Friday, November 30, 2012

Factor Groups Via Cycles- And Some Beautiful Math

Having seen some basic group theory, I'd like to go to factor groups which disclose some really beautiful mathematics which has wide applicability. There are different ways to approach these groups but one I've found very useful is via plane, geometric (actually topological cycles). First - a simple example of a factor group: [4] = {4 + 5j: j Î Z}= {4, 9, -1, 14, -6}

Yielding five equivalence classes in all for I = (5), and:

Zn = Z/ (n)

Yielding coset: 0 + 5Z, 1 + 5Z, 2 + 5Z, 3 + 5Z, 4 + 5Z

So the factor group is: Z/ 5Z

Whence: S/I = Z/ 5Z and S/I is a factor group since

Z/ 5Z is a factor group.

In advanced physics algebraic cycles are popular because they make it easier to deal with higher dimensional space. This is accomplished by first converting the higher dimensional entity into flat, two-dimensional configuration, then assigning algebraic symbols to each 'dimension' (chain). Let's consider a relatively simple example: the basic torus pattern shown in the diagram above (copy and paste the image into a Microsoft Word or Paint field to see it better. )

If one were now to fold the left side over to the right side, so they join, ABA-left to ABA-right, s/he would be well on the way to re-forming the torus. Taping the two sides together, for example, would form a cylinder or straight tube. To complete the process, one simply joins the oppositely situated circles, ADA-top to ADA-bottom.

In the sketch, four basic regions are shown - defined as separate homology spaces. Each is of a different ‘equivalence class’ as well. Starting with the lower left corner ‘space’ and going around clockwise, these can be written:





with each one representing an individual rectangular region within the whole. In a comprehensive approach, arrows are used to define consistent directions, and either numbers or Greek letters are assigned to the box sides. This is for ease of identification of the particular equivalence classes.

For example, arrows assigned to segments AB and BA on both sides of the rectangle shown above are made to point in the same direction, say top to bottom. The same direction implies two sides have to seamlessly fit together when connected. A similar consideration applies to the bottom ends (AD + DA) when joined, so that the arrow from A to D on top would match an arrow direction from A to D on the bottom.

Thus, for the ‘top’ side of the 2D-torus:

A ---->-----D ----->------ A

and, for the ‘bottom’ side:

A ---->-----D -----> ------ A

One could go one step further, as I indicated, and assign Greek letters to the different segments. For example:

a:  A ---->-----D -----> ------ A

b:  A ---->-----D -----> ------ A

We now have a one-dimensional homology space (H1) denoted by:

H1 = (a  +   b)

The same applies to the complementary homology space (H1') that runs vertically so as to join the left and right sides, which we might denote by:

H1' = (d  + g))

These are not just homological spaces but cycles - that are themselves not boundaries. For example, one large cycle would be made by going around the outermost ‘space’ in a clockwise sense, starting from the ‘A’ in the upper left corner. We would have:

(A-D-A) -> (A-B-A)-> [-(A-D-A)] -> [-(A-B-A)]

where the minus signs precede the last two terms and help to distinguish their direction from opposite the ‘positive’ space- defined above. This could also be written in a shorthand form:

H1 + (H1') - H1 - (H1')

For which it can clearly be seen that the ‘boundary’ vanishes, since both pairs of sides (H1, H1') cancel out (having opposite signs for opposite directions). This can, of course, be written to include the ‘space’ elements:

[(a + b)] + [(d + g) ]- [(a + b)] - [(d + g )]

whence we clearly see mutually cancelling space elements
Note here that 1-cycles in a triangulated space can be generated by closed curves of the space formed by the edges of the triangulation. One can thereby form the factor group:

H1 = Z1/ B1

which amounts (roughly) to counting the closed curves that appear in the space (which are not there simply by virtue of being the boundary of a 2-dimensional segment)

Re-posing the factor group by dimension (dim):

dim H1 = dim Z1 - dim B1

where dim Z1 = [b + 1 - n]

for any connected complex and: b = branches, n = nodes (nodes are the corners of the rectangle ABCD)

For the torus: b = 4 and n = 4

so: dim Z1 = 4 + 1 - 4 = 1 = dim H1

And dim B1 = dim Z1 - dim H1 = 1 - 1 = 0

Well, what does all this gain us? Where might we be going? The beauty of this branch of math is that higher

dimensionality can be represented with simpler, lower dimensional configurations.

If we desired, we could place the points referencing the nodes into a set, which we’d call Co :  the space of “zero-chains”.  The “dim” or dimension measures, are very important, and we need some of these to proceed to basic relations later:
dim Co = the number of nodes = n

dim C1 = the number of branches = b

dim H1 = the number of connected components of a complex

dim Z1 = the number of cycles = [b + 1 – n]

We can see fairly easily that if the set Co is basically the set of nodes, then the set C1 is basically the figure (rectangle ABCD) itself, comprising: a) the set of nodes or zero-dimensional points A, B, C, and D, and b) the set of branches or 1-dimensional objects: ab,  dg etc.

Meanwhile the space of 1-chains can be represented: k = {k_a , k_b ……} within which each branch can be specified as a vector such that: a, = [1, 0, 0,……]T, with the next:

b = [0, 1, 0,……]T etc. .

Thus, each branch has prescribed values corresponding to the order for the Greek letter assigned, such that ‘1’ always commences for a, with all other places occupied by 0, extending to the full number of nodes, n.  Such specification is very useful when we seek to describe the “boundary map” say from C1 to  Co  . This is annotated in such as way that the segment or partial boundary, call it (branch) = (node)final – (node) initial

Problems: For the figure given in the inset graphic

1) Show that: dim Ho - dim Z1 = dim Co - dim C1

2) Show that: dim Z1 = dim  C1    -  dim Bo  

3) Show that: dim Ho   =   dim Co  - dim Bo  

4) Find: a,  b

5) Thence find, given (4):    a  -   ¶b

Is it true that k =0 for the WHOLE figure?

No comments: