COMPUTING HOMOLOGY GROUPS
We are given the following (model of) topological space X via its ∆–complex We
S
T
a
a
b
b V
c
V
VV
Figure 1. Model of the space X
decompose X into the sequence of cells:
X0 = {V }, X1 = {a, b, c}, X2 = {T , S}, Xn = ∅, n ≥ 3.
The chain complex C∗(X) is given by
Ci =< Xi >, i ≥ 0
as every group Ci is an abelian group ( =Z module) finitely genrated by the i–cells.
As such, we can see Ci as a kind of ”vector space” and we can describe the chain
homomorphisms ∂i in terms of matrices. Let us describe ∂2: We have ∂2(T ) = 2b − c,
∂2(S) = 2a − c, hence in our ordering
∂2 =
2 0 −1
0 2 −1
We can see that the kernel of ∂2 is 0. Then it is rather easy to see that 0 = Z2(X) =
H2(X).
In order to compute H1(X), we first see that ker ∂1 = C1 =< a, b, c >. (because the
matrix is the zero matrix).
The group H1 is a factor group of C1 where the relations that definie factoring are
2b − c = 0 and 2a − c = 0.
I am not aware of any very easy algorithmic way how to compute the factoring now
(there is in fact a way using the Smith normal form, but I deem it unnecessary). We
have to use our wit and some dirty tricks:
(1) we observe that c depends on both a and b and so we have
< a, b, c|2a = −c, 2b = −c >=< a, b|2a = 2b >
Date: March 12, 2013.
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2 COMPUTING HOMOLOGY GROUPS
(2) From our ”vector space” idea we can change our basis from (a, b) to (a − b, a).
Let us denote d = a − b. Then
H1 =< a, d|2d = 0 >∼= Z ⊕ Z2.
This concludes the calculation. Good luck with the homework. And by the way what is
really our space X geometrically? (HINT: First think just about one of the triangles).