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I had heard from a professor saying that "Polyakov and ’tHooft discover the magnetic monopoles in $SU(2)$ gauge theory with scalar fields [Georgi-Glashow model]." And he cited two references:

’t Hooft G., Nucl.Phys. B79 (1974), 276.

A. Polyakov, Pis’ma v ZhETF 20 (1974), 430.

My question is that isnt that $$\pi_1(SU(2))=0$$ so in the $SU(2)$ gauge theory , there is no magnetic object (magnetic monopole as the cut end of ’t Hooft line or Dirac string)? So why/how/what do Polyakov and ’t Hooft discover the magnetic monopoles in $SU(2)$ gauge theory?

p.s. Maybe providing some of your personal view on ’t Hooft and Polyakov works will be nice.

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    $\begingroup$ Where did the $\pi_1$ come from? Aren't monopoles classified by $\pi_2(G/H)$ for a $G\to H$ symmetry breaking, so in this case $\pi_2(SU(2)/U(1))=\pi_1(U(1))=\mathbb Z$? $\endgroup$
    – Nihar Karve
    Jul 20, 2021 at 14:41
  • $\begingroup$ $𝜋_1$ is associated with the magnetic line operator, which is the ’tHooft magnetic line operator. And the magnetic monopole as the cut end of ’tHooft line or Dirac string. $\endgroup$
    – ann marie cœur
    Jul 20, 2021 at 14:48
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    $\begingroup$ possible duplicate: physics.stackexchange.com/q/512570/84967 $\endgroup$
    – AccidentalFourierTransform
    Jul 20, 2021 at 14:57
  • $\begingroup$ The magnetic monopoles in the Georgi-Glashow model are monopoles with respect to the unbroken $U(1)$ subgroup. There are no $SU(2)$ monopoles. $\endgroup$
    – fewfew4
    Jul 20, 2021 at 15:36
  • $\begingroup$ @fewfew4, yes that clarifies. It counts a partial answer. $\endgroup$
    – ann marie cœur
    Jul 20, 2021 at 16:30

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When a gauge group $G$ is Higgsed down to a subgroup $H\subset G$, there are different notions of magnetic charges that we should consider.

  • The first notion is the one you mentioned, particles which carry non-trivial $\pi_1(G)$ charge. These are called $G$-monopoles.

  • Then there are $H$-monopoles, which don't have $\pi_1(G)$ charge, but carry non-trivial $\pi_1(H)$ charge. First, we note that one can always find group homomorphism $f:\pi_1(H)\to\pi_1(G)$. Even if $\pi_1(G)=0$, there can still be $H$-monopoles, as long as its image onto $\pi_1(G)$ is trivial. Physically this means the magnetic charge looks non-trivial with respect to $H$, but is trivial when looking at the full group $G$. The type of elements of $\pi_1(H)$ which satisfy this make up $\text{ker}f$. Making contact with @NiharKarve 's comment, we note that $\text{ker}f=\pi_2(G/H)$. Also note that $\pi_1(G)$ does not need to be trivial for this definition to work.

In the case of the Georgi-Glashow model, $\pi_2(SU(2)/U(1))=\pi_1(U(1))=\mathbb{Z}$, as @NiharKarve mentions. $U(1)$ monopoles are allowed to exist in this theory, and it was shown by 't Hooft that they do.

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