In a previous blog I looked at how we can work out transpositions, and how permutations fit into this. See: http://www.brane-space.blogspot.com/2012/12/extending-permutation-groups.html
What I'd now like to do is extend the permutation basis to three-dimensional geometry (specifically solids) and look at the tetrahedron(Fig. 1, below). Since the interface in the main blogger frame doesn't allow the adequate use of symbols such as I'll be using - I will insert some the working in the image (Fig. 1) and already known alternative symbols in the text.
The main proposition we're going to be proving here is that for a 3D figure like the tetrahedron, we can reduce it to an algebraic complex then show the "boundary of a boundary" is zero, by applying the permutation principle.
Now, consider the ordered tetrahedron (vertices ordered by number) shown in Fig. 1. Call the ordering '1234'. In terms of signage (sign rules - e.g. for (+) or (-) being followed, it's important to note that a segment (1 2) induces orientation (+1) in the associated complex, but a segment (2 1) induces (-1). This is how differing segments acquire negative signage in the complex.
The boundary of the tetrahedron, in terms of its four faces can than be written:
- (1 2 3) - (1 3 4) + (1 2 4) + (1 3 4)
And the calculation of the boundary is shown in Fig. 1 using the appropriate boundary symbols for the respective faces.
Leading to the result which we saw earlier, i.e. http://www.brane-space.blogspot.com/2012/11/factor-groups-via-cycles-and-some.html , that the boundary of a boundary is zero, or ¶ ¶= 0 (Given as Delta Delta in the Figure)
By definition, the factor group: Hr = Zr/ Br
Then, in our case, Br = B2 (for the boundary) while: H2 = Z2
where: Z2 = a(1 2 3) + b(1 2 4) + c (1 3 4) + d( 2 3 4)
The careful reader who's followed the earlier blog entries on groups, factor groups should be able to show this.