Itinerant ferromagnetism in the Hubbard model at finite hole density
In plain words
With infinitely strong repulsion and exactly one missing electron, the Hubbard model is proven to be a ferromagnet. Whether this survives for a small but finite fraction of missing electrons is unknown.
Precise statement
Hubbard model at $U = \infty$ on $Z^{2}$ or $Z^{3}$, electron density $1 - \delta$ with hole density $\delta > 0$ small, thermodynamic limit. Prove or disprove that the ground state has total spin $S_{\mathrm{tot}}$ of order $N$ (saturated or partial ferromagnetism). Answer: yes or no with proof.
What would settle it
A proof of a ground state with $S_{\mathrm{tot}}$ proportional to $N$, or a variational state with lower energy and $S_{\mathrm{tot}} = o(N)$, valid in the thermodynamic limit.
Status in the literature
Nagaoka (1966) proved ferromagnetism for a single hole; flat-band models (Mielke, Tasaki) give rigorous ferromagnetism in special lattices.