Classification of topolgical phases when eigenstates belong to complex Grassmannian

Question

I want to understand the paper which belongs to Ludwig (I put it below). I do not understand why exactly he got the new space $U(m+n)/U(m) \times U(n)$. My understanding from Grassmannian Manifold is that I think because we have m positive eigenvalues which are identified and n minus eigenvalues which are identified also we should mod out these two. Maybe it is wrong!

Source: Topological phases: Classification of topological insulators and superconductors of non-interacting fermions, and beyond Below equation 38

Answer 1

You have filled all the negative energy one-particle states $|i\rangle$ and the many particle state you get is the wedge product $$ \Psi= |1\rangle\wedge |2\rangle\wedge \ldots \wedge |n\rangle $$ This state depends only on the subspace of $ {\mathcal H}^{n+m}$ spanned by the states $|1\rangle, |2\rangle, \ldots, |n\rangle$ and this is left alone by the $U(n)\times U(m)$ that transforms only the occupied space and the unoccupied space, but does not take any state from to the other. The set of all maps that takes one-Slater-determinant states to one-Slater determiant states is $U(n+m)$. The set of distinct one-Slater-determinant states is therefore the coset $U(n+m)/(U(n)\times U(m))$.