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I am trying to understand algebraic invariants in topological insulators and topological superconductors through homotopy. But I encounter kind of a conceptual question. Let's say we have a second quantized Hamiltonian given by: $$ \hat{\mathcal{H}}=\sum_{I,J}\hat{\psi}^{\dagger}_{I}H_{IJ}\hat{\psi}_{J} $$ where $I,J$ are compound indices: $I=(i,\sigma_i,\cdots)$ and $J=(j,\sigma_j,\cdots)$. I know that the time reversal (T), particle-hole (C) and chiral symmetries (S), in first-quantised form are given by: \begin{align*} T&:\lbrace U_{T}K,H\rbrace=0\\ C&:\lbrace U_{C}K,H\rbrace=0\\ S&:[U_{S},H]=0 \end{align*} And the ten symmetry classes are determined by a 3-tuple in the following set: $$\Big\{(T,C,S):T,C\in\{-1,0,1\}\mathrm{\ and\ }S=\lvert TC\rvert\Big\}\cup\{0,0,1\}$$

Let's say I am considering a translationally-invariant system with $N_{sites}\to\infty$. Then, the first Brillouin zone becomes continuous and we can write \begin{align*} \hat{\mathcal{H}}&=\sum_{\alpha,\beta}\int\frac{\mathrm{d}^{n}\boldsymbol{k}}{(2\pi)^{n}}\hat{\psi}^{\dagger}_{\boldsymbol{k}\alpha}H_{\alpha\beta}(\boldsymbol{k})\hat{\psi}_{\boldsymbol{k}\beta} \end{align*} where $\alpha,\beta$ are summed over the additional degrees of freedom beyond site indices. Let's say I consider the symmetry class $A$ with $(T,C,S)=(0,0,0)$, i.e. there is no symmetry at all. Then, I know that \begin{align*} H_{0}=\begin{pmatrix} I_{n} & 0\\ 0 & -I_{m} \end{pmatrix}\mathrm{\ and\ }H_{1}=\begin{pmatrix} I_{n+1} & 0\\ 0 & -I_{m-1} \end{pmatrix} \end{align*} where $n+m=N_{f}$ is the number of degrees of freedom, are two valid Hamiltonians since they are Hermitian. So, if these two Hamiltonians can be continuously deformed into each other, there exists a continuous map $H:T^{n}\times [0,1]\to GL(N_{f},\mathbb{C})\cap\mathfrak{u}(N_{f})$ such that $$ H(\boldsymbol{k},0)=H_{0}\mathrm{\ and\ }H(\boldsymbol{k},1)=H_1 $$ i.e. $H$ is a homotopy from $H_0$ to $H_1$.

Note: I am using physics convention that Lie algebra of unitary group is the space of hermitian matrices so $GL(N_{f},\mathbb{C})\cap\mathfrak{u}(N_{f})$ means the set of invertible $N_{f}$-by-$N_f$ hermitian matrices.

Explicitly, I can consider the $(n+1,n+1)$ entry of the homotopy: $$ H_{n+1,n+1}(\boldsymbol{k},t)=\begin{cases} -1 &, t=0\\ 1 &, t=1 \end{cases} $$ Since this should be a continuous map, for every $\boldsymbol{k}\in T^{n}$, intermediate value theorem tells us that there exists $t(\boldsymbol{k})\in(0,1)$ such that $$ H_{n+1,n+1}(\boldsymbol{k},t(\boldsymbol{k}))=0 $$ Hence, by definition, any homotopy between $H_0$ and $H_1$ must encounter a gapless Hamiltonian. Thus, as long as the number of filled/empty bands is distinct, the two Hamiltonians must describe different topological phases. Since this proof is independent of dimension $n$, it would seem that the symmetry class $A$ is always topologically non-trivial.

However, of course this is not what we see from the tenfold way. Are we fixing the number of filled bands in momentum space such that the above two Hamiltonians simply cannot co-exist?

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One is allowed to "stack" an arbitrary trivial insulator on top of the target system you are studying. This means one can add arbitrary number of topologically trivial valance and conduction bands.

if one is not allowed to do so, one can get some exotic topological insulators which are not stable under such stacking. This is the fragile topological insulator.

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