11Manifolds and Poincare duality

III Algebraic Topology



11.4 Applications
We go through two rather straightforward applications, before we move on to
bigger things like the intersection product.
Signature
We focus on the case where
d
= 2
n
is even. Then we have, in particular, a
non-degenerate bilinear form
h · , · i : H
n
(M; R) H
n
(M; R) R.
Also, we know
ha, bi = (1)
n
2
hb, ai = (1)
n
hb, ai.
So this is a symmetric form if
n
is even, and skew-symmetric form if
n
is odd.
These are very different scenario. For example, we know a symmetric matrix is
diagonalizable with real eigenvalues (if
R
=
R
), but a skew-symmetric form does
not have these properties.
So if
M
is 4
k
-dimensional, and
Z
-oriented, then in particular
M
is
R
-oriented.
Then the map
h · , · i
:
H
2k
(
M
;
R
)
H
2k
(
M
;
R
)
R
can be represented by a
symmetric real matrix, which can be diagonalized. This has some real eigenvalues.
The eigenvalues can be positive or negative.
Definition
(Signature of manifold)
.
Let
M
be a 4
k
-dimensional
Z
-oriented
manifold. Then the signature is the number of positive eigenvalues of
h · , · i : H
2k
(M; R) H
2k
(M; R) R
minus the number of negative eigenvalues. We write this as sgn(M).
By Sylvester’s law of inertia, this is well-defined.
Fact.
If
M
=
W
for some compact 4
k
+ 1-dimensional manifold
W
with
boundary, then sgn(M) = 0.
Example. CP
2
has
H
2
(
CP
2
;
R
)
=
R
, and the bilinear form is represented by
the matrix (1). So the signature is 1. So
CP
2
is not the boundary of a manifold.
Degree
Recall we defined the degree of a map from a sphere to itself. But if we have a
Z
-oriented space, we can have the fundamental class [
M
], and then there is an
obvious way to define the degree.
Definition
(Degree of map)
.
If
M, N
are
d
-dimensional compact connected
Z-oriented manifolds, and f : M N, then
f
([M]) H
d
(N, Z)
=
Z[N].
So f
([M]) = k[N] for some k. This k is called the degree of f, written deg(f).
If
N
=
M
=
S
n
and we pick the same orientation for them, then this recovers
our previous definition.
By exactly the same proof, we can compute this degree using local degrees,
just as in the case of a sphere.
Corollary.
Let
f
:
M N
be a map between manifolds. If
F
is a field and
deg(f) 6= 0 F, then then the induced map
f
: H
(N, F) H
(M, F)
is injective.
This seems just like an amusement, but this is powerful. We know this is in
fact a map of rings. So if we know how to compute cup products in
H
(
M
;
F
),
then we can do it for H
(N; F) as well.
Proof. Suppose not. Let α H
k
(N, F) be non-zero but f
(α) = 0. As
h · , · i : H
k
(N, F) H
dk
(N, F) F
is non-singular, we know there is some β H
dk
(N, F) such that
hα, βi = (α ^ β)[N] = 1.
Then we have
deg(f) = deg(f) · 1
= (α ^ β)(deg(f)[N])
= (α ^ β)(f
[M])
= (f
(α) ^ f
(β))([M])
= 0.
This is a contradiction.