Consider the Lie subalgebra of $\mathfrak{u}(n)$ given by $L = \{A \in \mathfrak{u}(n): \sum\_{j=1}^n A\_{ij} = 0 \text{ for all } i \in [n]\}$. What is its dimension? What does the corresponding Lie subgroup of $U(n)$ look like? 


Edit: I believe the dimension is $(n-1)^2$, by explicitly determining the number of independent constraints. I highly suspect the subgroup is $U(n-1)$, acting on the subspace $\{z\_1 + \ldots + z\_n = 0\}$.



An innocent looking subgroup of $U(n)$
Yup. Your condition is $M \vec v = 0$ for $\vec v = [1 1 1\ldots 1]$. Exponentiating, that's $\exp(M) \vec v = \vec v$.


