$\newcommand{\d}{\mathrm{d}}\newcommand{\U}{\mathrm{U}}\newcommand{\b}[1]{\overline{#1}}\newcommand{\C}{\mathbb{C}}\newcommand{\ex}[1]{\mathrm{e}^{#1}}\newcommand{\i}{\mathrm{i}}$ Consider a free theory of a complex $\U(1)$ gauge-field, $a\in\Omega^1(X;\C)$, $$S[a] :=\frac12\int_X\b{\d a}\wedge\star\d a.$$ The gauge invariance of this theory, $\U(1)_\C$, is a complexified version of $\U(1)$ where a gauge transformation is $\delta_\lambda a = \d \lambda$, with a complex parameter $\lambda\in\Omega^0(X;\C)$.

This theory has, moreover, a real $\U(1)$ zero-form global symmetry acting on the gauge field as $$a\mapsto \ex{\i\varphi} a, \qquad \varphi\in\mathbb{R}.$$ One might be tempted to background-gauge it by allowing $\varphi$ for some curvature and compensating by coupling the theory minimally to a real flat $\U(1)$-connection $B$, as $$S[a,B]:=\frac12\int_X \b{\left(\d a - B\wedge a\right)}\wedge\star \left(\d a - B\wedge a\right),$$ where $B$ transforms under $\U(1)$ as $B\mapsto B+\i\,\d\varphi$, and looking at the resulting partition function as a functional of $B$ $$ Z[B]:= \int\frac{\mathrm{D}a}{\text{vol}(\Omega^0(X;\C))}\ex{-S[a,B]}.$$

However one can quickly realise that $Z[B]$ doesn't make sense, as the coupling to $B$ broke gauge-invariance $\U(1)_\C$. Indeed, $S[a,B]$ contains a term that behaves like a mass term for $a$, thus violating gauge invariance. For example, if I took the connection $B$ to be a constant (which I can, since $B$ is just a background field), it would be exactly a mass term for $a$. Moreover, I can't see a way to say make $B$ transform under $\U(1)_\C$ in a way that cancels the unwanted terms. Nor can I imagine adding extra terms in $B$ and $a$ that could cancel it.

In some ways, this looks a lot like a mixed anomaly. The theory simply doesn't allow me to gauge $\U(1)$. But in other ways, it seems a bit more complicated. Usually, with mixed anomalies, the background-gauged theory makes perfect sense; it is just the gauge transformations of the background-gauged symmetry that don't make sense. After all, the symmetry is still global. Here, not even the background-gauged partition function makes sense. You might say, oh but that's because $\U(1)_\C$ is already gauged. That's probably right, but, if we momentarily think of $S[a]$ as just the kinetic terms for the gauging of a global $\U(1)_\C$ symmetry, in the background-gauged $\U(1)_\C\times\U(1)$ version of the theory, the anomaly is not there at all, as it only appears through the kinetic terms of $\U(1)_\C$ and in the background-gauged version $a$ would be flat. So the theory wouldn't warn me that I can't gauge both symmetries until it's too late.

So my question is, is this actually a mixed anomaly? Has it been studied in the literature? (it seems too simple to not have been studied) If so, where does it fall into, classification-wise? Or have I just missed something and everything is fine and the theory makes sense?


1 Answer 1


Okay, I'm an idiot. Gauge invariance is still there and life still makes sense. The solution is that once $B$ is introduced, $a$ no longer transforms as $\delta_\lambda a = \mathrm{d}\lambda$, but rather $\delta_\lambda a = (\mathrm{d}-B)\lambda$ (and $\delta_\lambda B = 0$). Since $B$ is a flat one-form, the operator $\mathrm{d}_B:=\mathrm{d}-B\wedge$ is still nilpotent so $\delta_\lambda\mathrm{d}_B\; a = \mathrm{d}_B\,\delta_\lambda a = 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.