= Solution
{c}
The <periodic boundary condition> requires $e^{ikM}=1$, so there are exactly $M$ independent <wavevectors> in a <Brillouin zone>:
$$
\boxed{k=\frac{2\pi l}{M},\qquad l=0,\ldots,M-1\quad(\bmod\ 2\pi).}
$$
The inverse <discrete Fourier transform>, with the sign convention of the paper, is
$$
c^\dagger_{\sigma,j}=\frac1{\sqrt M}\sum_k e^{-ikj}\widetilde c^\dagger_{\sigma,k},\qquad
c_{\sigma,j}=\frac1{\sqrt M}\sum_k e^{ikj}\widetilde c_{\sigma,k}.
$$
<Fourier orthogonality> gives the <tight binding> <dispersion relation> $\epsilon_k=-2t\cos k$. For the interaction, first use the <canonical anticommutation relations> at distinct sites to write $n_{\sigma,j}n_{\sigma,j+1}=c^\dagger_{\sigma,j}c^\dagger_{\sigma,j+1}c_{\sigma,j+1}c_{\sigma,j}$. Summing its <discrete Fourier transform> over $j$ enforces <crystal momentum conservation>, giving
$$
\boxed{H=\sum_{\sigma,k}(-2t\cos k)\widetilde c^\dagger_{\sigma,k}\widetilde c_{\sigma,k}
+\frac VM\sum_{\sigma,k,p,q}e^{iq}\widetilde c^\dagger_{\sigma,k+q}\widetilde c^\dagger_{\sigma,p-q}\widetilde c_{\sigma,p}\widetilde c_{\sigma,k}.}
$$
Every <wavevector> is reduced modulo $2\pi$. The phase $e^{iq}$ comes from the displacement of the second site by one lattice spacing. Equivalently replace $e^{iq}$ by $\cos q$: exchanging the two incoming and the two outgoing <fermions> changes $q$ to $-q$ with two cancelling <fermionic signs>. Thus the usual interaction vertex is $2V\cos q$ accompanied by the factor $1/(2M)$.
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