(p1 ’ p2 )ρ . (23)

»µνρ

On the other hand, by commuting γ5 in (22), we notice that “ (3) is formally a

symmetric function of the three sets of external arguments. A divergence, being

proportional to (23), which is not symmetric, breaks the symmetry between

external arguments. Therefore, a symmetric regularization, of the kind we adopt

in the ¬rst calculation, leads to a ¬nite result. The result is not ambiguous

because a possible ambiguity again is proportional to (23).

Similarly, if the regularization is consistent with gauge invariance, the vector

current is conserved:

(3)

p1µ “»µν (k; p1 , p2 ) = 0 .

Applied to a possible divergent contribution, the equation implies

’p1µ p2ρ = 0,

»µνρ

which cannot be satis¬ed for arbitrary p1 , p2 . Therefore, the sum of the two

diagrams is ¬nite. Finite ambiguities must also have the form (23) and thus

are also forbidden by gauge invariance. All regularizations consistent with gauge

invariance must give the same answer.

Therefore, there are two possibilities:

(3)

(i) k» “»µν (k; p1 , p2 ) in a regularization respecting the symmetry between

the three arguments vanishes. Then both “ (3) is gauge invariant and the axial

current is conserved.

188 J. Zinn-Justin

(3)

(ii) k» “»µν (k; p1 , p2 ) in a symmetric regularization does not vanish. Then it

is possible to add to “ (3) a term proportional to (23) to restore gauge invariance

but this term breaks the symmetry between external momenta: the axial current

is not conserved and an anomaly is present.

4.2 Explicit Calculation

Momentum Regularization. The calculation can be done using one of the

various gauge invariant regularizations, for example, Momentum cut-o¬ regular-

ization or dimensional regularization with γ5 being de¬ned as in dimension four

and thus no longer anticommuting with other γ matrices. Instead, we choose

a regularization that preserves the symmetry between the three external argu-

ments and global chiral symmetry, but breaks gauge invariance. We modify the

fermion propagator as

(q)’1 ’’ (q)’1 ρ(µq 2 ),

where µ is the regularization parameter (µ ’ 0+ ), ρ(z) is a positive di¬erentiable

function such that ρ(0) = 1, and decreasing fast enough for z ’ +∞, at least

like 1/z.

Then, as we have argued, current conservation and gauge invariance are com-

(3)

patible only if k» “»µν (k; p1 , p2 ) vanishes.

It is convenient to consider directly the contribution C (2) (k) of order A2 to

5

k» J» (k) , which sums the two diagrams:

d4 q

(2) 2 4 4

ρ µ(q + k)2

C (k) = e d p1 d p2 Aµ (p1 )Aν (p2 ) 4

(2π)

p1 +p2 +k=0

—ρ µ(q ’ p2 )2 ρ µq 2 tr γ5 k(q + k)’1 γµ (q ’ p 2 )’1 γν q ’1 ,

because the calculation then suggests how the method generalizes to arbitrary

even dimensions.

We ¬rst transform the expression, using the identity

k(q + k)’1 = 1 ’ q(q + k)’1 . (24)

Then,

(2) (2)

C (2) (k) = C1 (k) + C2 (k)

with

d4 q

(2) 2 4 4

ρ µ(q + k)2

C1 (k) =e d p1 d p2 Aµ (p1 )Aν (p2 ) 4

(2π)

p1 +p2 +k=0

—ρ µ(q ’ p2 )2 ρ µq 2 tr γ5 γµ (q ’ p 2 )’1 γν q ’1

and

d4 q

(2)

= ’e 2 4 4

ρ µ(q + k)2

C2 (k) d p1 d p2 Aµ (p1 )Aν (p2 ) 4

(2π)

p1 +p2 +k=0

—ρ µ(q ’ p2 )2 ρ µq 2 tr γ5 q(q + k)’1 γµ (q ’ p 2 )’1 γν q ’1 .

Chiral Anomalies and Topology 189

In C2 (k) we use the cyclic property of the trace and the commutation of γν q ’1

(2)

and γ5 to cancel the propagator q ’1 and obtain

d4 q

(2)

C2 (k) = ’e2 d4 p1 d4 p2 Aµ (p1 )Aν (p2 ) ρ µ(q + k)2

4

(2π)

p1 +p2 +k=0

—ρ µ(q ’ p2 )2 ρ µq 2 tr γ5 γν (q + k)’1 γµ (q ’ p 2 )’1 .

We then shift q ’ q + p2 and interchange (p1 , µ) and (p2 , ν),

d4 q

(2)

C2 (k) = ’e2 ρ µ(q ’ p2 )2

d4 p1 d4 p2 Aµ (p1 )Aν (p2 ) 4

(2π)

p1 +p2 +k=0

—ρ µq ρ µ(q + p1 )2 tr γ5 γµ (q ’ p2 )’1 γν q ’1 .

2

(25)

(2) (2)

We see that the two terms C1 and C2 would cancel in the absence of regula-

tors. This would correspond to the formal proof of current conservation. How-

ever, without regularization the integrals diverge and these manipulations are

not legitimate.

Instead, here we ¬nd a non-vanishing sum due to the di¬erence in regulating

factors:

d4 q

ρ µ(q ’ p2 )2 ρ µq 2

(2) 2 4 4

C (k) = e d p1 d p2 Aµ (p1 )Aν (p2 ) 4

(2π)

p1 +p2 +k=0

— tr γ5 γµ (q ’ p 2 )’1 γν q ’1 ρ µ(q + k)2 ’ ρ µ(q + p1 )2 .

After evaluation of the trace, C (2) becomes (using (11))

d4 q

C (2) (k) = ’4e2 ρ µ(q ’ p2 )2 ρ µq 2

d4 p1 d4 p2 Aµ (p1 )Aν (p2 ) 4

(2π)

p1 +p2 +k=0

p2ρ qσ

— ρ µ(q + k)2 ’ ρ µ(q + p1 )2 .

µνρσ 2

q (q ’ p )2

2

Contributions coming from ¬nite values of q cancel in the µ ’ 0 limit. Due to

the cut-o¬, the relevant values of q are of order µ’1/2 . Therefore, we rescale q

accordingly, qµ1/2 ’ q, and ¬nd

√

d4 q

C (2) (k) = ’4e2 ρ (q ’ p2 µ)2

d4 p1 d4 p2 Aµ (p1 )Aν (p2 )

(2π)4

p1 +p2 +k=0

√ √

ρ (q + k µ)2 ’ ρ (q + p1 µ)2

p2ρ qσ

√ √

—ρ q µνρσ 2

2

.

q (q ’ p2 µ)2 µ

Taking the µ ’ 0 limit, we obtain the ¬nite result

C (2) (k) = ’4e2 d4 p1 d4 p2 Aµ (p1 )Aν (p2 )Iρσ (p1 , p2 )

µνρσ

p1 +p2 +k=0

with

d4 q

Iρσ (p1 , p2 ) ∼ p q ρ2 (q 2 )ρ (q 2 ) [2q» (k ’ p1 )» ] .

4 q 4 2ρ σ

(2π)

190 J. Zinn-Justin

The identity

d4 q q± qβ f (q 2 ) = 1 δ±β d4 q q 2 f (q 2 )

4

transforms the integral into

µd4 q 2 2

Iρσ (p1 , p2 ) ∼ ’ 1 p2ρ (2p1 ρ (q )ρ (q 2 ).

+ p2 )σ

2 (2π)4 q 2

The remaining integral can be calculated explicitly (we recall ρ(0) = 1):

∞

d4 q 1 1

qdq ρ2 (q 2 )ρ (q 2 ) = ’

ρ2 (q 2 )ρ (q 2 ) = ,

(2π)4 q 2 8π 2 48π 2

0

and yields a result independent of the function ρ. We ¬nally obtain

e2

=’

5

d4 p1 d4 p2 p1µ Aν (p1 )p2ρ Aσ (p2 )

k» J» (k) (26)

µνρσ

12π 2

and, therefore, from the de¬nition (21):

e2

(3)

k» “»µν (k; p1 , p2 ) = µνρσ p1ρ p2σ .

6π 2

This non-vanishing result implies that any de¬nition of the determinant det D

breaks at least either axial current conservation or gauge invariance. Since gauge

invariance is essential to the consistency of a gauge theory, we choose to break

axial current conservation. Exchanging arguments, we obtain the value of

e2

(3)

p1µ “»µν (k; p1 , p2 ) = »νρσ kρ p2σ .

6π 2

Instead, if we had used a gauge invariant regularization, the result for “ (3) would

have di¬ered by a term δ“ (3) proportional to (23):

(3)

’ p2 )ρ .

δ“»µν (k; p1 , p2 ) = K »µνρ (p1

The constant K then is determined by the condition of gauge invariance

(3) (3)

p1µ “»µν (k; p1 , p2 ) + δ“»µν (k; p1 , p2 ) = 0 ,

which yields

e2

(3)

=’ 2 ’ K = e2 /(6π 2 ).

p1µ δ“»µν (k; p1 , p2 ) »νρσ kρ p2σ

6π

This gives an additional contribution to the divergence of the current

e2

(3)

k» δ“»µν (k; p1 , p2 ) = µ»ρσ p1ρ p2σ .

3π 2

Chiral Anomalies and Topology 191

Therefore, in a QED-like gauge invariant ¬eld theory with massless fermions, the

axial current is not conserved: this is called the chiral anomaly. For any gauge

invariant regularization, one ¬nds

e2 2±

(3)

≡

k» “»µν (k; p1 , p2 ) = µνρσ p1ρ p2σ , (27)

2π 2 π

where ± is the ¬ne stucture constant. After Fourier transformation, (27) can be

rewritten as an axial current non-conservation equation:

±

‚» J» (x) = ’i

5

µνρσ Fµν (x)Fρσ (x) . (28)

4π

Since global chiral symmetry is not broken, the integral over the whole space

of the anomalous term must vanish. This condition is indeed veri¬ed since the

anomaly can immediately be written as a total derivative:

µνρσ Fµν Fρσ = 4‚µ ( µνρσ Aν ‚ρ Aσ ).

The space integral of the anomalous term depends only on the behaviour of the

gauge ¬eld at the boundaries, and this property already indicates a connection

between topology and anomalies.

Equation (28) also implies