The difference is whether you consider the symmetries of the *single-particle* or *many-particle* (Fock) Hilbert spaces.
Consider, for example, class A. An example of a free-fermion Hamiltonian in this symmetry class, acting in the Fock space, would be
$$H_{mb} = \sum_{i,j \in \mathcal{A}} t_{i,j} \hat{c}_i^{\dagger} \hat{c}_j$$
where $\hat{c}_i^{\dagger}, \hat{c}_i, i \in \mathcal{A}$ are the fermionic creation and annihilation operators, and $\mathcal{A}$ is a set indexing all the fermionic modes. It is readily seen that this Hamiltonian has a $U(1)$ symmetry, since the total particle number $\sum_i c_i^{\dagger} c_i$ is conserved.
However, because the Hamiltonian $H_{mb}$ is non-interacting, it is equivalent to consider the single-particle Hamiltonian
$$H_{sp} = \sum_{i,j \in \mathcal{A}} t_{i,j} |i\rangle \langle j |,$$
which acts in the single-particle Hilbert space of dimension $|\mathcal{A}|$. The eigenstates of $H_{mb}$ are just indexed by the occupation numbers $n_1, n_2, \cdots$ of the single-particle orbitals (eigenstates of $H_{sp}$). This single-particle Hamiltonian indeed does not have a non-trivial $U(1)$ symmetry, since the particle number in this space is just a scalar, namely 1.
Similarly, in class AII, the single-particle Hamiltonian $H_{mb}$ has an anti-unitary symmetry $\mathbb{T}$ such that $\mathbb{T}^2 = -1$. In the many-body Hilbert space, the corresponding statement is that $\mathbb{T}^2 = (-1)^{\hat{N}_f}$, where $\hat{N}_f$ is the fermion number. The reason is that the sector of Fock space with a fixed particle number $n$ is basically built out out of $n$ tensor products of the single-particle Hilbert space (subject to anti-symmetrization), and therefore you get $n$ factors of $-1$.
Class D and the other ones with "particle-hole symmetry" are more subtle. In many-body language, the statement is that we allow ourselves to add superconducting pairing terms which violate particle number conservation, for example
$$ \sum_{i,j} \Delta_{i,j} c_i c_j + h.c. $$
Thus, in the many-body language, class D has only fermion parity symmetry, because the $U(1)$ gets broken down. It might not be obvious at this point why this corresponds to particle-hole symmetry in the single-particle language. (Indeed, it is rather surprising that *reducing* the symmetry in the many-body language could *increase* the symmetry in the single-particle picture!) I don't have a great intuitive understanding here, but mathematically it comes about from the fact that, to convert a many-body (but non-interacting, i.e. quadratic) Hamiltonian containing pairing terms to a single-particle one, we first have to get rid of the pairing terms through a Bogoliubov transformation. This involves introducing redundant degrees of freedom, and as a consequence it turns out that the resulting Hamiltonian always has particle-hole symmetry. You perhaps shouldn't think of this as a "real" symmetry, it's more like a "gauge" symmetry in the sense that it results from a redundancy in a certain description of the system.
In answer to question (2), I think that once you go to interacting systems (as in the second reference in the paper), there is nothing that distinguishes the AZ symmetry classes from any other symmetries. Which is why the classification of interacting SPTs is much richer than free-fermion SPTs.
Q&A (4871)
