また位相幾何学

The next homology group and Betti number to be examined is the second homology group of the cube (H2(cube)). The only two 2-cycles on a cube are null and ABCDEF (where these are the six faces). These are not homologous, as the only two-boundary (by convention) is null. So H2(cube) consists of two elements. Since ABCDEF+ABCDEF = null, we can prove that this group is isomorphic to the two-element cyclic group C2. The second Betti number h2 is one, but I am not sure how this is computed. It seems that, as was the case for h0, all members of the group can be generated from the single element ABCDEF, so this single element would be the basis. C2 also has rank of one, it seems.
The next complex to compute homology groups for is the torus. In this case, every k-chain (k=1,2) has a null boundary. This seems difficult to comprehend without viewing the diagram of the torus in the plane (pg 135). In this case, a square is drawn with all four vertices labelled "P". Thus, any two points bounding a two-chain are the same point, so their sum (P+P) is null. In the case of two chains, there is only one face (A) which is bound by two "a" edges and two "b" edges. a+a = null and b+b=null, so the boundary is null here as well. As the boundaries of all one and two-chains are null, all k chains (k=1,2) are cycles. As the two-boundary is null, all boundaries on this complex are null. This means that x and y are homologous if and only if (x+y) = null. In other words only if x equals y (as this is the definition of when the sum is null in the case of chain group addition). So homology and equality are equivalent. This means that the homology groups are the same as the chain groups.
H0 consists of P and null (P+P). Again, this group is isomorphic to C2. H2 contains null and A, and is also isomorphic to C2. H1 contains, a, b, and null (a+a or b+b). a and b are not homologous, so a+b is also present. Thus, there are four elements. Although I am unsure of the proof, this group is apparently isomorphic to Klein's four group. Its table of addition would be:
null a b a+b
null null a b a+b
a a null a+b b
b b a+b null a
a+b a+b b a null