As requested, here’s a concrete example; I have no idea whether the numbers are at all plausible, but they at least illustrate the computation.
First let me correct the expression; it’s
$$Purity=\frac1{N_{cm}}\sum_{i=1}^{N_{cm}}\max_{1\le j\le K}\frac{n_{ij}}{n_i}\;,$$
where $N_{cm}$ is the number of communities, $n_{ij}$ is the number of nodes belonging to topic $j$ and community $i$, and $n_i$ is the number of members in community $i$. $K$ is the number of topics.
Suppose that $K=3$, so that there are three topics, and the algorithm detects $N_{cm}=2$ communities, which I’ll call $C_1$ and $C_2$. Suppose that $C_1$ has $n_1=4$ members, and $C_2$ has $n_2=7$ members. Finally, suppose that the numbers $n_{ij}$ are given by the following array:
$$\begin{array}{c|cc}
i\backslash j&1&2&3\\ \hline
1&0&2&3\\
2&7&2&1
\end{array}$$
Then
$$\begin{align*}
\sum_{i=1}^2\max_{1\le j\le 3}\frac{n_{ij}}{n_i}&=\max_{1\le j\le 3}\frac{n_{1j}}4+\max_{1\le j\le 3}\frac{n_{2j}}7\\
&=\max\left\{\frac{n_{11}}4,\frac{n_{12}}4,\frac{n_{13}}4\right\}+\max\left\{\frac{n_{21}}7,\frac{n_{22}}7,\frac{n_{23}}7\right\}\\
&=\frac14\max\{n_{11},n_{12},n_{13}\}+\frac17\max\{n_{21},n_{22},n_{23}\}\\
&=\frac14\max\{0,2,3\}+\frac17\max\{7,2,1\}\\
&=\frac34+\frac77\\
&=\frac74\;,
\end{align*}$$
and $Purity=\frac12\cdot\frac74=\frac78$.
It is permissible to add or subtract summations when the indices start and end at the same values. So you can do this:
$$\sum_{i=1}^{n}a_i\pm\sum_{i=1}^{n}b_i=\sum_{i=1}^n(a_i\pm b_i).$$
Multiplications are possible, but not so easy: you have to change one of the sum variables to avoid confusion:
$$\left(\sum_{i=1}^na_i\right)\left(\sum_{i=1}^nb_i\right)=
\left(\sum_{i=1}^na_i\right)\left(\sum_{j=1}^nb_j\right)=\sum_{i=1}^n\sum_{j=1}^na_ib_j.$$
This works because of the distributive law. Division I wouldn't even worry about. The AND or OR operations would both be possible if the $a_i$ and $b_i$ are boolean variables - otherwise what would be the meaning of it? In that case you would have to use the distributive laws of boolean algebra carefully, and you would have to define what the "sum" meant.
Best Answer
You seem to be coming from the TensorFlow documentation, as am I. After half an hour of searching for commas in Sigma Notation, I have realized that the comma is separating the subscript $i$ and $j$ of $\mathbf W$, not any part of the whole sum.
It is a little more clear here: $$\sum_{j} \mathbb W_{i,j} x_j+b_i $$
$W_{i,j}$ is the weight for class $i$ at pixel $j$.