An unshellable triangulation of a tetrahedron
An unshellable triangulation of a tetrahedron
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四面体的不可壳三角剖分
DOI：
10.1090/s0002-9904-1958-10168-8
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发表时间：
1958
影响因子：
1.3
通讯作者：
M. Rudin
中科院分区：
文献类型：
--
作者：
M. Rudin
对于四面体T，如果四面体Ku••••，Kn (K)可以如此有序，使得KJUK*-+iVJ•••••u<e:1> cn同纯于T (i= 1，•••，n)，则四面体T的三角剖分是可壳的。数学。Soc。[vol. 8 (1957) p. 917]已经证明，如果它是一个四面体的欧几里得三角剖分，那么K的一个细分K'是可壳的；他提出了一个四面体的欧几里得三角形是否存在的问题，这个四面体是不可壳的。这里将描述这种三角测量。设T是一个边长为1的四面体。我们将描述一个非平凡欧几里得三角形K (T)，如果R是K的任意四面体，则（T - R）的闭包不同态于T。K的构造：设Xi， X2， X$和X*是T的顶点。字母i和j的可能值分别为1,2,3和4，涉及i或j的加法以4为模。对于每个i，设F*表示与Xit相对的T的面，设Ui为区间X，-X；+2的中点。观察U\ = U%和
A triangulation if of a tetrahedron T is shellable if the tetrahedra Ku • • é , Kn of K can be so ordered that KJUK*-+iVJ • • • UüCn is homeomorphic to Tfor i=l, • • • , n. Sanderson [Proc. Amer. Math. Soc. vol. 8 (1957) p. 917] has shown that, if if is a Euclidean triangulation of a tetrahedron then there is a subdivision K' of K which is shellable; and he raises the question of the existence of a Euclidean triangulation of a tetrahedron which is unshellable. Such a triangulation will be described here. Let T be a tetrahedron each of whose edges has length 1. We will describe a nontrivial Euclidean triangulation K of T such that, if R is any tetrahedron of K, then the closure of (T — R) is not homeomorphic to T. I. Construction of K: Let Xi, X2, X$, and X* be the vertices of T. The possible values for the letters i and j are 1,2,3, and 4 and addition involving i or j will be modulo 4. For each i, let F* denote the face of T opposite Xit and let Ui be the midpoint of the interval X,-X;+2. Observe that U\ = U% and