The bottom of the spectrum of a Riemannian covering.
The bottom of the spectrum of a Riemannian covering.
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黎曼覆盖谱的底部。
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发表时间：
1985
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通讯作者：
R. Brooks
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作者：
R. Brooks
设M1是完备的黎曼流形，M2是M1的黎曼覆盖空间，即存在映射f：M2 -> Mt，它是局部等距的。我们假设Mi具有“有限拓扑型”，即它在拓扑上是多个单形的并。一个基本的问题是要理解L-谱A0的底部在这样的覆盖下是如何表现的。我们在[2]（另见[11]）中在M1是紧的（因此A0（M1）= 0）并且M2 = M1是Mi的泛覆盖的特殊情况下考虑了这个问题。在这种情况下，我们完全解决了这个问题的以下：定理（[2]）。>10（M1）= 0当且仅当ni（M1）是顺从群。仔细检查证明表明，我们把M2取为M1并不重要。一个一致的一般版本是假设π1（M2）是π1（M1）的正规子群，从而定义群π1（M1）/π1（M2）。然后在定理的陈述中用π1（M1）/π1（M2）代替π1（A/1）。通过考虑π1（M1）在陪集空间π1（M1）/π1（M2）中的作用，可以得到一个完全一般的陈述，但我们在这里不这样做。在[5]中，我们在Ml具有有限体积的特殊情况下重新考虑了这个问题。在这种情况下，我们发现我们可以利用一个“尖点等周不等式”来得到定理的结论，这在带的情况下是很标准的。本文考虑一般非紧流形的问题。很容易看出，当Mx和M2大于s时，我们必须有A0（M2）^A0（M1）。为了证明这一点，我们利用经典的观察，即对于任何完备黎曼流形，A0由正的I 0-调和函数（不一定是L）表示，如果λ是存在正的Λ 0-调和函数的任何数，则λ0^>λ（参见[17]或[8]）。然后，将M1上的正A0-调和函数提升到M2，以显示λ0（M2）^λ0（M1）。
Let Ml be a complete Riemannian manifold, and let M2 be a Riemannian covering space of Ml—i.e. there is a map/: M2 —> Mt which is a local isometry. We will assume that Mi has "finite topological type," that is, it is topologically the union of finitely many simplices. A fundamental question is to understand how the bottom of the L-spectrum A0 behaves under such coverings. We considered this problem in [2] (see also [11]) in the special case where M1 is compact (so that A0(M1) = 0) and M2 = M1 is the universal covering of Mi. In this case, we resolved the question completely by the following: Theorem ([2]). >10(M1) = 0 if and only if ni(Ml) is an amenable group. A close examination of the proof shows that it was not important that we take M2 to be M1. An agreeably general version would be to assume only that π1(Μ2) is a normal subgroup of π1(Μ1), so that the group π1(Μ1)/π1(Μ2) is defined. One then replaces π1(Α/1) with π1(Μ1)/π1(Μ2) in the Statement of the theorem. A completely general Statement could be achieved by considering the action of π1(Μ1) οη the coset space π1(Μ1)/π1(Μ2), but we will not do that here. In [5], we were led to reconsider this question in a special case in which Ml had fmite volume. In that case, we found that we could achieve the conclusion of the theorem by making use of an "isoperimetric inequality in the cusps" which in the case at band was quite Standard. In this paper, we consider the problem for general noncompact manifolds. It is easy to see that when Mx and M2 are s above, we must have A0(M2)^A0(M1). To see this, we make use of the classical observation that for any complete Riemannian manifold, A0 is represented by a positive I0-harmonic function (which need not be L) and if λ is any number for which there exists a positive Λ,-harmonic function, then λ0^>λ (see [17] or [8]). One then lifts a positive A0-harmonic function on M1 to M2 to show λ0(Μ2)^λ0(Μ1).