Metallic to insulating spin density wave transition in Hubbard model For a half-filled Hubbard model with weak on-site Coulomb interaction ($U/t<<1$), it's quite intuitive that very likely the system will be in metallic phase. However, there is also such a statement that "the commensurability of the Fermi wavelength with the lattice can initiate a transition to an insulating spin density wave state characterized by a small quasi-particle energy gap".
What exactly does this statement mean? I'm asking for a qualitative explanation for the transition if there is one.
 A: The half-filling case is very special because you can easily see that the Fermi surface of the non-interacting system ($U=0$) is a square. As a result, a wave vector $Q = (\pi,\pm\pi)$ will connect opposite sides. This commensurability is the so called "nesting effect" and it induces instability of the system. If you calculate the magnetic susceptibility $\chi(k)$, you will get \begin{equation} \chi(q) \propto \sum_k \frac{n(\epsilon_{k+q}) -n(\epsilon_k)}{\epsilon_{k }-\epsilon_{k+q}}.\end{equation} Because of the commensurability, there will be a significant density of states with  $|\epsilon_{k }-\epsilon_{k+Q}| \ll W $, the bandwidth. The result is an instability at this wave vector $Q$. This is a special case of spin density wave, anti-ferromagnetic phase. 
Using mean field theory by assuming magnetic order, you can get that an energy gap $\Delta \propto e^{-\sqrt{12t/U}}$, which is nonzero even for $U/t \ll 1$. This is what the statement says. For for information, see Chapter 4 in Interacting Electrons and Quantum Magnetism and Chapter 3 in Field Theories of Condensed Matter Physics. 
