Limits of rational transformations and multiplicative amplification

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Status and scope

The statements below are proved limitations of two proposed ways to settle the density question. They supply no density conclusion for a diagonal value set or for $S_r$. A family of rational transformations whose complexity grows with the input scale is not controlled by these bounds.

Throughout, $r\ge3$ and

$$F(\mathbf x)=\sum_{i=1}^r a_i x_i^r,\qquad a_i\in\mathbb Q_{>0}.$$

Summable multiplier sets preserve density zero

Let $A,M\subseteq\mathbb N$, with $1\in M$ and $S=\sum_{m\in M}1/m<\infty$. Then

$$\overline d(MA)\le S\,\overline d(A). \tag{1}$$

Indeed, for any fixed $K$,

$$\#(MA\cap[1,X]) \le\sum_{\substack{m\in M\\m\le K}}\#(A\cap[1,X/m]) +X\sum_{\substack{m\in M\\m>K}}\frac1m.$$

Divide by $X$, take the limsup, and then let $K\to\infty$. As $A\subseteq MA$, the set $MA$ has density zero if and only if $A$ does. This includes the full set of $r$-powerful multipliers: its reciprocal sum is at most $\prod_p(1+p^{-r}/(1-p^{-1}))<\infty$.

Thus multiplicative saturation cannot turn a density-zero value set into one of positive upper density, even if the original set has a strong near-linear counting lower bound. This does not decide whether the original set has density zero.

Integer inputs of a fixed nonpolynomial rational map

Lemma. Let $T\in\mathbb Q(x_1,\ldots,x_r)^s$ have at least one coordinate which is not a polynomial. For every $\varepsilon>0$,

$$\#\{\mathbf x\in[1,N]^r\cap\mathbb Z^r: T\text{ is defined at }\mathbf x,\ T(\mathbf x)\in\mathbb Z^s\} \ll_{T,\varepsilon}N^{r-1+\varepsilon}. \tag{2}$$

Choose a nonpolynomial coordinate $P/Q$ in reduced form with $P,Q\in\mathbb Z[\mathbf x]$. Let $Q_0$ be a primitive irreducible nonconstant factor of $Q$, and choose $x_j$ on which $Q_0$ depends. Over $\mathbb Q(\widehat{\mathbf x})[x_j]$, the polynomials $P,Q_0$ are coprime, by Gauss's lemma. Hence their resultant

$$R(\widehat{\mathbf x})=\operatorname{Res}_{x_j}(P,Q_0)$$

is nonzero. The resultant identity, after clearing fixed constants if necessary, implies that an integral value of $P/Q$ satisfies

$$Q_0(\mathbf x)\mid cR(\widehat{\mathbf x})$$

for a fixed nonzero integer $c$.

Discard parameter slices where $R$ or the leading $x_j$-coefficient of $Q_0$ vanishes. These lie on fixed nonzero polynomial zero sets and number $O_T(N^{r-2})$; including all $x_j$ costs $O_T(N^{r-1})$. On any other slice, $|cR|\ll_T N^{O_T(1)}$. There are $O_{T,\varepsilon}(N^\varepsilon)$ signed divisors of $cR$, and each equation $Q_0(x_j)=d$ has at most $\deg_{x_j}Q_0$ roots. There are $O(N^{r-1})$ slices. This proves (2).

Fixed rational self-maps cannot create typical multiplicity

Proposition. Suppose $T\in\mathbb Q(\mathbf x)^r$ satisfies $F\circ T=F$ identically. Unless $T$ is a positive monomial symmetry of $F$, only $o_{F,T}(N^r)$ inputs in $[1,N]^r\cap\mathbb Z^r$ have a defined image in $\mathbb N^r$.

If a coordinate is nonpolynomial, apply (2) with $\varepsilon<1$. Otherwise $T$ is polynomial. At every counted input, positivity and the identity imply $0<T_i(\mathbf x)\ll_F N$ for each $i$. If some coordinate has degree $d>1$, with nonzero leading homogeneous part $G$, then, uniformly for $\mathbf u\in[0,1]^r$,

$$N^{-d}T_i(N\mathbf u)=G(\mathbf u)+O_T(N^{-1}).$$

The counted normalized inputs lie in $|G(\mathbf u)|\ll_{F,T}N^{-1}$. For any fixed $\eta>0$, Riemann sums bound their upper proportion by the volume of $\{|G|\le\eta\}\cap[0,1]^r$. Letting $\eta\downarrow0$ gives zero, since a nonzero polynomial zero set has measure zero.

It remains to consider $T=A\mathbf x+\mathbf b$. Highest-degree terms give $F(A\mathbf x)=F(\mathbf x)$. The matrix $A$ is invertible: a nonzero $v\in\ker A$ would make the directional derivative $\sum_i r a_i v_i x_i^{r-1}$ identically zero, which is impossible. The Hessian determinant of $F$ is a nonzero constant times $(x_1\cdots x_r)^{r-2}$. The transformation rule and unique factorization force each row of $A$ to be a scalar coordinate form. Comparing degree-$(r-1)$ terms then gives $\mathbf b=0$. If any scalar is negative, no positive input has a positive image; otherwise this is precisely a positive monomial symmetry. This proves the proposition.

Such symmetries merely permute the actual summands $a_i x_i^r$ and form a finite group. Consequently no fixed finite list of rational involutions or self-maps forces unbounded multiplicity on a positive proportion of inputs. The constants in (2), and the polynomial small-neighborhood argument, are not uniform over varying maps.

The multiplier classes of linear similitudes

Let $G=\mathbb Q_{>0}^{\times}/\mathbb Q_{>0}^{\times r}$, and let $\mathcal A$ be the multiset of classes of $a_1,\ldots,a_r$. The positive multiplier classes of rational linear similitudes

$$F(T\mathbf x)=\lambda F(\mathbf x), \qquad T\in\operatorname{GL}_r(\mathbb Q),\quad\lambda>0,$$

are exactly

$$H=\{h\in G:h\mathcal A=\mathcal A\},\qquad |H|\mid r. \tag{3}$$

The Hessian argument just given makes $T$ monomial, proving necessity of multiset stabilization. Conversely, a matching of the coefficient classes supplies the rational coordinate scalings, which can be chosen positive. This proves the characterization.

The stabilizer is finite: for a fixed coefficient class $g$, the map $h\mapsto hg$ injects it into the finite support of $\mathcal A$. Its action on $G$ is free. Each orbit in the coefficient multiset has $|H|$ distinct classes with equal multiplicities. Summing their multiplicities proves $|H|\mid r$.

In particular, two distinct primes cannot both be similitude multipliers: their classes generate a subgroup of order $r^2$. More generally, distinct primes with multipliers $p_j^{e_j}$ require $\prod_j r/\gcd(r,e_j)\mid r$. Each class in $H$ has a unique positive integer representative $c_h$ with prime exponents in $\{0,\ldots,r-1\}$. Its positive integer members are exactly $c_h m^r$, $m\ge1$. Thus integer similitude multipliers have a convergent reciprocal sum. By (1), even allowing every such output dilation cannot amplify a density-zero set to positive upper density. Rational coordinate changes may impose further integrality restrictions; ignoring them only enlarges this saturation.

These results rule out fixed rational amplification and a nonsummable family of linear dilation symmetries. They do not estimate a family of transformations of growing complexity, or the actual primitive fibers of $F$.

No composition law linear in one input

There is no identity

$$F(B(\mathbf x,\mathbf y))=F(\mathbf x)F(\mathbf y)$$

in which $B$ is rational in $\mathbf y$ and linear in $\mathbf x$. This statement holds for all nonzero rational coefficients, without positivity, when $r\ge3$.

Indeed, over $K=\mathbb Q(\mathbf y)$ write $B=A\mathbf x$. The matrix $A$ is invertible: otherwise a kernel vector would give a nonzero direction along which $F$ has identically zero derivative. The same Hessian argument over $K$ makes $A$ monomial. Comparing any one coefficient then gives

$$F(\mathbf y)=c\,g(\mathbf y)^r, \qquad c\in\mathbb Q^\times,\quad g\in\mathbb Q(\mathbf y)^\times.$$

The polynomial $F$ is absolutely irreducible, as proved in the fixed-norm argument. Taking the valuation at its prime divisor gives $1=r v_F(g)$, a contradiction.

In particular, a reduced-norm construction cannot supply such a bilinear composition law for a diagonal form. This excludes that construction, not nonlinear rational composition laws or positive density of the value set.