Amenability, bilipschitz equivalence, and the von Neumann conjecture
Amenability, bilipschitz equivalence, and the von Neumann conjecture
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顺从性、bilipschitz 等价和冯诺依曼猜想
DOI：
10.1215/s0012-7094-99-09904-0
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发表时间：
1999
影响因子：
2.5
通讯作者：
K. Whyte
中科院分区：
文献类型：
--
作者：
K. Whyte
我们确定当一个准等距离散空间之间的bilipschitz映射在有界距离。由此，我们证明了一个几何版本的冯诺依曼猜想顺从。我们还得到了几何群论中的一些例子，这些例子表明欧拉特征线的符号不是粗糙不变量。最后得到了一致有限同调的一些一般结果，这些结果将在以后的文章中应用于流形。在他的基本文件的顺从性，冯诺依曼证明，如果Γ是一个无约束生成，非顺从组，那么Γ有一个子群是免费的两个发电机。虽然这个猜想被证明是错误的一般[O]，它是真实的许多类的群体。鉴于最近的兴趣研究群体通过几何方法，似乎很自然地要求一个几何版本的猜想。如果Γ包含一个自由子群，那么它的陪集将把Γ划分为自由群的拷贝。自由群在几何学上是正规的四价树。我们的“几何冯诺依曼猜想”是：定理1如果X是有界几何的一致离散空间（特别是，对于X是紧致流形的一个叶的叶中的一个网或一个群生成的群），X是不服从的当且仅当它允许一个具有与正则4-叶树一致等价的bilipschitz块的划分。证明是通过构造X × {0，1}与X在投影映射附近的一个bilipschitz等价来实现的。我们的主要技术成果是一般1
We determine when a quasi-isometry between discrete spaces is at bounded distance from a bilipschitz map. From this we prove a geometric version of the Von Neumann conjecture on amenability. We also get some examples in geometric groups theory which show that the sign of the Euler characteristic is not a coarse invariant. Finally we get some general results on uniformly we finite homology which we will apply to manifolds in a later paper. In his fundamental paper on amenability, Von Neumann conjectured that if Γ is a finitely generated, non-amenable group, then Γ has a subgroup which is free on two generators. While this conjecture proved to be false in general [O], it is true for many classes of groups. Given the recent interest in studying groups via geometric methods, it seems natural to ask for a geometric version of the conjecture. If Γ does contain a free subgroup then its cosets will partition Γ into copies of the free group. The free group is, geometrically, the regular 4-valent tree. Our " geometric Von Neumann conjecture " , is: Theorem 1 If X is a uniformly discrete space of bounded geometry (in particular, for X a finitely generated group or a net in a leaf of a foliation of a compact manifold), X is non-amenable iff it admits a partition with pieces uniformly bilipschitz equivalent to the regular 4-valent tree. The proof depends on constructing a bilipschitz eqivalence of X × {0, 1} and X near the projection map. Our main technical result is a general 1