Graphics Reference
In-Depth Information
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[]
S=
s
.
2 simplex
-
s
in K
For example, [
v
0
v
2
v
8
] is one oriented simplex that appears in the sum S.
Claim 1.
S is a 2-cycle.
To see that ∂
2
(S) = 0, consider an arbitrary 1-simplex
v
i
v
j
in K. If
v
i
v
j
is a face of
the two 2-simplices s and s¢, then the coefficients of [
v
i
v
j
] in ∂
2
([s]) is the negative of
the coefficient of [
v
i
v
j
] in ∂
2
([s¢]). It follows that the coefficient of [
v
i
v
j
] in ∂
2
(S) is zero.
Claim 2.
[S] generates Z
2
(K).
To prove Claim 2 let z be a 2-cycle. Let s and s¢ be an arbitrary pair of 2-simplices
in K that meet in an edge
v
i
v
j
. It is easy to see that the coefficient of [
v
i
v
j
] will vanish
in ∂
2
(z) if and only if [s] and [s¢] appear in z with the same multiplicity. Since this is
true for all pairs of adjacent 2-simplices in K, we must have that z = aS for some
integer a.
Claim 1 and 2 prove that
()
=
()
=
ZZ
HK ZK
S
.
2
2
To determine H
1
(K), define two 1-cycles u and v by
=
[
]
+
[
]
+
[
]
=
[
]
+
[
]
+
[
]
u
vv
vv
vv
and
v
vv
vv
vv
40
.
01
12
20
03
34
Claim 3.
H
1
(K) is generated by [u] and [v].
To prove Claim 3 observe that
any
1-chain z is homologous to a 1-chain z
1
of the
form
[
]
+
[
]
+
[
]
+
[
]
+
[
]
z
=
a
vv
a
vv
a
vv
a
vv
a
vv
1
1 03
2 34
3 40
4 01
5 12
[
]
+
[
]
+
[
]
+
[
]
+
[
]
+
a
vv
a
vv
a
vv
a
vv
a
vv
,
for some a
i
Œ
Z
.
620
736
865
948
087
We leave this as an exercise for the reader. For example, one can start by first replac-
ing any appearing [
v
2
v
3
] in z by [
v
2
v
6
] + [
v
6
v
3
], then replacing any [
v
2
v
6
] by [
v
2
v
1
] +
[
v
1
v
6
], and so on. Each of these replacements produces a new chain that is homolo-
gous to the previous one, so that the final chain z
1
is homologous to z. This is similar
to what we did in Example 7.2.1.6. If we were to start with a 1-cycle z, then z
1
will
also be a 1-cycle. However, for an element like z
1
to be a 1-cycle, it must satisfy two
other properties. First, a
i
must be 0 for i = 7, 8, 9, 10; otherwise, one or more of the
vertices
v
5
,
v
6
,
v
7
, and
v
8
would appear in ∂
1
(z
1
) with a nonzero coefficient. Second,
a
1
= a
2
= a
3
and a
4
= a
5
= a
6
; otherwise, ∂
1
(z
1
) would not be zero. This shows that any
1-cycle z is homologous to a 1-cycle of the form au + bv, where a, b Œ
Z
and Claim 3
is proved.
Claim 4.
The homology classes [u] and [v] are linearly independent.
