gpt 6 astra ultra attempt on Math-488

Zariski Dense Orbit Conjecture

Definitions and mathematical statement

Let $X$ be an irreducible quasiprojective variety over an algebraically closed characteristic-zero field $K$, and let $f:X\dashrightarrow X$ be dominant. The invariant rational functions form \[ K(X)^f=\{g\in K(X):g\circ f=g\}. \] Assume $K(X)^f=K$. The conjecture asks for a point $x\in X(K)$ whose entire forward orbit is defined and Zariski dense: \[ f^n(x)\notin\operatorname{Indet}(f)\ (n\ge0), \qquad\overline{\{f^n(x):n\ge0\}}^{\rm Zar}=X. \] The hypothesis rules out invariant rational fibrations detected by a nonconstant rational function. The field need not be uncountable, so an argument simply avoiding a countable union of proper closed sets is not sufficient in the full scope. The conclusion is existence of one point, not a dense set of such points.

Formulation and concept references: [S1].

Short English statement

If a dominant rational self-map has no nonconstant invariant rational function, must it have a well-defined Zariski-dense forward orbit?

Research attempt

Scope and source audit (27 September 2026).

The field is algebraically closed of characteristic zero and may be countable, including $\overline{\mathbb Q}$. A $K$-point is required, not merely a generic point over a larger transcendental extension. Dominance does not make a rational map everywhere defined. In dimension zero the irreducible variety is a point and the conclusion is immediate; the discussion below concerns positive dimension.

The supplied Bell--Ghioca paper [S1, Conjecture 1.1] states the general dichotomy and records the established uncountable-field case, as well as several special families. Its proposed strengthening for birational maps of dynamical degree one is not an assumption available in this problem. The recent Matsuzawa--Xie paper [E1] proves a birational case for projective varieties over $\overline{\mathbb Q}$ when the first dynamical degree exceeds the third, including projective threefolds over that field with first dynamical degree greater than one. Its more general arithmetic-degree result is not an unconditional dense-orbit theorem for every rational self-map. Those statements were checked directly in the primary sources; their deep proofs are not reproduced here.

The present attempt makes the countable-field existence step explicit in two families: translations on a split torus and simultaneous power maps. The proofs use finite algebraic identities rather than choosing a point outside a countable union. The counterexample track tests several tempting extensions: an arbitrary starting point, a highest possible height-growth rate, and an assertion that finite avoidance automatically extends to infinite avoidance. These tests identify failures of methods, not counterexamples to the catalogue conjecture.

A complete torus-translation criterion.

Let $T=(\mathbb G_m)^r$, $r\ge1$, and fix $a=(a_1,\ldots,a_r)\in T(K)$. Consider the everywhere-defined automorphism \[ f_a(x_1,\ldots,x_r)=(a_1x_1,\ldots,a_rx_r). \] For $u=(u_1,\ldots,u_r)\in\mathbb Z^r$ write $x^u=\prod x_i^{u_i}$ and $a^u=\prod a_i^{u_i}$. Define \[ \Lambda_a=\{u\in\mathbb Z^r:a^u=1\}. \] The condition $\Lambda_a=\{0\}$ is multiplicative independence. It also excludes a nontrivial monomial taking a root-of-unity value: raising that relation to the order of the root of unity would produce a nonzero element of $\Lambda_a$.

Lemma 484.1. For $f_a$ the following conditions are equivalent: $\Lambda_a=0$; $K(T)^{f_a}=K$; and every point of $T(K)$ has a Zariski-dense forward orbit. If $\Lambda_a\ne0$, no such orbit is dense.

Proof of density and the obstruction. If $0\ne u\in\Lambda_a$, the nonconstant regular function $x^u$ is invariant. An orbit starting at $b$ lies in the proper closed hypersurface $x^u=b^u$, so it is not dense.

Now assume $\Lambda_a=0$. A nonzero Laurent polynomial has the form \[ P(x)=\sum_{j=1}^N c_jx^{u_j}, \qquad c_j\ne0,\quad u_j\text{ distinct}. \] Its values on the orbit of $b\in T(K)$ are \[ P(f_a^n(b))=\sum_{j=1}^N(c_jb^{u_j})\lambda_j^n, \qquad \lambda_j=a^{u_j}. \] The $\lambda_j$ are nonzero and pairwise distinct, because an equality would give $u_i-u_j\in\Lambda_a$. If all orbit values vanished, using $n=0,\ldots,N-1$ would give a homogeneous system with Vandermonde determinant \[ \prod_{1\le i<j\le N}(\lambda_j-\lambda_i)\ne0. \] It would force every $c_jb^{u_j}$ to be zero, a contradiction. Thus no nonzero Laurent polynomial vanishes on the orbit. Since $T$ is affine with coordinate ring $K[x_1^{\pm1},\ldots,x_r^{\pm1}]$, this is exactly Zariski density. The same argument works for any consecutive $N$ iterates, after multiplying the coefficients by nonzero powers of the $\lambda_j$.

Proof of the invariant-field assertion. Suppose $g=P/Q$ is invariant, with coprime nonzero Laurent polynomials $P,Q$ in that unique factorization domain. The identity \[ P(ax)Q(x)=P(x)Q(ax) \] implies $P(x)\mid P(ax)$, by coprimality of $P,Q$. Write $P(ax)=H(x)P(x)$. For each coordinate, the largest exponent in a product is the sum of the largest exponents of its factors, and the same is true of the smallest exponent. This follows by treating the Laurent polynomial as a one-variable Laurent polynomial over the integral domain generated by the other coordinates. The exponent ranges of $P(ax)$ and $P(x)$ coincide, so the largest and smallest exponents of $H$ in every coordinate are both zero. Hence $H$ is a nonzero constant $c$.

Comparing coefficients in $P(ax)=cP(x)$ shows that every exponent $u$ in the support of $P$ satisfies $a^u=c$. Independence implies that this support contains only one exponent. Applying the same argument to $Q$ and then the invariance identity gives $Q(ax)=cQ(x)$ with the same $c$. Thus $P$ and $Q$ are monomials of the same exponent, and $g$ is constant. If $\Lambda_a\ne0$, the monomial already exhibited is a nonconstant invariant. This proves all the assertions.$\square$

For example $a=(2,3,5,\ldots,p_r)$ is independent over every characteristic-zero field: equality of a rational product of distinct primes to one forces every exponent to vanish by unique factorization in $\mathbb Q$. The starting point $(1,\ldots,1)$ therefore works over $\mathbb Q$ itself, and also over every permitted $K$. Countability causes no gap in this construction.

A second family with an explicit arithmetic starting point.

Fix an integer $d\ge2$ and consider \[ F_d:T\longrightarrow T,\qquad F_d(x_1,\ldots,x_r) =(x_1^d,\ldots,x_r^d). \] Let $q_1,\ldots,q_r$ be distinct positive rational primes and put $b=(q_1,\ldots,q_r)$. The forward orbit is $F_d^n(b)=(q_1^{d^n},\ldots,q_r^{d^n})$; it is defined for every $n$.

Lemma 484.2. This orbit is Zariski dense in $T$ over every algebraically closed characteristic-zero field $K$. More strongly, a nonzero Laurent polynomial over $K$ vanishes on only finitely many points of this orbit.

Proof. For $P=\sum_{j=1}^N c_jx^{u_j}$ with nonzero coefficients and distinct exponents, set $\beta_j=\prod_iq_i^{u_{ji}}$. Unique factorization shows that the $\beta_j$ are distinct positive rational numbers. Evaluating gives \[ P(F_d^n(b))=\sum_{j=1}^N c_j\beta_j^{d^n}. \] The coefficients lie in a subfield $k\subset K$ finitely generated over $\mathbb Q$. Such a field embeds into $\mathbb C$: send a transcendence basis to algebraically independent complex numbers and extend the map over the remaining finite algebraic extension. This embedding is used only for the finitely many coefficients of this one polynomial; there is no assertion that the entire field $K$ embeds into $\mathbb C$.

After applying the embedding, order the rational bases as $0<\beta_1<\cdots<\beta_N$. Division by $\beta_N^{d^n}$ gives \[ \beta_N^{-d^n}P(F_d^n(b)) =c_N+\sum_{j<N}c_j(\beta_j/\beta_N)^{d^n}\longrightarrow c_N\ne0. \] Consequently the expression is nonzero for all sufficiently large $n$. Injectivity of the embedding transfers this conclusion back to $K$. Every proper closed subset of the affine torus is annihilated by a nonzero Laurent polynomial, so the orbit meets it only finitely often. In particular the orbit and all its tails are dense.$\square$

This also verifies the invariant-field hypothesis in this family. If $g=P/Q\in K(T)$ satisfies $g\circ F_d=g$, the denominator $Q$ is nonzero on all sufficiently late orbit points by Lemma 484.2. On that tail the rational identity can be evaluated at each consecutive pair, so $g$ has a constant value $c$. Density of the tail implies $P-cQ=0$ as a Laurent polynomial, and $g=c$. This argument checks denominator avoidance rather than silently evaluating a rational function at its base points.

The proof is insensitive to whether the largest base is below or above one: it is the ratios $\beta_j/\beta_N<1$ that matter. Negative Laurent exponents therefore cause no problem. In dimension one it reduces to the simple fact that the sequence $q^{d^n}$ is infinite and hence dense in $\mathbb G_m$.

A maximal-growth orbit still need not be dense.

Even in the family just solved, the hypothesis does not make every starting point suitable. For $F_2$ on $(\mathbb G_m)^2$, the orbit of $(2,2)$ is contained in the diagonal $x_1=x_2$, whereas the orbit of $(2,3)$ is dense by Lemma 484.2. Both have the same exponential growth rate for logarithmic projective height. With $h(x_1,x_2)=h_{\mathbb P^1}(x_1)+h_{\mathbb P^1}(x_2)$ on rational positive coordinates, \[ h(F_2^n(2,2))=2^{n+1}\log2, \qquad \lim_{n\to\infty}\max\{1,h(F_2^n(2,2))\}^{1/n}=2. \] On $\mathbb P^1\times\mathbb P^1$ the map pulls back each ruling divisor to twice itself, so its first dynamical degree is also two. Thus even attaining the largest predicted height-growth rate does not, by itself, imply a dense orbit. The additional geometry in the recent results cited above cannot be replaced by that implication.

Roots of unity give a second useful stress test. On $\mathbb G_m$ the map $x\mapsto x^2$ has trivial invariant rational-function field, but every root of unity has a finite forward orbit. There are infinitely many such roots, and an infinite subset of an irreducible curve is Zariski dense. Consequently no nonempty Zariski-open subset consists entirely of dense-orbit points in this example. The conjecture asks for existence of one point; it does not assert that deleting a single algebraic exceptional set leaves only successful starting points.

Finite avoidance is valid; countable avoidance is an extra step.

Let $f:X\dashrightarrow X$ be an arbitrary dominant rational self-map. For each integer $N\ge0$, let $U_N$ be the set of points for which the first $N$ applications of $f$ are successively defined. This is a dense open subset. Indeed $U_0=X$ and the induction step restricts a dominant iterate to the inverse image of the nonempty open domain of $f$; that inverse image is nonempty open because the iterate is dominant. A finite intersection of these domains is still nonempty by irreducibility.

Since $K$ is algebraically closed, each nonempty open $U_N$ has a $K$-point. This proves \[ \text{for every }N\text{ there exists }x_N\in X(K) \text{ whose orbit is defined through time }N. \] It does not prove existence of a single $x$ that works for all $N$. The required order of quantifiers is reversed, and there is no compactness of the set of $K$-points in the Zariski topology that repairs that reversal.

For a model of the failure over a countable field, enumerate $\overline{\mathbb Q}=\{a_1,a_2,\ldots\}$. The subsets \[ V_N=\mathbb A^1(\overline{\mathbb Q})\setminus\{a_1,\ldots,a_N\} \] are nonempty open and decreasing, but their intersection has no $\overline{\mathbb Q}$-point. These particular sets are not claimed to be the orbit domains of a rational self-map. They disprove the proposed general countable-avoidance principle, which would otherwise be used without any dynamical justification.

Similarly, over a countable algebraically closed field the union of all closed points is a countable union of proper closed subsets covering $\mathbb A^1(K)$. Passing to an uncountable extension and choosing a point outside that union does not descend the point to $K$. All varieties and maps can have finitely many defining coefficients while the property of avoiding every iterate or every proper orbit closure remains an infinite collection of conditions on the starting point.

Why a finite orbit computation does not close the argument.

Noetherianity says descending chains of closed subsets stabilize; it does not say that ascending chains of finite orbit prefixes stabilize. The closed subsets $\{1,\ldots,N\}\subset\mathbb A^1$ are an explicit ascending nonstationary chain. Thus taking larger finite prefixes, computing their algebraic relations, and appealing to Noetherianity cannot certify the final orbit closure without an independent degree bound or recurrence argument.

The Vandermonde proof succeeds for torus translations because every Laurent polynomial with $N$ terms is ruled out by $N$ consecutive iterates, and the argument works symbolically for arbitrary $N$. The power-map proof instead uses an asymptotic separation of finitely many distinct positive bases, separately for every polynomial. Neither proof gives these properties for an arbitrary rational map. The lack of invariant rational functions has not been converted here into an equally concrete mechanism excluding every lower-dimensional orbit closure.

One possible next statement would be a dynamical avoidance lemma: from $K(X)^f=K$, construct a $K$-point with an infinite defined orbit and prove that every proper closed set misses at least one point of that orbit. The substantive missing input is a means of selecting that point over the original countable field while controlling all iterates. It cannot be replaced by the finite nonempty-open argument just given. An alternative would provide effective algebraic restrictions on all possible proper orbit closures, strong enough that finitely many explicitly verified conditions exclude them. No such restriction is established in general.

Exact checks and outcome.

The companion script computes an exact Vandermonde certificate for the translation $(x,y)\mapsto(2x,3y)$ on the 16 monomials $x^iy^j$ with $0\le i,j\le3$. It compares the product-of-differences determinant with an independent fraction-free elimination. A second exact evaluation matrix tests the first nine iterates of $(2,3)$ under the squaring map against the nine monomials with $0\le i,j\le2$. Nonzero determinants exclude relations in those finite spaces; they are not a computational proof of density by themselves. The all-polynomial density assertions are proved above. All starting points and coefficients in the finite certificates are rational, so no generic transcendental point is hidden.

Outcome: UNRESOLVED, with complete elementary families. The attempt proves the torus-translation criterion, constructs explicit dense power-map orbits over the permitted countable fields, and identifies specific failures of arbitrary starting-point selection, growth-only tests, and finite-to-infinite avoidance. The first remaining gap is a construction of one original-field point whose defined orbit excludes every proper closed closure for an arbitrary dominant rational map with trivial invariant field. No general proof or counterexample is supplied.

Sources

• S1 Jason Bell and Dragos Ghioca, A conjecture strengthening the Zariski dense orbit problem for birational maps of dynamical degree one. https://personal.math.ubc.ca/~dghioca/papers/conjecture_automorphisms.pdf. Formulation source.
• E1 Yohsuke Matsuzawa and Junyi Xie, Arithmetic degree and its application to Zariski dense orbit conjecture, Journal of the London Mathematical Society 112 $2025$, e70282; author manuscript arXiv:2409.06160v2, 17 September 2024. https://doi.org/10.1112/jlms.70282; https://arxiv.org/html/2409.06160v2. Publisher abstract and manuscript introduction inspected 27 September 2026. Birational and dynamical-degree hypotheses retained; the height-growth theorem is not used as a general dense-orbit theorem.