It is a well-known fact that a massive Dirac fermion minimally coupled to a gauge field $A_\mu$ induces a Chern-Simons term when integrating out the fermion: \begin{align} i\bar{\psi}\gamma^\mu(\partial_\mu + A_\mu)\psi + m\bar{\psi} \psi \rightarrow \frac{i\operatorname{sign}(m)}{4\pi}\epsilon_{\mu\nu\gamma}A_\mu\partial_\nu A_\gamma \end{align}

What happens in the case that $A_\mu$ is no longer gauged, but is just a vector field? Would it simply generate a non-quantized Chern-Simons term, or is gauging necessary for a Chern-Simons term at all? As far as I can tell, it seems that perturbative calculations at leading order are insensitive to whether or not $A_\mu$ is gauged.


1 Answer 1


The (2+1)-dimensional QED, after integrating out the fermions, is still a theory with a gauge symmetry. This is not obvious from the expressions that are given in your question, however, because what appears in the question are the Lagrangian densities for the respective theories. The gauge symmetry is not a symmetry of the Lagrangian density, but only of the integrated action, $$ S=\int d^{3}\,{\cal L}.$$ The Chern-Simons term proportional to $\epsilon_{\mu\nu\rho}A^{\mu}\partial^{\nu}A^{\rho}$ is not gauge invariant, but it changes by a total derivative under a gauge transformation. Under $A^{\mu}\rightarrow A^{\mu}+\partial^{\mu}\Lambda$, the Chern-Simons term undergoes $$\epsilon_{\mu\nu\rho}A^{\mu}\partial^{\nu}A^{\rho}\rightarrow\epsilon_{\mu\nu\rho}A^{\mu}\partial^{\nu}A^{\rho}+\epsilon_{\mu\nu\rho}\partial^{\mu}\Lambda\partial^{\nu}A^{\rho}+\epsilon_{\mu\nu\rho}A^{\mu}\partial^{\nu}\partial^{\rho}\Lambda=\epsilon_{\mu\nu\rho}A^{\mu}\partial^{\nu}A^{\rho}+\partial^{\mu}\left(\epsilon_{\mu\nu\rho}\Lambda\partial^{\nu}A^{\nu}\right).$$

That means that the integrate action does not change under the gauge transformation, $S\rightarrow S$. Consequently, the equations of motion of the theory still possess gauge invariance. In spite of the explicit appearance of the potential $A^{\mu}$ in ${\cal L}$, the theory is still a gauge theory.

  • $\begingroup$ My question is not asking whether or not Chern-Simons theory is gauge invariant, but whether to what extent the explicit gauging of the theory, namely enforcing the gauge symmetry of the theory, affects the renormalization flow. Does the Chern-Simons term still dominate even in the IR limit if the theory is not gauged? $\endgroup$
    – Aaron
    Apr 3, 2020 at 14:35
  • $\begingroup$ @Aaron You seem to be trying to make a distinction that does not really exist, between an "explicit" gauge symmetry and a theory that merely possesses a gauge symmetry. Any theory with a gauge-invariant action is really a gauge theory, unless the matter sector is such that the regularization cannot preserve the gauge symmetry (that is, when there is a gauge anomaly), in which case the theory is generally inconsistent. $\endgroup$
    – Buzz
    Apr 4, 2020 at 2:05
  • $\begingroup$ The reason why I am trying to explicitly differentiate between gauge theory vs (ungauged) local symmetry is that it is commonly said that the massless Dirac fermion has a parity anomaly enforced on it because the regularization (via a mass term) must preserve gauge invariance. On the face, I agree with you that it doesn't matter if the theory is gauged or not, but relevant terms such as a $mA^2$ become allowed if the theory is not gauged, so I don't know the fate of the IR theory. $\endgroup$
    – Aaron
    Apr 4, 2020 at 4:34
  • $\begingroup$ @Aaron My point is that, because the theory is still gauge invariant, those terms automatically do not appear. $\endgroup$
    – Buzz
    Apr 4, 2020 at 15:54

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