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In Wikipedia, it states about the magnetic monopole of the gauge theory is determined by the fact:

This argument for monopoles is a restatement of the lasso argument for a pure U(1) theory. It generalizes to $d + 1$ spacetime dimensions with $d \geq 2$ in several ways. One way is to extend everything into the extra dimensions, so that U(1) monopoles become sheets of dimension $d − 3$. Another way is to examine the type of topological singularity at a point with the homotopy group $$ \pi_{d −2}(G). $$


  • My goal and my questions are trying to get generic definitions and classifications with reasonable assumptions. Do you have one?

  • Can you explain the Wik quote above?

  • Can I understand the above statement from Wikipedia as follows:


My attempt: The magnetic monopole's wordline is $d+1-3=d-2$ dimensions (which generally has a worldvolume of $d-2$ dimensions). Thus the magnetic monopole itself is $d-3$ dimension (as a constant time slice of $d-2$ dimensions).

  • magnetic monopole's wordline is the so-called 't Hooft operator.

Consider the full spacetime as $S^{d+1}$. Cutting out the ${d-2}$ dimensional worldvolume as the time trajectory of the magnetic monopole. We can consider the cut with a tubular neighborhood around this, thus we have $$D^{d-3} \times S^{1},$$ where the closed time trajectory is $S^{1}$. I think what I should really do about the surgery is $$ S^{d+1} \smallsetminus (D^{2} \times D^{d-3} \times S^{1}) = ? $$

But all I know so far is that $$ S^{d+1} \smallsetminus D^{d} \times S^{1} = S^{d-1} \times D^{2}. $$

So I can consider the covering charts over $S^{d-1}$. I suppose that minimal overlapping of the charts form a trajectory of $S^{d-2}$ (True of false?).

Thus consider the homotopy group (for the map from $S^{d-2}$ to the gauge group $G$) $$ \pi_{d −2}(G) $$ seems reasonable. Is my argument above correct, or is it incorrect?

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    $\begingroup$ I'm voting to close this question as off-topic because it is a general policy of mathoverflow to reject proof checking questions. $\endgroup$ – Jan-Christoph Schlage-Puchta Aug 9 '18 at 9:48
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    $\begingroup$ @Jan-ChristophSchlage-Puchta: isn't part of what constitutes a good question on MO to include what one has already tried? $\endgroup$ – M.G. Aug 9 '18 at 12:08
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    $\begingroup$ Thanks M.G. It is not about checking my answer only. There is a statement I am trying to know its general form. Any other approach is fine. I am only showing my attempt. Because of some other people closing questions, for the opposite reason --- showing no attempts to answer your own question by yourself. $\endgroup$ – wonderich Aug 9 '18 at 14:01

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