Using Fujikawa's path integral treatment of the triangle diagram, one can show that

$$\mathrm{Tr} \gamma^5 = \int d^4 x\ \partial_{\mu}j^{\mu} $$

Where $j^{\mu}$ is the Noether current of $U(1)_A$. Thus, the $U(1)_A$ anomaly can be traced to to fact that the trace of $\mathrm{Tr} \gamma^5 \neq 0$ at loop order. My question is, why isn't $SU(2)_A$ anomalous too? I fail to understand why this applies only to $U(1)_A$.

  • $\begingroup$ Somewhat related. $\endgroup$ Commented Jul 25, 2018 at 22:36
  • $\begingroup$ How on earth did you get the impression chiral charges are never anomalous? $\endgroup$ Commented Jul 26, 2018 at 0:04

2 Answers 2


In general the chiral non-Abelian anomaly does not vanish. It is proportional to the three dimensional symmetric tensor

$$ d_{ABC} = \mathrm{tr}(T_A\{ T_B T_C\})$$

This tensor vanishes in the particular case of $SU(2)$, but it is nonvanishing for $SU(N)$, for $N>2$.

It should be also mentioned, however, that in the special case of $SU(2)$ there is a special anomaly called the Witten's $SU(2)$ global anomaly (please see the following lecture note by: Roberto Catenacci). This anomaly vanishes when the number of doublets is even.

In addition if the gauge group is $SU(2)_L \times U(1)$. Then the $SU(2)$ axial anomaly does not vanish either for a single doublet, because of the triangle diagram with two outcoming photons. However, in the standard model this anomaly cancels because the contribution to this diagram is proportional to the square of the electric charge times the isospin. It is easy to see that for each generation, the total coefficient of a single generation vanishes:

$$3(\frac{4}{9}-\frac{1}{9}) -1=0$$

where the factor $3$ counts the number of colors.


The existing answer is correct, but seems to be answering a different question than the one asked. As stated, the anomaly for three $SU(2)$ currents vanishes because all anomaly coefficients for $SU(2)$ representations are zero. But the anomaly asked about here is not that one; in particular the OP's $SU(2)_A$ is not the $SU(2)$ factor in the SM gauge group.

This question is about a triangle diagram involving an $SU(2)_A$ current and two $U(1)_V$ currents. The OP is completely correct: the logic essentially goes through unchanged and produces a nonzero $SU(2)_A$ anomaly. This is the original historical motivation for anomalies: if this were not true then the decay $\pi^0 \to \gamma \gamma$ should proceed much more slowly than it does because it's a Goldstone boson associated with spontaneous breaking of $SU(2)_A$.

Caveat: $SU(2)_A$ is not really a group, but a coset $SU(2)_L \times SU(2)_R / SU(2)_V$. Strictly speaking I should talk about the $U(1)$ subgroup of this coset associated with the $\pi^0$.

  • $\begingroup$ But... those coset generators are of course anomalous! They are the celebrated "good" chiral flavor anomalies: non-gauge (global) ones. Low energy physics all but runs on them, and their effective theory, the 4D WZW term, summarizes their effects. $\endgroup$ Commented Jul 25, 2018 at 22:21
  • $\begingroup$ @CosmasZachos Absolutely! But the direct statement of this basic fact seems to be missing from several popular textbooks. $\endgroup$
    – knzhou
    Commented Jul 25, 2018 at 22:28
  • $\begingroup$ I'll take your word for it about the textbooks... How do they introduce the WZW term and how it works and how our low energy effective physics would be inconsistent nonsense, then? Ah, never mind... what is the use of rhetorical questions? $\endgroup$ Commented Jul 25, 2018 at 22:31

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