I'm reading Chapter 10.4 on the 't Hooft-Polyakov monopoles in Ryder's Quantum Field Theory.

On page 412 he explains why magnetic monopoles cannot appear in the Weinberg-Salam model. I'm I right by saying that he shows that the electromagnetic gauge group $U(1)_{em}$ is not compactly embedded into the $U(1)\times U(1)$ subgroup of $SU(2)\times U(1)$?

He then immediately concludes that the first fundamental group of the unbroken symmetry, which is $H=U(1)_{em}$, $\pi_1(H)$ must be trivial or doesn't exists. Could someone refer me why?

Comment: I know that in the $SU(2)\times U(1)$ ones must consider the second homotopy group from $S^2$ to the orbit $G/H=SU(2)\times U(1)/U(1)$, where $H$ is the isotropy group of a vacuum state, after symmetry breaking. But the second homotopy group of a quotient can be related through a exact series to the kernel of the map from $\pi_1(H)$ into $\pi_1(G)$.

What I do not understand is by which theorem for $H$ having a non-compact covering group $\pi_1(H)$ must be trivial or non-existing (???)?

  • 1
    $\begingroup$ I don't own Ryder so I am not sure what you are asking, but you might be interested in this answer. $\endgroup$
    – Hunter
    Commented May 4, 2014 at 9:37
  • $\begingroup$ Although it is in the same line, it doesn't answer my question. I've also looked into the references but they don't seem to use the same argument. $\endgroup$ Commented May 4, 2014 at 10:47
  • $\begingroup$ Does Ryder mention the following isomorphism result:$$ \pi_2 ( G / H ) \cong \pi_1 (H) $$ provided $G$ is compact, connected and simply connected. Does that answer your question? $\endgroup$
    – Hunter
    Commented May 4, 2014 at 11:12
  • $\begingroup$ Btw. (not that it is essential to OP's question), note that Ryder in Chap. 10 writes $O(3)$ in many places where he should have written $SO(3)$. $\endgroup$
    – Qmechanic
    Commented May 4, 2014 at 12:23
  • $\begingroup$ Have you checked E. Weinberg's Classical Solutions in QFT text? $\endgroup$
    – JamalS
    Commented May 12, 2014 at 11:05

2 Answers 2


I) Consider a Yang-Mills type theory with gauge group $G$. In principle we can consider the same theory with its covering group $\tilde{G}$, with $\pi:\tilde{G}\to G$. The covering group is by definition simply connected: $$\tag{1} \pi_1(\tilde{G})~=~\{\bf 1\}.$$ Any representation $\rho$ of $G$ can naturally be viewed as a representation $\rho\circ \pi$ of $\tilde{G}$. (Warning: the opposite is not true.) Hence the fields of the theory also transform under $\tilde{G}$, and the theory is invariant under $\tilde{G}$. In other words, we could in principle from the very beginning view $\tilde{G}$ as the gauge group of the theory, cf. Section III below.

II) Next assume a spontaneous symmetry breaking of the gauge group $G \to H$ to a subgroup $H$. Define subgroup

$$\tag{2} \tilde{H}~:=~\pi^{-1}(H)~\subseteq~ \tilde{G}. $$

Concretely for the electroweak theory, the gauge group is

$$\tag{3} G~:=~ SU(2)_I\times U(1)_Y ~\supset~ U(1)_I\times U(1)_Y; $$

the covering group

$$\tag{4} \tilde{G}~=~SU(2)_I\times (\mathbb{R},+)_Y $$

is non-compact; and the unbroken electromagnetic subgroup

$$\tag{5} H~:=~U(1)_Q~\subset~ U(1)_I\times U(1)_Y$$

is irregularly/non-compactly/incommensurably/non-topologically embedded, cf. e.g. my Phys.SE answer here. Here we have assume that tangent to the Weinberg angle $\tan\theta_W\in \mathbb{R}\backslash\mathbb{Q}$ is irrational. Now since the non-compact group $H$ is not a subgroup of $SU(2)_I$, the subgroup

$$\tag{6} \tilde{H}~=~(\mathbb{R},+)_Q$$

is also non-compact. It is also simply connected

$$\tag{7} \pi_1(\tilde{H})~\cong~\{\bf 1\}.$$

III) Let us now view the non-compact gauge group $\tilde{G}$ in eq. (4) as the gauge group of electroweak theory. The non-compact gauge group of electromagnetism (5) is then replaced by the non-compact subgroup (6). Standard monopole analysis then shows that there are no magnetic monopoles in the electroweak theory

$$\tag{8} \pi_2(\tilde{G}/\tilde{H})~\cong~{\rm Ker}\left(\pi_1(\tilde{H})\to \pi_1(\tilde{G})\right)~\stackrel{(1)}{\cong}~\pi_1(\tilde{H})~\stackrel{(7)}{\cong}~\{\bf 1\}, $$

cf. e.g. Ref. 2.

IV) Now let us return to OP's question. Ref. 1 considers the system from the perspective of $G$ and $H$ rather than $\tilde{G}$ and $\tilde{H}$, and recast the standard monopole analysis in this language. The problem is that a closed loop $\gamma\in G$ that wraps a $U(1)\subseteq G$ has a non-compact lift $\tilde{\gamma}\in \tilde{G}$ which is not a closed loop. Similarly, a closed loop $\gamma\in H$ that wraps a $U(1)\subseteq H$ has a non-compact lift $\tilde{\gamma}\in \tilde{H}$ which is not a closed loop.

Intuitively/heuristically, the relevant notion $\tilde{\pi}_1(\tilde{H})$ (of such non-compact "closed" paths) has more paths than just (7), and for the electroweak theory the relevant quantity

$$\tag{9} \pi_2(G/H)~\cong~\pi_2(\tilde{G}/\tilde{H})~\cong~{\rm Ker}\left(\tilde{\pi}_1(\tilde{H})\to \tilde{\pi}_1(\tilde{G})\right)~\cong~\{\bf 1\} $$

is trivial, in agreement with eq. (8). See Ref. 1 for more details.


  1. L.H. Ryder, Quantum Field Theory, 2nd eds., 1996; p. 412.

  2. S. Coleman, Aspects of Symmetry, 1985; p. 217 & 221.

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

  4. S. Weinberg, Quantum Theory of Fields, vol. 2, 1996; p. 443-445.

  5. J. Preskill, Magnetic Monopoles, Ann. Rev. Nucl. Part. Sci. 34 (1984) 461-530; Sections 4.2 & 4.3. The PDF file is available here.

  • $\begingroup$ I'm rather confused by the last equation. How do you know that the kernel is trivial, since you don't mention it except that it is different from $\pi_1(\tilde{H})=\mathbb{Z}$. And isn't it $\pi_1(H)=\mathbb{Z}$ while $\pi_1(\tilde{H})=1$? $\endgroup$ Commented May 12, 2014 at 9:06
  • $\begingroup$ Yes: $\pi_1(\tilde{H})\cong\{\bf 1\}$. $\endgroup$
    – Qmechanic
    Commented May 12, 2014 at 10:59
  • $\begingroup$ What is the difference between $\tilde{\pi}_1$ and $\pi_1$? $\endgroup$ Commented Feb 20, 2016 at 13:34
  • $\begingroup$ I'm sorry, I am still unclear of its definition, and why $\pi_2(\tilde{G}/\tilde{H})\cong Ker (\tilde{\pi}_1(\tilde{H})\rightarrow \tilde{\pi}_1(\bar{G}))$. Should it not be $\pi_2(\tilde{G}/\tilde{H})\cong Ker (\pi_1(\tilde{H})\rightarrow\pi_1(\tilde{G}))$? $\endgroup$ Commented Feb 21, 2016 at 6:55
  • $\begingroup$ Okay, I believe $\tilde{\pi}_1$ involves maps from $\mathbb{R}$ to a topological space instead of from $S^1$ to that topological space, but I am still unclear on the reason for the isomorphism. $\endgroup$ Commented Feb 21, 2016 at 7:03

(The answer of Qmechanics is indeed correct but I would like to spend some more words. I finally got the picture by reading Quantum Field Theory of Fields II by Weinberg.)

Weinberg adreses the question in his second Quantum Theory of fields whether when the involved fields do not belong to the extra representation of the covering group $\tilde{G}$, if we should consider the gauge to be the non-simply connected group $G$ or its covering group $\tilde{G}$.

This was my problem earlier I did not understood which group you had to consider $G$ or $\tilde{G}$. Does the results, i.e. the spectrum of monopoles, change by making either choice?

He then arguments that it does not. E.g. In a three dimensional world, i.e. with a $S^3$ boundary, the topologically stable monopole configuration is determined by the second homotopy group $\pi_2(G/H)$. This as mentioned by Qmechanics this can be computed as the kernel of the map $$\pi_1(H) \rightarrow \pi_1(G)$$ when $H$ is embedded in $G$.

However when we replace $G$ by its covering group $\tilde{G}$, we must also change $H$ its pre-image $\tilde{H} = \Pi ^{-1}(H)$ by the canonical projection $\Pi: \tilde{G}\rightarrow G$ . This is because the loops in $H$ do not necessarily 'close back' when $H$ is embedded in $\tilde{G}$. These are the loops that do not become trivial when $H$ is embedded into the covering group $\tilde{G}$. We can therefore identify $$\pi_2(G/H)=\pi_1(\tilde{H}) $$.

But this means that it does noet matter, for the monopoles in any case, if we consider the gauge group to $\tilde{G}$ instead of $G$.

An example: the Georgi-Glashow model. Here the gauge group can be identified as the doubly-connected group $SO(3)$. Its covering group is (the simply connected) $SU(2)$.

The unbroken symmetry are the rotation in the plane $SO(2)$ where the rotations that differ from $2\pi$ are identified. Hence path from the identity to $2\pi$ are loops in $\pi_1(SO(2))$. However once we embed $SO(3)$ in $SU(2)$ we have to go twice around, rotate by $4\pi$, to get back to the identity and close a loop: it excludes the loops that closes after a uneven multiple of $2\pi$ in $\pi_1(SO(2))$ and therefore $$ \pi_1(SO(2)) \neq \pi_1(\tilde{SO(2)}) = \pi_2(SO(3)/SO(2))).$$

The elements of $\pi_1(SO(2))$ that map to the trivial element of $\pi_1(SO(3))$ are those that go from the unit element to a rotation of $2\pi$: $$\pi_2(SO(3)/SO(2))\cong U(1) $$ where $U(1)$ is a subgroup of $SU(2)$, the covering group $\tilde{G}$ regardless whether we had considered $G$ or $\tilde{G}$ as the gauge group from the onset!

Quick answer: You can always consider the gauge group $G$ to be the simply connected covering group, such that you can use the result $\pi_2(G/H)=\pi_1(H)$.


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