Question

Over a spatial continuum, it is easy to see why some topological solitons like vortices and monopoles have to be stable. For similar reasons, Skyrmions also have to be stable, with a conserved topological density. The reason is nontrivial homotopy.

Surprisingly, in some phases, but not all phases, the analog of topological solitons, or at least what can be interpreted as them, also emerge over lattice models. Why is that? There is no nontrivial homotopy over a lattice. Why are there some phases of the XY-model with deconfined vortices and antivortices? Why are deconfined monopoles present in some 3D lattice models?

Answer 1

I was wondering about exactly the same question some days ago, reading the seminal paper of Mermin (Rev. Mod. Phys. 51, 591--648 (1979), The topological theory of defects in ordered media), where you find an introductory discussion for the example of spins within the two-dimensional plane. There you find a lot of plots with spins (depicted as arrows in the plane) arranged in a circle, showing the orientation of the order parameter as one performs a closed loop in real space. However, as you have pointed out, for spins on a lattice there is no nontrivial homotopy. Using the famous rubber band analogy, it is clear that for a continuous rubber band a deformation from a loop winding one time around the circle to a constant loop (while keeping the end points fixed) is not possible, because continuity ensures that you cannot move individual parts of your rubber band without affecting the adjacent parts (i.e. stretching the band). For a lattice, there is no continuity, such that without invoking energetic arguments, you can rotate your spins arbitrarily at each point of the lattice. It behaves like having a rubber band that is cut into pieces and you can just take each ragged piece to the constant loop position without problems.

However, when invoking energetic arguments, i.e. a preference for the adjacent spins to align, you recover the rubber band tension without invoking continuity. Remark that when trying to bring the n=1 configuration to the constant n=0 loop, you will try to rotate the spins on one half of the circle to the left and the ones on the other half to the right, such that at some point you will encounter an incompatibility. Of course, this argument is far from being rigorous but it makes sense to me from an intuitive perspective. Still I am not sure it is applicable to other settings.