Vojta’s refinement of the subspace theorem

Abstract:

Vojta’s refinement of the Subspace Theorem says that given linearly independent linear forms ${L_1}, \ldots , {L_n}$ in $n$ variables with algebraic coefficients, there is a finite union $U$ of proper subspaces of ${\mathbb {Q}^n}$, such that for any $\varepsilon > 0$ the points $\underline {\underline x} \in {\mathbb {Z}^n}\backslash \{ \underline {\underline 0} \}$ with (1) $|{L_1}(\underline {\underline x} ) \cdots {L_n}(\underline {\underline x} )|\; < \;|\underline {\underline x} {|^{ - \varepsilon }}$ lie in $U$, with finitely many exceptions which will depend on $\varepsilon$ . Put differently, if $X(\varepsilon )$ is the set of solutions of (1), if $\bar X(\varepsilon )$ is its closure in the subspace topology (whose closed sets are finite unions of subspaces) and if $\bar X\prime (\varepsilon )$ consists of components of dimension $> 1$ , then $\bar X\prime (\varepsilon ) \subset U$ . In the present paper it is shown that $\bar X\prime (\varepsilon )$ is in fact constant when $\varepsilon$ lies outside a simply described finite set of rational numbers. More generally, let $k$ be an algebraic number field and $S$ finite set of absolute values of $k$ containing the archimedean ones. For $\upsilon \in S$ let $L_1^\upsilon , \ldots ,L_m^\upsilon$ be linear forms with coefficients in $k$, and for $\underline {\underline x} \in {K^n}\backslash \{ \underline {\underline 0} \}$ with height ${H_k}(\underline {\underline x} ) > 1$ define ${a_{\upsilon i}}(\underline {\underline x} )$ by $|L_i^\upsilon (\underline {\underline x} )|_\upsilon /|\underline {\underline x} |_\upsilon = {H_k}{(\underline {\underline x} )^{ - {a_{\upsilon i}}(\underline {\underline x} )/{d_\upsilon }}}$ where the ${d_\upsilon }$ are the local degrees. The approximation set $A$ consists of tuples $\underline {\underline a} = \{ {a_{\upsilon i}}\} \;(\upsilon \in S,1 \leqq i \leqq m)$ such that for every neighborhood $O$ of $\underline {\underline a}$ the points $\underline {\underline x}$ with $\{ {a_{{v_i}}}\{ \underline {\underline x} )\} \in O$ are dense in the subspace topology. Then $A$ is a polyhedron whose vertices are rational points.