Graphics Reference
In-Depth Information
Before computing H
0
(K) it is convenient to introduce some more notation.
Definition.
Let K be a simplicial complex. If x, y ΠC
q
(K), then we shall say that
x is homologous to y
, or that
x and y are homologous
, and write x ~ y, provided that
x - y =∂
q+1
(w) for some w ΠC
q+1
(K). Also, if z ΠZ
q
(K), let [z] = z + B
q
(K) ΠH
q
(K).
The coset [z] is called the
homology class
in H
q
(K)
determined by z
.
Note.
Recall that if
v
is a vertex of K, then we use
v
to denote the element of C
0
(K)
rather than [
v
]. Therefore, [
v
] will always mean the homology class of
v
.
The relation ~ on C
q
(K) is clearly an equivalence relation and two q-cycles x, y Œ
Z
q
(K) Ã C
q
(K) determine the same homology class in H
q
(K) if and only if x ~ y.
Returning to the computation of H
0
(K) in Example 7.2.1.5, note that Z
0
(K) =
C
0
(K). Also,
vvv
0
~~
1
2
because
(
[
]
)
=-
(
[
]
)
=-
∂
vv
v
v
and
∂
vv
v
v
.
101
1 0
112
2 1
This proves that H
0
(K) is generated by the homology class [
v
0
]. On the other hand, if
n[
v
0
] = 0, then
()
n
v
0
=
∂
x
1
for some 1-chain
[
]
+
[
]
+
[
]
xa
=
vv
b
vv
c
vv
20
.
01
12
Applying ∂
1
to the element x and equating the coefficients of the vertices
v
0
,
v
1
, and
v
2
that one gets to the same coefficients in n
v
0
= n
v
0
+ 0
v
1
+ 0
v
2
easily shows that n =
0. It follows that the map from
Z
to H
0
(K) that sends k to k[
v
0
] is an isomorphism,
that is,
0
()
ª
Z
.
HK
This finishes Example 7.2.1.5.
7.2.1.6. Example.
To compute the homology groups of K =∂<
v
0
v
1
v
2
v
3
>. See Figure
7.4.
Solution.
First of all, since there are no q-simplices for q > 2,
q
()
=
H
K
0
,
if q
>
2
.
On the other hand,
