# Consistent and covariant anomalies for the abelian case

Consider the theory of left and right fermions, which interact with an abelian gauge field. Left and right sectors of the theory have the gauge anomaly: by defining the anomaly as the variation of the quantum effective action $\Gamma[A]$, obtained by integrating out the heavy fermions (the so-called consistent anomaly), we have $$\partial_{\mu}J^{\mu}_{L/R} = \pm \frac{1}{48\pi^2}F_{\mu\nu}\tilde{F}^{\mu\nu}.$$ I have met the statement that for the abelian case the above consistent anomaly is related to the so-called covariant anomaly, $$\partial_{\mu}J^{\mu}_{L/R} =\pm\frac{1}{16\pi^2}F_{\mu\nu}\tilde{F}^{\mu\nu},$$ by the factor one third. What I don't understand is that the formal difference between the covariant and consistent anomalies for the abelian case is absent, as I think, since the consistent anomaly is the covariant one as well as the covariant anomaly. So I don't understand how to obtain the factor one third from formal thinking, and what in fact the covariant anomaly is in the abelian case.

I believe the source of confusion is that in non-abelian gauge theories, one defines the consistent anomaly as obeying the Wess-Zumino consistency condition and the covariant anomaly as being gauge-covariant. We can see from your equations that in the abelian case both forms obey both of these conditions, so that is not a workable definition [edit: although the anomaly is covariant, the current for the consistent anomaly is not -- see the comments to this answer]. However, there are other properties that can distinguish the two forms even in abelian theories. Some of these are:

As you said, the consistent anomaly corresponds to the variation of the effective action obtained by integrating out the fermionic fields, that is, if $$e^{iW[A]}=\int \mathcal{D}\psi \mathcal{D}\bar{\psi}e^{iS[A,\psi,\bar{\psi}]},$$ under a gauge transformation we will have $\delta W=G$, where $\partial_\mu J^\mu=G$ is the consistent form of the anomaly. There is no way to obtain an effective action in terms of $A$ whose variation is the covariant form of the anomaly, although you can modify the current such that its divergence is in the covariant form.

A way to understand the factor of 1/3 is that it comes from Bose symmetrisation of the vertices. This might be what you mean when you talk about obtaining it by "formal thinking". Typically the covariant form appears for global symmetry currents on a gauge theory, where $J$ is the current for a symmetry distinct from the one being gauged. In this case, there is no symmetrisation between the vertices of the triangle diagram. When $J$ is the current for the symmetry $A$ is gauging, the labelling of internal fermionic lines should be symmetric among the external lines, giving a result which is three times smaller. See the discussion in Weinberg around equation (22.3.38), which applies to abelian and non-abelian theories alike.

Another way in which the covariant form appears, once again for global (i.e., spacetime independent) chiral anomalies, is as the variation of the quantum effective action. Under a chiral global transformation of $\psi$, the effective action $W$ as defined above is invariant. However we can calculate the effective action with sources, $$e^{iW[A,\chi,\bar{\chi}]}=\int \mathcal{D}\psi \mathcal{D}\bar{\psi}e^{iS[A,\psi,\bar{\psi}]+\int (\chi\psi+\bar{\psi}\bar{\chi})},$$ and then Legendre transform from $W$ to the quantum effective action for the fermions (see eg Srednicki ch. 21), $\Gamma[A,\psi,\bar{\psi}]$. Under a chiral transformation, $\delta\Gamma$ will be the covariant anomaly above. A good source here is section 3.6 of https://arxiv.org/abs/0802.0634.

• I read an article written by Bardeen and Zumino, in which they introduced the Bardeen-Zumino polynomial. Although they write that the consistent anomaly implies breaking of gauge covariance only for non-singlet gauge current, after this statement they derive (and state) that the consistent current isn't gauge covariant as long as the consistent anomaly is not zero, which is true even if the current corresponds to the singlet gauge group. Could You please comment this? – Name YYY Dec 12 '16 at 20:49
• I believe you will find in the singlet case the expression reduces to a gauge-covariant form. – user121664 Dec 14 '16 at 10:13
• @usee121664 : but the formula for the gauge current gauge variation they've derived explicitly says that the current isn't gauge covariant as long as the animaly doesn't vanish. – Name YYY Dec 14 '16 at 11:08
• What formula are you talking about exactly? – user121664 Dec 14 '16 at 14:53
• precisely, $$\delta_{\epsilon}J^{\mu}_{a} = [\epsilon, J_{\mu}]_{a} + \frac{\delta }{\delta A_{\mu}^{a}}\int d^{4}x \epsilon_{a}\text{A}_{a}(x),$$ where $A_{\mu}^{a}$ is the associated gauge field. This expression shows that even in the abelian case (not only the singlet non-abelian, but in the case of singlet abelian) the gauge current is not gauge invariant. – Name YYY Dec 14 '16 at 16:12