## Thursday, March 4, 2010

### TETRAHEDRAL TREATMENT via PERMUTATIONS

In a much earlier blog I looked at how we can work out transpositions, and how permutations fit into this. See:

http://brane-space.blogspot.com/2008/12/some-fun-with-transpositions.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). 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). (Note also that some algebraic homology goes into this - so interested readers may want to go back to earlier blogs where I dealt with it, e.g.:

http://brane-space.blogspot.com/2007/12/excursion-into-algebraic-homology.html

and

http://brane-space.blogspot.com/2009/11/back-to-fractals.html

(Last part- showing the Sierpinski gasket as an oriented oriented 2-simplex(see Fig. 2- top) for which we can write: a1a2a3 = a2a3a1 = a3a1a2 = -a1a3a2 = - a3a2a1 = -a2a1a3)

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 that the boundary of a boundary is zero, or 'delta delta = 0"

By definition, the factor group: H_r = Z_r/ B_r

Then, in our case, B_r = B_2 (for the boundary) while:

H_2 = Z_2

where:

Z_2 = 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 should be able to show this.