位相幾何学の続き

以前に立方体のzeroth homology groupを計算しました。　　今回はキューブのH1を考えます。
Points to consider when calculating H1(cube) are:
1. cycles are 1-boundaries (bound 2-chains)
2. There are 64 1-cycles on a cube. This point is a bit bemusing, as it seems that a 1-cycle would surround a face, and there are six faces on a cube. However, some one-cycles may span multiple faces, it seems. Another approach is to realize that there are eight vertices on a cube, and each vertex is on the boundary of three faces. So three one-cycles per vertex are apparent. In addition, for each face-cycle, one of the four edges can be replaced with three edges of another face. This yields 12 cycles per vertex, or 96 total cycles. Thus, some overlap must occur to reduce the count to 64.
3. Each one-cycle is homologous to the null cycle. a. For a cycle c1=ABCD, c1+null is a boundary (since c1 is a boundary). Thus the two are by definition homologous.
4. Thus, the homology group consists of the single element (call it e) and is isomorphic to the trivial cyclic group C1.
5. Given a null cycle, we have found all members (equivalent under homology) of H1(K). Why would the basis thus not be a single null cycle and have a single element? In fact, the rank of the first homology group for any complex on a sphere is defined to be zero. Consider the analog of points on a line. If having no points at all is equivalent to having any point on the line, then no point is needed to generate the entire set, it seems. So the rank is zero.
6. The first Betti number h1 is called the connectivity number of the surface. It provides the largest number of closed curves that can be drawn on the surface without dividing the surface into two or more pieces. This is zero for a sphere and two for a torus. On the torus, it seems that one curve might be drawn along the inside of the "hole of the doughnut" and one curve might be drawn along the exterior of the tube which is curved to form the doughnut. The surface would still be connected (from any point on the surface, any other point could be reached without crossing the curves.