Notation: What does $\sum_{i>j}$ mean? I'm currently writing a review of neutrino oscillation however there is one line that I don't understand:

More specifically, I do not understand the condition on the summation:
$$\sum_{i>j}$$
Does this actually mean sum over both i and j, using only values of i that satisfy i>j ?
 A: This notation often occurs in physics when we have a collection of, say $N$ interacting particles (or anything of the sort). Then we have a quantity which characterizes the interaction, i.e. let $e_{i,j}$ be the potential energy of interaction between the $i$-th and the $j$-th particle. Of course, and this is crucial, in a lot of cases (potential energy being one) the order of the indices is irrelevant and $e_{i,j}=e_{j,i}$. If we wanted to compute the total potential energy of the system, we would be inclined to write
$$\sum _{i=1}^{N }\sum _{j=1}^{N }e_{i,j}$$
But this would give a false result since we would be summing every interaction twofold, e.g. for $i=2$ and $j=3$ one term in the sum would be $e_{2,3}$ which is the same as for $j=3$ and $i=2$, as $e_{2,3}=e_{3,2}$. Therefore the given sum is actually equal to twice the actual potential energy and we sometimes write
$$E= \frac{1}{2} \sum _{i=1}^{N }\sum _{j=1}^{N }e_{i,j}$$
If you've taken courses in physics, you may remember the factor $\frac{1}{2}$ occurring often in interaction energy calculations. One way to avoid this is to sum using the condition $i>j$ as this would guarantee that every term would be present in the sum AND we won't have any duplication. One other thing, $e_{i,i}$ makes no sense as the particle cannot interact with itself (at least in the classical sense), and so the more precise notation would be 
$$E= \frac{1}{2} \sum _{i\neq j} e_{i,j}$$
But note that the condition $i>j$ excludes self-interaction.  For example, let $N=3$. Using the condition $i>j$ gives
$$E=\sum _{i>j} e_{i,j}=\sum _{i=1}^{3}\sum _{j=1}^{i-1}e_{i,j}=e_{2,1}+e_{3,1}+e_{3,2}$$
Compare this to the previous representation of $E$:
$$E=\frac{1}{2} \sum _{i\neq j} e_{i,j}=\frac{1}{2}(e_{1,2}+e_{1,3}+e_{2,3}+e_{2,1}+e_{3,1}+e_{3,2})=\frac{1}{2}(2e_{1,2}+2e_{1,3}+2e_{2,3})=\sum _{i>j} e_{i,j}$$
We have used the important property $e_{i,j}=e_{j,i}$.
If in your problem the variable $m_{i,j}$ satisfies $m_{i,j}=m_{j,i}$ and if those $U$ variables commute, as I think they do, since they seem like matrix entries, then the entire summand is symmetric to changing the indices and the summation $\sum _{i>j}$ has the said meaning.
A: If you're looking for unambiguous notation to use in your own papers, you can use one of:
$$\sum_{i,j \in \mathbb{N} \,\mid\, i > j} a_{ij} \qquad \sum_{i \in \mathbb{N} \,\mid\, i > j} a_{ij}$$
(which mean very different things).
Alternatively, use the Iverson Bracket. The above expressions become
$$\sum_{i,j \in \mathbb{N}} [i>j]a_{ij} \qquad \sum_{i \in \mathbb{N}} [i>j]a_{ij}$$
One further notation that technically works (but that most people won't understand) is
$$\sum_{(i,j) \in (>_\mathbb{N})} a_{ij} \qquad \sum_{i \in (>_\mathbb{N}\:\! j)} a_{ij}$$
A: It just means that your summing across all the elements/observations that fulfill that condition.
