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


1 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 $U(n+m)/(U(n)\times U(m))$.

  • $\begingroup$ Thank you. So since we want just negative or just positive many-body wave function we have to eliminate those which are redundant which are $U(n)\times U(m)$. I also read from another reference (Classification of topological quantum matter with symmetries) that since we have gauge redundancy: U(m) or U(n) rotations among occupied/unoccupied Bloch wave functions gives rise to the same ground state (Fermi sea) at a given k, that's why we mod out. So here since U(m) or U(n) gives us the filled negative or positive energies we have to get rid of all states which don't belong to U(m) or U(n)? $\endgroup$ May 13, 2021 at 23:42

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