11Manifolds and Poincare duality
III Algebraic Topology
11.3 Poincar´e duality
We now get to the main theorem — Poincar´e duality:
Theorem
(Poincar´e duality)
.
Let
M
be a
d
-dimensional
R
-oriented manifold.
Then there is a map
D
M
: H
k
c
(M; R) → H
d−k
(M; R)
that is an isomorphism.
The majority of the work is in defining the map. Afterwards, proving it is
an isomorphism is a routine exercise with Mayer-Vietoris and the five lemma.
What does this tell us? We know that
M
has no homology or cohomology
in negative dimensions. So by Poincar´e duality, there is also no homology or
cohomology in dimensions > d.
Moreover, if
M
itself is compact, then we know
H
0
c
(
M
;
R
) has a special
element 1. So we also get a canonical element of
H
d
(
M
;
R
). But we know there
is a special element of
H
d
(
M
;
R
), namely the fundamental class. They are in
fact the same, and this is no coincidence. This is in fact how we are going to
produce the map.
To define the map D
M
, we need the notion of the cap product
Definition (Cap product). The cap product is defined by
· _ · : C
k
(X; R) × C
`
(X; R) → C
k−`
(X; R)
(σ, ϕ) 7→ ϕ(σ|
[v
0
,...,v
`
]
)σ|
[v
`
,...,v
k
]
.
We want this to induce a map on homology. To do so, we need to know how
it interacts with differentials.
Lemma. We have
d(σ _ ϕ) = (−1)
d
((dσ) _ ϕ − σ _ (dϕ)).
Proof. Write both sides out.
As with the analogous formula for the cup product, this implies we get a
well-defined map on homology and cohomology, i.e. We obtain a map
H
k
(X; R) × H
`
(X; R) → H
k−`
(X; R).
As with the cup product, there are also relative versions
_ : H
k
(X, A; R) × H
`
(X; R) → H
k−`
(X, A; R)
and
_ : H
k
(X, A; R) × H
`
(X, A; R) → H
k−`
(X; R).
We would like to say that the cap product is natural, but since maps of spaces
induce maps of homology and cohomology in opposite directions, this is rather
tricky. What we manage to get is the following:
Lemma.
If
f
:
X → Y
is a map, and
x ∈ H
k
(
X
;
R
) And
y ∈ H
`
(
Y
;
R
), then
we have
f
∗
(x) _ y = f
∗
(x _ f
∗
(y)) ∈ H
k−`
(Y ; R).
In other words, the following diagram commutes:
H
k
(Y ; R) × H
`
(Y ; R) H
k−`
(Y ; R)
H
k
(X; R) × H
`
(Y ; R)
H
k
(X; R) × H
`
(X; R) H
k−`
(X; R)
_
f
∗
×id
id ×f
∗
_
f
∗
Proof.
We check this on the cochain level. We let
x
=
σ
: ∆
k
→ X
. Then we
have
f
#
(σ _ f
#
y) = f
#
(f
#
y)(σ|
[v
0
,...,v
`
]
)σ|
[v
`
,...,v
k
]
= y(f
#
(σ|
[v
0
,...,v
`
]
))f
#
(σ|
[v
`
,...,v
k
]
)
= y((f
#
σ)|
[v
0
,...,v
`
]
)(f
#
σ)|
[v
`
,...,v
k
]
= (f
#
σ) _ y.
So done.
Now if M is compact, then we simply define the duality map as
D
M
= [M] _ · : H
d
(M; R) → H
d−`
(M; R).
If not, we note that
H
d
c
(
M
;
R
) is the directed limit of
H
d
(
M, K
;
R
) with
K
compact, and each
H
d
(
M, L
;
R
) has a fundamental class. So we can define the
required map for each K, and then put them together.
More precisely, if
K ⊆ L ⊆ M
are such that
K, L
are compact, then we have
an inclusion map
(id, inc) : (M, M \ L) → (M, M \ K).
Then we have an induced map
(id, inc) : H
d
(M | L; R) → H
d
(M | K; R)
that sends the fundamental class
µ
L
to
µ
K
, by uniqueness of the fundamental
class.
Then the relative version of our lemma tells us that the following map
commutes:
H
`
(M | K; R) H
`
(M | L; R)
H
d−`
(M; R) H
d−`
(M; R)
(id,inc)
∗
µ
K
_ · µ
L
_ ·
id
Indeed, this is just saying that
(id)
∗
(µ
L
_ (id, inc)
∗
(ϕ)) = µ
K
_ ϕ.
So we get a duality map
D
M
= lim
−→
(µ
K
_ · ) : lim
−→
H
`
(M | K; R) → H
d−`
(M; R).
Now we can prove Poincar´e duality.
Proof.
We say
M
is “good” if the Poincar´e duality theorem holds for
M
. We
now do the most important step in the proof:
Claim 0. R
d
is good.
The only non-trivial degrees to check are
`
= 0
, d
, and
`
= 0 is straightforward.
For ` = d, we have shown that the maps
H
d
c
(R
d
; R) H
d
(R
d
| 0; R) Hom
R
(H
d
(R
d
| 0; R), R)
∼
UCT
are isomorphisms, where the last map is given by the universal coefficients
theorem.
Under these isomorphisms, the map
H
d
c
(R
d
; R) H
0
(R
d
; R) R
D
R
d
ε
corresponds to the map
Hom
K
(
H
d
(
R
d
|
0;
R
)
, R
)
→ R
is given by evaluating a
function at the fundamental class
µ
0
. But as
µ
0
∈ H
d
(
R
d
|
0;
R
) is an
R
-module
generator, this map is an isomorphism.
Claim 1. If M = U ∪ V and U, V, U ∩ V are good, then M is good.
Again, this is an application of the five lemma with the Mayer-Vietoris
sequence. We have
H
`
c
(U _ V ) H
`
c
(U) ⊕ H
d
c
(V ) H
`
c
(M) H
`+1
c
(U _ V )
H
d−`
(U _ V ) H
d−`
(U) ⊕ H
d−`
(V ) H
d−`
(M) H
d−`−1
(U _ V )
D
U_V
D
U
⊕D
V
D
M
D
U_V
We are done by the five lemma if this commutes. But unfortunately, it doesn’t.
It only commutes up to a sign, but it is sufficient for the five lemma to apply if
we trace through the proof of the five lemma.
Claim 2.
If
U
1
⊆ U
2
⊆ · · ·
with
M
=
S
n
U
n
, and
U
i
are all good, then
M
is
good.
Any compact set in M lies in some U
n
, so the map
lim
−→
H
`
c
(U
n
) → H
`
c
(U
n
)
is an isomorphism. Similarly, since simplices are compact, we also have
H
d−k
(M) = lim
−→
H
d−k
(U
n
).
Since the direct limit of open sets is open, we are done.
Claim 3. Any open subset of R
d
is good.
Any
U
is a countable union of open balls (something something rational
points something something). For finite unions, we can use Claims 0 and 1 and
induction. For countable unions, we use Claim 2.
Claim 4. If M has a countable cover by R
d
’s it is good.
Same argument as above, where we instead use Claim 3 instead of Claim 0
for the base case.
Claim 5. Any manifold M is good.
Any manifold is second-countable by definition, so has a countable open
cover by R
d
.
Corollary. For any compact d-dimensional R-oriented manifold M, the map
[M] _ · : H
`
(M; R) → H
d−`
(M; R)
is an isomorphism.
Corollary.
Let
M
be an odd-dimensional compact manifold. Then the Euler
characteristic χ(M) = 0.
Proof.
Pick
R
=
F
2
. Then
M
is
F
2
-oriented. Since we can compute Euler
characteristics using coefficients in F
2
. We then have
χ(M) =
2n+1
X
r=0
(−1)
i
dim
F
2
H
i
(M, F
2
).
But we know
H
i
(M, F
2
)
∼
=
H
2n+1−i
(M, F
2
)
∼
=
(H
2n+1−i
(M, F
2
))
∗
∼
=
H
2n+1−i
(M, F
2
)
by Poincar´e duality and the universal coefficients theorem.
But the dimensions of these show up in the sum above with opposite signs.
So they cancel, and χ(M) = 0.
What is the relation between the cap product and the cup product? If
σ ∈ C
k+`
(X; R), ϕ ∈ C
k
(X; R) and ψ ∈ C
`
(X; R), then
ψ(σ _ ϕ) = ψ(ϕ(σ|
[v
0
,...,v
k
]
)σ|
[v
k
,...,v
k+`
]
)
= ϕ(σ|
[v
0
,...,v
k
]
)ψ(σ|
[v
k
,...,v
k+`
]
)
= (ϕ ^ ψ)(σ),
Since we like diagrams, we can express this equality in terms of some diagram
commuting. The map
h
:
H
k
(
X
;
R
)
→ Hom
R
(
H
k
(
X
;
R
)
, R
) in the universal
coefficients theorem is given by
[ϕ] 7→ ([σ] 7→ ϕ(σ)).
This map exists all the time, even if the hypothesis of the universal coefficients
theorem does not hold. It’s just that it need not be an isomorphism. The formula
ψ(σ _ ϕ) = (ϕ ^ ψ)(σ)
then translates to the following diagram commuting:
H
`
(X; R) Hom
R
(H
`
(X; R), R)
H
k+`
(X; R) Hom
R
(H
`+k
(X; R), R)
ϕ^ ·
h
( · _ϕ)
∗
h
Now when the universal coefficient theorem applies, then the maps
h
are isomor-
phisms. So the cup product and cap product determine each other, and contain
the same information.
Now since Poincar´e duality is expressed in terms of cap products, this
correspondence gives us some information about cupping as well.
Theorem.
Let
M
be a
d
-dimensional compact
R
-oriented manifold, and consider
the following pairing:
h · , · i : H
k
(M; R) ⊗ H
d−k
(M, R) R
[ϕ] ⊗ [ψ] (ϕ ^ ψ)[M]
.
If
H
∗
(
M
;
R
) is free, then
h · , · i
is non-singular, i.e. both adjoints are isomor-
phisms, i.e. both
H
k
(M; R) Hom(H
d−k
(M; R), R)
[ϕ] ([ψ] 7→ hϕ, ψi)
and the other way round are isomorphisms.
Proof. We have
hϕ, ψi = (−1)
|ϕ||ψ|
hψ, ϕi,
as we know
ϕ ^ ψ = (−1)
|ϕ||ψ|
ψ ^ ϕ.
So if one adjoint is an isomorphism, then so is the other.
To see that they are isomorphsims, we notice that we have an isomorphism
H
k
(M; R) Hom
R
(H
k
(M; R), R) Hom
R
(H
d−k
(M; R), R)
[ϕ] ([σ] 7→ ϕ(σ)) ([ψ] 7→ ϕ([M] _ ψ))
UCT
D
∗
m
.
But we know
ϕ([M] _ ψ) = (ψ ^ ϕ)([M]) = hψ, ϕi.
So this is just the adjoint. So the adjoint is an isomorphism.
This is a very useful result. We have already seen examples where we can
figure out if a cup product vanishes. But this result tells us that certain cup
products are not zero. This is something we haven’t seen before.
Example. Consider CP
n
. We have previously computed
H
∗
(CP
n
, Z) =
(
Z ∗ = 2i, 0 ≤ i ≤ n
0 otherwise
Also, we know
CP
n
is
Z
-oriented, because it is a complex manifold. Let’s forget
that we have computed the cohomology ring structure, and compute it again.
Suppose for induction, that we have found
H
∗
(CP
n−1
, Z) = Z[x]/(x
n
).
As
CP
n
is obtained from
CP
n−1
by attaching a 2
n
cell, the map given by inclusion
i
∗
:
H
2
(
CP
n
, Z
)
→ H
2
(
CP
n−1
, Z
) is an isomorphism. Then the generator
x ∈ H
2
(CP
n−1
, Z) gives us a generator y ∈ H
2
(CP
n
, Z).
Now if we can show that
y
k
∈ H
2k
(
CP
n
, Z
)
∼
=
Z
is a generator for all
k
, then
H
∗
(CP
n
, Z)
∼
=
Z[y]/(y
n+1
).
But we know that
y
n−1
generates
H
2n−2
(
CP
n
, Z
), since it pulls back under
i
∗
to x
n−1
, which is a generator. Finally, consider
H
2
(CP
2
, Z) ⊗ H
2n−2
(CP
2
, Z) Z
y ⊗ y
n−1
y
n
[CP
n
].
Since this is non-singular, we know y
n
∈ H
2n
(CP
n
, Z) must be a generator.
Of course, we can get H
∗
(RP
n
, F
2
) similarly.