Open nonnegatively curved 3-manifolds with a point of positive curvature-猫眼课题宝

具有正曲率点的开非负弯曲 3 流形

DOI：

10.1090/s0002-9939-1979-0529221-3

发表时间：

1979

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作者：

Doug Elerath

设M是完备开非负曲黎曼3-流形，其上有一点的截面曲率均为正，并设M含有极点。则M在任何紧集的补集上都不是平坦的。注意，这对于2-流形显然是错误的。在这篇短文中，我将证明如下：设M3是一个完全开的非负曲3-流形，其上有一点所有的截面曲率都是正的。进一步假设M包含一个极点。则M在任何紧集上都不是平坦的。尽管这一点很容易得到证明，但我认为，有几个理由证明这一正式介绍是有道理的。其中，这些问题包括：（1）结果可能是反直觉的--在二维空间中它显然是错误的;（2）它在更高的维度中提出了类似的问题，而这些问题是不容易回答的;（3）最后，如果对极点的需要代表了证据的缺陷，而不是真实的限制，我相信情况就是这样，这将导致少数情况之一，其中一点的几何条件意味着全局几何（而不是拓扑）结果。注释和初步评论。(1)M总是表示一个完备的黎曼流形。(2)M中的点p称为极点（见[2]），如果指数映射expp：MpM是淹没或在所有Mp上具有最大秩。如果p是极点，则如果M是单连通的，则expp是一个单同态。(3)如果M3是一个开的非负曲3-流形，且有一个正曲率点，则（见[1]）M与R3同构;特别是它是单连通的。(4)Tr（A）表示关于集合AcM的半径为r的开管状邻域，Sr（A）表示Tr（A），即关于A的半径为r的球面。N.B. Sr（A）存在于M中，不存在于TM中。(5)如果x E M和y是M中的正常测地线，y（0）= x，则HX（y）将表示由x和y确定的互补半空间;即，对于TC中的X和Y，Hx（y）= M \ {y E Mld（y（t），y）。(7)我们把值为p-形式的q-形式称为（q，p）-形式。关于（q，p）-形式的微积分的恢复，例如参见[3]。这似乎是以下两个引理，虽然肯定知道在某些方面，没有出现在文献中。因此，我提供了简短的证明。LEMMA 1.设M是完备开非负曲流形，p ∈ M是极点.如果X ∈ Sr（p），y是从p发出并穿过x的射线，则令N（x）= y ′（x）确定Sr（p）的定向矢量场N。则IIN在Sr（p）上是半正定的。证据设x和y与引理的陈述相同，并假设y（0）= p，y（a）= x。然后YI [a，o.）是源自x的射线，因此可以构造互补半空间Hx（y）。显然Tr（p）c Hx（y），所以Hx（y）在x处的支撑平面是Tr（p）在x处的支撑平面。因为Hx（y）至少是局部凸的，引理如下。C1注释。使用相同的基本思想，人们可以很容易地给出一个初等的，如果稍微长一点，证明这个引理只使用劳赫比较定理。LEMMA 2.设M'是完备开非负曲流形.设CcM是一个具有方向向量场N的方向余维为1的子流形.通过单位速度测地线将N扩展到C的邻域;即，如果对于x ∈ C，y = expxtoN（x），则令N（y）=（expxtN（x））× ′（to）。进一步假设IIN在C上是半正定的。设Ci = {y ′ i = expxtN（x），对于某个x ∈ C）。则C上的（d/dt）（f c det IIN）t =p 0，对于所有0（Y）=。然后我们可以写fcdet“N = Jc A IIn 'N 1，其中是（0，1）-形式，因此A II。'N1是（（n1），n）-形式，因此在C上可积。此外，J nAIIr-f AII.7 =f|其中Dt，e = Tt（C）| T1，（C）.但一个简单的计算产生fd（AII）n-I fln+（n1）fAON AII Vn 2 Die Dee Die其中ON是（2，1）-形式RN（X，Y）（Z）=（Z）= .由于VNN = 0，因此在Dte上IIn = 0。此外，如果X1，. . ., Xn是IIN的特征向量的局部正交基，特征值为X1，. .,并且XI = N，如果K，i =，则本内容于2016年6月19日星期日05：30：32 UTC从 157.55.39.78下载所有使用http://about.jstor.org/terms约束

Let M be a complete open nonnegatively curved Riemannian 3-manifold with a point at which all sectional curvatures are positive, and suppose that M contains a pole. Then M is not flat on the complement of any compact set. Note that this is clearly false for 2-manifolds. In this short note I will prove the following: Let M3 be a complete open nonnegatively curved 3-manifold with a point at which all sectional curvatures are positive. Suppose further that M contains a pole. Then M is not flat off any compact set. Despite the ease with which this is proven, I feel that this formal presentation is warranted on several grounds. Among others, these include the following: (1) the result is, perhaps, counterintuitive-it is clearly false in two dimensions; (2) it opens up similar questions in higher dimensions which cannot be so easily answered; (3) finally, if the need for a pole represents a defect in the proof and not a real restriction, which I believe to be the case, it would lead to one of the few instances in which a geometric condition at one point implies a global geometric (not topological) result. Notation and preliminary remarks. (1) M will always denote a complete Riemannian manifold. (2) A point p in M is called a pole (see [2]) if the exponential map expp: MpM is a submersion or has maximal rank on all Mp. If p is a pole, expp will be a diffeomorphism if M is simply connected. (3) If M3 is an open nonnegatively curved 3-manifold with a point of positive curvature, then (see [1]) M is diffeomorphic to R3; in particular it is simply connected. (4) Tr(A) will denote the open tubular neighborhood of radius r about the set A c M, and Sr (A) will denote a Tr(A), the sphere of radius r about A. N.B. Sr(A) is contained in M, not TM. (5) If x E M and y is a normal geodesic ray in M with y(0) = x, then HX(y) will denote the complementary half-space determined by x and y; i.e. Hx(y) = M \ {y E Mld(y(t), y) for X and Y in TC. (7) We call a q-form with values in p-forms a (q, p)-form. For a resume of the calculus of (q, p)-forms, see for example [3]. It would seem that the following two lemmas, although surely known in some quarters, do not appear in the literature. Hence I have included brief proofs. LEMMA 1. Let M be a complete open nonnegatively curved manifold, and let p E M be a pole. If X E Sr(p) and y is the ray originating at p and passing through x, let N (x) = y'(x) determine an orienting vector field N for Sr(p). Then IIN is positive semidefinite on Sr(p). PROOF. Let x and y be as in the statement of the lemma, and suppose that y(0) = p, y(a) = x. Then YI [a,o.) is a ray originating at x, and so the complementary half space Hx(y) may be constructed. Clearly Tr(p) c Hx(y), and so the support plane for Hx(y) at x is a support plane for Tr(p) at x. Since Hx(y) is at least locally convex, the lemma follows. C1 REMARK. Using the same underlying idea one could easily give an elementary, if somewhat longer, proof of this lemma using only the Rauch comparison theorem. LEMMA 2. Let M' be a complete open nonnegatively curved manifold. Let C c M be an oriented codimension one submanifold with orientation vector field N. Extend N to a neighborhood of C by unit speed geodesics; i.e., if y = expx toN (x) for x E C, let N (y) = (expx tN (x))'(to). Furthermore suppose that IIN is positive semidefinite on C. Set C, = { y'I = expxtN (x) for some x E C) . Then (d/dt)(f c det IIN)t =p 0 on C, for all 0 (Y)= . Then we can write fcdet "N = Jc A IIn'N 1, where is a (0, 1)-form, and thus A II.'N1 is an ((n 1), n)-form, and thus is integrable over C. Furthermore, J nAIIr-f AII.7 =f| d( A IIln) where Dt,e = Tt(C) \ TI>,(C). But an easy computation yields f d( AII )n-I f ln + (n 1)f AON AII Vn2 Die Dee Die where ON is the (2, 1)-form RN(X, Y)(Z) = (Z) = . Since VNN = 0, IIn = 0 on Dte. Furthermore, if X1, . . ., Xn is a local orthonormal basis of eigenvectors of IIN with eigenvalues X1,.. ., and XI = N, and if K,i = , then This content downloaded from 157.55.39.78 on Sun, 19 Jun 2016 05:30:32 UTC All use subject to http://about.jstor.org/terms

Open nonnegatively curved 3-manifolds with a point of positive curvature

具有正曲率点的开非负弯曲 3 流形

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