7Simplicial homology
II Algebraic Topology
7.3 Homology calculations
We’ll get back to topology and compute some homologies.
Here, K is always a simplicial complex, and C
·
= C
·
(K).
Lemma. Let
f
:
K → L
be a simplicial map. Then
f
induces a chain map
f
·
: C
·
(K) → C
·
(L). Hence it also induces f
∗
: H
n
(K) → H
n
(L).
Proof.
This is fairly obvious, except that simplicial maps are allowed to “squash”
simplices, so
f
might send an
n
-simplex to an (
n −
1)-simplex, which is not in
D
n
(L). We solve this problem by just killing these troublesome simplices.
Let
σ
be an oriented
n
-simplex in
K
, corresponding to a basis element of
C
n
(K). Then we define
f
n
(σ) =
(
f(σ) f(σ) is an n-simplex
0 f(σ) is a k-simplex for k < n
.
More precisely, if σ = (a
0
, · · · , a
n
), then
f
n
(σ) =
(
(f(a
0
), · · · , f(a
n
)) f(a
0
), · · · , f(a
n
) spans an n-simplex
0 otherwise
.
We then extend f
n
linearly to obtain f
n
: C
n
(K) → C
n
(L).
It is immediate from this that this satisfies the chain map condition, i.e.
f
·
commutes with the boundary operators.
Definition (Cone). A simplicial complex is a cone if, for some v
0
∈ V
k
,
|K| = St
K
(v
0
) ∪ Lk
K
(v
0
).
v
0
We see that a cone ought to be contractible — we can just squash it to the point
v
0
. This is what the next lemma tells us.
Lemma. If
K
is a cone with cone point
v
0
, then inclusion
i
:
{v
0
} → |K|
induces
a chain homotopy equivalence i
·
: C
n
({v
0
}) → C
n
(K). Therefore
H
n
(K) =
(
Z n = 0
0 n > 0
The homotopy inverse to
i
·
is given by
r
·
, where
r
:
k 7→ v
0
is the only map.
It is clear that
r
·
◦ i
·
=
id
, and we need to show that
i
·
◦ r
·
' id
. This chain
homotopy can be defined by
h
n
:
C
n
(
K
)
→ C
n+1
(
K
), where
h
n
associates to
any simplex σ in Lk
K
(v
0
) the simplex spanned by σ and v
0
. Details are left to
the reader.
Corollary. If ∆
n
is the standard
n
-simplex, and
L
consists of ∆
n
and all its
faces, then
H
k
(L) =
(
Z k = 0
0 k > 0
Proof. K is a cone (on any vertex).
What we would really like is an example of non-trivial homology groups.
Moreover, we want them in higher dimensions, and not just the examples we got
for fundamental groups. An obvious candidate is the standard n-sphere.
Corollary. Let
K
be the standard (
n −
1)-sphere (i.e. the proper faces of
L
from above). Then for n ≥ 2, we have
H
k
(K) =
Z k = 0
0 0 < k < n − 1
Z k = n − 1
.
Proof. We write down the chain groups for K and L.
0 C
0
(L) C
1
(L) · · · C
n−1
(L) C
n
(L)
0 C
0
(K) C
1
(K) · · · C
n−1
(K) C
n
(K) = 0
=
d
L
1
=
d
L
n−1
=
d
L
n
d
K
1
d
K
n−1
For
k < n −
1, we have
C
k
(
K
) =
C
k
(
L
) and
C
k+1
(
K
) =
C
k+1
(
L
). Also, the
boundary maps are equal. So
H
k
(K) = H
k
(L) = 0.
We now need to compute
H
n−1
(K) = ker d
K
n−1
= ker d
L
n−1
= im d
L
n
.
We get the last equality since
ker d
L
n−1
im d
L
n
= H
n−1
(L) = 0.
We also know that
C
n
(
L
) is generated by just one simplex (
e
0
, · · · , e
n
). So
C
n
(
L
)
∼
=
Z
. Also
d
L
n
is injective since it does not kill the generator (
e
0
, · · · , e
n
).
So
H
n−1
(K)
∼
=
im d
L
n
∼
=
Z.
This is very exciting, because at last, we have a suggestion for a non-trivial
invariant of
S
n−1
for
n >
2. We say this is just a “suggestion”, since the simplicial
homology is defined for simplicial complexes, and we are not guaranteed that if
we put a different simplicial complex on
S
n−1
, we will get the same homology
groups. So the major technical obstacle we still need to overcome is to see that
H
k
are invariants of the polyhedron
|K|
, not just
K
, and similarly for maps.
But this will take some time.
We will quickly say something we’ve alluded to already:
Lemma (Interpretation of
H
0
).
H
0
(
K
)
∼
=
Z
d
, where
d
is the number of path
components of K.
Proof.
Let
K
be our simplicial complex and
v, w ∈ V
k
. We note that by definition,
v, w
represent the same homology class in
H
0
(
K
) if and only if there is some
c
such that
d
1
c
=
w − v
. The requirement that
d
1
c
=
w − v
is equivalent to saying
c
is a path from
v
to
w
. So [
v
] = [
w
] if and only if
v
and
w
are in the same path
component of K.
Most of the time, we only care about path-connected spaces, so
H
0
(
K
)
∼
=
Z
.