In J, you can square a vector and get a vector of the squared entries. So "sum under square" really is doing "sqrt(sum(square(vector)))", and more generally "(a under b) x" does "b^-1(a(b(x)))".
Even without the mapping wrinkle, though, this is definitely a deviation from the examples the article provides: instead of unsquaring the vector we initially squared, we unsquare the sum.