0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.