Brill-Noether Existence on Graphs via R-Divisors, Polytopes and Lattices

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Problem: Reconstructed conjecture: for every connected loop-free multigraph $G$ of genus $g$, and nonnegative real $r,d$ with $d\le 2g-2$, if

$$\rho(g,r,d)=g-(r+1)(g-d+r)\ge0,$$

then $G$ has an $\mathbb R$-divisor of degree at most $d$ and rank exactly $r$. The source paper’s full statement includes $d\ge0$; the counterexample below satisfies this, so the ambiguity is immaterial.

Result: The conjecture is false.

Let $G$ have vertices $1,2,3$, two parallel edges between $1,2$, and two parallel edges between $2,3$. Then $G$ is connected and loop-free, with $m=4,n=3$, hence genus

$$g=m-n+1=2.$$

Take

$$r=\frac12,\qquad d=\frac76.$$

Then $d\le 2g-2=2$, and

$$\rho\!\left(2,\frac12,\frac76\right) =2-\frac32\left(2-\frac76+\frac12\right) =2-\frac32\cdot\frac43 =0.$$

We show no $\mathbb R$-divisor of degree $\le 7/6$ has rank $1/2$.

For this graph the Laplacian lattice is

$$L_G=\langle (2,-2,0),(0,-2,2)\rangle=2A_2.$$

The acyclic-orientation representatives $\operatorname{outdeg}-1$ are

$$(-1,1,1),\ (1,-1,1),\ (-1,3,-1),\ (1,1,-1),$$

all congruent modulo $2A_2$. Thus

$$\mathcal N_G=\{\,\nu\in\mathbb Z^3:\nu_i\text{ odd for all }i,\ \nu_1+\nu_2+\nu_3=1\,\}.$$

Using the standard degree-plus rank formula for $\mathbb R$-divisors,

$$r(D)=\min_{\nu\in\mathcal N_G}\deg^+(D-\nu)-1.$$

Let $D=(x_1,x_2,x_3)$ with $s=x_1+x_2+x_3\le 7/6$. Put

$$a_i=\frac{x_i-1}{2}.$$

Then $\sum_i a_i=(s-3)/2\le -11/12$. Write $a_i=q_i+f_i$, where $q_i\in\mathbb Z$, $0\le f_i<1$, and set $N=-1-\sum_i q_i\ge0$. Choose nonnegative integers $\ell_i$ summing to $N$, killing the largest $N$ fractional parts if $N\le2$, and all fractional parts if $N\ge3$. Let $k_i=q_i+\ell_i$, so $\sum k_i=-1$, and define $\nu_i=2k_i+1$. Then $\nu\in\mathcal N_G$, and

$$\deg^+(D-\nu)=2\sum_i \max(0,f_i-\ell_i).$$

If $N=0$, this is at most $2\cdot \frac1{12}<\frac32$.
If $N=1$, it is at most $2\cdot \frac23\cdot \frac{13}{12}=\frac{13}{9}<\frac32$.
If $N=2$, it is at most $2\cdot \frac13\cdot \frac{25}{12}=\frac{25}{18}<\frac32$.
If $N\ge3$, it is $0$.

Therefore for every $D$ of degree $\le7/6$,

$$r(D)<\frac12.$$

So $G$ satisfies the hypotheses but has no $\mathbb R$-divisor of degree at most $d$ and rank $r$. This refutes the conjecture.

Citation: Definitions and the rank formula are from Manjunath, “Brill-Noether Existence on Graphs via $\mathbb R$-Divisors, Polytopes and Lattices,” Selecta Math. 28 (2022), arXiv:1911.11514.