Temporal Central Limit Theorem for a Multidimensional Adding Machine

Abstract Let p 1 ,..., p s +1 be distinct primes and let T pi be the von Neumann-Kakutani adding machine (1 ≤ i ≤ s ), T 𝒫 ( x )= ( T p 1 ( x 1 ),..., T ps ( x s )). Let y i ∈ (0, 1) be a p s +1 -rational (1 ≤ i ≤ s ), 𝟙 [0 , y) the indicator function of the box [0, y 1 )×…×[0, y s ). In this paper, we prove the following central limit theorem: <m:math xmlns:m="http://www.w3.org/1998/Math/MathML" display="block"> <m:mrow> <m:mfrac> <m:mrow> <m:munderover> <m:mo>∑</m:mo> <m:mrow> <m:mi>k</m:mi> <m:mo>=</m:mo> <m:mo>-</m:mo> <m:mi>n</m:mi> </m:mrow> <m:mrow> <m:mi>n</m:mi> <m:mo>-</m:mo> <m:mn>1</m:mn> </m:mrow> </m:munderover> <m:msub> <m:mo>𝟙</m:mo> <m:mrow> <m:mo>[</m:mo> <m:mn>0</m:mn> <m:mo>,</m:mo> <m:mi>y</m:mi> <m:mo>)</m:mo> </m:mrow> </m:msub> <m:mo>(</m:mo> <m:msup> <m:msub> <m:mi>T</m:mi> <m:mi>P</m:mi> </m:msub> <m:mi>k</m:mi> </m:msup> <m:mo>(</m:mo> <m:mi>x</m:mi> <m:mo>)</m:mo> <m:mo>-</m:mo> <m:mn>2</m:mn> <m:mi>n</m:mi> <m:msub> <m:mi>y</m:mi> <m:mn>1</m:mn> </m:msub> <m:msub> <m:mi>y</m:mi> <m:mn>2</m:mn> </m:msub> <m:mo>⋯</m:mo> <m:msub> <m:mi>y</m:mi> <m:mi>s</m:mi> </m:msub> <m:mo>)</m:mo> </m:mrow> <m:mrow> <m:msub> <m:mi>ℋ</m:mi> <m:mi>N</m:mi> </m:msub> <m:mo>(</m:mo> <m:mi>x</m:mi> <m:mo>)</m:mo> <m:mo>⋅</m:mo> <m:msup> <m:msub> <m:mi>log</m:mi> <m:mn>2</m:mn> </m:msub> <m:mrow> <m:mi>s</m:mi> <m:mo>/</m:mo> <m:mn>2</m:mn> </m:mrow> </m:msup> <m:mi>N</m:mi> </m:mrow> </m:mfrac> <m:mo>→</m:mo> <m:mrow> <m:mi>N</m:mi> <m:mo>(</m:mo> <m:mn>0</m:mn> <m:mo>,</m:mo> <m:mn>1</m:mn> <m:mo>)</m:mo> </m:mrow> </m:mrow> </m:math> \frac{\sum_{k=-n}^{\,n-1}\mathbb{1}_{[0,y)}\!\left(T_{P}^{k}(\mathbf{x})- 2n\,y_{1}y_{2}\cdots y_{s}\right)}{\mathcal{H}_{N}(\mathbf{x})\,\log_{2}^{\,s/2} N}\xrightarrow{w} \mathcal{N}(0,1) when n is sampled uniformly from {1,..., N }, ℋ N ( x )∈[υ 1 ,υ 2 ] with some υ 1 ,υ 2 &gt; 0, for almost all x ∈[0, 1) s . The main tool in the proof is the S -unit theorem and the theorem on linear forms in the logarithm.