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I am currently studying topological insulators and repeatedly found the claim (e.g. here), that the "basic hamiltonian" of a topological system in $d$ spatial dimensions can be written using the elementary representation matrices of a Clifford algebra $$ H(\vec{k}) = \vec{h}(\vec{k}) \cdot \vec{\gamma} = \sum_{i=0}^d h_i(\vec{k}) \gamma_i $$ where the $\gamma_i$ are generators of a Clifford algebra ($\left\{ \gamma_i, \gamma_j \right\} = 2 \delta_{ij}$).

For $d=0$ I understand that the $\gamma_i$ represent the presence / absence of time reversal symmetry $T$ and a particle number symmetry $Q$ and one sets $\gamma_1 = T$ and $\gamma_2 = QT$ and then wants to find another Clifford generator $\gamma_0 = \tilde{H}$, where $\tilde{\cdot}$ denotes spectral flattening. (See Kitaevs Paper on the details)

Now my question is: What happens for $d > 0$? Where do I get the additional $d-1$ generators from? And as a bonus question: What happens to $\gamma_2 = QT$ in the zero-dimensional case?

I guess that somehow the extra clifford generators come from the additional translational symmetries in $d$ space dimensions, but these are symmetries commuting with the hamiltonian instead of anti-commuting.

Edit: If I go to higher spatial dimensions, I get another symmetry that commutes with the hamiltonian, the translation $P_i$ by one unit cell in each spatial direction. If I set a $\gamma_{i+1} = T P_i$ I get another term that anti-commutes with the hamiltonian. But how do I also get it to anti-commute with $T$? Or does it already anti-commute with $T$?

Edit 2: Just figured out, how to extend a Clifford algebra, if I already have an even number of generatrs and a new operator that commutes with all of them and squares to 1.

Assume $\{\gamma_i, \gamma_j\} = 2 \delta_{ij}$ for $i,j=1...2n$. And we also have an operator $g$ s.t. $[g, \gamma_i] = 0$ for $i=1...2n$ and $\gamma^2$. Then we can set $\gamma_{2n+1} = g \gamma_1 \cdots \gamma_{2n}$ and easily verify that

$$ \{\gamma_{2n+1}, \gamma_i\} = g \gamma_1 \cdots \gamma_{2n} \gamma_i + \gamma_i g \gamma_1 \cdots \gamma_{2n}\\ = g \gamma_1 \cdots \gamma_{2n} \gamma_i - g \gamma_1 \cdots \gamma_{2n} \gamma_i = 0 $$

and $\gamma_{2n+1}^2 = \pm 1$ (might need to throw a factor of $i$ in to make this a +, but that doesn't change the general idea). But this only works if the Clifford algebra I start with has an even number of generators.

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