Abstract

Let X X be a variety defined over an algebraically closed field k k of characteristic 0 0 , let N ∈ N N\in \mathbb {N} , let g : X ⇢ X g:X\dashrightarrow X be a dominant rational self-map, and let A : A N ⟶ A N A:\mathbb {A}^N\longrightarrow \mathbb {A}^N be a linear transformation defined over k ( X ) k(X) , i.e., for a Zariski open dense subset U ⊂ X U\subset X , we have that for x ∈ U ( k ) x\in U(k) , the specialization A ( x ) A(x) is an N N -by- N N matrix with entries in k k . We let f : X × A N ⇢ X × A N f:X\times \mathbb {A}^N\dashrightarrow X\times \mathbb {A}^N be the rational endomorphism given by ( x , y ) ↦ ( g ( x ) , A ( x ) y ) (x,y)\mapsto (g(x), A(x)y) . We prove that if the determinant of A A is nonzero and if there exists x ∈ X ( k ) x\in X(k) such that its orbit O g ( x ) \mathcal {O}_g(x) is Zariski dense in X X , then either there exists a point z ∈ ( X × A N ) ( k ) z\in (X\times \mathbb {A}^N)(k) such that its orbit O f ( z ) \mathcal {O}_f(z) is Zariski dense in X × A N X\times \mathbb {A}^N or there exists a nonconstant rational function ψ ∈ k ( X × A N ) \psi \in k(X\times \mathbb {A}^N) such that ψ ∘ f = ψ \psi \circ f=\psi . Our result provides additional evidence to a conjecture of Medvedev and Scanlon.

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