Why endpoint weights do not close the surface approximation argument

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This is a retained earlier research note. The current reading guide gives the latest status and distinguishes completed proofs from remaining cases.

P699 research · Erdős problem 699 · The uniform cubic attempt

This note tests a Diophantine approximation theorem against the remaining cubic case. Arbitrary weights at the five endpoint points cannot give the strict height inequality needed in the range $\log j/\log n\to1$. Section 4 proves this for every big rational divisor on the indicated surface, without assuming symmetry or nefness. Thus the fact that all three residue groups must grow does not repair this application of the theorem. This is an obstruction to a proposed proof, not a proof of nonexistence or finiteness of bad triples.

Sections 1--3 retain the earlier exact calculations for symmetric nef divisors. They illustrate the failure numerically; the argument in Section 4 explains why nonsymmetric choices fail as well.

1. The surface and the inequality needed

Start with $n=2^h$ and the necessary odd-prime endpoint conditions. Let $X$ be the blowup of $\mathbf P^1_n\times\mathbf P^1_j$ at

$$(1,0),(1,1),(2,0),(2,1),(2,2).$$

Write $F$ for a fiber of the $n$ projection and $H$ for a fiber of the $j$ projection, and write $E_{rs}$ for the exceptional divisor above $(r,s)$. Thus $H^2=F^2=0$ and $H.F=1$. Put $Q=(\infty,\infty)$. Consider

$$L=aH+bF-x(E_{10}+E_{11})-y(E_{20}+E_{21}+E_{22}),\tag{1}$$

where $a,b,x,y>0$ are rational and

$$a\ge2x,\qquad a\ge3y,\qquad b\ge x+y.\tag{2}$$

These conditions make $L$ nef. For a curve other than a vertical fiber, the sums of its multiplicities on each of the two rows are at most its intersection with a vertical fiber; this proves nonnegative intersection with (1). The two special vertical fibers give the first two inequalities in (2). Moreover

$$V:=L^2=2ab-S>0,\qquad S:=2x^2+3y^2.$$

For a closed subscheme $Y$, use

$$\beta(L,Y)=\liminf_{m\to\infty} \frac{\sum_{r\ge1}h^0(X,mL\otimes\mathcal I_Y^r)} {m h^0(X,mL)}.$$

The main theorem of Huang–Levin–Xiao permits a divisor and a point on it as a nested regular chain. Applied to $F_\infty\supset Q$ at infinity and $F_0$ at $2$, it gives the relevant inequality with contributions $\beta(L,Q)\log n$ and $\beta(L,F)\log n$ when $\log j/\log n\to1$. The endpoint gcds give

$$h_L(n,j)=(a+b-x-y)\log n+o(\log n).$$

Consequently, to exclude this remaining range by this application, one would need

$$\beta(L,Q)+\beta(L,F)>a+b-x-y.\tag{3}$$

At the finite place one may complete $F_0$ to a regular chain with an arbitrary point on it and discard the additional nonnegative term. For fixed odd $u$ in $n=u2^h$, enlarging the fixed set of places gives the same asymptotic test. The theorem's exceptional algebraic curves do not repair failure of the strict numerical inequality.

2. Exact beta constants

Put $z=(2x+y-a)_+$, where $t_+=\max(t,0)$, and define

$$C_F= \begin{cases} \dfrac23x^3+2xy^2+\dfrac13y^3+\dfrac16z^3,&x\ge y,\\[3pt] y^3+2x^2y+\dfrac16z^3,&x\le y, \end{cases}$$ $$T= \begin{cases} 2x^3+3xy^2+\dfrac52y^3,&2x\ge3y,\\[3pt] \dfrac{14}{9}x^3+2x^2y+4y^3,&2x\le3y. \end{cases}$$

Then

$$V\beta(L,F)=ab^2-Sb+C_F,\tag{4}$$ $$V\beta(L,Q)=ab^2+(a^2-S)b-aS+T.\tag{5}$$

Here are derivations which also specify the geometry behind the formulas.

For (4), fix a sufficiently divisible $m$. At $n=1$, the Taylor coefficient of order $r$ belongs to the subspace of polynomials in $j$ divisible by $[j(j-1)]^{(xm-r)_+}$; at $n=2$, the corresponding condition is divisibility by $[j(j-1)(j-2)]^{(ym-s)_+}$. These are two filtrations of $k[j]_{\le am}$, so they admit a simultaneous constant basis $e_i(j)$. The section module has basis

$$e_i(j)(n-1)^{r_i}(n-2)^{s_i}.$$

Thus its dimension at $n$-degree at most $bm$ is $\sum_i(bm-r_i-s_i+1)_+$. The intersection of the two filtration spaces has dimension

$$[am-2\max((xm-r)_+,(ym-s)_+)-(ym-s)_++1]_+.$$

After scaling, the first moment of the total cost $r_i+s_i$ is $S/2$, and its second moment is $C_F$. Integrating $\sum_i(b-t-r_i-s_i)_+$ over $t$ gives (4). In particular, the mixed second moment follows by integrating the joint tail

$$2\min(p,q)+(2p+q-a)_+, \qquad 0\le p\le x,\quad0\le q\le y.$$

For (5), blow up $Q$ as well and denote its exceptional divisor by $E_Q$. Put $A_1=a-2x$, $A_2=a-3y$, and $\tau=a+b-x-y$. For $0\le t\le\tau$, the negative part of $L-tE_Q$ is supported on the strict transforms of $F_\infty,H_\infty,F_1,F_2$. Its coefficients are

$$\delta=(t-a)_+,\quad \gamma,\quad r_1=\frac{(\gamma-A_1)_+}{2},\quad r_2=\frac{(\gamma-A_2)_+}{3},$$

where $\gamma-r_1-r_2=(t-b)_+$. These are obtained by setting the intersections of the remaining divisor with the active curves to zero. The remainder is nef and becomes zero at $t=\tau$. One can check nefness on the exceptional curves, the vertical and horizontal fibers through the six points, and the four diagonal curves through $Q$ and an unequal-$j$ pair from the two rows. These are all the negative curves: the row and fiber intersection bounds, combined with adjunction, force any other negative nonfiber curve to have bidegree $(1,1)$; the indicated four are exactly those possibilities.

Writing

$$s(\gamma)=\gamma-\frac{(\gamma-A_1)_+}{2} -\frac{(\gamma-A_2)_+}{3},$$

integration of this Zariski decomposition gives

$$V\beta(L,Q)=ab^2+(a^2-S)b-\frac{a^3}{3} +\int_0^a s(\gamma)^2\,d\gamma.$$

The elementary integral is $a^3/3-aS+T$, proving (5).

3. The strict inequality always has the wrong sign

Write $p=x+y$. Multiplying the difference between the two sides of (3) by $V$ gives

$$G(b)=-[a^2-2ap+S]b+C_F+T-Sp.$$

Its coefficient of $b$ is negative, since

$$a^2-2a(x+y)+2x^2+3y^2 =2(x-a/2)^2+3(y-a/3)^2+a^2/6>0.$$

Thus $G$ is largest at $b=p$. There it equals

$$G(p)=C_F+T-p(a^2-2ap+2S).$$

Its derivative with respect to $a$ is

$$2p(p-a)-\frac12(2x+y-a)_+^2<0,$$

because (2) implies $p\le5a/6$. It therefore suffices to set $a=\max(2x,3y)$. The resulting expressions are

$$G= \begin{cases} -\dfrac43x^3+3xy^2-3y^3,&2x\ge3y,\\[3pt] -\dfrac{22}{9}x^3+6x^2y-3xy^2-4y^3,&x\le y,\ 2x\le3y,\\[3pt] -\dfrac49x^3+3xy^2-6y^3,&y\le x\le3y/2. \end{cases}$$

All three are negative. In the first case, putting $r=y/x\le2/3$ gives $G/x^3=-4/3+3r^2-3r^3\le-8/9$. In the second case, putting $r=x/y\le1$ and using $6r^2\le6r$ gives $G/y^3\le-1$. In the last case, $x/y\le3/2$ gives $G/y^3\le-3/2$.

We have proved, throughout (2),

$$\boxed{\beta(L,Q)+\beta(L,F)<a+b-x-y.}$$

For example, on the entire boundary family $L=6H+bF-3(E_{10}+E_{11})-2(E_{20}+E_{21}+E_{22})$, $b\ge5$, the difference is exactly $-(b-1)/(2b-5)$.

4. Arbitrary endpoint weights still cannot give strictness

Suppose $(n,j)$ runs through an unbounded sequence of bad cubic pairs with $\log j/\log n\to1$. Let

$$D_i=\gcd(n-1,j-i)\quad(i=0,1),$$

and let $C_0,C_1,C_2$ be the three residue groups at $n-2$. Then

$$D_0D_1=(n-1)/\lambda,\qquad C_0C_1C_2=(n-2)/2,$$

where $\lambda\in\{1,3\}$. Pass to a subsequence on which the five normalized logarithms converge, and write their limits as

$$\alpha_i=\lim\frac{\log D_i}{\log n},\qquad \gamma_r=\lim\frac{\log C_r}{\log n}.$$

All five numbers are nonnegative, and

$$\alpha_0+\alpha_1=1,\qquad \gamma_0+\gamma_1+\gamma_2=1.\tag{6}$$

For each pair $(i,r)$, the coprime integers $D_i,C_r$ both divide

$$j-(r-i)n-(2i-r).$$

This affine expression is nonzero in $4\le j<n/2$, and its absolute value is less than $2n$. Hence

$$\alpha_i+\gamma_r\le1 \qquad(i=0,1;\ r=0,1,2).\tag{7}$$

The polytope defined by (6)--(7) has exactly five vertices:

$$\begin{aligned} &(\alpha_0,\alpha_1)=(1/3,2/3)\text{ or }(2/3,1/3), && (\gamma_0,\gamma_1,\gamma_2)=(1/3,1/3,1/3),\\ &(\alpha_0,\alpha_1)=(1/2,1/2), && (\gamma_0,\gamma_1,\gamma_2) \text{ is a permutation of }(1/2,1/2,0). \end{aligned}\tag{8}$$

For verification, put $m=\min(\alpha_0,\alpha_1)$. The inequalities say $1/3\le m\le1/2$ and $0\le\gamma_r\le m$. At fixed $m$, the vertices of the latter triangle are the permutations of $(m,m,1-2m)$. Each lies on a segment between the corresponding two points in (8). This proves the assertion about the whole polytope.

A weighted Bezout inequality

For any nonzero affine polynomial $f(n,j)$ of total degree $d$, we have

$$\alpha_0\operatorname{mult}_{(1,0)}f +\alpha_1\operatorname{mult}_{(1,1)}f +\sum_{r=0}^2\gamma_r\operatorname{mult}_{(2,r)}f \le d.\tag{9}$$

It is enough to check an irreducible factor and the five mass vectors in (8). If that factor is one of the lines $n=1$ or $n=2$, its weighted multiplicity sum is exactly one, equal to its degree. Otherwise Bezout with those two lines gives

$$m_{10}+m_{11}\le d,\qquad m_{20}+m_{21}+m_{22}\le d,$$

and each individual multiplicity is at most $d$. At a vertex with all nonzero masses equal to $1/2$, average the two row inequalities and discard the unselected multiplicity in the second row. At a vertex with one $\alpha_i=2/3$, average the two row inequalities and the bound $m_{1i}\le d$, with coefficient $1/3$ on each. These are exactly (9) at the five vertices. Convexity and additivity over factors prove (9) in general. This argument also shows that the divisor consisting of a line minus these five weighted exceptional divisors is nef on the blowup of $\mathbf P^2$ at the endpoint points.

The resulting upper bound for both beta constants

Return to the surface $X$ of Section 1 and take any big rational divisor

$$L=aH+bF-x_0E_{10}-x_1E_{11}-\sum_{r=0}^2y_rE_{2r}.$$

Symmetry and nefness are not assumed. A section of $mL$, for sufficiently divisible $m$, is represented by a polynomial of degree at most $mb$ in $n$ and at most $ma$ in $j$, with the required endpoint multiplicities. Write this polynomial as $n^k g(n,j)$, where $n\nmid g$. Its vanishing order along $F_0$ is $k$. At $Q=(\infty,\infty)$ its vanishing order is

$$m(a+b)-\deg(n^kg)=m(a+b)-k-\deg g.$$

The factor $n^k$ is a unit at all five endpoint points. Apply (9) to $g$, using the multiplicities required by $mL$. Every nonzero section therefore satisfies

$$\operatorname{ord}_{F_0}(s)+\operatorname{ord}_{Q}(s) \le m\left(a+b-x_0\alpha_0-x_1\alpha_1 -\sum_{r=0}^2y_r\gamma_r\right).\tag{10}$$

The two filtrations of $H^0(X,mL)$ by vanishing at $F_0$ and at $Q$ admit a common adapted basis. Sum (10) over that basis, divide by $m h^0(X,mL)$, and let $m$ tend to infinity. By the definition of beta,

$$\boxed{ \beta(L,F_0)+\beta(L,Q) \le a+b-x_0\alpha_0-x_1\alpha_1 -\sum_{r=0}^2y_r\gamma_r.}\tag{11}$$

For the bad pairs in question, the right side of (11) is precisely $h_L(n,j)/\log n+o(1)$. At infinity the nested chain $F_\infty\supset Q$ contributes $\beta(L,Q)\log n+o(\log n)$. At two the single divisor $F_0$ contributes at most $\beta(L,F_0)\log n+O(1)$, with equality to leading order for fixed odd part of $n$. Thus (11) prevents this application of the theorem from exceeding the available height. Allowing the odd part of $n$ to vary cannot improve the inequality.

This conclusion concerns this choice of approximation data: the chain at infinity and the divisor $F_0$ at two, with arbitrary weights at the five endpoint points. It does not rule out using additional information at two, other subschemes, or a different Diophantine theorem. Nor does it cover a sequence with $\log j/\log n$ bounded away from one. Finally, $C_r\to\infty$ does not imply $\gamma_r>0$; even positivity of all three limiting masses would not turn (11) into the required strict inequality.

Source and provenance

Keping Huang, Aaron Levin and Zheng Xiao, A New Diophantine Approximation Inequality on Surfaces and Its Applications, arXiv:2406.18879v1 (2024), Theorem 1.4 and Definition 2.14. The theorem was checked in the primary TeX source Surface_20240510.tex, at the theorem labeled BetaMain and the definition labeled regulardef.

The calculations (1)–(5) and the general obstruction (6)–(11) are deductions here, not claims made in that paper.