Graphics Reference
In-Depth Information
Figure 7.4.
A triangulating complex for a sphere.
v
3
v
2
v
0
v
1
K = ∂ <v
0
v
1
v
2
v
3
>
|K| = boundary of tetrahedron
()
=
[
]
≈
[
]
≈
[
]
≈
[
]
CK
C
Zvvv
Zvvv
Zvvv
Zvvv
,
2
012
013
123
023
()
=
[
]
≈
[
]
≈
[
]
≈
[
]
≈
[
]
≈
[
]
K
Zv v
Zvv
Zv v
Zv v
Zv v
Zvv
,
and
1
0
1
1
2
0
2
0
3
3
2
1
3
()
=≈≈≈
CK
Zv
Zv
Zv
Zv
3
.
0
0
1
2
Now B
2
(K) = 0 and ∂
0
= 0. Therefore, H
2
(K) = Z
2
(K) and Z
0
(K) = C
0
(K). To compute
the group Z
2
(K), let
[
]
+
[
]
+
[
]
+
[
]
z =
a
vvv
b
vvv
c
vvv
d
vvv
012
013
123
0 23
be a 2-cycle. By definition, ∂
2
(z) = 0. Computing the coefficients of the 1-simplices in
∂
2
(z) and setting them equal to 0 implies that a =-b =-c = d. In other words,
()
=
(
[
]
≈
[
]
≈
[
]
≈
[
]
)
Z
2
Z vvv
vvv
vvv
vvv
,
0 21
013
123
032
and so
2
()
ª
Z
.
HK
We could compute H
1
(K) by calculating Z
1
(K) and B
1
(K) using arguments as
before, but the approach to calculating homology groups by explicitly determining
the group of cycles and the group of boundaries would become very tedious as spaces
get more complicated. To simplify computations it is helpful to use certain tricks and
shortcuts. The point is that we are looking for homology classes. Any representative
cycle for such a class will do, and so we are free to replace any such cycle by a homol-
ogous one.
Let z be a 1-chain. The equation
[
]
=
[
]
+
[
]
+
(
[
]
)
vv
vv
vv
∂
vvv
23
21
13
2
123
shows that the chains [
v
2
v
3
] and [
v
2
v
1
] + [
v
1
v
3
] are homologous. Therefore, if the ori-
ented 1-simplex [
v
2
v
3
] appears in z with some nonzero coefficient, then we can replace
[
v
2
v
3
] by [
v
2
v
1
] + [
v
1
v
3
] to get a homologous 1-chain z
1
in which [
v
2
v
3
] does not appear.
