According to Mukhanov's Physical Foundations of Cosmology,

Homotopy groups give us a useful unifying description of topological defects. Maps of the $n$-dimensional sphere $\mathbb{S}^n$ into a vacuum manifold $\mathcal{M}$ are classified by the homotopy group $\pi_n(\mathcal{M})$. This group counts the number of topologically inequivalent maps from $\mathbb{S}^n$ into $\mathcal{M}$ that cannot be continuously deformed into each other.

For example, cosmic strings correspond to the homotopy group $\pi_1(\mathcal{M})=\pi_1(\mathbb{S}^1)=\mathbb{Z}$, which describes the maps of a one-dimensional sphere $\mathbb{S}^1$ into itself in a ${\rm U}(1)$ theory.

Question In this map, one of the two $\mathbb{S}^1$ spaces (between which we consider the inequivalent maps) is the vacuum manifold of ${\rm U}(1)$ given by $$|\phi|^2=\phi_1^2+\phi^2_2=v^2\tag{1}$$ where $\phi=\phi_1+i\phi_2$ and $v$ is a constant corresponding to the vacuum expectation value.

What is the other $\mathbb{S}^1$ in this case?


The source $S^1$ is, as is usual in the homotopical classification of defects, coming from considering a loop around a defect line (in this case a cosmic string).

If no such nontrivial loops would exist, then you can always deform the string to a "pointlike" object, i.e. it is not a stringy (one-dimensional) defect.

  • $\begingroup$ Okay. I guess this is a loop because the defect (cosmic string) is a one-dimensional curve. If the defect were pointlike or zero-dimensional, like the monopole, the appropriate shape to consider around it will be a sphere $\mathbb{S}^2$. Right? @NDewolf $\endgroup$ – SRS Dec 6 '20 at 10:08
  • $\begingroup$ Exactly. A simple introduction can be found here: lassp.cornell.edu/sethna/OrderParameters/… $\endgroup$ – NDewolf Dec 6 '20 at 10:26

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