ
˙
X
and (e
iθ
ξ
A
)(e
−iθ
¯
ξ
˙
X
)=ξ
A
¯
ξ
˙
X
.
We call e
iθ
the phase factor of the corresponding member of the family.
Exercise 3.5.4 Show that, conversely, if ψ is a spin vector with the same
flagpole as ξ,thenψ
A
= e
iθ
ξ
A
for some θ ∈ R. Hint:Writeoutv
1
,v
2
,v
3
and
v
4
in terms of ξ
A
and ψ
A
, show that ψ
1
= e
iθ
1
ξ
1
and ψ
0
= e
iθ
0
ξ
0
and then
show θ
1
= θ
0
+2nπ for some n =0, ±1,....
Our geometrical representation of ξ must therefore contain more than just
the flagpole if it is to distinguish spin vectors which differ only by a phase
factor. To determine this additional element in the picture we now observe
that ξ also determines a symmetric spinor φ of valence
00
20
defined by
φ
AB
= ξ
A
ξ
B
.
As we saw in the discussion following (3.5.9), φ
AB
gives rise to a spinor of
valence
00
22
defined by
F
A
˙
XB
˙
Y
=
AB
¯
φ
˙
X
˙
Y
+ φ
AB
¯
˙
X
˙
Y
,
which satisfies (3.5.4)and(3.5.5) and which, in turn, determines a bivector
˜
F giv
en
by F
ab
= σ
a
A
˙
X
σ
b
B
˙
Y
F
A
˙
XB
˙
Y
, i.e.,
F
ab
= σ
a
A
˙
X
σ
b
B
˙
Y
(
AB
¯
ξ
˙
X
¯
ξ
˙
Y
+ ξ
A
ξ
B
¯
˙
X
˙
Y
). (3.5.10)
To simplify (3.5.10) we select a spin vector η which,
tog
ether with ξ,forma
spin frame {ξ,η} with
<η,ξ>= ξ
A
η
A
=1=−ξ
A
ηA.
Exercise 3.5.5 Show that
ξ
A
η
B
− ξ
B
η
A
=
AB
(3.5.11)
3.5 Bivectors and Null Flags 183
and
¯
ξ
˙
X
¯η
˙
Y
−
¯
ξ
˙
Y
¯η
˙
X
=¯
˙
X
˙
Y
. (3.5.12)
Substitute (3.5.11)and(3.5.12)into(3.5.10)toobtain
F
ab
= σ
a
A
˙
X
σ
b
B
˙
Y
[(ξ
A
η
B
− ξ
B
η
A
)
¯
ξ
˙
X
¯
ξ
˙
Y
+(
¯
ξ
˙
X
¯η
˙
Y
−
¯
ξ
˙
Y
¯η
˙
X
)ξ
A
ξ
B
]
= σ
a
A
˙
X
σ
b
B
˙
Y
ξ
A
η
B
¯
ξ
˙
X
¯
ξ
˙
Y
− σ
a
A
˙
X
σ
b
B
˙
Y
ξ
B
η
A
¯
ξ
˙
X
¯
ξ
˙
Y
+ σ
a
A
˙
X
σ
b
B
˙
Y
¯
ξ
˙
X
¯η
˙
Y
ξ
A
ξ
B
− σ
a
A
˙
X
σ
b
B
˙
Y
¯
ξ
˙
Y
¯η
˙
X
ξ
A
ξ
B
=
σ
a
A
˙
X
ξ
A
¯
ξ
˙
X
σ
b
B
˙
Y
η
B
¯
ξ
˙
Y
+ σ
b
B
˙
Y
ξ
B
¯η
˙
Y
−
σ
b
B
˙
Y
ξ
B
¯
ξ
˙
Y
σ
a
A
˙
X
η
A
¯
ξ
˙
X
+ σ
a
A
˙
X
ξ
A
¯η
˙
X
= v
a
σ
b
B
˙
Y
(η
B
¯
ξ
˙
Y
+ ξ
B
¯η
˙
Y
) − v
b
σ
a
A
˙
X
(η
A
¯
ξ
˙
X
+ ξ
A
¯η
˙
X
).
Now define a spinor of valence
00
11
by
W
A
˙
X
= η
A
¯
ξ
˙
X
+ ξ
A
¯η
˙
X
and observe that W
A
˙
X
is Hermitian since
¯
W
A
˙
X
= W
˙
AX
= (¯η
˙
A
ξ
X
+
¯
ξ
˙
A
η
X
)=
η
A
¯
ξ
˙
X
+ ξ
A
¯η
˙
X
= W
A
˙
X
. Consequently (Theorem 3.4.2), we may define a cov-
ector w
∗
∈M
∗
by
w
a
= −σ
a
A
˙
X
W
A
˙
X
= −σ
a
A
˙
X
(η
A
¯
ξ
˙
X
+ ξ
A
¯η
˙
X
).
Thus, our expression for F
ab
now becomes
F
ab
= v
b
w
a
− v
a
w
b
. (3.5.13)
Notice that, by (3.4.22),
v · w = −V
A
˙
X
W
A
˙
X
= −ξ
A
¯
ξ
˙
X
(η
A
¯
ξ
˙
X
+ ξ
A
¯η
˙
X
)
= −ξ
A
η
A
(
¯
ξ
˙
X
¯
ξ
˙
X
) −
¯
ξ
˙
X
¯η
˙
X
(ξ
A
ξ
A
)
= −(−1)(0) − (−1)(0)
=0.
Thus, w is orthogonal to v.Sincev is null, w is spacelike.
Exercise 3.5.6 Show that, in fact, w · w =2.
Thus far we have found that the spin vector ξ determines a future-directed
null vector v (its flagpole) and a bivector F
ab
= v
b
w
a
− v
a
w
b
,wherew is a
spacelike vector orthogonal to v. However, w is not uniquely determined by ξ
since our choice for the “spinor mate” η for ξ is not unique. We now examine
the effect on w of making a different selection ˜η for η (still with < ˜η, ξ > =1).