7Simplicial homology

II Algebraic Topology



7.8 Rational homology, Euler and Lefschetz numbers
So far, we have been working with chains with integral coefficients. It turns out
we can use rational coefficients instead. In the past,
C
n
(
K
) was an abelian group,
or a
Z
-module. If we use rational coefficients, since
Q
is a field, this becomes a
vector space, and we can use a lot of nice theorems about vector spaces, such as
the rank-nullity theorem. Moreover, we can reasonably talk about things like
the dimensions of these homology groups.
Definition (Rational homology group). For a simplicial complex
K
, we can de-
fine the rational
n
-chain group
C
n
(
K, Q
) in the same way as
C
n
(
K
) =
C
n
(
K, Z
).
That is,
C
n
(
K, Q
) is the vector space over
Q
with basis the
n
-simplices of
K
(with a choice of orientation).
We can define d
n
, Z
n
, B
n
as before, and the rational nth homology group is
H
n
(K; Q)
=
Z
n
(K; Q)
B
n
(K; Q)
.
Now our homology group is a vector space, and it is much easier to work with.
However, the consequence is that we will lose some information. Fortunately,
the way in which we lose information is very well-understood and rather simple.
Lemma. If H
n
(K)
=
Z
k
F for F a finite group, then H
n
(K; Q)
=
Q
k
.
Proof. Exercise.
Hence, when passing to rational homology groups, we lose all information
about the torsion part. In some cases this information loss is not very significant,
but in certain cases, it can be. For example, in RP
2
, we have
H
n
(RP
2
)
=
Z n = 0
Z/2 n = 1
0 n > 1
If we pass on to rational coefficients, then we have lost everything in
H
1
(
RP
2
),
and RP
2
looks just like a point:
H
n
(RP
2
; Q)
=
H
n
(; Q).
Example. We have
H
n
(S
n
; Q)
=
(
Q k = 0, n
0 otherwise
.
We also have
H
n
g
, Q)
=
Q k = 0, 2
Q
2g
k = 1
0 otherwise
.
In this case, we have not lost any information because there was no torsion part
of the homology groups.
However, for the non-orientable surfaces, one can show that
H
k
(E
n
) =
Z k = 0
Z
n1
× Z/2 k = 1
0 otherwise
,
so
H
k
(E
n
; Q) =
Q k = 0
Q
n1
k = 1
0 otherwise
.
This time, this is different from the integral coefficient case, where we have
an extra Z
2
term in H
1
.
The advantage of this is that the homology “groups” are in fact vector spaces,
and we can talk about things like dimensions. Also, maps on homology groups
are simply linear maps, i.e. matrices, and we can study their properties with
proper linear algebra.
Recall that the Euler characteristic of a polyhedron is defined as “faces
edges + vertices”. This works for two-dimensional surfaces, and we would like to
extend this to higher dimensions. There is an obvious way to do so, by counting
higher-dimensional surfaces and putting the right signs. However, if we define it
this way, it is not clear that this is a property of the space itself, and not just a
property of the triangulation. Hence, we define it as follows:
Definition (Euler characteristic). The Euler characteristic of a triangulable
space X is
χ(X) =
X
i0
(1)
i
dim
Q
H
i
(X; Q).
This clearly depends only on the homotopy type of
X
, and not the triangula-
tion. We will later show this is equivalent to what we used to have.
More generally, we can define the Lefschetz number.
Definition (Lefschetz number). Given any map
f
:
X X
, we define the
Lefschetz number of f as
L(f) =
X
i0
(1)
i
tr(f
: H
i
(X; Q) H
i
(X; Q)).
Why is this a generalization of the Euler characteristic? Just note that the
trace of the identity map is the number of dimensions. So we have
χ(X) = L(id).
Example. We have
χ(S
n
) =
(
2 n even
0 n odd
We also have
χ
g
) = 2 2g, χ(E
n
) = 2 n.
Example. If
α
:
S
n
S
n
is the antipodal map, we saw that
α
:
H
n
(
S
n
)
H
n
(S
n
) is multiplication by (1)
n+1
. So
L(α) = 1 + (1)
n
(1)
n+1
= 1 1 = 0.
We see that even though the antipodal map has different behaviour for
different dimensions, the Lefschetz number ends up being zero all the time. We
will soon see why this is the case.
Why do we want to care about the Lefschetz number? The important thing
is that we will use this to prove a really powerful generalization of Brouwer’s
fixed point theorem that allows us to talk about things that are not balls.
Before that, we want to understand the Lefschetz number first. To define the
Lefschetz number, we take the trace of
f
, and this is a map of the homology
groups. However, we would like to understand the Lefschetz number in terms of
the chain groups, since these are easier to comprehend. Recall that the homology
groups are defined as quotients of the chain groups, so we would like to know
what happens to the trace when we take quotients.
Lemma. Let
V
be a finite-dimensional vector space and
W V
a subspace.
Let
A
:
V V
be a linear map such that
A
(
W
)
W
. Let
B
=
A|
W
:
W W
and C : V/W V /W the induced map on the quotient. Then
tr(A) = tr(B) + tr(C).
Proof. In the right basis,
A =
B A
0
0 C
.
What this allows us to do is to not look at the induced maps on homology,
but just the maps on chain complexes. This makes our life much easier when it
comes to computation.
Corollary. Let f
·
: C
·
(K; Q) C
·
(K; Q) be a chain map. Then
X
i0
(1)
i
tr(f
i
: C
i
(K) C
i
(K)) =
X
i0
(1)
i
tr(f
: H
i
(K) H
i
(K)),
with homology groups understood to be over Q.
This is a great corollary. The thing on the right is the conceptually right
thing to have homology groups are nice and are properties of the space itself,
not the triangulation. However, to actually do computations, we want to work
with the chain groups and actually calculate with chain groups.
Proof. There is an exact sequence
0 B
i
(K; Q) Z
i
(K; Q) H
i
(K; Q) 0
This is since
H
i
(
K, Q
) is defined as the quotient of
Z
i
over
B
i
. We also have
the exact sequence
0 Z
i
(K; Q) C
i
(K; Q) B
i1
(K; Q) 0
d
i
This is true by definition of
B
i1
and
Z
i
. Let
f
H
i
, f
B
i
, f
Z
i
, f
C
i
be the various
maps induced by f on the corresponding groups. Then we have
L(|f|) =
X
i0
(1)
i
tr(f
H
i
)
=
X
i0
(1)
i
(tr(f
Z
i
) tr(f
B
i
))
=
X
i0
(1)
i
(tr(f
C
i
) tr(f
B
i1
) tr(f
B
i
)).
Because of the alternating signs in dimension, each
f
B
i
appears twice in the sum
with opposite signs. So all f
B
i
cancel out, and we are left with
L(|f|) =
X
i0
(1)
i
tr(f
C
i
).
Since
tr
(
id |
C
i
(K;Q)
) =
dim
Q
C
i
(
K
;
Q
), which is just the number of
i
-simplices,
this tells us the Euler characteristic we just defined is the usual Euler character-
istic, i.e.
χ(X) =
X
i0
(1)
i
number of i-simplices.
Finally, we get to the important theorem of the section.
Theorem (Lefschetz fixed point theorem). Let
f
:
X X
be a continuous
map from a triangulable space to itself. If L(f ) 6= 0, then f has a fixed point.
Proof.
We prove the contrapositive. Suppose
f
has no fixed point. We will show
that L(f ) = 0. Let
δ = inf{|x f(x)| : x X}
thinking of
X
as a subset of
R
n
. We know this is non-zero, since
f
has no fixed
point, and X is compact (and hence the infimum point is achieved by some x).
Choose a triangulation L : |K| X such that µ(K) <
δ
2
. We now let
g : K
(r)
K
be a simplicial approximation to
f
. Since we picked our triangulation to be so
fine, for x σ K, we have
|f(x) g(x)| <
δ
2
since the mesh is already less than
δ
2
. Also, we know
|f(x) x| δ.
So we have
|g(x) x| >
δ
2
.
So we must have
g
(
x
)
6∈ σ
. The conclusion is that for any
σ K
, we must have
g(σ) σ = .
Now we compute L(f ) = L(|g|). The only complication here is that g is a map
from
K
(r)
to
K
, and the domains and codomains are different. So we need to
compose it with
s
i
:
C
i
(
K
;
Q
)
C
i
(
K
(r)
;
Q
) induced by inverses of simplicial
approximations to the identity map. Then we have
L(|g|) =
X
i0
(1)
i
tr(g
: H
i
(X; Q) H
i
(X; Q))
=
X
i0
(1)
i
tr(g
i
s
i
: C
i
(K; Q) C
i
(K; Q))
Now note that
s
i
takes simplices of
σ
to sums of subsimplices of
σ
. So
g
i
s
i
takes every simplex off itself. So each diagonal terms of the matrix of
g
i
s
i
is
0! Hence the trace is
L(|g|) = 0.
Example. If
X
is any contractible polyhedron (e.g. a disk), then
L
(
f
) = 1 for
any
f
:
X X
, which is obvious once we contract the space to a point. So
f
has a fixed point.
Example. Suppose
G
is a path-connected topological group, i.e.
G
is a group
and a topological space, and inverse and multiplication are continuous maps.
If g 6= 1, then the map
r
g
: G G
γ 7→ γg
has no fixed point. This implies
L
(
r
g
) = 0. However, since the space is path
connected,
r
g
' id
, where the homotopy is obtained by multiplying elements
along the path from e to r
g
. So
χ(G) = L(id
G
) = L(r
g
) = 0.
So if G 6= 1, then in fact χ(G) = 0.
This is quite fun. We have worked with many surfaces. Which can be
topological groups? The torus is, since it is just
S
1
× S
1
, and
S
1
is a topological
group. The Klein bottle? Maybe. However, other surfaces cannot since they
don’t have Euler characteristic 0.