7Simplicial homology
II Algebraic Topology
7.1 Simplicial homology
For now, we will forget about simplicial approximations and related fluff, but
just note that it is fine to assume everything can be considered to be simplicial.
Instead, we are going to use this framework to move on and define some new
invariants of simplicial complexes
K
, known as
H
n
(
K
). These are analogous
to
π
0
, π
1
, · · ·
, but only use linear algebra in the definitions, and are thus much
simpler. The drawback, however, is that the definitions are slightly less intuitive
at first sight.
Despite saying “linear algebra”, we won’t be working with vector spaces most
of the time. Instead, we will be using abelian groups, which really should be
thought of as
Z
-modules. At the end, we will come up with an analogous theory
using
Q
-modules, i.e.
Q
-vector spaces, and most of the theory will carry over.
Using
Q
makes some of our work easier, but as a result we would have lost some
information. In the most general case, we can replace
Z
with any abelian group,
but we will not be considering any of these in this course.
Recall that we defined an
n
-simplex as the collection of
n
+ 1 vertices that
span a simplex. As a simplex, when we permute the vertices, we still get the
same simplex. What we want to do is to remember the orientation of the simplex.
In particular, we want think of the simplices (a, b) and (b, a) as different:
a
b b
a
Hence we define oriented simplices.
Definition (Oriented
n
-simplex). An oriented
n
-simplex in a simplicial complex
K
is an (
n
+ 1)-tuple (
a
0
, · · · , a
n
) of vertices
a
i
∈ V
k
such that
ha
0
, · · · , a
n
i ∈ K
,
where we think of two (
n
+ 1)-tuples (
a
0
, · · · , a
n
) and (
a
π(0)
, · · · , a
π(n)
) as the
same oriented simplex if π ∈ S
n
is an even permutation.
We often denote an oriented simplex as
σ
, and then
¯σ
denotes the same
simplex with the opposite orientation.
Example. As oriented 2-simplices, (
v
0
, v
1
, v
2
) and (
v
1
, v
2
, v
0
) are equal, but
they are different from (
v
2
, v
1
, v
0
). We can imagine the two different orientations
of the simplices as follows:
v
0
v
2
v
1
v
0
v
1
v
2
One and two dimensions are the dimensions where we can easily visualize
the orientation. This is substantially harder in higher dimensions, and often we
just work with the definition instead.
Definition (Chain group
C
n
(
K
)). Let
K
be a simplicial complex. For each
n ≥ 0, we define C
n
(K) as follows:
Let
{σ
1
, · · · , σ
`
}
be the set of
n
-simplices of
K
. For each
i
, choose an
orientation on
σ
i
. That is, choose an order for the vertices (up to an even
permutation). This choice is not important, but we need to make it. Now when
we say σ
i
, we mean the oriented simplex with this particular orientation.
Now let
C
n
(
K
) be the free abelian group with basis
{σ
1
, · · · , σ
`
}
, i.e.
C
n
(
K
)
∼
=
Z
`
. So an element in C
n
(K) might look like
σ
3
− 7σ
1
+ 52σ
64
− 28σ
1000000
.
In other words, an element of C
n
(K) is just a formal sum of n-simplices.
For convenience, we define
C
`
(
K
) = 0 for
` <
0. This will save us from
making exceptions for n = 0 cases later.
For each oriented simplex, we identify −σ
i
with ¯σ
i
, at least when n ≥ 1.
In this definition, we have to choose a particular orientation for each of our
simplices. If you don’t like making arbitrary choices, we could instead define
C
n
(K) as some quotient, but it is slightly more complicated.
Note that if there are no
n
-simplices (e.g. when
n
=
−
1), we can still
meaningfully talk about C
n
(K), but it’s just 0.
Example. We can think of elements in the chain group
C
1
(
X
) as “paths” in
X
.
For example, we might have the following simplex:
v
0
v
1
v
2
σ
1
σ
2
σ
3
σ
6
Then the path
v
0
v
1
v
2
v
0
v
1
around the left triangle is represented
by the member
σ
1
− σ
2
+ σ
3
+ σ
1
= 2σ
1
− σ
2
+ σ
3
.
Of course, with this setup, we can do more random things, like adding 57 copies
of
σ
6
to it, and this is also allowed. So we could think of these as disjoint union
of paths instead.
When defining fundamental groups, we had homotopies that allowed us to
“move around”. Somehow, we would like to say that two of these paths are
“equivalent” under certain conditions. To do so, we need to define the boundary
homomorphisms.
Definition (Boundary homomorphisms). We define boundary homomorphisms
d
n
: C
n
(K) → C
n−1
(K)
by
(a
0
, · · · , a
n
) 7→
n
X
i=0
(−1)
i
(a
0
, · · · , ˆa
i
, · · · , a
n
),
where (
a
0
, · · · , ˆa
i
, · · · , a
n
) = (
a
0
, · · · , a
i−1
, a
i+1
, · · · , a
n
) is the simplex with
a
i
removed.
This means we remove each vertex in turn and add them up. This is clear if
we draw some pictures in low dimensions:
v
0
v
1
−v
0
v
1
If we take a triangle, we get
v
0
v
2
v
1
v
0
v
2
v
1
An important property of the boundary map is that the boundary of a boundary
is empty:
Lemma. d
n−1
◦ d
n
= 0.
In other words, im d
n+1
⊆ ker d
n
.
Proof.
This just involves expanding the definition and working through the
mess.
With this in mind, we will define the homology groups as follows:
Definition (Simplicial homology group
H
n
(
K
)). The
n
th simplicial homology
group H
n
(K) is defined as
H
n
(K) =
ker d
n
im d
n+1
.
This is a nice, clean definition, but what does this mean geometrically?
Somehow,
H
k
(
K
) describes all the “
k
-dimensional holes” in
|K|
. Since we
are going to draw pictures, we are going to start with the easy case of
k
= 1.
Our
H
1
(
K
) is made from the kernel of
d
1
and the image of
d
2
. First, we give
these things names.
Definition (Chains, cycles and boundaries). The elements of
C
k
(
K
) are called
k
-chains of
K
, those of
ker d
k
are called
k
-cycles of
K
, and those of
im d
k+1
are
called k-boundaries of K.
Suppose we have some
c ∈ ker d
k
. In other words,
dc
= 0. If we interpret
c
as a “path”, if it has no boundary, then it represents some sort of loop, i.e. a
cycle. For example, if we have the following cycle:
e
0
e
2
e
1
We have
c = (e
0
, e
1
) + (e
1
, e
2
) + (e
2
, e
0
).
We can then compute the boundary as
dc = (e
1
− e
0
) + (e
2
− e
1
) + (e
0
− e
2
) = 0.
So this c is indeed a cycle.
Now if
c ∈ im d
2
, then
c
=
db
for some 2-chain
b
, i.e.
c
is the boundary of
some two-dimensional thing. This is why we call this a 1-boundary. For example,
suppose we have our cycle as above, but is itself a boundary of a 2-chain.
v
0
v
2
v
1
We see that a cycle that has a boundary has been “filled in”. Hence the “holes”
are, roughly, the cycles that haven’t been filled in. Hence we define the homology
group as the cycles quotiented by the boundaries, and we interpret its elements
as k-dimensional “holes”.
Example. Let
K
be the standard simplicial 1-sphere, i.e. we have the following
in R
3
.
e
0
e
1
e
2
Our simplices are thus
K = {he
0
i, he
1
i, he
2
i, he
0
, e
1
i, he
1
, e
2
i, he
2
, e
0
i}.
Our chain groups are
C
0
(K) = h(e
0
), (e
1
), (e
2
)i
∼
=
Z
3
C
1
(K) = h(e
0
, e
1
), (e
1
, e
2
), (e
2
, e
0
)i
∼
=
Z
3
.
All other chain groups are zero. Note that our notation is slightly confusing
here, since the brackets
h · i
can mean the simplex spanned by the vertices, or
the group generated by certain elements. However, you are probably clueful
enough to distinguish the two uses.
Hence, the only non-zero boundary map is
d
1
: C
1
(K) → C
0
(K).
We can write down its matrix with respect to the given basis.
−1 0 1
1 −1 0
0 1 −1
We have now everything we need to know about the homology groups, and we
just need to do some linear algebra to figure out the image and kernel, and thus
the homology groups. We have
H
0
(K) =
ker(d
0
: C
0
(K) → C
−1
(K))
im(d
1
: C
1
(K) → C
0
(K))
∼
=
C
0
(K)
im d
1
∼
=
Z
3
im d
1
.
After doing some row operations with our matrix, we see that the image of
d
1
is
a two-dimensional subspace generated by the image of two of the edges. Hence
we have
H
0
(K) = Z.
What does this
H
0
(
K
) represent? We initially said that
H
k
(
K
) should represent
the
k
-dimensional holes, but when
k
= 0, this is simpler. As for
π
0
,
H
0
just
represents the path components of
K
. We interpret this to mean
K
has one
path component. In general, if
K
has
r
path components, then we expect
H
0
(
K
)
to be Z
r
.
Similarly, we have
H
1
(K) =
ker d
1
im d
2
∼
=
ker d
1
.
It is easy to see that in fact we have
ker d
1
= h(e
0
, e
1
) + (e
1
, e
2
) + (e
2
, e
0
)i
∼
=
Z.
So we also have
H
1
(K)
∼
=
Z.
We see that this
H
1
(
K
) is generated by precisely the single loop in the triangle.
The fact that
H
1
(
K
) is non-trivial means that we do indeed have a hole in the
middle of the circle.
Example. Let L be the standard 2-simplex (and all its faces) in R
3
.
e
0
e
1
e
2
Now our chain groups are
C
0
(L) = C
0
(K)
∼
=
Z
3
∼
=
h(e
0
), (e
1
), (e
2
)i
C
1
(L) = C
1
(K)
∼
=
Z
3
∼
=
h(e
0
, e
1
), (e
1
, e
2
), (e
2
, e
0
)i
C
2
(L)
∼
=
Z = h(e
0
, e
1
, e
2
)i.
Since
d
1
is the same as before, the only new interesting boundary map is
d
2
. We
compute
d
2
((e
0
, e
1
, e
2
)) = (e
0
, e
1
) + (e
1
, e
2
) + (e
2
, e
0
).
We know that
H
0
(
L
) depends only on
d
0
and
d
1
, which are the same as for
K
.
So
H
0
(L)
∼
=
Z.
Again, the interpretation of this is that
L
is path-connected. The first homology
group is
H
1
(L) =
ker d
1
im d
2
=
h(e
0
, e
1
) + (e
1
, e
2
) + (e
2
, e
0
)i
h(e
0
, e
1
) + (e
1
, e
2
) + (e
2
, e
0
)i
∼
=
0.
This precisely illustrates the fact that the “hole” is now filled in L.
Finally, we have
H
2
(L) =
ker d
2
im d
3
= ker d
2
∼
=
0.
This is zero since there aren’t any two-dimensional holes in L.
We have hopefully gained some intuition on what these homology groups
mean. We are now going to spend a lot of time developing formalism.