III Algebraic Topology - Vector bundles

10Vector bundles

III Algebraic Topology

10.3 The Thom isomorphism theorem

We now get to the main theorem about vector bundles.

Theorem

(Thom isomorphism theorem)

Let

E → X

be a

-dimensional

vector bundle, and {ε

}

x∈X

be an R-orientation of E. Then

(i) H

(E, E

; R) = 0 for i < d.

(ii)

There is a unique class

∈ H

(

E, E

;

) which restricts to

on each

fiber. This is known as the Thom class.

(iii) The map Φ given by the composition

(X; R) H

(E; R) H

i+d

(E, E

; R)

∗

−^u

is an isomorphism.

Note that (i) follows from (iii), since H

(X; R) = 0 for i < 0.

Before we go on and prove this, we talk about why it is useful.

Definition

(Euler class)

Let

E → X

be a vector bundle. We define the

Euler class e(E) ∈ H

(X; R) by the image of u

under the composition

(E, E

; R) H

(E; R) H

(X; R)

∗

This is an example of a characteristic class, which is a cohomology class

related to an oriented vector bundle that behaves nicely under pullback. More

precisely, given a vector bundle

E → X

and a map

Y → X

, we can form

a pullback

∗

E E

Y X

∗

Since we have a fiberwise isomorphism (

∗

)

∼

f(y)

, an

-orientation for

induces one for

∗

, and we know

∗

(

) =

∗

by uniqueness of the Thom

class. So we know

e(f

∗

(E)) = f

∗

e(E) ∈ H

(Y ; R).

Now the Euler class is a cohomology class of

itself, so we can use the Euler

class to compare and contrast different vector bundles.

How can we think of the Euler class? It turns out the Euler class gives us an

obstruction to the vector bundle having a non-zero section.

Theorem.

If there is a section

X → E

which is nowhere zero, then

(

) =

0 ∈ H

(X; R).

Proof.

Notice that any two sections of

E → X

are homotopic. So we have

e ≡ s

∗

. But since

∈ H

(

E, E

;

), and

maps into

, we have

∗

Perhaps more precisely, we look at the long exact sequence for the pair

(E, E

), giving the diagram

(E, E

; R) H

(E; R) H

; R)

(X; R)

∗

Since

and

are homotopic, the diagram commutes. Also, the top row is exact.

∈ H

(

E, E

;

) gets sent along the top row to 0

∈ H

(

;

), and thus

∗

sends it to 0

∈ H

(

;

). But the image in

(

;

) is exactly the Euler

class. So the Euler class vanishes.

Now cohomology is not only a bunch of groups, but also a ring. So we can

ask what happens when we cup u

with itself.

Theorem. We have

^ u

= Φ(e(E)) = π

∗

(e(E)) ^ u

∈ H

∗

(E, E

; R).

This is just tracing through the definitions.

Proof. By construction, we know the following maps commute:

(E, E

; R) ⊗ H

(E, E

; R) H

(E, E

; R)

(E; R) ⊗ H

(E, E

; R)

∗

⊗id

We claim that the Thom class

⊗ u

∈ H

(

E, E

;

)

⊗ H

(

E, E

;

) is sent

to π

∗

(e(E)) ⊗ u

∈ H

(E; R) ⊗ H

(E, E

; R).

By definition, this means we need

∗

= π

∗

(e(E)),

and this is true because π

∗

is homotopy inverse to s

∗

and e(E) = s

∗

So if we have two elements Φ(c), Φ(d) ∈ H

∗

(E, E

; R), then we have

Φ(c) ^ Φ(d) = π

∗

c ^ u

^ π

∗

d ^ u

= ±π

∗

c ^ π

∗

d ^ u

^ u

= ±π

∗

(c ^ d ^ e(E)) ^ u

= ±Φ(c ^ d ^ e(E)).

(

) is precisely the information necessary to recover the cohomology ring

∗

(E, E

; R) from H

∗

(X; R).

Lemma.

E → X

is a

-dimensional

-module vector bundle with

odd,

then 2e(E) = 0 ∈ H

(X; R).

Proof.

Consider the map

E → E

given by negation on each fiber. This then

gives an isomorphism

∗

: H

(E, E

; R) H

(E, E

; R).

∼

This acts by negation on the Thom class, i.e.

∗

) = −u

as on the fiber

, we know

is given by an odd number of reflections, each

of which acts on

(

, E

;

) by

−

1 (by the analogous result on

). So we

change

by a sign. We then lift this to a statement about

by the fact that

is the unique thing that restricts to ε

for each x.

But we also know

a ◦ s

= s

which implies

∗

)) = s

∗

Combining this with the result that a

∗

) = −u

, we get that

2e(E) = 2s

∗

) = 0.

This is a disappointing result, because if we already know that

(

;

) has

no 2-torsion, then e(E) = 0.

After all that fun, we prove the Thom isomorphism theorem.

Proof of Thom isomorphism theorem.

We will drop the “

” in all our diagrams

for readability (and also so that it fits in the page).

We first consider the case where the bundle is trivial, so

X × R

. Then

we note that

∗

, R

\ {0}) =

(

R ∗ = d

0 ∗ 6= d

In particular, the modules are free, and (a relative version of) K¨unneth’s theorem

tells us the map

× : H

∗

(X) ⊗ H

∗

, R

\ {0}) H

∗

(X × R

, X × (R

\ {0}))

∼

is an isomorphism. Then the claims of the Thom isomorphism theorem follow

immediately.

(i)

For

i < d

, all the summands corresponding to

(

X × R

, X ×

(

\ {

}

))

vanish since the H

∗

, R

\ {0}) term vanishes.

(ii) The only non-vanishing summand for H

(X × R

, X × (R

\ {0}) is

(X) ⊗ H

, R

\ {0}).

Then the Thom class must be 1

⊗ u

, where

is the object corresponding

to ε

∈ H

, E

) = H

, R

\ {0}), and this is unique.

(iii) We notice that Φ is just given by

Φ(x) = π

∗

(x) ^ u

= x × u

which is an isomorphism by K¨unneth.

We now patch the result up for a general bundle. Suppose

E → X

is a

bundle. Then it has an open cover of trivializations, and moreover if we assume

our

is compact, there are finitely many of them. So it suffices to show that

U, V ⊆ X

are open sets such that the Thom isomorphism the holds for

restricted to U, V, U ∩ V , then it also holds on U ∪ V .

The relative Mayer-Vietoris sequence gives us

d−1

(E|

U∩V

, E

U∩V

) H

(E|

U∪V

, E

U∪V

)

(E|

, E

) ⊕ H

(E|

, E

) H

(E|

U∩V

, E

U∩V

∂

We first construct the Thom class. We have

∈ H

(E|

, E

), u

∈ H

(E|

, E

We claim that (

, u

)

∈ H

(

, E

)

⊕ H

(

, E

) gets sent to 0

∗

− i

∗

. Indeed, both the restriction of

and

U ∩ V

are Thom

classes, so they are equal by uniqueness, so the difference vanishes.

Then by exactness, there must be some

U∪V

∈ H

(

U∪V

, E

U∪V

) that

restricts to

and

respectively. Then this must be a Thom

class, since the property of being a Thom class is checked on each fiber. Moreover,

we get uniqueness because

d−1

(

U∩V

, E

U∩V

) = 0, so

and

must

be the restriction of a unique thing.

The last part in the Thom isomorphism theorem come from a routine appli-

cation of the five lemma, and the first part follows from the last as previously

mentioned.