It follows from γ
2
2n+1
= 1 that the eigenvalues of γ
2n+1
are restricted to be ±1,
which we call chirality.
Exercise 12.10. Use the Clifford algebra to show that
γ
2
2n+1
= 1 {γ
µ
,γ
2n+1
}=0.
The set of sections (M,)for an even k is not an irreducible representation
of SPIN(k) but can be decomposed into two subspaces according to the chirality
as
(M,) = (M,
+
) ⊕ (M,
−
) (12.167)
where ψ
±
∈ (M,
±
) satisfy γ
2n+1
ψ
±
=±ψ
±
. We assign the fermion
number F = 0 to sections in (M,
+
) while F = 1 for those in (M,
−
).
Then the defined in (12.164) can be written as
= (−1)
F
. (12.168)
It is clear that the operator Q flips the chirality and hence {Q,}=0.
Let Q be the Dirac operator on M and let = γ
2n+1
. In fact, it follows from
exercise 12.11 that {Q,γ
2n+1
}=0andγ
2n+1
is identified with (−1)
F
.When
is diagonalized as in (12.164), the chirality eigensections are expressed as
6
ψ
+
=
ψ
+
0
ψ
−
=
0
ψ
−
. (12.169)
It should be then clear that
: (M,
+
) → (M,
−
) and
†
: (M,
−
) →
(M,
+
) are identified with D and D
†
, respectively, in (12.79). Accordingly,
the index of the Dirac operator is defined as
ind Q = dim ker D − dim ker D
†
. (12.170)
Physicists often call the sections in ker D and ker D
†
zero modes. Then, the
index of the Dirac operator is the difference between the number of positive and
negative chirality zero modes. This index has a path integral expression as we see
in the next subsection.
12.10.2 Path integral and index theorem
Let us consider a Dirac operator Q on a 2n-dimensional spin manifold M.We
employ Euclidean time (t →−it) from now on.
Let H = (iQ)
2
=
1
2
g
µν
p
µ
p
ν
be the Hamiltonian corresonding to Q.Then
the index of the Dirac operator has a path integral expression
ind Q = Tr e
−β H
= Tr(−1)
F
e
−β H
=
PBC
x ψ e
−
β
0
dtL
(12.171)
6
Note the slight abuse of notations. The symbols ψ
±
have been used to denote sections in (M, S)
as well as those in (M,
±
).