Questions about Kosterlitz–Thouless (KT) transition Why we extend $\theta$ from $(0,2\pi)$ to $(-\infty, \infty)$? I mean we cannot measure $\theta$ in experiment, can we? 
Secondly,the feature of vortex solution (at least in KT transition) can be summarized as the following: have singularity and multi-valued.
I'm wondering is this the definition of vortex in mathematics? And why do we say that vortex is topological effect? I cannot see any topology from these two features.
 A: It is not fully clear what is meant by $\theta$ - assuming that it is the angle between spins, extending it from $(0,2\pi)$ to $(-\infty, +\infty)$ is a matter of convenience: it does not change anything physically, as enters the Hamiltonian only via trigonometric functions; on the other hand, it may significantly simplify calculations - both analytical and numerical, since one does not have to take care of taking module of $\theta$ every time it exceeds $2\pi$ or falls below $0$.
Vortex may mean different things in different fields. Thus, the definition used in this or that branch of mathematics might be an unreliable guideline. Yet, singularities definitely have topological meaning. The simplest examples come from the complex analysis:


*

*$1/z$ is a singularity and integrating around it results in a finite residue, unlike integration along a closed contour not enclosing this point.

*$\log z = \log r + i\phi + 2\pi n $ is a multi-valued function: going around the point $z=0$ brings us to a sheet with different $n$.

A: We extend $\theta$ from $(0, 2 \pi)$ to $(-\infty, +\infty)$ exactly because it makes it possible to see the multi-valued nature of $\theta$.
Imagine a simple vortex with the associated complex field $\psi(r, \varphi) = |\psi(r, \varphi)| e^{i \theta(r, \varphi)} = r e^{i \varphi}$.
In that case, $\psi(r=1, \varphi=0) = 1$, but what about $\theta$? In principle, you could choose $\theta = 0$ (and if you would restrict $\theta$ from $0$ to $2\pi$ that would be the only choice). But if you allow $\theta$ to run between $-\infty$ to $+\infty$, and you want $\theta(r, \phi)$ to be locally smooth (continuous), $\theta$ has to be multivalued.
This is simply because if you want $\theta$ to be continuous, then you need to have $\theta(r, \varphi + d\varphi) = \theta(r, \varphi) + d\varphi$ (you cannot "jump" by $2 \pi$ at once). But in that case, it is easy to see that you must have "$\theta(r, 2\pi) = \theta(r, 0) + 2\pi = \theta(r, 0)$" so $\theta$ is multivalued. Of course, you would still be able to see this with $\theta \in (0, 2 \pi)$ by integrating the change in $\theta$ along a circle, but that would require extra-care (especially at the points where $\theta$ is discontinuous).

Now, why is this topological behavior? One way to understand "topological" is to say that the property is robust to small deformations (probably not a rigourous definition, but that's the idea). This is to say that for any infinitesimal change $d \psi$ in the wavefunction $\psi$, some property of the new wavefunction remains unchanged. Here such a property would be for instance the winding number, which is a fancy way of saying "by how many multiples of $2 \pi$ does $\theta$ change when going in a circle around some point" (here, for any circle centered around $r=0$, the winding number would be $+1$). I think (but I am not an expert), that this is equivalent to saying that the values of the winding number/topological number are isolated (for any winding number $\omega_1$, there exists $\epsilon > 0$ such that there is no winding number $\omega_2 \neq \omega_1$ in the interval $(\omega_1 - \epsilon, \omega_1 + \epsilon)$.
For the winding number, this is obviously the case, as by definition it takes values over the integers $\mathbb{Z}$. Intuitively, you can convince yourself that an infinitesimal change in a continuous field $\psi$ cannot possibly result in a discrete jump for a quantity that can be expressed as a function of the field.
As for why a vortex is always associated with a singularity (which is to say that if the winding number over a closed curve is non-zero, there must be a point inside the region delimited by the curve for which $\psi(r_{\mathrm{singularity}}, \varphi_{\mathrm{singularity}}) = 0$, this is also a consequence of the "robustness" of the phase winding. Just as the same way an infinitesimal change in $\psi$ cannot change the winding number, an infinitesimal deformation of the curve cannot change the winding number. This is true as long as $\psi \neq 0$, because then $\theta$ is ill-defined. 
So imagine that there is no point inside your curve for which $\psi = 0$. You could continuously shrink the curve without changing the winding number because everything is continuous. This would  mean that, as the curve becomes smaller and smaller, $\theta$ would still take every values from $0$ to $2 \pi$ at least once along the curve. But if you keep shrinking the curve, it will eventually become a single point $(r, \varphi)$. But what is the value of $\theta$ here? If $\psi \neq 0$, then $\theta$ is well-definded an unique up to a multiple of $2 \pi$. At the same time, because the point is defined as the limit of the shrinking curve, and the winding number was non-zero all along the shrinking process, $\theta$ must also take every value between $0$ and $2 \pi$ at a single point, which is clearly impossible. The only way out of this situation is to relax the assumption that $\psi \neq 0$ inside the initial curve. Then there is a point for which $\theta$ is ill-defined and even an infinitesimal deformation of the curve can lead to a discrete change in winding number if the curve happens to cross the point for which $\psi = 0$.
