---

Source. Daniel W. Cranston and Landon Rabern, The fractional chromatic number of the plane, arXiv:1501.01647 (2015); later Combinatorica 37 (2017), 837–861, doi:10.1007/s00493-016-3380-3. Theorem 2 on p. 12 of arXiv v1 (7 January 2015), the edition named on the source card; the journal's labels and pagination may differ.

Notation (pp. 1–2). $\chi_f(\mathbb R^2)$ is the fractional chromatic number of the graph whose vertices are the points of the plane, two points adjacent when their distance is $1$: the least total weight of a nonnegative weighting of its independent sets under which every vertex lies in sets of total weight at least $1$.

Statement

Theorem 2 (p. 12). The paper states:

The fractional chromatic number of the plane is at least $\frac{76}{21}$, i.e., $\chi_f(\mathbb{R}^2) \ge \frac{76}{21}$.

Here $76/21\approx3.619047$. The previous best lower bound, recalled on p. 3, was $32/9\approx3.5556$.

Read depth. Claims checked: the statement, the construction of the graphs $G'_d$ and the weights were read clause by clause on the arXiv v1 PDF. The discharging proof (Claims 1–6) was read for structure; nothing here is independently reviewed.

Proof pointer

Section 3.3 (pp. 12–18), on the graphs $G'_d$ built at the end of Section 3.2 (pp. 11–12). The lower bound comes from a sequence of finite unit-distance graphs $G'_d$ and the fact that the total vertex weight divided by the largest weight of an independent set bounds $\chi_f$ below (p. 3).

• The graphs (pp. 3, 6, 11–12). The core $C_d$ is the part of the triangular lattice within distance $d$ of a fixed lattice vertex. Moser spindles are attached to the core along every diamond (two lattice vertices at distance $\sqrt3$ and their two common neighbours), now in six directions at each core vertex, so that each interior core vertex meets $12$ spindles; the second spindle on each pair is the reflection of the first across the perpendicular bisector, and the rotation angle is $\cos^{-1}(5/6)$. Spindle vertices that happen to coincide are merged and their weights added, which keeps the argument valid (p. 12).
• The weights (p. 12). Each core vertex gets $31/5$ and each spindle vertex $1/2$. With $M$ core vertices there are $M(18-o(1))$ spindle vertices as $d\to\infty$, so the total weight is $M(31/5+9-o(1))$.
• The bound (pp. 12–13). For an arbitrary independent set $I$, the weight in $I$ is moved onto the core so that core vertices end with average weight at most $21/5$ and spindles with weight at most $0$. The averaging is done tile by tile over the tiling of Lemma 1, in three discharging phases (rules R1–R6, p. 13), and Claims 1–6 (pp. 14–18) show that every tile ends with excess at most $0$ and every spindle with nonnegative weight. The ratio $(31/5+9-o(1))/(21/5)$ tends to $76/21$.

Remarks in the source

• p. 18: changing two discharging rules (the $1/3$ in R1 to $13/42$, the $3/10$ in R4 to $2/7$) improves the bound to $105/29\approx3.6207$; the authors state that the proof needs four further phases and an additional 5-spindle block, and do not present it. They ask for the value of $\lim_{d\to\infty}\chi_f(G'_d)$, and whether it exceeds $105/29$.
• p. 5: a linear program on a larger version of the Fisher–Ullman graph gave $\chi_f\ge 1732/481\approx3.6008$; the authors say they offer no proof of that bound beyond their code generating the LP.
• p. 19: since every $G'_d$ has its vertices in $\mathbb Q(\sqrt3,\sqrt{11})\times\mathbb Q(\sqrt3,\sqrt{11})$, the same argument gives $\chi_f\bigl(\mathbb Q(\sqrt3,\sqrt{11})\times\mathbb Q(\sqrt3,\sqrt{11})\bigr)\ge 76/21$.

Bears on

• Problem 508: the problem asks for the chromatic number $\chi(\mathbb R^2)$. Since $\chi(\mathbb R^2)\ge\chi_f(\mathbb R^2)$, the theorem gives $\chi(\mathbb R^2)\ge 76/21$, hence only $\chi(\mathbb R^2)\ge4$ for the integer $\chi$, which the Moser spindle already gives. The theorem is a bound on the fractional relaxation, not on the problem's quantity; the later fractional bound $4$ is on the Matolcsi–Ruzsa–Varga–Zsámboki card.