线性环面自同态、精确性、伯努利性与生成元
Linear toral endomorphisms, exactness, Bernoulliness and generators
AI总结：
本文用初等构造性方法重证Krzyżewski关于环面自同态精确性的定理，并证明非扩张矩阵对应的自同态不存在光滑有限生成元，且扩张性等价于单边伯努利移位同构。
AI中文摘要：
设 $T$ 为群 $\ttf^d\\,:=\rrf^d/\zzf^d$ 上的一个不可逆的满射连续自同态。则 $T$ 保持 Haar 测度，与某个具有整数系数的 $d \times d$ 矩阵 $A$ 相关联，并且是 $r$-对一的，其中 $r = |\det A| \ge 2$。在文献~\cite{Krzyzewski} 中，Krzyżewski 给出了 $T$ 为精确的充分必要条件，并证明了 $T$——作为保测映射——同构于某个（可能平凡的）环面自同构与某个精确环面自同态的直积。我们通过更初等且具有构造性的方法重新证明了这些结果，而 Krzyżewski 实际上处理的是紧致阿贝尔群的自同态，并依赖于非构造性定理。Krzyżewski 的精确性条件以及 Mihailescu 的论文~\cite{Mihailescu 2012} 和~\cite{Mihailescu 2013} 表明，$T$ 同构于 $\odc 0,r-1 \fdc^{\zzf\\_+}$ 上的单边均匀伯努利移位，当且仅当 $A$ 是扩张的（即每个特征值的模都 $>1$）。当 $A$ 不是扩张的且是双曲的（即所有特征值的模都不等于 $1$）时，Mihailescu~\cite{Mihailescu 2012} 证明了 $T$ 不可能具有生成性的 Rokhlin 划分。更一般地，当 $A$ 不是扩张的（无论是否双曲）时，我们证明了自同态 $T$ 不可能具有光滑的有限生成元（无论是否独立）。
英文摘要：
Let $T$ be a non-invertible surjective continuous endomorphism of the group $\ttf^d\,:= \rrf^d/\zzf^d$. Then $T$ preserves the Haar measure, is associated to some $d \times d$ matrix $A$ with integer coefficients and is $r$-to-one, where $r = |\det A| \ge 2$. In~\cite{Krzyzewski}, Krzyżewski gives a necessary and sufficient condition for $T$ to be exact, and shows that $T$ -- viewed as a measure-preserving map -- is isomorphic to the direct product of some (possibly trivial) toral automorphism by some exact toral endomorphism. We re-prove these results by more elementary and constructive methods, whereas Krzyżewski actually works with endomorphisms of compact Abelian groups and relies on non-constructive theorems. Krzyżewski's exactness condition and Mihailescu's papers ~\cite{Mihailescu 2012} and~\cite{Mihailescu 2013} show that $T$ is isomorphic to the one-sided uniform Bernoulli shift on $\odc 0,r-1 \fdc^{\zzf\_+}$ if and only if $A$ is expanding (i.e. the modulus of each eigenvalue is $>1$). When $A$ is not expanding and hyperbolic (i.e. all eigenvalues have modulus different from $1$), Mihailescu~\cite{Mihailescu 2012} shows that $T$ cannot have a generating Rokhlin partition. More generally, when $A$ is not expanding (whether hyperbolic or not), we show that the endomorphism $T$ cannot have a smooth finite generator (whether independent or not).
