11Manifolds and Poincare duality
III Algebraic Topology
11.5 Intersection product
Recall that cohomology comes with a cup product. Thus, Poincar´e duality gives
us a product on homology. Our goal in this section is to understand this product.
We restrict our attention to smooth manifolds, so that we can talk about
the tangent bundle. Recall (from the example sheet) that an orientation of the
manifold is the same as an orientation on the tangent bundle.
We will consider homology classes that come from submanifolds. For con-
creteness, let
M
be a compact smooth
R
-oriented manifold, and
N ⊆ M
be
an
n
-dimensional
R
-oriented submanifold. Let
i
:
N → M
be the inclusion.
Suppose
dim M
=
d
and
dim N
=
n
. Then we obtain a canonical homology class
i
∗
[N] ∈ H
n
(M; R).
We will abuse notation and write [
N
] for
i
∗
[
N
]. This may or may not be zero.
Our objective is to show that under suitable conditions, the product of [
N
1
] and
[N
2
] is [N
1
∩ N
2
].
To do so, we will have to understand what is the cohomology class Poinacr´e
dual to [
N
]. We claim that, suitably interpreted, it is the Thom class of the
normal bundle.
Write
ν
N⊆M
for the normal bundle of
N
in
M
. Picking a metric on
T M
, we
can decompose
i
∗
T M
∼
=
T N ⊕ ν
N⊆M
,
Since
T M
is oriented, we obtain an orientation on the pullback
i
∗
T M
. Similarly,
T N is also oriented by assumption. In general, we have the following result:
Lemma.
Let
X
be a space and
V
a vector bundle over
X
. If
V
=
U ⊕ W
, then
orientations for any two of U, W, V give an orientation for the third.
Proof.
Say
dim V
=
d
,
dim U
=
n
,
dim W
=
m
. Then at each point
x ∈ X
, by
K¨unneth’s theorem, we have an isomorphism
H
d
(V
x
, V
#
x
; R)
∼
=
H
n
(U
x
, U
#
x
; R) ⊗ H
m
(W
x
, W
#
x
; R)
∼
=
R.
So any local
R
-orientation on any two induces one on the third, and it is
straightforward to check the local compatibility condition.
Can we find a more concrete description of this orientation on
ν
N⊆M
? By
the same argument as when we showed that H
i
(R
n
| {0})
∼
=
H
i
c
(R
d
), we know
H
i
(ν
N⊆M
, ν
#
N⊆M
; R)
∼
=
H
i
c
(ν
N⊆M
; R).
Also, by the tubular neighbourhood theorem, we know
ν
N⊆M
is homeomorphic
to an open neighbourhood U of N in M . So we get isomorphisms
H
i
c
(ν
N⊆M
; R)
∼
=
H
i
c
(U; R)
∼
=
H
d−i
(U; R)
∼
=
H
d−i
(N; R),
where the last isomorphism comes from the fact that
N
is homotopy-equivalent
to U .
In total, we have an isomorphism
H
i
(ν
N⊆M
, ν
#
N⊆M
; R)
∼
=
H
d−i
(N; R).
Under this isomorphism, the fundamental class [
N
]
∈ H
n
(
N
;
R
) corresponds to
some
E
N⊆M
∈ H
d−n
(ν
N⊆M
, ν
#
N⊆M
; R)
But we know
ν
N⊆M
has dimension
d − n
. So
E
N⊆M
is in the right dimension to
be a Thom class, and is a generator for
H
d−n
(
ν
N⊆M
, ν
#
N⊆M
;
R
), because it is a
generator for H
d−n
(N; R). One can check that this is indeed the Thom class.
How is this related to the other things we’ve had? We can draw the commu-
tative diagram
H
n
(N; R) H
n
(U; R) H
d−n
c
(U; R)
H
n
(M; R) H
d−n
(M; R)
∼ ∼
i
∗
extension by 0
∼
The commutativity of the square is a straightforward naturality property of
Poincar´e duality.
Under the identification
H
d−n
c
(
U
;
R
)
∼
=
H
d−n
(
ν
N⊆M
, ν
#
N⊆M
;
R
), the above
says that the image of [
N
]
∈ H
n
(
N
;
R
) in
H
d−n
c
(
U
;
R
) is the Thom class of the
normal bundle ν
N⊆M
.
On the other hand, if we look at this composition via the bottom map, then
[N] gets sent to D
−1
M
([N]). So we know that
Theorem.
The Poincar´e dual of a submanifold is (the extension by zero of) the
normal Thom class.
Now if we had two submanifolds
N, W ⊆ M
. Then the normal Thom classes
give us two cohomology classes of
M
. As promised, When the two intersect
nicely, the cup product of their Thom classes is the Thom class of [
N ∩ W
]. The
niceness condition we need is the following:
Definition
(Transverse intersection)
.
We say two submanifolds
N, W ⊆ M
intersect transversely if for all x ∈ N ∩ W , we have
T
x
N + T
x
W = T
x
M.
Example. We allow intersections like
but not this:
It is a fact that we can always “wiggle” the submanifolds a bit so that the
intersection is transverse, so this is not too much of a restriction. We will neither
make this precise nor prove it.
Whenever the intersection is transverse, the intersection
N ∩ W
will be a
submanifold of M, and of N and W as well. Moreover,
(ν
N∩W ⊆M
)
x
= (ν
N⊆M
)
x
⊕ (ν
W ⊆M
)
x
.
Now consider the inclusions
i
N
: N ∩ W → N
i
W
: N ∩ W → W.
Then we have
ν
N∩W ⊆M
= i
∗
N
(ν
N⊆M
) ⊕ i
∗
W
(ν
W ⊆M
).
So with some abuse of notation, we can write
i
∗
N
E
N⊆M
i
∗
W
E
W ⊆M
∈ H
∗
(ν
N∩W ⊆M
, ν
#
N∩W ⊆M
; R),
and we can check this gives the Thom class. So we have
D
−1
M
([N]) D
−1
M
([W ]) = D
−1
M
([N ∩ W ]).
The slogan is “cup product is Poincar´e dual to intersection”. One might be
slightly worried about the graded commutativity of the cup product, because
N ∩ W = W ∩ N as manifolds, but in general,
D
−1
M
([N]) D
−1
M
([W ]) 6= D
−1
M
([W ]) D
−1
M
([N]).
The fix is to say that
N ∩ W
are
W ∩ N
are not the same as oriented manifolds
in general, and sometimes they differ by a sign, but we will not go into details.
More generally, we can define
Definition
(Intersection product)
.
The intersection product on the homology
of a compact manifold is given by
H
n−k
(M) ⊗ H
n−`
(M) H
n−k−`
(M)
(a, b) a · b = D
M
(D
−1
M
(a) D
−1
M
(b))
Example. We know that
H
2k
(CP
n
, Z)
∼
=
(
Z 0 ≤ k ≤ n
0 otherwise
.
Moreover, we know the generator for these when we computed it using cellular
homology, namely
[CP
k
] ≡ y
k
∈ H
2k
(CP
n
, Z).
To compute [
CP
`
]
·
[
CP
`
], if we picked the canonical copies of
CP
`
, CP
k
⊆ CP
n
,
then one would be contained in the other, and this is exactly the opposite of
intersecting transversely.
Instead, we pick
CP
k
= {[z
0
: z
1
· · · z
k
: 0 : · · · : 0]}, CP
`
= {[0 : · · · : 0 : w
0
: · · · : w
`
]}.
It is a fact that any two embeddings of
CP
k
→ CP
n
are homotopic, so we can
choose these.
Now these two manifolds intersect transversely, and the intersection is
CP
k
∩ CP
`
= CP
k+`−n
.
So this says that
y
k
· y
`
= ±y
l+`−n
,
where there is some annoying sign we do not bother to figure out.
So if x
k
is Poincar´e dual to y
n−k
, then
x
k
x
`
= x
k+`
,
which is what we have previously found out.
Example. Consider the manifold with three holes
b
1
b
2
b
3
a
1
a
2
a
3
Then we have
a
i
· b
i
= {pt}, a
i
· b
j
= 0 for i 6= j.
So we find the ring structure of
H
∗
(
E
g
, Z
), hence the ring structure of
H
∗
(
E
g
, Z
).
This is so much easier than everything else we’ve been doing.
Here we know that the intersection product is well-defined, so we are free to
pick our own nice representatives of the loop to perform the calculation.
Of course, this method is not completely general. For example, it would be
difficult to visualize this when we work with manifolds with higher dimensions,
and more severely, not all homology classes of a manifold have to come from
submanifolds (e.g. twice the fundamental class)!