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In Schwartz's QFT book, it is stated that anomalies cannot exist in a theory without gauge fields. This is because anomalies always give equations like $$\partial_\mu j^\mu \sim F \tilde{F}$$ where the $F_{\mu\nu}$ on the right-hand side must be the field strength associated with some gauge field. So if there were no gauge fields involved, there would be "nothing for the current to diverge to".

I don't understand this argument. First off, in the typical triangle diagram computations used to establish anomalous Ward-Takahashi relations, we never use the fact that some of the vertices are associated with gauge currents. Suppose they weren't. While it's true that you can't convert these relations into operator equations of the form $\partial_\mu j^\mu \sim F \tilde{F}$, I don't see any reason why something else couldn't go on the right-hand side.

To clarify the question: can there be anomalies for global continuous symmetries in a theory with no gauge fields? If so, what kind of things can $\partial_\mu j^\mu$ be equal to?

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  • $\begingroup$ Just a comment which may help. Consider the anomalous chiral symmetry in QCD (the anomaly which would lead you to introducing che axion). The chiral rotation is absolutely global but the measure of the path integral is not invariant. The anomaly is proproprional to the SU(3) gauge filed strengths, or to $G\tilde{G}$. There is no gauge field associated to the chiral symmetry. $\endgroup$
    – apt45
    Commented Jul 26, 2018 at 14:40
  • $\begingroup$ @apt45 I'm not asking "can a non-gauged symmetry be anomalous", rather I am asking "can a global symmetry be anomalous in a theory with no gauge fields at all". $\endgroup$
    – knzhou
    Commented Jul 26, 2018 at 14:41
  • $\begingroup$ So I think the answer has been already posted below. $\endgroup$
    – apt45
    Commented Jul 26, 2018 at 14:51

2 Answers 2

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I would defer to the extant answer, only to adapt its item 2 to your potentially narrower question. Matt S is somewhat glib here, in that there would be something in theories without [manifest, overt] gauge fields. In low energy effective hadron theories with anomalous chiral s.broken symmetry, the (4d) WZW sigma model term does the trick of providing the r.h.side to the divergence of the anomalous current.

If a Martian landed here and talked to a low energy data-analyzer without the benefit of a theorist (who'd know about anomaly consistency conditions and the like), and were curious about couplings such as KKπππ inaccessible through the standard chiral model, he might be then be served the WZW term of the effective theory that provides it. Where? he might ask.

The earthling would point to the leading term of the WZW term of low energy physics, silent about how it was produced, lacking manifest gauge fields, $$ \frac{2N_c}{15\pi^2f^5}\int_{S^4} d^4x ~ \epsilon^{\mu\nu\kappa\lambda} \operatorname {Tr} (\pi \partial_\mu \pi \partial_\nu \pi \partial_\kappa \pi \partial_\lambda \pi), $$ the πs being are SU(3) flavor octet axial-goldstons, $\pi\equiv \pi^i \lambda^i/2$ with their adjoint eightfold-way indices compacted into flavor Gell-Mann matrices.

These 8 goldstons shift under the 8 sbroken axial transformations, $$ \delta_{\theta_A } \pi\propto \theta_A ~ v_{\chi SB} + ... $$ with higher order goldston pieces.

He'd immediately see, to leading order in goldston fields, axial invariance for a hyperspherical closed boundary, since the only surviving variation of the lagrangian would be the constant underived term inside the trace, so the lagrangian would be a surface term, vanishing on the closed hyperspherical boundary.

So... what is the axial current octet corresponding to these shifts? He'd compute an extra piece beyond the standard linear one, $$ \theta_A\cdot J_{A\mu}=\frac{-8N_c}{15\pi^2f^5} \epsilon^{\mu\nu\kappa\lambda} \operatorname {Tr}(\theta_A ~\pi \partial_\nu \pi \partial_\kappa \pi \partial_\lambda \pi ), $$ diverging to $$ \partial^\mu (\theta_A\cdot J_{A\mu})=\frac{-8N_c}{15\pi^2f^5} \epsilon^{\mu\nu\kappa\lambda} \operatorname {Tr}(\theta_A ~\partial_\mu\pi \partial_\nu \pi \partial_\kappa \pi \partial_\lambda \pi ). $$ The r.h.side is non-vanishing, without manifest gauge fields, and (behold!) a total divergence, so, then, a surface term. Without manifest extraneous gauge fields. It leads with a quadrilinear term, but of course higher orders cannot help comporting with the pattern. That is, as the answer above indicates, the anomaly is a feature of the functional integral transformation structure and not of the extraneous handles that latch on to the relevant currents.

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  • $\begingroup$ Nice! Do you have a book reference for WZW terms? I haven't seen them in any of the standard introductory QFT texts. $\endgroup$
    – knzhou
    Commented Jul 26, 2018 at 19:08
  • $\begingroup$ Yes, they do , section VII-3, p 196 et seq, "the wess-zumino-witten anomaly action".... $\endgroup$ Commented Jul 26, 2018 at 19:34
  • $\begingroup$ @knzhou You may want to check Weinberg, Vol.II (chapters 19 and 22). See also Francesco et al, Conformal field theory (chapter 15), $\endgroup$ Commented Jul 26, 2018 at 22:36
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There are three types of anomalies:

  1. 't Hooft anomaly: This is a potential anomaly. The symmetry is exact, global, but connot be consistently gauged. This type of anomaly occurs for example in the flavor symmetry $SU(N_f)_L \times SU(N_f)_R \times U(1)_V$. This anomaly is important because it effects the low energy content of the theory due to the 't Hooft anomaly matching condition which states that the anomaly should be the same when computed at high or low energies.

  2. The chiral anomaly. In this case a vector symmetry is made local and the corresponding current couples to a quantized gauge field. In this case due to $VVA$ triangle diagram, there is an anomaly in the axial current, which will not be conserved, but since the axial current is not coupled itself to a gauge field, it does not need to be conserved and the theory is consistent.

  3. The chiral gauge anomaly. In this case an axial symmetry is gauged and the corresponding axial curent is coupled to a quantized gauge field. In this case the $AAA$ triangle diagram results an anomaly in the axial current and it will not be conserved; but it is gauged thus must be conserved in order to couple to a gauge field. This is the dangerous case where the theory becomes inconsistent. This is the case in the standard model where the axial current is coupled to the vector bosons. Here the anomaly must be canceled for the theory to be consistent. The cancelation can be taken care of by an educated selection of the fermion species representations as in the standard model, or the cancelation of the bulk anomaly with the edge anomaly as in the quantum Hall effect.

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    $\begingroup$ Thank you for the answer, but I'm not sure it quite addresses what I'm asking about. I'm not asking whether a global symmetry can be anomalous, but whether a global symmetry can be anomalous when there is no gauge symmetry anywhere in the theory. $\endgroup$
    – knzhou
    Commented Jul 26, 2018 at 14:27
  • $\begingroup$ This is just the first case that I addressed: the t'Hoof anomaly. Here the axial symmetry is global, there is no gauge field, but it is still anomalous; not because the corresponding current is not conserved, but because it can potentially become nonconserved had the vector symmetry been gauged. This is different from a nonanomalous symmetry because of the consequences stemming from the 't Hooft anomaly matching condition: The potential anomaly should be the same in the IR and the UV. $\endgroup$ Commented Jul 26, 2018 at 14:34
  • $\begingroup$ Ah, I see. But I think I want not just a potentially nonzero $\partial_\mu j^\mu$ if something were to be gauged, but a nonzero $\partial_\mu j^\mu$ if nothing is gauged at all. (In other words, an anomaly rather than a potential anomaly.) $\endgroup$
    – knzhou
    Commented Jul 26, 2018 at 14:40
  • $\begingroup$ The term potential anomaly is used for the sake of explanation. A potential anomaly is still an anomaly. This is why it is called 't Hooft anomaly. A nonzero $\partial_{\mu}j^{\mu}$ is a sufficient condition for an anomaly but not necessary. $\endgroup$ Commented Jul 26, 2018 at 14:46
  • $\begingroup$ How would you define an anomaly besides by $\partial_\mu j^\mu \neq 0$? $\endgroup$
    – knzhou
    Commented Jul 26, 2018 at 14:46

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