vortex anti-vortex configuration to ground state i'm studying kosterlitz transition. I'm reading this: 
https://assets.nobelprize.org/uploads/2018/06/advanced-physicsprize2016-1.pdf?_ga=2.51324009.1302372948.1538119052-1605759177.1538119052
Now  at page 5 it says :
"The right panel shows a vortex anti-vortex configuration which can be smoothly transformed to the ground state " where the ground state is the state where all the spins are aligned.
Now i'm trying to figure out this thing.

Pay attention in my picture the vectors represent the velocities, and not the phase $\theta$ as in figure 3 at page 5, like in the picture below

The link between this image and the figure 3 at page 6 is that $v \propto \nabla \theta$
 A: Say $d$ is the distance vortex-antivortex. When $d\rightarrow 0$ the vortex and antivortex annihilate each other. Here's a gif I found on google which makes it clear.

When the vortex and antivortex are at the same position, you get back the ground state of the system (a ferromagnet here if the arrows represent spins).
A: I think it may be worth describing this in a mathematical way, rather than trying to give sketches, although the animation referred to in the previous answer, and also in my answer to your previous question, helps give a good picture.
You can find plenty of discussion of the Kosterlitz-Thouless transition online, for instance here.
The angular field around a $+1$ defect located at position $(-a,0)$ has the form
$$
\theta^+(x,y) = \tan^{-1}\left(\frac{y}{x+a}\right)
$$
and around a $-1$ defect at position $(+a,0)$
$$
\theta^-(x,y) = -\tan^{-1}\left(\frac{y}{x-a}\right)
$$
where $\tan^{-1}$ is the arctangent function.
If you combine these two (simply add them) you get
$$
\theta^{\pm}(x,y) = \tan^{-1}\left(\frac{2ay}{a^2-x^2-y^2}\right)
$$
for the field around a pair of opposite defects separated
by $2a$ on the $x$ axis.
This form is given here
and you can derive it yourself using an identity for the difference between two arctangent functions.
Now you can plot the relevant fields in any form you like,
either as the gradient $\nabla\theta^{\pm}(x,y)$ or as the 
direction field $(\cos\theta^{\pm}(x,y),\sin\theta^{\pm}(x,y))$.
If you vary the
separation $2a$, allowing $a\rightarrow 0$ 
you should be able to convince yourself
that the separated $+1$ and $-1$ defects can be smoothly
converted back to the ground state, when all the angles are the same.
There is a practical issue here (the ambiguity of the $\tan^{-1}$ function with respect to which quadrant the result
is in) which you may need to tackle.
One possible approach is to do the calculation 
numerically on a grid,
using the above formula,
but employing a version of "arctan" which allows you to supply
the numerator and denominator separately.
Many programming languages provide this function,
for exactly this reason.
Also, as usual, any constant may be added uniformly to the
angle field without changing its validity.
Anyway, I hope this will help you sort out what is going on.
