In TASI 2003 Lectures on Anomalies (section 1.6) Jeffrey A. Harvey present arguments, why chiral anomaly is IR effect (in contrast to calculation, where UV regulator was used):

  1. Only massless particles can contribute to the anomaly. Massive particles can be regularised using the Pauli-Villars method, and so would not be anomalous. So massive particles (i.e. UV d.o.f.) doesn't contribute to anomaly.

  2. The anomaly can be understood as a statement about the analytic structure of current correlation functions.

For me, such arguments are unsatisfactory.

Is some more rigorous arguments, which clarify that anomaly is IR effect?

Could somebody present example, when anomaly will change under change of topology of manifold or something other IR deformation?


1 Answer 1


First of all, both your question and the discussion in the lecture notes, are more general than the chiral anomaly. They apply to all 't Hooft anomalies$^{(*)}$.

The claim is not that the anomaly is an IR effect. In fact if you carry on reading the lecture notes you will arrive at the 't Hoot anomaly matching conditions, which guarantee that the anomaly is an invariant of the RG flow.

The claim is that the anomaly controls the behaviour of the IR of the theory. In fact it tells you that the IR cannot be trivially gapped due to the anomaly. It must contain some interesting physics. Either some topological degrees of freedom, or a gapless phase, or the (anomalous) symmetry should have broken spontaneously at some point before reaching the deep IR.

The most elegant way to see this is through anomaly inflow, where a $d$-dimensional anomalous QFT$_d$ cannot be consistently defined in $d$ dimensions, but rather must be accompanied by a Symmetry Protected Topological phase in one higher dimension (SPT$_{d+1}$) carrying edge modes which exactly compensate for the anomaly.

Now, suppose that somewhere you calculated the anomaly. Equivalently you found the non-trivial SPT$_{d+1}$ phase which captures the anomaly. Now keeping in mind the anomaly matching conditions, you flow with the RG all the way to the IR. With you flows also the SPT$_{d+1}$ phase. Any non-trivial SPT$_{d+1}$ phase can't have a unique ground state. Translated back to QFT this means that the IR of your QFT can't be trivially gapped.

These said, I must mention that I too find these arguments confusing and unsatisfactory. Also these notes are from 2003. Since then there has been tremendous progress on the understanding of anomalies, so these notes are severely outdated in my opinion.

$^{(*)}$ strictly speaking the chiral anomaly as discussed in these lecture notes is not an 't Hooft anomaly but rather an ABJ anomaly. It is, however an 't Hooft anomaly if you consider the gauge field non-dynamical, i.e. just forget about $\int \mathrm{D}A$

  • $\begingroup$ Thank you! I know this facts about t'Hooft anomaly. But Harvey said that Chiral anomaly determined by IR physics, right? Ant I think it's very important, that this is ABJ anomaly, i.e. is not real symmetry of quantum theory. If I fogrt about $\int DA$, I will whatever have ABJ anomaly, it is not related to dynamics of gauge field. $\endgroup$
    – Nikita
    Apr 4, 2020 at 19:11
  • $\begingroup$ 'determined by' with 'determines' here sounds like /toˈmeɪ.to/ /toʊˈmɑː.toʊ/ (tomato tomato, for non IPA readers), doesn't it? If you forget the dynamics it will most definitely be an 't Hooft anomaly and not an ABJ anomaly. See also the lecture notes by Tachikawa at CERN 2008, where he discusses this exact issue (member.ipmu.jp/yuji.tachikawa/lectures/2018-cern-rikkyo) $\endgroup$ Apr 4, 2020 at 19:14
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    $\begingroup$ So, fact that massive particles (i.e. UV d.o.f.) doesn't contribute to anomaly is an accident and haven't any meaning? $\endgroup$
    – Nikita
    Apr 4, 2020 at 19:25

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