We known that the ABJ anomaly for non-abelian gauge fields with gauge group containing $SU(2)$ as a subgroup has a topological argument from the Euclidean path integral. Through studying the Euclidean path integral we can find the BPST instanton. In presence of the BPST instanton, the number of left-handed chiral fermioinc zero-modes and right-handed chiral fermionic zero-modes is not equal. One can then interpret this fact as chiral violation.

However, the ABJ anomaly exists for abelian gauge field as well. And we know in this case, there is no Euclidean instantons anymore due to the gauge group structure. But we can still derive the anomaly by studying the triangle diagram. So it seems that the instanton argument is not general?


1 Answer 1


In the case of non-Abelian gauge theories, instantons classify the principal $SU(N)$ bundles $P$ by means of the second Chern Class $c_2(P)$ (For other gauge groups, we might need additional topological invariants for the classification). They are self-dual solutions to the Yang-Mills equations but their topological class, i.e. other configurations belonging to the same bundle, are in general not self-dual and not solutions of the equations of motion. The topological term gives a common weight for all these configurations belonging to the same bundle in the path integral.

In the case of U(1) gauge theories over a compact $4-$manifold $\mathcal{M}$, the self-dual fields, although exist in general, but they do not characterize principal bundles, and in addition, both the Maxwell term and the theta term vanish for a pure self dual(anti-self dual) configuration.

However, there can be non-trivial U(1) bundles, thus a topological theta term does give different weights to different bundles in the path integral.

Principal U(1) bundles $Q$ are classified by the first Chern class. Since the first Chern class is represented by an Abelian gauge field, the theta term has the form:

$$\mathcal{L}_{\theta} = \theta \int F \wedge F = \theta \int c_1(Q) \wedge c_1(Q)$$

Thus, for example, The Abelian topological term does not detect an element of the fourth cohomology group $H^4(\mathcal{M})$ which cannot be decomposed as a wedge product of two elements of $H^2(\mathcal{M})$.

Since on a compact manifold the Dirac's quantization condition must be satisfied: $$\frac{q}{2\pi \hbar} \int_{\sigma} F = m(Z) \in \mathbb{Z} $$ where, $Z$ is two dimensional cycle on $M$. ($m(Z)$ counts the units of flux through the surface of $Z$), then decomposing $F$ as a linear combination of integral two forms, the topological term becomes: $$(\frac{q}{2\pi \hbar})^2 \int_{\mathcal{M}} F \wedge F = \sum_{i, j = 1}^{b_2} m(Z_i) Q_{ij }m(Z_j)$$ (Please see the following work by Olive (equation (3)).

The integer matrix $Q_{ij}$ is the reciprocal intersection matrix, i.e., its reciprocal counts the number of points of the intersection of the cycles $Z_i$ and $Z_j$.

However, the self-dual fields do enter the Dirac's index, but they do so due to the interaction with the background gravitational field of the manifold $\mathcal{M}$. The index of the Dirac-Weyl operator on a compact $4-$ dimensional manifold is given by: $$\mathrm{ind}(D_A) = (\frac{q}{2\pi \hbar})^2 \int_{\mathcal{M}} F \wedge F - \frac{\eta}{8} $$ where $\eta$ is Hirzebruch signature which is the difference between the number of Harmonic self dual and Harmonic anti-self dual two forms on $\mathcal{M}$. (Please see Olive and Luis Alvarez-Gaumé.

  • $\begingroup$ Thank you very much for your answer. I am not familiar with some of the mathematical concepts you mentioned here. So, in conclusion, do you mean that $F\tilde{F}$ in U(1) gauge theory is still a topological term (first Chern class) though it does not have an interpretation of the fourth cohomology group or, more commonly like in non-abelian gauge theory, the homotopy group $\pi_3(S^3)$? $\endgroup$
    – Wein Eld
    Jun 8, 2018 at 12:26
  • $\begingroup$ But how can we see directly the chiral violation in U(1) theory as we saw in non-abelian gauge theory (via the A-S index theorem), i.e., see directly the difference of left-handed fermioinc zero modes and right-handed fermionic zero modes? $\endgroup$
    – Wein Eld
    Jun 8, 2018 at 12:29
  • 2
    $\begingroup$ $\int F\wedge F$ is indeed topological even in the $U(1)$ case. It depends on the first Chern class which is a topological invariant. Homotopy arguments are rather restrictive, the apply only for spherical space-time manifolds. Cohomology on the other hand does not possess this restriction. The given index is the difference between the Left handed and right handed Weyl zero modes even if we are considering a four dimensional compact manifold. $\endgroup$ Jun 10, 2018 at 13:11

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