INTRODUCTION TO ALGEBRAIC TOPOLOGY
MARTIN ˇCADEK
8. Vector bundles and Thom isomorphism
In this section we introduce the notion of vector bundle and define its important
algebraic invariants Thom and Euler classes. The Thom class is involved in so called
Thom isomorphism. Using this isomorphism we derive the Gysin exact sequence which
is an important tool for computing cup product structure in cohomology.
8.1. Fibre bundles. A fibre bundle structure on a space E, with fiber F, consists
of a projection map p : E → B such that each point of B has a neighbourhood U for
which there is a homeomorphism h : p−1
(U) → U × F such that the diagram
p−1
(U)
h //
p
##FFFFFFFFF
U × F
pr1
||xxxxxxxxx
U
commutes. Here pr1 is the projection on the first factor. h is called a local trivialization,
the space E is called the total space of the bundle and B is the base space.
A subbundle (E , B, p ) of a fibre bundle (E, B, p) has the total space E ⊆ E,
the fibre F ⊆ F, p = p/E and local trivializations in E are restrictions of local
trivializations of E.
A vector bundle is a fibre bundle (E, B, p) whose fiber is a vector space (real or
complex). Moreover, we suppose that for each b ∈ B the fiber p−1
(b) over b is a vector
space and all local trivializations restricted to p−1
(b) are linear isomorphisms. The
dimension of a vector bundle is the dimension of its fiber. For p−1
(U) where U ⊆ B
we will use notation EU . Further, E0
U will stand for EU without zeroes in vector spaces
Ex = p−1
(x) for x ∈ U.
8.2. Orientation of a vector space. Let V be a real vector space of dimension n.
The orientation of V is the choice of a generator in Hn
(V, V − {0}; Z) = Z. If R is
a commutative ring with a unit, the R-orientation of V is the choice of a generator
in Hn
(V, V − {0}; R) = R. For R = Z we have two possible orientations, for R = Z2
only one.
8.3. Orientation of a vector bundle. Consider a vector bundle (E, B, p) with fiber
Rn
. The R-orientation of the vector bundle E is a choice of orientation in each vector
1
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space p−1
(b), b ∈ B, i. e. a choice of generators tb ∈ Hn
(Eb, E0
b ; R) = R such that for
each b ∈ B there is a neighbourhood U and an element
tU ∈ Hn
(EU , E0
U ; R)
with the property
i∗
x(tU ) = tx
for each x ∈ U and the inclusion ix : Ex → EU .
If a vector bundle has an R-orientation, we say that it is R-orientable. An R-oriented
vector bundle is a vector bundle with a chosen R-orientation. Talking on orientation
we will mean Z-orientation.
Example. Every vector bundle (E, B, p) is Z2-orientable. After we have some knowledge
of fundamental group, we will be able to prove that vector bundles with π1(B) = 0
are orientable.
8.4. Thom class and Thom isomorphism. The Thom class of a vector bundle
(E, B, p) of dimension n is an element t ∈ Hn
(E, E0
; R) such that i∗
b(t) is a generator
in Hn
(Eb, E0
b ; R) = R for each b ∈ B where ib : Eb → E is an inclusion.
It is clear that any Thom class determines a unique orientation. The reverse statement
is also true.
Theorem (Thom Isomorphism Theorem). Let (E, B, p) be an R-oriented vector bundle
of real dimension n. Then there is just one Thom class t ∈ Hn
(E, E0
; R) which
determines the given R-orientation. Moreover, the homomorphism
τ : Hk
(B; R) → Hk+n
(E, E0
; R), τ(a) = p∗
(a) ∪ t
is an isomorphism (so called Thom isomorphism).
Remark. Notice that Thom Isomorphism Theorem is a generalization of the K¨unneth
Formula 7.5 for (Y, A) = (Rn
, Rn
− {0}). We use it in the proof.
Proof. (1) First suppose that E = B × Rn
. Then according to Theorem 7.5
H∗
(E, E0
; R) = H∗
(B × Rn
, B × (Rn
− {0}); R) = H∗
(B; R) ⊗ H∗
(Rn
, Rn
− {0}); R)
∼= H∗
(B; R)[e]/ e2
where e ∈ Hn
(Rn
, Rn
− {0}); R) is the generator given by the orientation of E. Now,
there is just one Thom class t = 1 × e and
τ(a) = p∗
(a) ∪ t = a × e
is an isomorphism.
(2) If U is open subset of B, then the orientation of (E, B, p) induces an orientation of
the vector bundle (EU , U, p). Suppose that U and V are two open subsets in B such
that the statement of Theorem is true for EU , EV and EU∩V with induced orientations.
Denote the corresponding Thom classes by tU , tV and tU∩V . The uniqueness of tU∩V
implies that the restrictions of both classes tU and tV on Hn
(EU∩V , E0
U∩V ; R) are tU∩V .
We will show that Theorem holds for EU∪V .
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Consider the Mayer-Vietoris exact sequence 5.13 for A = EU , B = EV , C = E0
U ,
D = E0
V together with the Mayer-Vietoris exact sequence for A = U, B = V and C =
D = ∅. Omitting coefficients these exact sequences together with Thom isomorphisms
τU , τV and τU∩V yield the following diagram where DEU stands for the pair (EU , E0
U )
δ∗
// Hk+n
(DEU∪V )
(j∗
U ,j∗
V )
// Hk+n
(DEU ) ⊕ Hk+n
(DEV )
i∗
U −i∗
V
// Hk+n
(DEU∩V ) //
δ∗
// Hk
(U ∪ V )
(j∗
U ,j∗
V )
//
τU∪V
OO


Hk
(U) ⊕ Hk
(V )
i∗
U −i∗
V
//
τU ⊕τV
OO
Hk
(U ∩ V )
τU∩V
OO
//
(At the moment we do not need commutativity.) From the first row of this diagram
we get that
Hi
(EU∪V , E0
U∪V ) = 0 for i < n
and that there is just one class tU∪V ∈ Hn
(EU∪V , E0
U∪V ) such that
(j∗
U , j∗
V )(tU∪V ) = (tU , tV ).
This is the Thom class for EU∪V and we can define the homomorphism τU∪V : Hk
(U ∪
V ) → Hk+n
(EU∪V , E0
U∪V ) by
τU∪V (a) = p∗(a) ∪ tU∪V .
Complete the diagram by this homomorphism. When we check the commutativity
of the completed diagram, it suffices to apply Five Lemma to show that τU∪V is an
isomorphism.
To prove the commutativity we have to go into the cochain level from which the
Mayer-Vietoris sequences are derived in natural way. Let tU and tV be cocycles representing
the Thom classes tU and tV . We can choose them in such a way that
i∗
U tU = i∗
V tV = tU∩V
where tU∩V represents the Thom class tU∩V . Consider the diagram where the rows are
the short exact sequences inducing the Mayer-Vietoris exact sequences above.
0 // C∗
0 (EU + EV )
(j∗
U ,j∗
V )
// C∗
0 (EU ) ⊕ C∗
0 (EV )
i∗
U −i∗
V
// C∗
(EU∩V ) // 0
0 // C∗
(U + V )
(j∗
U ,j∗
V )
//
τU∪V
OO


C∗
(U) ⊕ C∗
(V )
τU ⊕τV
OO
i∗
U −i∗
V
// C∗
(U ∩ V ) //
τU∩V
OO
0
Here we use the following notation: C∗(U + V ) is the free Abelian group generated
by simplices lying in U and V , C∗
(U + V ) = HomR(C∗(U + V ), R). C∗
0 (EU + EV )
are the cochains from C∗
(EU + EV ) which are zeroes on simplices from C∗(E0
U + E0
V ).
τU (a) = p∗
(a) ∪ tU . (According to Lemma in 3.12 the cohomology of C∗
0 (EU + EV ) is
H∗
(EU∪V , E0
U∪V ; R).)
There is just one cocycle tU∪V representing the Thom class tU∪V such that
(j∗
U , j∗
V )(tU∪V ) = (tU , tV ).
If we show that τU , τV , τU∩V and τU∪V are cochain homomorphisms which make the
diagram commutative, then the diagram with the Mayer-Vietoris exact sequences will
be also commutative. To prove the commutativity of the cochain diagram above is
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not difficult and left to the reader. Here we prove that τU is a cochain homomorhism.
(The proof for the other τ is the same.)
Let a ∈ Ck
(U). Since tU is cocycle we get
δτU (a) = δ(p∗
(a) ∪ tU ) = δ(p∗
(a)) ∪ tU + (−1)k
p∗
(a) ∪ δ(tU ) = p∗
(δ(a)) ∪ tU = τU δ(a).
(3) Let B be compact (particullary a finite CW-complex). Then there is a finite open
covering U1, U2, . . . , Um such that EUi
is homeomorphic to Ui × Rn
. So according to
(1) the statement of Theorem holds for all EUi
. Using (2) and induction we can show
that Theorem holds for E = m
i=1 EUi
as well.
(4) The proof for the other base spaces B needs a limit transitions in cohomology and
the fact that for any B there is always a CW-complex X and a map f : B → X
inducing isomorphism in cohomology. Here we omit this part.
8.5. Euler class. Let ξ = (E, B, p) be oriented vector bundle of dimension n with
the Thom class tξ ∈ Hn
(E, E0
; Z). Consider the standard inclusion j : E → (E, E0
).
Since p : E → B is a homotopy equivalence, there is just one class e(ξ) ∈ Hn
(B; Z),
called the Euler class of ξ, such that
p∗
(e(ξ)) = j∗
(tξ).
For R-oriented vector bundles we can define the Euler class e(ξ) ∈ Hn
(B; R) in
the same way. Particulary, any vector bundle ξ = (E, B, p) has an Euler class with
Z2-coefficients called the n-th Stiefel-Whitney class wn(ξ) ∈ Hn
(B; Z2).
8.6. Gysin exact sequence. The following theorem gives us a useful tool for computation
of the ring structure of singular cohomology of various spaces.
Theorem (Gysin exact sequence). Let ξ = (E, B, p) be an R-oriented vector bundle
of dimension n with the Euler class e(ξ) ∈ Hn
(B; R). Then there is a homomorphism
∆∗
: H∗
(E0
; R) → H∗
(B; R) of modules over H∗
(B; R) such that the sequence
. . .
p∗
−→ Hk+n−1
(E0
; R)
∆∗
−→ Hk
(B; R)
∪e(ξ)
−−−→ Hk+n
(B; R)
p∗
−→ Hk+n
(E0
; R)
∆∗
−→ . . .
is exact.
Proof. The definition of ∆∗
and the exactness follows from the following cummutative
diagram where we have used the long exact sequence for the pair (E, E0
) and the
Thom isomorphism τ:
Hk+n−1
(E0
)
δ∗
//
∆∗
((PPPPPP
Hk+n
(E, E0
)
j∗
// Hk+n
(E)
i∗
// Hk+n
(E0
)
Hk
(B)
τ∼=
OO
∪e(ξ)
//____ Hk+n
(B)
p∗∼=
OO
p∗
88qqqqqqqqqq
The right action of b ∈ H∗
(B) on H∗
(E0
) is given by
x · b = x ∪ i∗
p∗
(b), x ∈ H∗
(E0
).
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Using the definition of the connecting homomorphism and the properties of cup product
one can show that
∆∗
(x · b) = ∆∗
(x) ∪ b.
The details are left to the reader.
Example. Consider the canonical one dimensional vector bundle γ = (E, RPn
, p)
where
E = {(l, v) ∈ RPn
× Rn+1
; v ∈ l},
the elements of RPn
are identified with lines in Rn+1
and p(l, v) = l. The space E0 is
equal to Rn+1
− {0} and homotopy equivalent to Sn
.
Using the Gysin exact sequence with Z2-coefficients and the fact that Hk
(RPn
; Z2) =
Z2 for 0 ≤ k ≤ n, we get successively that the first Stiefel-Whitney class w1(γ) ∈
H1
(RPn
; Z2) is different from zero and that
H∗
(RPn
); Z2) ∼= Z2[w1(γ)]/ w1(γ)n+1
.
Exercise. Using the Gysin exact sequence show that
H∗
(CPn
; Z) ∼= Z[x]/ xn+1
where x ∈ H2
(CPn
; Z).
CZ.1.07/2.2.00/28.0041
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