11Manifolds and Poincare duality

III Algebraic Topology



11.7 Lefschetz fixed point theorem
Finally, we are going to prove the Lefschetz fixed point theorem. This is going
to be better than the version you get in Part II, because this time we will know
how many fixed points there are.
So let
M
be a compact
d
-dimensional manifold that is
Z
-oriented, and
f
:
M M
be a map. Now if we want to count the number of fixed points,
then we want to make sure the map is “transverse” in some sense, so that there
aren’t infinitely many fixed points.
It turns out the right condition is that the graph
Γ
f
= {(x, f(x)) M × M} M × M
has to be transverse to the diagonal. Since Γ
f
is exactly the fixed points of
f, this is equivalent to requiring that for each fixed point x, the map
D
x
D
x
F : T
x
M T
x
M T
x
M T
x
M
is an isomorphism. We can write the matrix of this map, which is
I I
D
x
f I
.
Doing a bunch of row and column operations, this is equivalent to requiring that
I 0
D
x
f I D
x
f
is invertible. Thus the condition is equivalent to requiring that 1 is not an
eigenvalue of D
x
f.
The claim is now the following:
Theorem
(Lefschetz fixed point theorem)
.
Let
M
be a compact
d
-dimensional
Z
-oriented manifold, and let
f
:
M M
be a map such that the graph Γ
f
and
diagonal intersect transversely. Then Then we have
X
xfix(f)
sgn det(I D
x
f) =
X
k
(1)
k
tr(f
: H
i
(M; Q) H
k
(M; Q)).
Proof. We have
f
] · [∆(M)] H
0
(M × M; Q).
We now want to calculate ε of this. By Poincar´e duality, this is equal to
(D
1
M×M
f
] ^ D
1
M×M
[∆(M)])[M × M] Q.
This is the same as
(D
1
M×M
[∆(M)])([Γ
f
]) = δ(F
[M]) = (F
δ)[M],
where F : M M × M is given by
F (x) = (x, f(x)).
We now use the fact that
δ =
X
(1)
|a
i
|
a
i
b
i
.
So we have
F
δ =
X
(1)
|a
i
|
a
i
f
b
i
.
We write
f
b
i
=
X
C
ij
b
j
.
Then we have
(F
δ)[M] =
X
i,j
(1)
|a
i
|
C
ij
(a
i
b
j
)[M] =
X
i
(1)
|a
i
|
C
ii
,
and C
ii
is just the trace of f
.
We now compute this product in a different way. As Γ
f
and ∆(
M
) are
transverse, we know Γ
f
∆(
M
) is a 0-manifold, and the orientation of Γ
f
and
∆(M) induces an orientation of it. So we have
f
] · [∆(m)] = [Γ
f
∆(M)] H
0
(M × M; Q).
We know this Γ
f
∆(
M
) has
| fix
(
f
)
|
many points, so
f
∆(
M
)] is the sum
of
| fix
(
f
)
|
many things, which is what we’ve got on the left above. We have to
figure out the sign of each term is actually sgn det(I D
x
f), and this is left as
an exercise on the example sheet.
Example.
Any map
f
:
CP
2n
CP
2n
has a fixed point. We can’t prove this
using the normal fixed point theorem, but we can exploit the ring structure of
cohomology to do this. We must have
f
(x) = λx H
2
(CP
2
; Q) = Qx.
for some λ Q. So we must have
f
(x
i
) = λ
i
x
i
.
We can now very easily compute the right hand side of the fixed point theorem
X
k
(1)
k
tr(f
: H
k
H
k
) = 1 + λ + λ
2
+ · · · + λ
2n
.
and this cannot be zero.