Primitive cubic multiplicity and low-degree geometry

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Status

The statements below are proved. They isolate, but do not estimate, the primitive large fibers in the cubic multiplicity problem. No bounded logarithmic moment or positive-density conclusion was obtained.

Removing common dilations costs bounded entropy

Fix positive integral coefficients and $F(\mathbf x)=\sum_{i=1}^3a_ix_i^3$. Let $r(n)$ count all positive representations and $r_*(n)$ those with $\gcd(x_1,x_2,x_3)=1$. Define, with $0\log0=0$,

$$A(X)=\sum_{n\le X}r(n)\log r(n),\qquad A_*(X)=\sum_{n\le X}r_*(n)\log r_*(n).$$

Then uniformly for sufficiently large $X$,

$$0\le A(X)-\sum_{g\ge1}A_*(X/g^3)\ll_F X. \tag{1}$$

To prove this, choose an input uniformly from the finite region $\{\mathbf x\in\mathbb N^3:F(\mathbf x)\le X\}$, of size $T(X)\asymp_F X$. Put $G=\gcd(\mathbf x)$ and $Y=F(\mathbf x)$. The representations of a fixed $n$ split into gcd classes of sizes $r_*(n/g^3)$, where $g^3\mid n$. The definition of conditional Shannon entropy therefore gives exactly

$$A(X)-\sum_g A_*(X/g^3)=T(X)H(G\mid Y).$$

Also $\Pr(G=g)\ll_F g^{-3}$, by the box bound for inputs of gcd $g$. Consequently $\mathbb E\log G$ is bounded. Comparing this probability law with $q_g=g^{-2}/\zeta(2)$ gives $H(G)\le\log\zeta(2)+2\mathbb E\log G\ll_F1$; this is the elementary nonnegativity of relative entropy. Since $0\le H(G\mid Y)\le H(G)$, (1) follows.

In particular, the uniform bounds $A(X)\ll_F X$ and $A_*(X)\ll_F X$ are equivalent. One implication uses $r_*(n)\le r(n)$; the other uses (1) and $\sum_g g^{-3}<\infty$. Bounds at small $X$ can be absorbed into the constant. This is not a statement about transferring a bound given only along one unspecified sparse sequence of scales.

For every fixed $1<p<3$, there is also the equivalence

$$\sum_{n\le X}r(n)^p\ll_{F,p} X \quad\Longleftrightarrow\quad \sum_{n\le X}r_*(n)^p\ll_{F,p} X. \tag{2}$$

Indeed $r(n)=\sum_{g^3\mid n}r_*(n/g^3)$, so Minkowski's inequality bounds the $\ell^p$ norm of $r$ by a constant times $X^{1/p}\sum_g g^{-3/p}$. The converse is immediate. Thus common dilations do not account for the missing logarithmic or fractional-moment estimate.

Low-height projected collision planes have negligible input mass

Let $E_H(N)$ be the triples in $[1,N]^3$ lying on a rational plane through the origin whose primitive integral normal has maximum norm at most $H$. Then

$$|E_H(N)|\ll N^2H^2+NH^3. \tag{3}$$

For a primitive normal $\mathbf b$ of height $B$, relabel coordinates so $|b_1|=B$. Set $g=\gcd(b_2,B)$. The congruence $b_2x_2+b_3x_3\equiv0\pmod B$ forces $g\mid x_3$, since $\gcd(g,b_3)=1$. For each such $x_3$, there is one residue class for $x_2$ modulo $B/g$. Thus there are at most $N^2/B+N/g\le N^2/B+N$ possible pairs $(x_2,x_3)$, and the equation then determines $x_1$. There are $O(B^2)$ normals of height $B$; summing proves (3).

The fixed-box representation count satisfies $R_N(n)\le N^2$. For any fixed $0<\eta<1/2$, (3) consequently gives

$$\sum_{\mathbf x\in E_{N^{1/2-\eta}}(N)} \log R_N(F(\mathbf x)) \ll N^{3-2\eta}\log N=o(N^3). \tag{4}$$

Every fixed two-parameter homogeneous linear collision identity projects to a fixed rational plane (or a subspace of one) in the input coordinates. Equation (4) removes such identities, and all projected planes in the stated height range, from the logarithmic-moment problem.

This range cannot simply be increased to cover all inputs. Every triple of height at most $N$ has a nonzero integral orthogonal vector of size $O(\sqrt N)$: pigeonhole the dot products of vectors in $\{0,\ldots,L\}^3$, with $L$ a sufficiently large constant times $\sqrt N$, into their $O(NL)$ possible values, and subtract two equal ones. Dividing by the coordinate gcd makes the normal primitive. High-height planes therefore contain the entire remaining problem, not an automatically negligible exceptional set.

Degrees of curves on the special output surfaces

For $n\in\mathbb Q_{>0}$ put

$$S_n:\quad x^3+2y^3+4z^3=nw^3\subset\mathbb P^3_{\mathbb Q}, \qquad C=S_n\cap\{w=0\}.$$

Proposition. Every curve defined over $\mathbb Q$ on $S_n$ has degree divisible by $3$. In particular, $S_n$ contains no rational line or rational conic. No cube-freeness hypothesis on $n$ is required.

First every closed point of $C$ has degree divisible by $3$. If $C$ has a point over a finite extension $K/\mathbb Q_2$, let $e=v_K(2)$ with the valuation normalized integrally. Unless $3\mid e$, the nonzero terms in

$$x^3+2y^3+4z^3=0$$

have valuations in distinct classes $0,e,2e\pmod3$. Their minimum would occur uniquely, which is impossible. Hence $3\mid e$. For a closed point with residue number field $L$, apply this at every place $v$ above $2$. Each local degree $[L_v:\mathbb Q_2]=e_vf_v$ is divisible by $3$, and their sum is $[L:\mathbb Q]$.

If a curve $D$ on $S_n$ is not contained in $w=0$, its hyperplane intersection with $w=0$ is an effective zero-cycle over $\mathbb Q$ of degree $\deg D$, supported on $C$. The preceding result makes this degree a multiple of $3$. The smooth plane cubic $C$ itself has degree $3$; this also handles curves contained in the boundary and arbitrary curve cycles by additivity.

This excludes degree-one and degree-two curves on individual output surfaces. It does not exclude the split planes on the six-variable collision variety, whose outputs vary along the plane. Nor does it bound the number of higher-degree curves or integral points uniformly as $n$ varies. Primitive high multiplicity remains uncontrolled.

Three special target cube classes have negligible input mass

For $F=x^3+2y^3+4z^3$, let $E(N)$ be the positive triples in $[1,N]^3$ with $F(\mathbf x)=2^e m^3$ for some $e\in\{0,1,2\}$ and positive integer $m$. Then

$$|E(N)|\ll N^{\sqrt3}(1+\log N)^4=o(N^3). \tag{5}$$

This uses the same checked statement of Salberger's Corollary 0.7 as the coordinate-sharing estimate, not a new uniform surface theorem. For fixed $e$, consider the nonsingular diagonal cubic surface

$$x^3+2y^3+4z^3-2^e m^3=0.$$

A positive solution has no vanishing two-term subsum: a subsum involving only positive terms cannot vanish, and cancellation with the negative term would leave two strictly positive terms summing to zero. Also $m\le7^{1/3}N$, so the projective height is $O(N)$. The quoted bound gives $O(N^{\sqrt3}(1+\log N)^4)$ primitive points. Summing common dilations gives the same order, since $\sum_{g\ge1}g^{-\sqrt3}<\infty$; the additive constant in each primitive bound contributes only $O(N)$. Summing over the three values of $e$ proves (5).

As $R_N(n)\le N^2$, deleting these inputs costs at most $2|E(N)|\log N=o(N^3)$ in the logarithmic-multiplicity sum. This removes only three fixed rational cube classes of targets. It supplies no bound for the remaining integer fibers and no positive-density conclusion.