In the paper A Duality Web in 2+1 Dimensions and Condensed Matter Physics by Seiberg, Senthil, Wang, and Witten, they studied the particle-vortex dualities in $2+1$ dimensions.

On page 20, section 3.1, they considered phase transitions of the $2+1$ dimensional theory


where $(D_{b})_{\mu}=\partial_{\mu}+ib_{\mu}$, and $(D_{\hat{b}})_{\mu}=\partial_{\mu}+i\hat{b}_{\mu}$, and $f_{\mu\nu}=\partial_{\mu}b_{\nu}-\partial_{\nu}b_{\mu}$, and $\hat{f}_{\mu\nu}=\partial_{\mu}\hat{b}_{\nu}-\partial_{\nu}\hat{b}_{\mu}$.

This theory has two gauge redundancies $U(1)_{b}$ and $U(1)_{\hat{b}}$:

$$U(1)_{b}:\,\,\,b_{\mu}(x)\rightarrow b_{\mu}(x)-\partial_{\mu}\lambda(x),\quad \phi(x)\rightarrow e^{-i\lambda(x)}\phi(x)$$

$$U(1)_{\hat{b}}:\,\,\,\hat{b}_{\mu}(x)\rightarrow\hat{b}_{\mu}(x)-\partial_{\mu}\hat{\lambda}(x),\quad \hat{\phi}(x)\rightarrow e^{-i\hat{\lambda}(x)}\hat{\phi}(x)$$

(the BF-coupling $b\wedge d\hat{b}$ is invariant up to a total derivative under the above gauge transformations)

In addition, there are two generalized global symmetries (introduced by Gaiotto, Kapustin, Seiberg, and Willett) $U(1)_{f}$ and $U(1)_{\hat{f}}$ associated with the conservation of the topological currencies

$$j=\ast f,\quad\mathrm{and}\quad\hat{j}=\ast\hat{f},$$

where $f=db$, and $\hat{f}=d\hat{b}$ are the field strength. The conservation of these two topological currents follows trivially from the Bianchi identity.

On page 22, the author studied the consequence by adding a Dirac monopole operator $\mathcal{M}_{\hat{b}}(x)$ of gauge field $\hat{b}$ into the action. To be more specific, such an operator would break the conservation of $\hat{j}=\ast d\hat{b}$, and insert a Dirac monopole at the point $x$, which results in


In such a monopole configuration, the gauge field $\hat{b}$ is not globally defined, and the field strength $\hat{f}$ belongs to a non-trivial first Chern class. i.e.


The authors claimed that adding such a monopole operator into the Lagrangian explicitly breaks the generalized global symmetry $U(1)_{\hat{f}}$. I will explain such an explicit symmetry breaking in the following simpler example.

Let's consider the free Maxwell theory in $2+1$ dimensions

$$S[A]=-\frac{1}{2}\int F\wedge\ast F$$

where $F=dA$, and $A$ is a $U(1)$-gauge field.

  1. This theory has two generalized global symmetries $U(1)_{e}$ and $U(1)_{m}$ associated with the topological currents

$$J_{e}=F,\quad\mathrm{and}\quad J_{m}=\ast F.$$

Their conservation follows directly from the EOM and the Bianchi identity

$$d\ast J_{e}=d\ast F=0,\quad d\ast J_{m}=dF=0.$$

The Lagrangian can be converted into the dual photon description. First, one imposes the Bianchi identity by hand into the path integral

$$\mathcal{Z}=\int\mathcal{D}F\int\mathcal{D}\sigma \exp\left\{i\int\left(-\frac{1}{2}F\wedge\ast F+\sigma dF\right)\right\}$$

where $\sigma$ is an auxiliary field, whose integral produces the Bianchi identity $dF=0$. Integrating out the gauge invariant variable $F$, one obtains the dual theory

$$\mathcal{Z}=\int\mathcal{D}\sigma\exp\left\{i\int\frac{1}{2}d\sigma\wedge\ast d\sigma\right\}$$

This theory should be equivalent to the original one, and its only Abelian symmetry is the shift


where $\alpha\in\mathbb{R}$. The corresponding Noether current is


This symmetry should be identified with the global symmetry $U(1)_{e}$ or $U(1)_{m}$, depending on which one of $F$ or $\ast F$ is dualized.

The vaccua manifold of this theory can be identified with $\mathbb{R}$. Picking out one of its vacuum, the global symmetry is spontaneously broken.

  1. Next, one can add a Dirac monopole operator $\mathcal{M}(x)$ into the above theory. This can be achieved by imposing


into the path-integral. One has

$$\mathcal{Z}=\int\mathcal{D}F\int\mathcal{D}\sigma \exp\left\{i\int\left(-\frac{1}{2}F\wedge\ast F+\sigma(x)(dF-2\pi\delta(x))\right)\right\}$$

Integrating out $F$, one obtains

$$\mathcal{Z}=\int\mathcal{D}\sigma\exp\left\{i\int\left(\frac{1}{2}d\sigma\wedge\ast d\sigma-2\pi\sigma(x)\delta(x)\right)\right\}=\int\mathcal{D}\sigma e^{-2\pi i\sigma(0)}e^{\frac{i}{2}\int d\sigma\wedge\ast d\sigma}.$$

Therefore, one can define the monopole operator in the dual photon description by

$$\mathcal{M}(x)=e^{-2\pi i\sigma(x)}$$

and insert it into the path integral, and write

$$\mathcal{Z}=\int\mathcal{D}\sigma\mathcal{M}(0)e^{\frac{i}{2}\int d\sigma\wedge\ast d\sigma}.$$

Under the global $U(1)$ transformation, one has

$$\mathcal{M}(x)\rightarrow e^{-2\pi i\alpha}\mathcal{M}(x)$$

where $\alpha\in S^{1}$. Therefore the global $U(1)$ symmetry is broken to $\mathbb{Z}$.

On the other hand, I found something strange from David Tong's Lecture Notes on Gauge Theory. In section 8.2 page 377, he claimed that for the Abelian-Higgs model,

$$S=\int d^{3}x\left(-\frac{1}{4e^{2}}F_{\mu\nu}F^{\mu\nu}+|D_{\mu}\phi|^{2}-m^{2}|\phi|^{2}-\lambda|\phi|^{4}\right)$$

where $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$, the topological symmetry associated with $J=\ast F$ is unbroken in the Higgs phase when the Dirac monopole is present.

Can anybody help me understand why the generalized global symmetry in this case in unbroken even when the monopole is present?

  • $\begingroup$ To your Maxwell example: if this is a duality, the theory should have two $U(1)$-symmetries also in the $\sigma$-formulation. $\endgroup$ – Lorenz Mayer Jul 28 '19 at 15:12

In the Higgs phase, there is no photon field in the IR since the gauge symmetry is spontaneously broken. There are vortex of $\phi$ that are classical solutions characterized by the winding of the pure gauge configuration around a point in space. This vortex has a finite strictly positive energy, so it is an excitation around a vacuum state, and not the vacuum itself. The vortex is certainly not invariant under $U(1)_{top}$ since it is a non-trivial representation but the vacuum state is invariant under that symmetry.

The fact that the vacuum state at that phase is symmetric under $U(1)_{top}$ means that it has unbroken $U(1)_{top}$ symmetry. One way to see that this symmetry is unbroken is by noticing the absence of Goldstone bosons in that phase. There is no massless particles. The theory is gapped at the Higgs phase.

Note also that doing $\mathcal{M}(x)=e^{i\sigma(x)}$ in the path integral that you present in the first part of the question does not apply to the IR limit of the Abelian-Higgs model in the Higgs phase since the gauge symmetry is spontaneously broken, such that the degrees of freedom of $F_{mn}$ is frozen out once you go to the IR. If this representation of $\mathcal{M}(x)=e^{i\sigma(x)}$ was valid then the $U(1)_{top}$ would be spontaneously broken since the $U(1)_{top}$ acts nonlinearly as $\sigma\rightarrow\sigma+\alpha$.

Every time that you have $F_{mn}$ degrees of freedom in the IR description you can go to the dual photon description (in terms of $\sigma$) where the $U(1)_{top}$ acts non-linearly as $\sigma\rightarrow\sigma+\alpha$, so the symmetry will be broken spontaneously. When $F_{mn}$ is frozen at the IR we can't.

| cite | improve this answer | |

If we have an abelian gauge field

$$ F = \frac{1}{2} F_{\mu \nu}d x^\mu \wedge d x^\nu \ , $$

It always holds that $ d F= 0$. Or, equivalently. Equations like $dF = \delta(x)$ can be true with a grain of salt, for example if a spontaneous breaking of gauge symmetry is taking place, it might be true that there is some other two-form field $G$, and in the vacuum phase it holds that $G=F$; and that there is some closed two-manifold $S$ with

$$ \int_S G \neq 0 \ .$$

This should not be confused with $d G = \delta(x)$, and not even with $ \int_S F \neq 0$! This happens, for example, in (3+1) dimensions, when there is a spontaneous breakdown of gauge symmetry $SU(2) \rightarrow U(1)$. This can be found, e.g. in Weinbergs book, Vol II, Chapters 23.1.d) and 23.3. This book is also the reference i used to prepare this answer.

However, this is not what Tong seems to write about. He calls them 'Monopoles', but more usually they are called 'Vortices'. This is also explained in the middle of page 179 of the lecture notes you quoted. I don't have a full answer to your question but i can explain in a bit more detail how this 'monopole' operator looks like.

To understand what happens, one should go from a $3$d theory to a $(2+1)d$-theory and consider the Hamilton operator, for now just focussing on the matter part.

$$ H_\text{matter} = \int [ L_{\text{kin}}(\phi,D_i \phi) + V(\phi(x))] d^2 \mathbf{x} \ . $$

Here and in the following $\mathbf{x}$ is the spatial coordinate.

Assume there is some unique minumum value $\phi_c$ s.t.

$$V(\phi_c) = 0 , \quad V'(\phi_c) = 0 \ , \quad V''(\phi_c) > 0 \ ,$$

and $V \geq 0$.

That is, $\phi_c$ is a stable vev.

In order for some configuration to have finite energy, one needs that

$$ V'(\phi(x)) \rightarrow 0 \quad \text{as} \quad |\mathbf{x}| \rightarrow \infty \ .$$

Hence, denoting $\rho_c = |\phi_c|^2$,

$$ \phi(x) \rightarrow \sqrt{\rho_c} e^{i \alpha(x)} \quad \text{as} \quad |\mathbf{x}| \rightarrow \infty \ .$$

Here we see already that there is some winding number appearing; this is because a configuration defines a map

$$ S^1 \rightarrow S^1 \ , \quad \frac{\mathbf{x}}{|\mathbf{x}|} \mapsto \frac{\phi(\mathbf{x}) }{ |\phi(\mathbf{x})| } \ ,$$

which is well-defined at spatial infinity. The homotopy classes of such maps are captured by the Homotopy group $\pi_1(S^1)$, and which element one is in can be calculated by the winding number:

$$ n = \int_{S^1} d \arg{\phi} = \int_{S^1} d \alpha \in \mathbb{Z} \ . $$

Now what is the relation to the $F$-field? For that we can use the e.o.m.:

$$ \partial_\mu F^{\mu\nu} = J^\nu = \frac{1}{2 i} \left[ \overline{\phi(x)} \partial^\nu \phi(x) - \partial^\nu \overline{\phi(x)} \phi(x) \right] \rightarrow \sqrt{\rho_c} \partial^\nu \alpha(x) \quad \text{as} \quad |\mathbf{x}| \rightarrow \infty \ .$$

Hence we see that

$$ \mathbb{Z} \ni n = \int_{S^1} d \alpha = \frac{1}{\sqrt{\rho_c}} \int_{S^1} d \star F \ .$$

Note though that we cannot use Stokes here to write this as an integral over space, since if we move away from the sphere at infinity, $\rho_c$ will become space-dependent.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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