10Vector bundles
III Algebraic Topology
10.4 Gysin sequence
Now we do something interesting with vector bundles. We will come up with a
long exact sequence associated to a vector bundle, known as the Gysin sequence.
We will then use the Gysin sequence to deduce something about the base space.
Suppose we have a
d
-dimensional vector bundle
π
:
E → X
that is
R
-oriented.
We want to talk about the unit sphere in every fiber of
E
. But to do so, we
need to have a notion of length, and to do that, we want an inner product. But
luckily, we do have one, and we know that any two norms on a finite-dimensional
vector space are equivalent. So we might as well arbitrarily choose one.
Definition
(Sphere bundle)
.
Let
π
:
E → X
be a vector bundle, and let
h · , · i : E ⊗ E → R be an inner product, and let
S(E) = {v ∈ E; hv, vi = 1} ⊆ E.
This is the sphere bundle associated to E.
Since the unit sphere is homotopy equivalent to
R
d
\ {
0
}
, we know the
inclusion
j : S(E) E
#
is a homotopy equivalence, with inverse given by normalization.
The long exact sequence for the pair (
E, E
#
) gives (as before, we do not
write the R):
H
i+d
(E, E
#
) H
i+d
(E) H
i+d
(E
#
) H
i+d+1
(E, E
#
)
H
i
(X) H
i+d
(X) H
i+d
(S(E)) H
i+1
(X)
s
∗
0
j
∗
Φ
· ^e(E)
π
∗
p
∗
p
!
Φ
where
p
:
S
(
E
)
→ E
is the projection, and
p
!
is whatever makes the diagram
commutes (since
j
∗
and Φ are isomorphisms). The bottom sequence is the Gysin
sequence, and it is exact because the top row is exact. This is in fact a long
exact sequence of
H
∗
(
X
;
R
)-modules, i.e. the maps commute with cup products.
Example.
Let
L
=
γ
C
1,n+1
→ CP
n
=
Gr
1
(
C
n+1
) be the tautological 1 complex
dimensional vector bundle on
Gr
1
(
C
n+1
). This is
Z
-oriented as any complex
vector bundle is, because if we consider the inclusion
GL
1
(C) → GL
2
(R)
obtained by pretending
C
is
R
2
, we know
GL
1
(
C
) is connected, so lands in
the component of the identity, so has positive determinant. The sphere bundle
consists of
S(L) = {(V, v) ∈ CP
n
× C
n+1
: v ∈ V, |v| = 1}
∼
=
{v ∈ C
n+1
: |v| = 1}
∼
=
S
2n+1
,
where the middle isomorphism is given by
(V, v) v
(hvi, v) v
The Gysin sequence is
H
i+1
(S
2n+1
) H
i
(CP
n
) H
i+2
(CP
n
) H
i+2
(S
2n+1
)
p
!
^e(L)
p
∗
Now if
i ≤
2
n −
2, then both outer terms are 0. So the maps in the middle are
isomorphisms. Thus we get isomorphisms
H
0
(CP
n
) H
2
(CP
n
) H
4
(CP
n
) · · ·
Z · 1 Z · e(L) Z · e(L)
2
^e(L) ^e(L)
Similarly, we know that the terms in the odd degree vanish.
Checking what happens at the end points carefully, the conclusion is that
H
∗
(CP
n
) = Z[e(L)]/(e(L)
n+1
)
as a ring.
Example. We do the real case of the above computation. We have
K = γ
R
1,n+1
→ RP
n
= Gr
1
(R
n+1
).
The previous trick doesn’t work, and indeed this isn’t
Z
-orientable. However, it
is
F
2
-oriented as every vector bundle is, and by exactly the same argument, we
know
S(L)
∼
=
S
n
.
So by the same argument as above, we will find that
H
∗
(RP
n
, F
2
)
∼
=
F
2
[e(L)]/(e(L)
n+1
).
Note that this is different from the one we had before, because here
deg e
(
L
) = 1,
while the complex case had degree 2.