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The Universal Bundle

在文檔中 1.2. Classifying Vector Bundles (頁 31-35)

We will show that there is a special n dimensional vector bundle En

Gnwith the

property that all n dimensional bundles over paracompact base spaces are obtainable as pullbacks of this single bundle. When n= 1 this bundle will be just the canonical line bundle over RP, defined earlier. The generalization to n > 1 will consist in

replacing RP, the space of 1 dimensional vector subspaces ofR, by the space of n dimensional vector subspaces of R.

First we define the Grassmann manifold Gn(Rk) for nonnegative integers n≤ k.

As a set this is the collection of all n dimensional vector subspaces of Rk, that is, n dimensional planes in Rk passing through the origin. To define a topology on Gn(Rk) we first define the Stiefel manifold Vn(Rk) to be the space of orthonormal n frames in Rk, in other words, n tuples of orthonormal vectors in Rk. This is a subspace of the product of n copies of the unit sphere Sk−1, namely, the subspace of orthogonal n tuples. It is a closed subspace since orthogonality of two vectors can be expressed by an algebraic equation. Hence Vn(Rk) is compact since the product of spheres is compact. There is a natural surjection Vn(Rk)

Gn(Rk) sending an n frame to the subspace it spans, and Gn(Rk) is topologized by giving it the quotient topology with respect to this surjection. So Gn(Rk) is compact as well. Later in this section we will construct a finite CW complex structure on Gn(Rk) and in the process show that it is Hausdorff and a manifold of dimension n(k− n).

The inclusions Rk ⊂ Rk+1 ⊂ ··· give inclusions Gn(Rk) ⊂ Gn(Rk+1)⊂ ···, and we let Gn(R)=S

kGn(Rk) . This is the set of all n dimensional vector subspaces of the vector space R since if we choose a basis for a finite-dimensional subspace of R, each basis vector will lie in some Rk, hence there will be anRk containing all the basis vectors and therefore the whole subspace. We give Gn(R) the weak or direct limit topology, so a set in Gn(R) is open iff it intersects each Gn(Rk) in an open set.

There are canonical n dimensional vector bundles over Gn(Rk) and Gn(R) . Define En(Rk) = { (`, v) ∈ Gn(Rk)×Rk | v ∈ ` }. The inclusions Rk ⊂ Rk+1 ⊂ ···

give inclusions En(Rk)⊂ En(Rk+1)⊂ ··· and we set En(R)=S

kEn(Rk) , again with the weak or direct limit topology.

L

emma 1.15. The projection p : En(Rk)

Gn(Rk) , p(`, v)= `, is a vector bundle., both for finite and infinite k .

P

roof: First suppose k is finite. For `∈ Gn(Rk) , let π`:Rk

` be orthogonal projec-tion and let U`= { `0∈ Gn(Rk) ||π`(`0) has dimension n}. In particular, ` ∈ U`. We will show that U` is open in Gn(Rk) and that the map h : p−1(U`)

U`×` ≈ U`×Rn

defined by h(`0, v)= (`0, π`(v)) is a local trivialization of En(Rk) .

For U` to be open is equivalent to its preimage in Vn(Rk) being open. This preimage consists of orthonormal frames v1,··· , vn such that π`(v1),··· , π`(vn) are independent. Let A be the matrix of π` with respect to the standard basis in the domain Rk and any fixed basis in the range ` . The condition on v1,··· , vn is then that the n×n matrix with columns Av1,··· , Avn have nonzero determinant.

Since the value of this determinant is obviously a continuous function of v1,··· , vn, it follows that the frames v1,··· , vn yielding a nonzero determinant form an open set in Vn(Rk) .

It is clear that h is a bijection which is a linear isomorphism on each fiber. We need to check that h and h−1 are continuous. For `0∈ U` there is a unique invertible linear map L`0:Rk

Rk restricting to π` on `0 and the identity on `= Ker π`. We claim that L`0, regarded as a k×k matrix, depends continuously on `0. Namely, we can write L`0 as a product AB−1 where:

— B sends the standard basis to v1,··· , vn, vn+1,··· , vk with v1,··· , vn an or-thonormal basis for `0 and vn+1,··· , vk a fixed basis for `.

— A sends the standard basis to π`(v1),··· , π`(vn), vn+1,··· , vk.

Both A and B depend continuously on v1,··· , vn. Since matrix multiplication and matrix inversion are continuous operations (think of the ‘classical adjoint’ formula for the inverse of a matrix), it follows that the product L`0 = AB−1 depends continuously on v1,··· , vn. But since L`0 depends only on `0, not on the basis v1,··· , vnfor `0, it follows that L`0 depends continuously on `0 since Gn(Rk) has the quotient topology from Vn(Rk) . Since we have h(`0, v) = (`0, π`(v))= (`0, L`0(v)) , we see that h is continuous. Similarly, h−1(`0, w) = (`0, L−1`0 (w)) and L−1`0 depends continuously on

`0, matrix inversion being continuous, so h−1 is continuous.

This finishes the proof for finite k . When k= ∞ one takes U` to be the union of the U`’s for increasing k . The local trivializations h constructed above for finite k then fit together to give a local trivialization over this U`, continuity being automatic

since we use the weak topology. tu

We will mainly be interested in the case k= ∞ now, and to simplify notation we will write Gn for Gn(R) and En for En(R) . As earlier in this section, we use the notation [X, Y ] for the set of homotopy classes of maps f : X

Y .

T

heorem 1.16. For paracompact X , the map [X, Gn]

Vectn(X) , [f ]

,

f(En) ,

is a bijection.

Thus, vector bundles over a fixed base space are classified by homotopy classes of maps into Gn. Because of this, Gnis called the classifying space for n dimensional vector bundles and En

Gn is called the universal bundle.

As an example of how a vector bundle could be isomorphic to a pullback f(En) , consider the tangent bundle to Sn. This is the vector bundle p : E

Sn where E =

{ (x, v) ∈ Sn×Rn+1| x ⊥ v }. Each fiber p−1(x) is a point in Gn(Rn+1) , so we have a map Sn

Gn(Rn+1) , x

,

p−1(x) . Via the inclusionRn+1

>

R we can view this as a map f : Sn

Gn(R)= Gn, and E is exactly the pullback f(En) .

P

roof of 1.16: The key observation is the following: For an n dimensional vector bundle p : E

X , an isomorphism E≈ f(En) is equivalent to a map g : E

R that

is a linear injection on each fiber. To see this, suppose first that we have a map f : X

Gn and an isomorphism E≈ f(En) . Then we have a commutative diagram

−−→ −−→ −−−−−→ −−−−−→

E

E f E

−−−−−→

−−−−−→

X f G

f p

π

( )

n n

n R

where π (`, v)= v . The composition across the top row is a map g : E

R that is

a linear injection on each fiber, since both ef and π have this property. Conversely, given a map g : E

R that is a linear injection on each fiber, define f : X

Gn by

letting f (x) be the n plane g(p−1(x)) . This clearly yields a commutative diagram as above.

To show surjectivity of the map [X, Gn]

-→

Vectn(X) , suppose p : E

X is an

n dimensional vector bundle. Let {Uα} be an open cover of X such that E is trivial over each Uα. By Lemma 1.21 in the Appendix to this chapter there is a countable open cover {Ui} of X such that E is trivial over each Ui, and there is a partition of unity i} with ϕi supported in Ui. Let gi: p−1(Ui)

Rn be the composition of a trivialization p−1(Ui)

Ui×Rn with projection onto Rn. The map (ϕip)gi, v

,

ϕi(p(v))gi(v) , extends to a map E

Rn that is zero outside p−1(Ui) . Near each point of X only finitely many ϕi’s are nonzero, and at least one ϕi is nonzero, so these extended (ϕip)gi’s are the coordinates of a map g : E

(Rn)= R that is

a linear injection on each fiber.

For injectivity, if we have isomorphisms E ≈ f0(En) and E ≈ f1(En) for two maps f0, f1: X

Gn, then these give maps g0, g1: E

R that are linear injections on fibers, as in the first paragraph of the proof. We claim g0 and g1 are homotopic through maps gt that are linear injections on fibers. If this is so, then f0 and f1 will be homotopic via ft(x)= gt(p−1(x)) .

The first step in constructing a homotopy gt is to compose g0with the homotopy Lt:R

R defined by Lt(x1, x2,···) = (1 − t)(x1, x2,···) + t(x1, 0, x2, 0,···). For each t this is a linear map whose kernel is easily computed to be 0 , so Lt is injective.

Composing the homotopy Lt with g0 moves the image of g0 into the odd-numbered coordinates. Similarly we can homotope g1 into the even-numbered coordinates. Still calling the new g ’s g0 and g1, let gt = (1 − t)g0+ tg1. This is linear and injective on fibers for each t since g0 and g1 are linear and injective on fibers. tu

Usually [X, Gn] is too difficult to compute explicitly, so this theorem is of limited use as a tool for explicitly classifying vector bundles over a given base space. Its importance is due more to its theoretical implications. Among other things, it can reduce the proof of a general statement to the special case of the universal bundle.

For example, it is easy to deduce that vector bundles over a paracompact base have inner products, since the bundle En

Gn has an obvious inner product obtained by restricting the standard inner product inR to each n plane, and this inner product on En induces an inner product on every pullback f(En) .

The preceding constructions and results hold equally well for vector bundles over C, with Gn(Ck) the space of n dimensional C linear subspaces of Ck, and so on. In particular, the proof of Theorem 1.16 translates directly to complex vector bundles, showing that VectnC(X)≈ [X, Gn(C)] .

There is also a version of Theorem ?? for oriented vector bundles. Let eGn(Rk) be the space of oriented n planes in Rk, the quotient space of Vn(Rk) obtained by identifying two n frames when they determine the same oriented subspace of Rk. Forming the union over increasing k we then have the space eGn(R) . The universal oriented bundle eEn(R) over eGn(R) consists of pairs (`, v)∈ eGn(R)×R with v ∈ `. In other words, eEn(R) is the pullback of En(R) via the natural projection Gen(R)

Gn(R) . Small modifications in the proof that Vectn(X) ≈ [X, Gn(R)]

show that Vectn+(X)≈ [X, eGn(R)] .

Both Gn(R) and eGn(R) are path-connected since Vectn(X) and Vectn+(X) have a single element when X is a point. The natural projection eGn(R)

Gn(R)

ob-tained by ignoring orientations is two-to-one, and readers familiar with the notion of a covering space will have no trouble in recognizing that this two-to-one projection map is a covering space, using for example the local trivializations constructed in Lemma

??. In fact eGn(R) is the universal cover of Gn(R) since it is simply-connected, be-cause of the triviality of Vectn+(S1)≈ [S1, eGn(R)] . One can also observe that a vector bundle E≈ f(En(R) is orientable iff its classifying map f : X

Gn(R) lifts to a map ef : X

Gen(R) , and in fact orientations of E correspond bijectively with lifts ef .

在文檔中 1.2. Classifying Vector Bundles (頁 31-35)

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