Vector Bundles

Math 225A Discussion Session Week 5 Notes November 1, 2019

Per a request, we’ll spend today’s discussion reviewing some material on vector bundles. Much (but not all) of this will be a repeat of material from lecture. Another relatively readable source is Conlon’s Diﬀerentiable Manifolds (particularly sections 3.3 and 3.4).

Vector bundles Deﬁnition. Let M be a smooth n-manifold, E a smooth (n + k)-manifold, and π : E → M a smooth map. We will call π : E → M a vector bundle of rank k if:

−1 (1) For each p ∈ M, Ep := π (p) has the structure of a k-dimensional vector space over R.

(2) There is a locally trivializing cover of M for E. That is, an open cover {Uα}α∈A of M with diﬀeomorphisms −1 k ψα : π (Uα) → Uα × R

−1 such that p1 ◦ ψα = π|π (Uα), where p1 is projection onto the ﬁrst factor.

k (3) For each α ∈ A and p ∈ Uα, ψα|Ep : Ep → {p} × R is a vector space isomorphism.

We call E the total space, M the base space, and call each (Uα, ψα) a trivializing neighborhood.

k Remark. Notice that E = M × R is a vector bundle of rank k, with the obvious projection map and local trivialization. The deﬁnition of vector bundles is meant to generalize this. A vector bundle has the local appearance of a product, but may have nontrivial global topology.

Let us deﬁne a notion of equivalence for vector bundles.

Deﬁnition. Let πi : Ei → M be a vector bundle of rank k, for i = 0, 1. A vector bundle isomorphism between π0 : E0 → M and π1 : E1 → M is a diﬀeomorphism ϕ: E1 → E2 such that the following diagram commutes: ϕ E0 E1 , π0 π1 M −1 −1 and such that ϕ: π0 (p) → π1 (p) is a vector space isomorphism, for each p ∈ M.

1 1 Example. Let M = S and consider the trivial bundle E = S × R, with π given by projection ∂ onto the first factor. We have a vector bundle isomorphism ϕ: E → TM defined as follows. Let ∂θ be a nonvanishing vector field on S1. Then we can define ∂ ϕ(p, c) = c | ∈ T M. ∂θ p p Check that ϕ is indeed a vector bundle isomorphism. Note that tangent bundles aren’t usually trivial — when TM is trivial, we say that M is parallelizable.

1 Cocycles As we said above, a vector bundle has the local appearance of a product — that is, just a trivial bundle — and each trivializing neighborhood gives us a way of identifying this product structure. But what happens where the trivializing neighborhoods overlap? Say we have p ∈ Uα ∩ Uβ. Then k we can identify Ep with R using either ψα or ψβ:

k ψα ψβ k {p} × R Ep {p} × R .

−1 k k So we have a vector space isomorphism ψβ ◦ ψα : {p} × R → {p} × R , which we can consider as k an element ψαβ(p) ∈ GL(k). This tells us how to take a vector in {p}×R , as identiﬁed by ψα, and −1 produce its coordinates in the trivialization ψβ. Notice that, by construction, ψβα(p) = (ψαβ(p)) . What happens on triple intersections? This time we have a diagram of the form

ψα ψγ ψβ ,

k k k R R R where we’ve suppressed the ﬁrst coordinate in the second row. Now ψαγ(p) ∈ GL(k) is obtained from −1 −1 −1 ψγ ◦ ψα = (ψγ ◦ ψβ ) ◦ (ψβ ◦ ψα ), so ψαγ(p) = ψβγ(p)ψαβ(p). Note that we recover the previous property of ψαβ from this one by taking γ = α.

Deﬁnition. We call the data of the open cover {Uα}α∈A of M and the maps {ψαβ : Uα ∩ Uβ → GL(k)} the structure cocycle for the vector bundle π : E → M. More generally, a GL(k)-cocycle on M is an open cover {Vα}α∈A of M along with maps ϕαβ : Vα ∩ Vβ → GL(k) satisfying the cocycle condition ϕαγ = ϕβγϕαβ, for all α, β, γ ∈ A.

So we obtain a GL(k)-cocycle on M from any vector bundle of rank k on M (namely, the structure cocycle). Let us now consider the converse problem of constructing a vector bundle from a GL(k)-cocycle.

Let the open cover of our GL(k)-cocycle be given by {Vα}α∈A, with transition maps given by ϕαβ : Vα ∩ Vβ → GL(k). Consider the disjoint union

˜ a k E := Vα × R , α∈A and deﬁne an equivalence relation on E˜ by

k k (p, v) ∈ Vα × R ∼ (p, w) ∈ Vβ × R if ϕαβ(p) · v = w.

2 (Check that this is an equivalence relation.) Of course we have a projection mapπ ˜ : E˜ → M, k deﬁned on each Vα × R as projection onto the ﬁrst component. This plays nicely with ∼, and thus we have π : E → M. It remains to check that this is a vector bundle of rank k on M. Notice that   −1 a k π (p) =  {p} × R  / ∼ . Vα3p

k Each copy of {p} × R in the disjoint union has a natural vector space structure, and ∼ identiﬁes all of these as a single vector space. Checking that we have a locally trivializing cover is left to you.

1 2 2 2 Example. Consider the following pair of GL(1)-cocycles on M = S = {x + y = 1} ⊂ R . First, take the open sets 1 Uα = Uγ = {(x, y) ∈ S |y > −} and 1 Uβ = Uδ = {(x, y) ∈ S |y < }, for some small > 0. We will construct one GL(1)-cocycle with open sets Uα and Uβ, and another with open sets Uγ and Uδ. We deﬁne ϕαα, ϕαβ, ϕβα, and ϕββ to all be identically 1. Then the vector bundle constructed from the cocycle

{Uα,Uβ; ϕαα, ϕαβ, ϕβα, ϕββ} is certainly trivial — it consists of two vector bundles over intervals (necessarily trivial) glued together trivially. On the other hand, we can deﬁne ϕγδ according to x ϕ (x, y) = , γδ |x| deﬁning ϕγγ, ϕδγ, and ϕδδ in the obvious way. Then the vector bundle constructed from

{Uγ,Uδ; ϕγγ, ϕγδ, ϕδγ, ϕδδ} is isomorphic to the M¨obiusband. This time we take a pair of line bundles over the interval and glue them together with a twist. We will see below that this vector bundle is not isomorphic to 1 S × R.

We can now identify the collection of vector bundles of rank k over M with the collection of GL(k)-cocycles on M, once we identify cocycles which lead to the same vector bundle.

Deﬁnition. We say that a pair of GL(k)-cocycles on M is equivalent if there is a common GL(k)- cocycle on M in which both are contained. (i.e., the common GL(k)-cocycle contains every set and map which appears in each of the ﬁrst two.) Note that in this (somewhat strange) formulation, the open sets in a GL(k)-cocycle need not be distinct.

Here’s a lemma which is left for you to prove:

3 Lemma 1. Equivalence of GL(k)-cocycles on M is an equivalence relation.

Hint: Show that if two GL(k)-cocycles on M contain a common GL(k)-cocycle, then they are contained in a common GL(k)-cocycle.

Here’s another fact that’s not so diﬃcult to prove:

Theorem 2. The isomorphism class of a vector bundle constructed from a GL(k)-cocycle depends only on the equivalence class of the cocycle.

From this it follows that there is a canonical bijective correspondence between isomorphism classes of vector bundles of rank k on M and equivalence classes of GL(k)-cocycles on M.

Example. Consider the GL(1)-cocycles on S1 constructed in the previous example. We suspect that the nontrivial GL(1)-cocycle determines a nontrivial vector bundle, but how do we prove this? One thing we can do is this: if the bundle is trivial, then the GL(1)-cocycle is equivalent to the trivial GL(1)-cocycle deﬁned above. In particular, the two GL(1)-cocycles are contained in a common cocycle. If this is the case, then the sign of ϕαγ must be constant, since Uα ∩ Uγ is connected. Similarly, the sign of ϕβδ is constant. The cocycle condition tells us that

ϕαγ = ϕβγϕαβ = ϕδγϕβδϕαβ.

But then ϕαγ(1, 0) = ϕδγ(1, 0)ϕβδ(1, 0)ϕαβ(1, 0) = ϕβδ(1, 0), while ϕαγ(−1, 0) = ϕδγ(−1, 0)ϕβδ(−1, 0)ϕαβ(−1, 0) = −ϕβδ(−1, 0), contradicting our assertion that the signs of ϕαγ and ϕβδ are constant. So these GL(1)-cocycles are inequivalent, and thus the corresponding vector bundles are not isomorphic. Perhaps less trivial is checking that every GL(1)-cocycle on S1 is equivalent to one of these two.

Reduction of the structure group The GL(k)-cocycle (equivalence class) associated to a rank k vector bundle π : E → M provides open sets of M over which E is trivial, as well as transition data telling us how to patch together the various local pieces. The map

ϕαβ : Uα ∩ Uβ → GL(k) should be thought of as a “change of coordinates,” telling us how to translate from the coordinates on Ep given by ψα into those given by ψβ, for each p ∈ Uα ∩ Uβ.

The fact that ϕαβ takes values in GL(k) corresponds to our requirement that the change of coordinates maps (on the ﬁbers Ep) be vector space isomorphisms. We can sometimes ask more of our vector bundles, insisting that this transition data be valued in some subgroup of GL(k).

4 Deﬁnition. Let G ⊆ GL(k) be a subgroup and let π : E → M be a vector bundle of rank k on M.A reduction of the structure group of E to G is a GL(k)-cocycle representing the isomorphism class of E, all of whose transition maps are valued in G. We also call this cocycle a G-cocycle for E. Remark. If G is a proper subgroup, there is a priori no reason for π : E → M to admit a reduction of its structure group to G.

Example. Consider G = GL+(k) ⊂ GL(k), the subgroup of maps with positive determinant. We say that π : E → M is orientable if E admits a GL+(k)-cocycle. In this case, we may use each of k k our local trivializations ψi : Ui → Ui × R to make an oriented identiﬁcation of Ep with R , and the transition data respect these identiﬁcations. We say that the manifold M is orientable precisely when the vector bundle TM is orientable.

By construction, trivial vector bundles are always orientable (we can construct them from GL(k)-cocycles whose maps are constant, valued at the identity map). It follows that TS1 is orientable. Show that the nontrivial vector bundle on S1 constructed above is not orientable. Example. Consider the subgroup O(k) ⊂ GL(k) consisting of orthogonal matrices. Remember k that these are precisely the linear maps which preserve the natural inner product structure on R . Thus if a vector bundle π : E → M admits a reduction of its structure group to O(k), we can endow each of its ﬁbers with an inner product, and the transition maps will respect this. An n-manifold M whose tangent bundle TM has seen its structure group reduced to O(n) is called a Riemannian manifold, and we say that M has been given a Riemannian structure. Unlike orientations, every smooth manifold can be given a Riemannian structure1. There is an important point to be made about equivalence of G-cocycles. Namely, a pair of G-cocycles may be equivalent when realized as GL(k)-cocycles (because there is a common GL(k)- cocycles containing them), but inequivalent as G-cocycles. Indeed, suppose M is orientable. Then there are two orientations with which we may endow TM. The corresponding GL+(n)-cocycles are equivalent as GL(n)-cocycles (because they both correspond to TM), but not as GL+(n)- cocycles, since any common reﬁnement of these cocycles will contain a transition map with negative determinant. Example. Another important example of structure group reduction which we won’t seriously pursue here is given by almost complex structures. We may write GL(n, C) ⊂ GL(2n) by realizing each matrix A + iB ∈ GL(n, C) as AB ∈ GL(2n). −BA We say that a smooth 2n-manifold M admits an almost complex structure if the structure group of TM can be reduced to GL(n, C). In this case we can give each tangent space TpM the structure of a complex vector space, and the transition data will respect these structures. Certainly if M admits the structure of a complex n-manifold, then M admits an almost complex structure. But in fact an almost complex structure need not arise in this way.

1This is not obvious.