A nilpotent Roth theorem
A nilpotent Roth theorem
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DOI：
10.1007/s002220100179
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发表时间：
2002-02
影响因子：
3.1
通讯作者：
V. Bergelson;A. Leibman
中科院分区：
文献类型：
--
作者：
V. Bergelson;A. Leibman
设T和S是保概率测度空间(X，ℬ，μ)变换的可逆测度。证明了如果T和S生成的群是幂零的，则$\lim_{N\ras\infty}\frac{1}{N}\sum_{n=1}^{N}u(T^{n}x)v(S^{n}x)$存在于安余的L2-范数中，v∈L∞(X，ℬ，μ)。我们还证明了对于具有∈ℬ(A)>；0的μ(A)>；0，有$\lim_{N\ras\inty}\frac{1}{N}\sum_{n=1}^{N}\Mu(A\capT^{-n}A\Cap S^{-n}A)>；0$。作为对比，我们举例说明：如果保测变换T，S生成一个不可解群，则(I)上述极限不一定存在；(Ii)双递归性失效，即对于SOMEA∈ℬ，μ(A)>；0，对所有的μ(A∩T-NA∩∈ℕ-NA)=0。最后，我们证明了当T和S生成一个幂零群时，$\lim_{N\ras\infty}\frac{1}{N}\sum_{n=1}^{N}u(T^{n}x)v(S^{n}x)=IntUd\Mu\IntVd$inL2(X)，v≤∞(X)当且仅当Ift×S是遍历Onx×X，且生成的群t-1s，T-2S2，…，T-cc遍历Onx。
LetTandSbe invertible measure preserving transformations of a probability measure space (X, ℬ, μ). We prove that if the group generated byTandSis nilpotent, then $\lim_{N\ras\infty}\frac{1}{N}\sum_{n=1}^{N}u(T^{n}x)v(S^{n}x)$ exists inL2-norm for anyu,v∈L∞(X, ℬ, μ). We also show that forA∈ℬ with μ(A)>0 one has $\lim_{N\ras\infty}\frac{1}{N}\sum_{n=1}^{N} \mu(A\cap T^{-n}A\cap S^{-n}A)>0$. By the way of contrast, we bring examples showing that if measure preserving transformationsT,Sgenerate a solvable group, then (i) the above limits do not have to exist; (ii) the double recurrence property fails, that is, for someA∈ℬ, μ(A)>0, one may have μ(A∩T-nA∩S-nA)=0 for alln∈ℕ. Finally, we show that whenTandSgenerate a nilpotent group of class ≤c, $\lim_{N\ras\infty}\frac{1}{N}\sum_{n=1}^{N}u(T^{n}x)v(S^{n}x) =\int ud\mu\int vd\mu$ inL2(X) for allu,v∈L∞(X) if and only ifT×Sis ergodic onX×Xand the group generated byT-1S,T-2S2,...,T-cScacts ergodically onX.