Wednesday, October 7, 2020

Looking Again At Basic Algebraic Homology

 Algebraic homology is a branch of topology that is used to analyze higher-dimensional structures. This is accomplished by first converting them into flat, two-dimensional configurations, then assigning algebraic symbols to each 'dimension' (chain). Let's consider a relatively simple example: the basic torus pattern shown below:

If one were now to fold over the left and right sides 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.

Arrows are used to define consistent directions, and either numbers or Greek letters can be 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 shape shown above are made to point in the same direction, say top to bottom. The same direction implies two sides have to blend 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 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 +   )

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' = ( +   g)

1 –Simplex:

A 1-simplex can be represented as follows:

a1 ---->-----a: -----> ------ a2  

Then the boundary of a 1-simplex is given by:

1 a  =    1 (a1 a2) =      a2  - a1

 That is, the formal difference of the end point and the initial point.  Similarly, we define the boundary of a 0-simplex to be the empty simplex which we denote by “0” or:

 0 (a1) = 0    which defines a “node”.


2 –Simplex:

Likewise the boundary of a 2-simplex, e.g.


Is defined by: 2 (a1 a2 a3 ) = a2 a3 – a1 a3 + a1 a2

But the 2-simplex is ordinarily depicted with greater generality as:


So that, the sum of k-chains, for a complex C 2  can be expressed :

C 2  =  åk   x i 2  i = 1

Where   x i 2   denotes a directed line segment, e.g. a2 a3   and the superscript denotes the dimension.  So the applicable boundary operator is specified:

2 (a1 a2 a3 ) = a2 a3 – a1  a3 + a1 a

Or:   C 2 =   x21   +  x2 2  +  x23  

Where:    x2= a2 a3,   x22  =     a1 a3,     x23    =   a1 a

Which is easily remembered by defining  DD 2  as the formal sum of terms obtained by dropping each a i  in succession from the 2-simplex, e.g.  a1 a2 a3 and taking the sign to be + if the first term is omitted and (-) if the second term is omitted, + is the third term is omitted.  Referring to the preceding diagram this corresponds to going around the boundary in the direction given by the orientation arrow. 

 3-simplex:   Now, apply the same reasoning to the tetrahedron as a 3-simplex.  Consider the ordered tetrahedron (vertices ordered by number) shown below:


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.    Note the segments here play the same role as the a,    b,  d,    g   etc. in the previous example of the 2D torus.

The boundary of the tetrahedron, in terms of its four faces can then be written:



- (1 2 3) - (1 3 4) + (1 2 4) + (1 3 4)

Leading to the result that the boundary of a boundary is zero, or  D = 0

We can formulate the boundary of the generic (non-ordered)  3 –simplex.

3 ( a1 a2 a3 a4 ) = a2 a3 a4 – a1 a3 a4 + a1 a2 a4 – a1 a2 a3

Homology Group:

 The factor group H n (X) = Z n  (X) / B n (X) is the n-dimensional homology group of X.

For example, take n = 1 for a 1-simplex then we can form the quotient space:

H1 = Z1 / B 1 

And re-posing the factor group by dimension (dim):

dim H1 = dim  Z1 - dim B 1 

where dim 
Z1 = [b + 1 - n]

for any connected complex and: b = branches, n = nodes
 

If we desired, we could place the points referencing the nodes into a set, which we might 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 C o = the number of nodes = n

dim C 1  = the number of branches = b  (space of 1-chains)

dim H 1  = the number of connected components of a complex

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

(For any connected complex)


Definition:  The group  C n (X ) of oriented n-chains of X is the free abelian group generated by the n-simplexes of X.

Consider again the 1-simplex:

a1 ---->-----a: -----> ------ a2 

Whence:  1 (a)  =     a2   - a1

Then we can generalize so that:

n (a) Î  C n-1 (X )  for n = 1, 2, 3

Similarly, we saw:  ¶ o (a1) = 0    which defines a “node”

Then, since n = 0 :  C -1 (X )  = {0}

Defines the trivial group of one element, or alternatively:

o (a) Î  C -1 (X ) 

Since C n (X ) is free abelian we can specify the homomorphism of such a group by giving its values on generators. For example, for n = 2,

2 (a) Î  C 1 (X ) 

Or:  C 2  =  åk   x i 2 i = 1

  i.e.:  ¶ 2 (a1 a2 a3 ) = a2 a3 – a1  a3 + a1 a

Thus, n  gives a unique boundary homomorphism denoted by n    - for which we expect mapping C n (X )  onto C n-1 (X )  for n = 1, 2, 3.

 By extension to the preceding it should also be clear that one can extract the group of (n – 1) boundaries  and n-cycles. (Why not (n – 1) cycles?)

Thus, the image of  the group of (n – 1) boundaries   consists precisely of those (n – 1) chains which are boundaries of n-chains.  We denote this group by:

B n-1 (X ) .  

And bear in mind every boundary is a cycle.

Problem for the Math Whiz:

Consider the simplicial complex shown below:

 Given  S 1  = { 1 11 2 ,   1 3 }   

a) Find:  dim C o ,  dim C 1  ,  dim H1 ,  dim  Z1 , dim 1  

Thence or otherwise, find:  H1 = Z/ B 1 

b)Write expressions for each branch (or chain, or 1 - simplex) for the figure.

c) This (triangle space )group we can denote by B 1 (D ).

 Which also implies the group of n-cycles or  Z n (D )  

Write an expression for the 1-cycle   Z 1  and thence show we may write:

1 (Z 1)  =    a  +   b    +   g

 d) Let each node be expressed:  A =   0 1         B =   0 2         C =   0 3          

 Then rewrite each as vectors of the chain space  C o 

e)  Repeat for the 1-simplexes (chains)  as vectors related to the chain space C 1 :

 

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