III Algebraic Topology - Vector bundles

10Vector bundles

III Algebraic Topology

10.1 Vector bundles

We now end the series of random topics, and work on a more focused topic. We

are going to look at vector bundles. Intuitively, a vector bundle over a space

is a continuous assignment of a vector space to each point in

. In the first

section, we are just going to look at vector bundles as topological spaces. In the

next section, we are going to look at homological properties of vector bundles.

Definition

(Vector bundle)

Let

be a space. A (real) vector bundle of

dimension

over

is a map

E → X

, with a (real) vector space structure

on each fiber

−1

(

{x}

), subject to the local triviality condition: for each

x ∈ X

, there is a neighbourhood

and a homeomorphism

−1

(U) → U × R

such that the following diagram commutes

U × R

and for each

y ∈ U

, the restriction

ϕ|

→ {y}× R

is a linear isomorphism

for each y ∈ U . This maps is known as a local trivialization.

We have an analogous definition for complex vector bundles.

Definition

(Section)

A section of a vector bundle

E → X

is a map

s : X → E such that π ◦ s = id. In other words, s(x) ∈ E

for each x.

Definition

(Zero section)

The zero section of a vector bundle is

X → E

given by s

(x) = 0 ∈ E

Note that the composition

E X E

is homotopic to the identity map on id

, since each E

is contractible.

One important operation we can do on vector bundles is pullback:

Definition

(Pullback of vector bundles)

Let

E → X

be a vector bundle,

and f : Y → X a map. We define the pullback

∗

E = {(y, e) ∈ Y × E : f(y) = π(e)}.

This has a map

∗

E → Y

given by projecting to the first coordinate. The

vector space structure on each fiber is given by the identification (

∗

)

f(y)

It is a little exercise in topology to show that the local trivializations of

E → X

induce local trivializations of f

∗

π : f

∗

E → Y .

Everything we can do on vector spaces can be done on vector bundles, by

doing it on each fiber.

Definition

(Whitney sum of vector bundles)

Let

E → F

and

F → X

be vector bundles. The Whitney sum is given by

E ⊕ F = {(e, f) ∈ E × F : π(e) = ρ(f)}.

This has a natural map

π ⊕ ρ

E ⊕ F → X

given by (

π ⊕ ρ

)(

e, f

) =

(

) =

(

This is again a vector bundle, with (

E ⊕ F

)

⊕ F

and again local

trivializations of E and F induce one for E ⊕ F .

Tensor products can be defined similarly.

Similarly, we have the notion of subbundles.

Definition

(Vector subbundle)

Let

E → X

be a vector bundle, and

F ⊆ E

a subspace such that for each x ∈ X there is a local trivialization (U, ϕ)

U × R

such that

takes

F |

U × R

, where

⊆ R

. Then we say

is a vector

sub-bundle.

Definition

(Quotient bundle)

Let

be a sub-bundle of

. Then

E/F

, given

by the fiberwise quotient, is a vector bundle and is given by the quotient bundle.

We now look at one of the most important example of a vector bundle. In

some sense, this is the “universal” vector bundle, as we will see later.

Example (Grassmannian manifold). We let

X = Gr

) = {k-dimensional linear subgroups of R

To topologize this space, we pick a fixed

V ∈ Gr

(

). Then any

-dimensional

subspace can be obtained by applying some linear map to

. So we obtain a

surjection

(R) → Gr

)

M 7→ M (V ).

So we can given Gr

) the quotient (final) topology. For example,

n+1

) = RP

Now to construct a vector bundle, we need to assign a vector space to each point

. But a point in

(

) is a vector space, so we have an obvious definition

E = {(V, v) ∈ Gr

) × R

: v ∈ V }.

This has the evident projection

E → X

given by the first projection. We

then have

= V.

To see that this is a vector bundle, we have to check local triviality. We fix a

V ∈ Gr

), and let

U = {W ∈ Gr

) : W ∩ V

⊥

= {0}}.

We now construct a map

→ U × V

∼

U × R

by mapping (

W, w

) to

(W, pr

(w)), where pr

: R

→ V is the orthogonal projection.

Now if

w ∈ U

, then

(

)

= 0 since

W ∩ V

⊥

{

}

. So

is a homeomor-

phism. We call this bundle γ

k,n

→ Gr

In the same way, we can get a canonical complex vector bundle

k,n

→

Example.

Let

be a smooth

-dimensional manifold, then it naturally has a

d-dimensional tangent bundle π : T M → M with (T M)|

= T

M ⊆ N

is a smooth submanifold, with

the inclusion map, then

T M

a subbundle of

∗

T N

. Note that we cannot say

T M

is a smooth subbundle of

T N

, since they have different base space, and thus cannot be compared without

pulling back.

The normal bundle of M in N is

M⊆N

∗

T N

T M

Here is a theorem we have to take on faith, because proving it will require

some differential geometry.

Theorem

(Tubular neighbourhood theorem)

Let

M ⊆ N

be a smooth sub-

manifold. Then there is an open neighbourhood

and a homeomorphism

M⊆N

→ U

, and moreover, this homeomorphism is the identity on

(where

we view M as a submanifold of ν

M⊆N

by the image of the zero section).

This tells us that locally, the neighbourhood of M in N looks like ν

M⊆N

We will also need some results from point set topology:

Definition

(Partition of unity)

Let

be a compact Hausdorff space, and

}

α∈I

be an open cover. A partition of unity subordinate to

}

is a

collection of functions λ

: X → [0, ∞) such that

(i) supp(λ

) = {x ∈ X : λ

(x) > 0} ⊆ U

(ii) Each x ∈ X lies in finitely many of these supp(λ

(iii) For each x, we have

α∈I

(x) = 1.

Proposition. Partitions of unity exist for any open cover.

You might have seen this in differential geometry, but this is easier, since we

do not require the partitions of unity to be smooth.

Using this, we can prove the following:

Lemma.

Let

E → X

be a vector bundle over a compact Hausdorff space.

Then there is a continuous family of inner products on

. In other words, there

is a map E ⊗ E → R which restricts to an inner product on each E

Proof.

We notice that every trivial bundle has an inner product, and since every

bundle is locally trivial, we can patch these up using partitions of unity.

Let {U

}

α∈I

be an open cover of X with local trivializations

: E|

→ U

× R

The inner product on

then gives us an inner product on

, say

h · , · i

We let

be a partition of unity associated to

}

. Then for

u ⊗ v ∈ E ⊗ E

we define

hu, vi =

α∈I

(π(u))hu, vi

Now if

(

) =

(

) is not in

, then we don’t know what we mean by

hu, vi

but it doesn’t matter, because

(

)) = 0. Also, since the partition of unity

is locally finite, we know this is a finite sum.

It is then straightforward to see that this is indeed an inner product, since a

positive linear combination of inner products is an inner product.

Similarly, we have

Lemma.

Let

E → X

be a vector bundle over a compact Hausdorff space.

Then there is some N such that E is a vector subbundle of X × R

Proof.

Let

}

be a trivializing cover of

. Since

is compact, we may wlog

assume the cover is finite. Call them U

, · · · , U

. We let

: E|

→ U

× R

We note that on each patch,

embeds into a trivial bundle, because it is a

trivial bundle. So we can add all of these together. The trick is to use a partition

of unity, again.

We define f

to be the composition

× R

Then given a partition of unity λ

, we define

f : E → X × (R

)

v 7→ (π(v), λ

(π(v))f

(v), λ

(π(v))f

(v), · · · , λ

(π(v))f

(v)).

We see that this is injective. If

v, w

belong to different fibers, then the first

coordinate distinguishes them. If they are in the same fiber, then there is some

with

(

))

= 0. Then looking at the

th coordinate gives us distinguishes

them. This then exhibits E as a subbundle of X × R

Corollary.

Let

E → X

be a vector bundle over a compact Hausdorff space.

Then there is some

F → X

such that

E ⊕ F

∼

X × R

. In particular,

embeds as a subbundle of a trivial bundle.

Proof.

By above, we can assume

is a subbundle of a trivial bundle. We can

then take the orthogonal complement of E.

Now suppose again we have a vector bundle

E → X

over a compact

Hausdorff

. We can then choose an embedding

E ⊆ X × R

, and then we get

a map

X → Gr

(

) sending

⊆ R

. Moreover, if we pull back the

tautological bundle along f

, then we have

∗

k,N

∼

So every vector bundle is the pullback of the canonical bundle

k,N

over a

Grassmannian. However, there is a slight problem. Different vector bundles

will require different

’s. So we will have to overcome this problem if we want

to make a statement of the sort “a vector bundle is the same as a map to a

Grassmannian”.

The solution is to construct some sort of

(

∞

). But infinite-dimensional

vector spaces are weird. So instead we take the union of all

(

). We note

that for each

, there is an inclusion

(

)

→ Gr

(

n+1

), which induces

an inclusion of the canonical bundle. We can then take the union to obtain

(

∞

) with a canonical bundle

. Then the above result shows that each

vector bundle over

is the pullback of the canoncial bundle

along some map

f : X → Gr

∞

Note that a vector bundle over

does not uniquely specify a map

X →

(

∞

), as there can be multiple embeddings of

into the trivial bundle

X × R

. Indeed, if we wiggle the embedding a bit, then we can get a new bundle.

So we don’t have a coorespondence between vector bundles

E → X

and maps

: X → Gr

∞

The next best thing we can hope for is that the “wiggle the embedding a

bit” is all we can do. More precisely, two maps

f, g

X → Gr

(

∞

) pull back

isomorphic vector bundles if and only if they are homotopic. This is indeed true:

Theorem. There is a correspondence







homotopy classess

of maps

f : X → Gr

∞

)







(

d-dimensional

vector bundles

π : E → X

)

[f] f

∗

] π

The proof is mostly technical, and is left as an exercise on the example sheet.