There are absolute constants C and 1≤β<4/3 such that every set of n points in the Euclidean plane determines at most Cnβ pairs at unit distance. The proof argues by contradiction: a configuration with n4/3−o(1) unit distances yields a degree-regular, expanding point-circle incidence piece, from which the paper extracts a large graph of multiplicatively organized pairs whose algebraic numbers have large heights, and an algebraic obstruction rules such a graph out. This improves the classical Spencer–Szemerédi–Trotter exponent 4/3 for the unit-distance problem by a power.
Where the new idea lives · 4 novel steps (one is the main theorem)
Proposition 2.4Incidence piece with degree and expansion control
From n=t3 points with at least t4−o(1) ordered unit pairs, one extracts algebraic point and center sets P,Q and a nonempty unit-incidence graph E⊆P×Q with ∣P∣=t1+2a+o(a), ∣Q∣=t2+a+o(a), ∣E∣=t2+2a+o(a). Degrees are uniform (d(p)=t1+o(a), d(q)=ta+o(a)), every marginal atom is at least γ/∣P∣ or γ/∣Q∣ with γ=27/50, and for all F⊆P, G⊆Q the edge mass satisfies PE(p∈F,q∈G)≤uv((uv)η+ξt) with η=1/25 and ξt→0 independent of the subsets.
Proposition 4.4Total variation S cannot stay bounded
For a sequence of the incidence pieces, no subsequence has the total variation quantity S bounded, so S→∞.
Proposition 8.1No algebraic configuration with large heights
No sequence exists of finite algebraic point sets Pn of size tn1+o(1) with a bipartite graph of tn2−o(1) edges, edge numbers λpr satisfying (zp−zr)(wp−wr)=2−λpr−λpr−1, heights h(λpr)≥cJn, rectangle cross-ratios of height o(Jn), and no edge inside a unit circle with at least tn9/10 points.