Which manifold is formed by the set of resolutions of the identity operator into orthogonal projectors? A resolution of the identity operator $I$ on $\mathbb{R}^n$ or $\mathbb{C}^n$ is a decomposition
$$I = \sum_{i=1}^n P_i,$$
where the $\{P_i\}$ are a set of orthogonal rank-one projection operators. What is the manifold (or Lie group) structure of the set of all resolutions of the identity, or equivalently sets $P_i$?
A naive guess is to express each $P_i$ as an outer product $|\psi_i\rangle \langle \psi_i|$ and identify every resolution of the identity with an orthonormal basis $\{|\psi_i\rangle\}$ - the set of which is diffeomorphic to $\mathrm{U}(n)$ (since you can rotate any orthonormal basis into another one via an appropriate unitary operator). But I don't think is quite right, because you could multiply any basis vector by a phase factor $e^{i \theta}$ (or in the real case, $-1$) without changing the corresponding projector $P_i$, so the orthonormal basis contains redundant information/degrees of freedom. (Permuting basis elements also leaves the decomposition unchanged, but we can ignore this possibility because it can't be done continuously.) Is the answer just $\mathrm{U}(n) / \mathrm{U}(1)^{\times n}$ to remove the $n$ redundant phase factors, or something more complicated? If so, is there a simpler expression for this quotient group?

Yes, it's just $U(n)/U(1)^n$. This is not a quotient group because $U(1)^n$ is not a normal subgroup. This manifold is known as the (complete) flag variety of $\mathbb{C}^n$, and much is known about it. 
