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I am a math student who is learning topological phases from this website.

Let's assume the fermi level is zero. For the graphene, the sublattice symmetry $\sigma_z H \sigma_z = -H$ makes the Hamiltonian look like

\begin{align*} H = \begin{bmatrix} 0 & H_{AB}\\ H_{AB}^\dagger & 0 \end{bmatrix} \end{align*}

I can see that the spectrum of $H$ is symmetric with respect to the fermi level, but I don't see why it is away from $0$. In my understanding, a Hamiltonian is a hermitian operator that may differ from questions to question. Therefore, for example, the trivial case $H_{AB} = 0$ clearly has $0$ in the spectrum; or when $H_{AB}$ has eigenvalue $0$, then so does $H$.

When physics people write down a Hamiltonian, is there any assumption or convention made without explicitly stated? For instance, $H_{AB}$ is assumed to be positive definite? Or the fact that the spectrum is away from $0$ is based on observation?

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When physics people write down a Hamiltonian, is there any assumption or convention made without explicitly stated? For instance, $H_{AB}$ is assumed to be positive definite?

There is no general assumption or convention. First, note that the page you link to provides an example of an $H_{AB}$ that is not positive definite, $$ H_{AB}=\pmatrix{0.47+0.62I&0.65+1.46I\\1.98-0.03I&-0.71+1.76I}. $$

Although there is no overall convention, it is often convenient to separately discuss the cases where $H$ is gapped and where $H$ is gapless. In many contexts, not least in discussions of topological states and invariants, the gapped case is more straight-forward, which is why it's often treated first. Some authors may find it convenient to consider gapped systems the default, but careful authors will specify whether a gap is present or absent. In the case of the resource you link to, the section on Sublattice symmetry follows the section on Topology and gapped quantum systems, which indicates that the notion of topology used in the rest of the page concerns gapped systems. They discuss topology in gapless systems elsewhere, specifically treating the case where $\det H_{AB}=0$.

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