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Chiral Current on a Compact Spacetime

+ 2 like - 0 dislike

This is a basic question I haven't see answered anywhere and I can't seem to figure out.

The usual statement of the 1+1D chiral anomaly Ward identity is that the divergence of the chiral current is the background field strength:

$\partial_\mu \langle j^\mu\rangle = \epsilon^{\mu \nu} F_{\mu \nu}/2\pi.$

I want to rewrite this in terms of the covariant chiral current $J^\mu = \epsilon^{\mu \nu}\langle j_\nu\rangle$. I believe it says $dJ = F/2\pi$. I am worried about this expression on a compact spacetime, however, since $F/2\pi$ may have a nonzero surface integral, while the integral of a divergence over a closed surface is zero. Must it be that somehow the covariant chiral current $J$ is not gauge invariant? I don't see a mechanism for this to happen though.


asked Feb 1 in Theoretical Physics by Ryan Thorngren (1,455 points) [ no revision ]

I don't remember all the story, but it is discussed, say, in Nakahara "Geometry, Topology and Physics". Check Ch. 13, and, in particular, formula (13.35).

I am aware of the index theorem and that reference, thanks. Maybe I should clarify that equation 13.35 is inconsistent with equation 13.33 because the integral of 13.33 is zero on the LHS.

Or rather, that whole part of Nakahara's discussion seems like nonsense because the integral of dj is zero unless j is somehow not gauge invariant. There is no gauge anomaly though, so this interpretation seems unlikely. Somehow I think that equation 13.33 only makes sense in flat (in particular contractible) space.

Hmm, as far as I understand, on the curved manifold the equation should have the form $D \star j=(d+\omega) \star j=\frac{1}{2 \pi} F$, where $\omega$ is a spin connection, because the Dirac operator must be modified, and generically $\int\limits_{M}D... \neq \int\limits_{\partial M}...$. Does it make sense?

P.S. Moreover, the right hand side must also be modified by the term quadratic in curvature. See, (45) here: https://arxiv.org/pdf/hep-th/0509097v1.pdf.

I agree those terms should be there, but adding the metric dependence doesn't really help. I can ask my question on a flat torus and have the same issue. The Chern number of the gauge bundle and the A-hat genus are independent.

@RyanThorngren Hmm, right. It seems that in order for this to be correct, $\star j$ must transform under the gauge transformation in the same way as $A$ does.

I am now thinking that the Ward identity as written only holds for flat space. All the derivations of it I can find use the LSZ techniques, which require asymptotic plane wave states and the Fourier transform. As you say, we would need $\star j$ to transform like A does, but there is no gauge anomaly. The classical current and the path integral measure are both invariant under vector gauge transformations. Thanks for your thoughts on this!

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