I was reading a paper (https://arxiv.org/pdf/1609.03891.pdf), and on the first page the author's write the following line: $$ \mu^\sigma = \frac{1}{n} \sum_{i = 1}^n \delta_{\left(\frac{2i}{n} - 1, \frac{2\sigma(i)}{n} -1\right)}. $$ Here, $\sigma$ is in the symmetric group on $n$ letters. The authors write that $\mu^\sigma$ is supposed to be a probability measure on $[-1, 1]^2$. I'm having trouble parsing the expression. Is the delta supposed to be the Kroenecker delta? I don't think so, because the subscript ${\left(\frac{2i}{n} - 1, \frac{2\sigma(i) }{n} -1\right)}$ doesn't seem to make sense (it would be equivalent to saying $i = \sigma(i)?$). Also, why is the domain $[-1, 1]^2$?
Could someone help me understand this expression?
Note that $$ {\left(\frac{2i}{n} - 1, \frac{2\sigma(i) }{n} -1\right)} $$ is a point in $[-1,1]^2$.
So $\delta$ with that subscript is a unit point mass at that point.
Add $n$ point masses and divide by $n$, we get a probability measure.