图的最大亏格与直径

摘要

设G是直径为4的简单图,若G不含3阶完全子图K3,则G的Betti亏数ξ（G）≤2,即G的最大亏格γM（G）≥1/2β（G）-1,并且不等式的下界是可达的。这种结合图的直径等条件的证明方法改进了相关结果。%Combined the condition of the diameter of a graph, the paper proves the following results; Let G be a simple graph with diameter four, if G does not contain the complete subgraph K3 of order three, then the Betti deficient number of G,ξ（G）≤2,and thus the maximurm genus of G,γM（G）≥1/2β（G）-1,And the lower bound is best possible. And some relative results are improved.