Graphics Reference
In-Depth Information
Figure 7.3.
A triangulating complex for a circle.
v
2
v
1
v
0
<v
0
v
1
v
2
>
|K| = boundary of triangle
K =
∂
()
=
()
=
()
=
H
K
Z
K
B
K
0
,
if q
>
0
,
and
q
q
q
()
=
()
=
()
ª
HK ZK CK
Z
0
0
0
7.2.1.5. Example.
To compute the homology groups of K =∂<
v
0
v
1
v
2
>. See Figure
7.3.
Solution.
In this case
q
()
=
C
K
0
,
if q
>
1
,
()
=
[
]
≈
[
]
≈
[
]
C
K
Zv v
Zvv
Zv v
,
and
1
0
1
1
2
2
0
()
=≈≈
CK
Zv
Zv
Zv
.
0
0
1
2
Trivially,
()
=
()
=
()
=
H
K
Z
K
B
K
0
,
if q
>
1
.
q
q
q
Next, assume that
[
]
+
[
]
+
[
]
xa
=
vv
b
vv
c
vv
01
12
20
is a 1-cycle. Then
()
0
=
∂ x
aa
1
(
)
+
(
)
+
(
)
=
vv
-
b b c c
v v vv
-
-
1
0
2
1
0
2
(
)
(
)
(
)
=-
ca
v
+-
ab
v
+-
bc
v
.
0
1
2
It follows that c - a = a - b = b - c = 0, which implies that a = b = c. Hence,
()
=
(
[
]
+
[
]
+
[
]
)
ª
ZK
1
Zvv
vv
vv
Z
.
0
1
1
2
2
0
But B
1
(K) = 0, and so
()
=
()
ª
Z
.
HK
ZK
1
1
