The text, see [1], compares the vortex solutions of a spontaneously broken symmetry $U(1) \rightarrow 1$ and $SU(2)\rightarrow U(1) \rightarrow \mathbb{Z}_2$. The vortices can be classified by elements in the first homotopy group which are respectively $\mathbb{Z}$ and $\mathbb{Z}_2$.

My questions

  1. What do they mean by an orientable vortex?

  2. Now this picture should show that the $SU(2)/\mathbb{Z}_2$ vortex is non-orientable:

One could try to label the quantum numbers of the vortices by arrows. But as the figure indicates, these arrows are unstable in the $SU(2)/\mathbb{Z}_2$ case. Proof

I don't understand how the arrows are associated to the quantum numbers: a single arrow is $-1$, a looping arrow is $+1$?! What does they mean with 'unstable'/'inconsistent'?

3 . I neither understand the physical consequence of this. The total flux of two $SU(2)\rightarrow U(1) \rightarrow \mathbb{Z}_2$ vortices is again a vacuum state (with zero flux). They conclude from this that there have been created `a pair of something that carries magnetic charge.' I don't see how they come to this conclusion.

This is corresponding pictorial representation:


Source: arxiv.org/abs/hep-th/0010225, p 20.


1 Answer 1


I) Ref. 1 is using the term orientable vortex in a specific situation without offering a general definition. However in the specific situation, Ref. 1 considers two cases:

  1. The vortices are labelled with additive quantum number $$n~\in~\mathbb{Z}.$$ ($n=0$ corresponds to no vortex.) Since the sign of $n$ makes physical sense, Ref. 1 calls the vortices orientable.

  2. The vortices are labelled with additive quantum number $$n~\in~ \mathbb{Z} \text{ mod } 2 ~\cong~\{0,1\}~\cong~\mathbb{Z}_2.$$ ($n=0\text{ mod } 2$ corresponds to no vortex.) Since the sign of $n$ makes no physical sense, Ref. 1 calls the vortices non-orientable.

As for how Ref. 1 at the end of Chapter 2 "concludes" that

[...] there must be magnetic monopoles,

it should likely not be read as a mathematical proof, but merely as an appetizer/advertisement for the next Chapter 3 titled Magnetic monopoles, where the mechanism is explained.

II) Let us here briefly summarize Chapter 3 as much as space permits. Consider classical static solutions to a $SU(2)$ Yang-Mills theory in 3+1 dimensions in temporal gauge $A^a_0=0$, and with a Higgs field $\phi^{\alpha}$ transforming in an $SU(2)$ irrep $R:G\to GL(2I+1,\mathbb{F})$. The Higgs field $\phi^{\alpha}$ has $2I+1$ components, where $\alpha=1, \ldots, 2I+1$. Let us call $I=\frac{1}{2}\mathbb{N}_0$ for the isospin. The gauge potential $A_i^a$ transforms in the adjoint representation $Ad$, i.e. it is $su(2)$-valued. Here $i=1,2,3$ is a spatial index, and $a=1,2,3$ is a Lie algebra index. There is also a Mexican hat potential for the Higgs to ensure a non-zero VEV. The $|D\phi|^2$ term must vanish asymptotically in order to have finite energy, which in turn strongly bind together the asymptotic behaviors of the gauge potential $A_i^a$ and the Higgs field $\phi^{\alpha}$.

We identify $eI_3$ as the generator of electric charge. We will only analyze the fields in the asymptotic region far away from the core.

III) Case of half-integer-isospin irreps. The irrep $R$ is complex and faithful. The minimal non-zero electric charge for half-integer irreps is $|e|/2$. The Dirac quantization condition states that magnetic charge must be a multiple of

$$\tag{A} g_m~=~\frac{2\pi}{|e|/2}~=~\frac{4\pi}{|e|}.$$

Next, the Higgs mechanism makes the full gauge potential massive and breaks all gauge symmetry

$$\tag{B} G~=~SU(2)~\to~ H~=~\{\bf 1\}$$

There are no monopoles

$$\tag{C} \pi_2(G/H)~\cong~ \{\bf 1\},$$

cf. e.g. this mathoverflow post. We will therefore not discuss this case further in this answer.

IV) Case of integer-isospin irreps. The irrep $R$ is real, i.e. the Higgs $\phi^{\alpha}\in \mathbb{R}$ is real-valued. The kernel of the irrep $R$ is

$$\tag{D} {\rm Ker}(R)~\cong~\{\pm {\bf 1}\}~\cong~\mathbb{Z}_2.$$

The minimal non-zero electric charge for integer irreps is $|e|$. The Dirac quantization condition states that magnetic charge must be a multiple of

$$\tag{E} g_m~=~\frac{2\pi}{|e|}. $$

Note that the center of $SU(2)$ is

$$\tag{F} Z(SU(2))~=~{\rm Ker}(Ad)~\cong~\mathbb{Z}_2.$$

This means that double-valued gauge transformations $\pm g\in SU(2)$ have a well-defined group action on the gauge potential $A_i^a$ as well as on the Higgs field $\phi^{\alpha}$ in the integer irrep $R$. So the gauge group is effectively$^1$

$$\tag{G} SU(2)/\mathbb{Z}_2~\cong~ SO(3)~=~G,$$

and we will assume this from now on.

Now apart from the central region $C\subset \mathbb{R}^3$ where possible magnetic monopoles are located, we can cover the rest of space $\mathbb{R}^3\backslash C$ with a "North" and a "South" coordinate chart, with a North and a South gauge potential, $A_{(N)i}^a$ and $A_{(S)i}^a$, respectively. The gauge transformation between the two charts in the equatorial overlap (which is homotopy equivalent to $S^1$) characterizes (the asymptotic features of) the physical multi-monopole configuration. Topologically, the equatorial gauge transformation is a map $S^1\to G$, and classified by the fundamental group $\pi_1(G)=\mathbb{Z}_2$.

V) Next, the Higgs is assumed to break the gauge symmetry

$$\tag{H} G~=~SO(3)~\to~ H~=~U(1),$$

so that only an Abelian gauge potential $A^3_i$ remains massless. [We assume isospin $I\neq 0$. For $I=1$ the isotropy group $H=U(1)$ is automatic. For higher integer-isospin, $H=U(1)$ only happens for special VEVs with enhanced symmetry, while $H=\{\bf 1\}$ is generic: There are no monopoles $\pi_2(G/H)\cong \{\bf 1\}$ for generic VEVs.] Topologically, the equatorial gauge transformation is then a map $S^1\to H$, and classified by the fundamental group $\pi_1(H)=\mathbb{Z}$. Without background vortices, the possible configurations of the Higgs are classified by

$$\tag{I} 2\mathbb{Z}~\cong~\pi_2(G/H)~\cong~ {\rm Ker}\left(\pi_1(H)\to \pi_1(G)\right) ~\subseteq~ \pi_1(H)~\cong~\mathbb{Z},$$

cf. Ref. 2. Thus only even multiples of the magnetic charge in eq. (E) is possible. In the 2+1 dimensional picture from Chapter 2, we allow background integer vortices in the $x^3$-direction, who are not constrained to live in the even part of eq. (I).

To make contact with Chapter 2, note that Chapter 2 is considering classical static solutions to a $U(1)$ Yang-Mills theory (aka. EM) in 2+1 dimensions in temporal gauge $A^3_0=0$, and with a complex$^2$ Higgs scalar $\phi^3$. The fields do not depend on the $x^3$-direction. In particular, the 2-dimensional $A^3_i$-vortex (2.6) should be identified with a equatorial tubular chart in the 3-dimensional picture. Vortices can be viewed as fat 1-dimensional strings, while magnetic monopoles behave more like particles.

Without additional symmetry breaking of the $U(1)$ symmetry, the above picture corresponds to the orientable vortices (1) above.

VI) Finally we imagine that we additionally break

$$\tag{J} U(1)~\to~ \{\bf 1\}.$$

Then the magnetic monopoles disappears $\pi_2(G/\{\bf 1\})\cong \{\bf 1\}$, and the vortices becomes the non-orientable vortices (2) above, cf. $\pi_1(G)=\mathbb{Z}_2$.

Depending on energy scales for the two symmetry breakings, the orientable vortices (1) could be quasi-stable before they break down to the stable non-orientable vortices (2), i.e. two vortices can snap, cf. Fig 2.7 and Fig 2.8. The remains of the two vortices constitute two quasi-stable magnetic monopoles, who has a net inflow or outflow of magnetic flux, respectively.


  1. G. 't Hooft and F. Bruckmann, Monopoles, Instantons and Confinement, arXiv:hep-th/0010225.

  2. F.A. Bais, To be or not to be? Magnetic monopoles in non-abelian gauge theories, arXiv:hep-th/0407197. (Hat tip: Hunter.)


$^1$ Later in Section 3.6, there is introduced fermionic matter, which transforms in the fundamental of $SU(2)$, and which hence distinguishes between $SO(3)$ and $SU(2)$.

$^2$ To compare with Chapter 3, which takes $\phi^{\alpha}$ as a real field, we pick $\phi^3$ to be a real field, cf. footnote on p. 15, aka. unitary gauge.


Your Answer

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