# Meaning of Delta Notation

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.
• @MJD So $\mu^\sigma$ places $n$ masses of $1/n$ in the square $[-1, 1]^2$ and assigns measure $0$ to every other point. Is that the right way to interpret the expression? Commented Dec 27, 2018 at 0:01
• Yes. A measure is a function on sets. The pdf is a distribution. That's the Dirac delta in $\mathbb{R}^2$ : $\delta(x,y) = \delta(x)\delta(y)$ the latter being the usual Dirac delta in $\mathbb{R}$ @Probably Commented Dec 27, 2018 at 1:00