11Manifolds and Poincare duality
III Algebraic Topology
11.6 The diagonal
Again, let
M
be a compact
Q
-oriented
d
-dimensional manifold. Then
M × M
is
a 2
d
-dimensional manifold. We can try to orient it as follows — K¨unneth gives
us an isomorphism
H
d+d
(M × M; Q)
∼
=
H
d
(M; Q) ⊗ H
d
(M; Q),
as
H
k
(
M
;
Q
) = 0 for
k > d
. By the universal coefficients theorem, plus the fact
that duals commute with tensor products, we have an isomorphism
H
d+d
(M × M; Q)
∼
=
H
d
(M; Q) ⊗ H
d
(M; Q).
Thus the fundamental class [
M
] gives us a fundamental class [
M × M
] =
[M] ⊗ [M].
We are going to do some magic that involves the diagonal map
∆ : M M × M
x (x, x).
This gives us a cohomology class
δ = D
−1
M×M
(∆
∗
[M]) ∈ H
d
(M × M, Q)
∼
=
M
i+j=d
H
i
(M, Q) ⊗ H
j
(M, Q).
It turns out a lot of things we want to do with this
δ
can be helped a lot by
doing the despicable thing called picking a basis.
We let
{a
i
}
be a basis for
H
∗
(
M, Q
). On this vector space, we have a
non-singular form
h · , · i : H
∗
(M, Q) ⊗ H
∗
(M; Q) → Q
given by
hϕ, ψi 7→
(
ϕ ψ
)([
M
]). Last time we were careful about the degrees
of the cochains, but here we just say that if
ϕ ψ
does not have dimension
d
,
then the result is 0.
Now let {b
i
} be the dual basis to the a
i
using this form, i.e.
ha
i
, b
j
i = δ
ij
.
It turns out δ has a really nice form expressed using this basis:
Theorem. We have
δ =
X
i
(−1)
|a
i
|
a
i
⊗ b
i
.
Proof. We can certainly write
δ =
X
i,
C
ij
a
i
⊗ b
j
for some
C
ij
. So we need to figure out what the coefficients
C
ij
are. We try to
compute
((b
k
⊗ a
`
) δ)[M × M] =
X
C
ij
(b
k
⊗ a
`
) (a
i
⊗ b
j
)[M × M]
=
X
C
ij
(−1)
|a
`
||a
i
|
(b
k
a
i
) ⊗ (a
`
b
i
)[M] ⊗ [M]
=
X
C
ij
(−1)
|a
`
||a
i
|
(δ
ik
(−1)
|a
i
||b
k
|
)δ
j`
= (−1)
|a
k
||a
`
|+|a
k
||b
k
|
C
k`
.
But we can also compute this a different way, using the definition of δ:
(b
k
⊗ a
`
δ)[M × M] = (b
k
⊗ a
`
)(∆
∗
[M]) = (b
k
a
`
)[M] = (−1)
|a
`
||b
k
|
δ
k`
.
So we see that
C
k`
= δ
k`
(−1)
|a
`
|
.
Corollary. We have
∆
∗
(δ)[M] = χ(M),
the Euler characteristic.
Proof. We note that for a ⊗ b ∈ H
n
(M × M), we have
∆
∗
(a ⊗ b) = ∆
∗
(π
∗
1
a π
∗
2
b) = a b
because π
i
◦ ∆ = id. So we have
∆
∗
(δ) =
X
(−1)
|a
i
|
a
i
b
i
.
Thus
∆
∗
(δ)[M] =
X
i
(−1)
|a
i
|
=
X
k
(−1)
k
dim
Q
H
k
(M; Q) = χ(M).
So far everything works for an arbitrary manifold. Now we suppose further
that
M
is smooth. Then
δ
is the Thom class of the normal bundle
ν
M⊆M×M
of
M → M × M.
By definition, pulling this back along ∆ to
M
gives the Euler class of the normal
bundle. But we know
ν
M⊆M×M
∼
=
T M
, because the fiber at
x
is the cokernel of
the map
T
x
M T
x
M ⊕ T
x
M
∆
and the inverse given by
T
x
M ⊕ T
x
M → T
x
M
(v, w) 7→ v − w
gives us an isomorphism
T
x
M ⊕ T
x
M
∆T
x
M
∼
=
T
x
M.
Corollary. We have
e(T M)[M] = χ(M).
Corollary. If M has a nowhere-zero vector field, then χ(M) = 0.
More generally, this tells us that
χ
(
M
) is the number of zeroes of a vector
field M → T M (transverse to the zero section) counted with sign.
Lemma.
Suppose we have
R
-oriented vector bundles
E → X
and
F → X
with
Thom classes u
E
, u
F
. Then the Thom class for E ⊕ F → X is u
E
u
F
. Thus
e(E ⊕ F ) = e(E) e(F ).
Proof. More precisely, we have projection maps
E ⊕ F
E F
π
E
π
F
.
We let U = π
−1
E
(E
#
) and V = π
−1
F
(F
#
). Now observe that
U ∪ V = (E ⊕ F )
#
.
So if dim E = e, dim F = f , then we have a map
H
e
(E, E
#
) ⊗ H
f
(F, F
#
) H
e
(E ⊕ F, U ) ⊗ H
f
(E ⊕ F, V )
H
e+f
(E ⊕ F, (E ⊕ F )
#
)
π
∗
E
⊗π
∗
F
^
,
and it is easy to see that the image of
u
E
⊗ u
F
is the Thom class of
E ⊕ F
by
checking the fibers.
Corollary. T S
2n
has no proper subbundles.
Proof.
We know
e
(
T S
2n
)
6
= 0 as
e
(
T S
2n
)[
S
2
] =
χ
(
S
2n
) = 2. But it cannot be a
proper cup product of two classes, since there is nothing in lower cohomology
groups. So
T S
2n
is not the sum of two subbundles. Hence
T S
2n
cannot have a
proper subbundle
E
or else
T S
2n
=
E ⊕E
⊥
(for any choice of inner product).