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In-Depth Information
Similarly, we may replace any occurrence of [
v
1
v
2
] in z
1
by [
v
1
v
0
] + [
v
0
v
2
]. This will
give us a 1-chain z
2
, which is homologous to the original z and which contains neither
[
v
2
v
3
] nor [
v
1
v
2
]. If we assume further that z actually was a 1-cycle, then z
2
will be a
1-cycle and [
v
0
v
2
] cannot appear in z
2
either, otherwise the coefficient of
v
2
in ∂
1
(z
2
)
would not vanish. An argument similar to the one used in the computation of H
1
(K)
in Example 7.2.1.5 now shows that z
2
must in fact be a multiple of
x =
[
]
+
[
]
+
[
]
vv
vv
vv
30
.
01
13
In other words, H
1
(K) is generated by [x]. On the other hand, x =∂
2
([
v
0
v
1
v
3
]) and so
[x] = 0 and we have proved that
()
=
HK
1
0
.
Finally, the group H
0
(K) can be computed like in Example 7.2.1.5. The first step
is to show that H
0
(K) is generated by [
v
0
]. Next, one shows that n[
v
0
] = 0 implies that
n = 0. We again get that
0
()
ª
Z
.
HK
This finishes Example 7.2.1.6.
7.2.1.7. Example.
The simplicial complex K in Figure 7.5 triangulates the torus. We
want to compute the homology groups of K.
Solution.
Again, since there are no q-simplices for q > 2,
q
()
=
H
K
0
,
if q
>
2
.
To compute H
2
(K), orient the 2-simplices s in K as indicated by the circular arrows
in Figure 7.5. In this example, the notation [s] will denote the 2-simplex s together
with that preferred orientation. Let S denote the element in C
2
(K) that is the sum of
these oriented 2-simplices, that is,
v
v
3
v
4
v
0
v
0
v
2
v
2
v
6
v
8
u
u
v
1
v
1
v
5
v
7
v
0
v
3
v
4
v
0
v
Figure 7.5.
A triangulating complex for the
torus.
|K| = S
1
¥ S
1
