# Quantized periods in electromagnetic duality path integral

In John McGreevy's notes (page 64 of https://mcgreevy.physics.ucsd.edu/w21/2021W-239-lectures.pdf), he describes a path integral derivation of electromagnetic duality for $$p$$-form gauge fields. The first step in the duality involves introducing a Lagrange multiplier field $$A^\vee$$: $$\int [dA] e^{-\frac{1}{2g} \int dA \wedge \star dA} = \int [dA \, dB \, dA^\vee] e^{-\frac{1}{2g} \int (F-B) \wedge \star (F-B) + i \int B \wedge dA^\vee}.$$ Here $$A$$ is a $$p$$-form gauge field with field strength $$F$$, $$A^\vee$$ is a $$(D-p-2)$$-form gauge field, $$B$$ is a $$(p+1)$$-form field, and $$D$$ is the spacetime dimension. In the second term, we also have a redundancy: $$A\rightarrow A+\Lambda, \, B \rightarrow B + d\Lambda$$.

I understand how performing the functional integration over the Lagrange multiplier $$A^\vee$$ will impose the constraint $$dB = 0$$: essentially by integrating by parts we can get a delta function

$$\delta(dB) \sim \int [dA^\vee] e^{\pm i \int dB \wedge A^\vee}$$

However, he makes another claim, which I have seen elsewhere in the literature (e.g Section 2.2 in Witten's "On S-Duality in Abelian Gauge Theory"): that this integration also forces the field $$B$$ to have integral periods, such that $$\oint_S B \in 2\pi \mathbb{Z},$$ where $$S$$ is any (closed) cycle in the spacetime manifold.

• How does the Lagrange multiplier also enforce integer periods of $$B$$?

• Furthermore, if $$B$$ is closed ($$dB=0$$) and has integer periods, how can we set $$B=0$$ through a gauge transformation $$B \rightarrow B + d \Lambda$$?

Naively it seems to me like this gauge transformation cannot change the cohomology class of $$B$$ because it is shifting it by an exact form, yet it is claimed that we have the freedom to set $$B=0$$ (This claim is also repeated in Witten's article above - perhaps $$d\Lambda$$ is only locally exact?)

In my opinion, in the lecture notes an monopole operator is implicitly inserted. Otherwise, it cannot make any sense. When a Dirac monopole is inserted in space, then the flux $$B$$ must satisfy the quantization condition $$\int_{S^{2}}\frac{B}{2\pi}\in\mathbb{Z}.$$

More explicity, let's consider a simpler action at the moment: $$S[B]=-\frac{1}{2}\int B\wedge\star B.$$

This model has two generalized symmetries, whose Noether currents are: $$j_{e}=B,\quad\mathrm{and}\quad j_{m}=\star B.$$

They are conserved on-shell because of the Bianchi identity $$d\star j_{m}=dB=0$$, and the equation of motion $$d\star j_{e}=d\star B=0$$.

To see this is indeed a global symmetry, one can impose the Bianchi identity in the path-integral $$\mathcal{Z}=\int\mathcal{D}B\int\mathcal{D}\sigma\exp\left(-i\frac{1}{2}\int B\wedge\star B+i\int\sigma dB\right),$$

and integrate out $$B$$ field. Then, one obtains $$\mathcal{Z}=\int\mathcal{D}\sigma\exp\left(i\frac{1}{2}\int d\sigma\wedge\star d\sigma\right).$$

From the above dual description, one finds that the symmetry is actually a constant shift $$\sigma\rightarrow\sigma+a$$, where $$a$$ is an arbitrary constant.

In the lecture notes, as far as I understand, it makes sense only when a Dirac monopole $$dB=2\pi\delta(x)$$

is imposed. Only in this case, then the condition $$\int_{S^3}dB=\oint_{S^2}B\in 2\pi\mathbb{Z}$$

makes sense. One can insert this monopole operator in the following procedure:

By introducing an auxiliary field as before, the partition function becomes $$\mathcal{Z}=\int\mathcal{D}B\int\mathcal{D}\sigma\exp\left(-i\frac{1}{2}\int B\wedge\star B+i\int\sigma(dB-2\pi\delta(x))\right).$$

Integrating out the $$B$$ field, one gets $$\mathcal{Z}=\int\mathcal{D}\sigma\left(e^{2\pi i\sigma(0)}\right)\exp\left(i\frac{1}{2}\int d\sigma\wedge\star d\sigma\right).$$

That is, one inserted a monopole operator $$\mathcal{M}(x)\equiv e^{2\pi i\sigma(x)}$$

at $$x=0$$ in the partition function. By doing so, the original $$\mathbb{R}$$ symmetry is broken into $$\mathbb{Z}$$.

But whenever $$x\neq 0$$, $$dB=0$$ is satisfied, which has a local redundancy $$B\rightarrow B+d\Lambda.$$

Then, by Poincaré lemma, one can always find an open neighborhood $$U_{x}$$ of $$\forall x\neq 0$$, where $$B$$ is exact, say $$B=dC$$ for some $$p$$-form $$C$$. Then, by choosing $$\Lambda=-C$$, one has $$B=0$$ on $$U_{x}$$. The result is not true if one chooses an open neighborhood $$U_{x}$$ containing $$x=0$$.

• Thanks for the detailed reply: I agree with your final point about locally setting $B=0$, yet I think there must be something more than an implicit assumption of a monopole insertion. In your duality transformation, you only integrate over two fields in the intermediate step. Why then do McGreevy and Witten go through this complicated introduction of a third field (with an additional gauge redundancy) if the duality can be made directly as you have written it? I suspect it is because somehow this additional step also enforces integer periods - without monopole assumptions. Commented May 31, 2022 at 7:48