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Rational versus transcendental points on analytic Riemann surfaces

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Let (X, L) be a polarized variety over a number field K. We suppose that L is an hermitian line bundle. Let M be a non compact Riemann Surface and $$U\subset… Click to show full abstract

Let (X, L) be a polarized variety over a number field K. We suppose that L is an hermitian line bundle. Let M be a non compact Riemann Surface and $$U\subset M$$ be a relatively compact open set. Let $$\varphi :M\rightarrow X(\mathbf{C})$$ be a holomorphic map. For every positive real number T, let $$A_U(T)$$ be the cardinality of the set of $$z\in U$$ such that $$\varphi (z)\in X(K)$$ and $$h_L(\varphi (z))\le T$$ . After a revisitation of the proof of the sub exponential bound for $$A_U(T)$$ , obtained by Bombieri and Pila, we show that there are intervals of the reals such that for T in these intervals, $$A_U(T)$$ is upper bounded by a polynomial in T. We then introduce subsets of type S with respect of $$\varphi $$ . These are compact subsets of M for which an inequality similar to Liouville inequality on algebraic points holds. We show that, if M contains a subset of type S, then, for every value of T the number $$A_U(T)$$ is bounded by a polynomial in T. As a consequence, we show that if M is a smooth leaf of an algebraic foliation in curves defined over K then $$A_U(T)$$ is bounded by a polynomial in T. Let S(X) be the subset (full for the Lebesgue measure) of points which verify some kind of Liouville inequalities. In the second part we prove that $$\varphi ^{-1}(S(X))\ne \emptyset $$ if and only if $$\varphi ^{-1}(S(X))$$ is full for the Lebesgue measure on M.

Keywords: bounded polynomial; varphi; let; versus transcendental; transcendental points; rational versus

Journal Title: manuscripta mathematica
Year Published: 2021

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