Solutions To Algebraic Homology Problems

 1)  Given  S ₁  = { 1 ₁ ,  1 ₂ ,   1 ₃ }   

a) Find:  dim C _o ,   dim C _{1  ,}   dim  Z₁ , dim B ₁  

Thence or otherwise, find:  H₁ = Z₁ / B ₁ 

Solutions:

dim C _o = the number of nodes = n  =  3

dim C ₁  = the number of branches = b =  3

dim Z ₁ = the number of cycles = [b + 1 – n] =   3 +1 – 3 = 1

dim B ₁  = 1  (The number of boundaries for the given cycle, i.e. every boundary is a cycle.).    Then the quotient space is:

H₁ = Z₁ / B ₁   =   dim Z ₁  -  dim B ₁  =  1 - 1 = 0

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

¶ (B – A)  =  1₁

¶ (C – B)  =  1₂

¶ (A – C)  =  1₃

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

We may write for the one cycle:

Z ₁ =  AB + BC + CA    Or:

¶ ₁ (Z ₁)  =    (B – A)  +   (C – B) +   (A – C)  

Or:  ¶ ₁ (Z ₁)  =    a  +   b   +   g

Letting:    a  =   (B – A) ;  b  = (C – B);   g =  (A – C)  

Thereby confirming Z ₁ is a 1-cycle.

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 

Let us take the set of nodes  S _o    first for which elements applied to the previous triangle ABC we have::

A =   0 ₁         B =   0 ₂              C =   0 ₃          

Then each of these can alternatively be denoted by vectors of the chain space  C _o 

A =   0 ₁       =   [1, 0,  0] ^T

B =   0 ₂       =   [0, 1,  0] ^T

C =   0 ₃       =   [0, 0,  1] ^T

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

By the same token each of the simplexes can be specified by vectors related to the chain space C ₁ :

 C ₁   : 1₁   =  a  =  (B - A) =   [0, 1,  0] ^T   -  [1, 0,  0] ^{T      etc.}