THE GEOMETRY OF VECTOR BUNDLES AND

AN INTRODUCTION TO GAUGE THEORY

LECTURE 12

Professor Steven Bradlow Class Notes From Math 433

University of Illinois at Urbana-Champaign

February 18, 1998

We will need to be familiar with Grassmanians and bundles over them for our construction of the unviversal bundle.

Definition 1. The real Grassmanian is the set of all k−planes in ℝ^{n} and is denoted by G_{ℝ}(k, n). Similarly, the complex Grassmanian, G_{ℂ}(k, n) is the set of all complex k−planes in ℂ^{n}.

Example 1. G(1, n) is the set of all lines in ℝ^{n}. This should be familiar as real projective n − 1 space, ℝℙ^{n−1}. Similarly, G_{ℂ}(1, n) = ℂℙ^{n−1}.

We need to establish the following key properties of the Grassmanian. We derive the properties for G(k, n) = G_{ℝ}(k, n) but the same hold for G_{ℂ}(k, n). These are:

(a) G(k, n) is a smooth manifold of dimension k(n − k).

(b) Let γ_{k}^{n} be the subset of G(k, n)×ℝ^{n} consisting of {(V, v⃗) : v⃗ ∈ V }. Then under the natural projection map, π : γ_{k}^{n} → G(k, n) is a bundle with fiber a k−plane.

Recall that the trivial bundle is denoted by \underline{ℝ^{n}}. Then the universal quotient bundle, Q, over G(k, n) is defined by the exact sequence

0 → γ_{k}^{n} → \underline{ℝ^{n}} → Q → 0

We have that Q_{|V} ≅ ℝ^{n}/V, an n − k-plane.

We begin with the first property above. A useful was to specify a frame is by tge n×k matrix, [v⃗_{1}, …, v⃗_{k}] (where eacg v⃗_{i} is displayed as an n × 1 column). The linear independence condition is equivalent to the condition that [v⃗_{1}, …, v⃗_{k}] has rank k. We can interpret this matrix as an injective linear map, ℝ^{k} → ℝ^{n}.

Of course each k−plane admits may frames, so that the description is in no way unique. In fact, given any A ∈ GL(k), we get a new frame described by the matrix, [v⃗_{1}, …, v⃗_{k}] · A, or equivalently, the map

ℝ^{k} \overline{}^{A} ℝ^{k} \overline{}^{[v⃗_{1},...,v⃗_{k}]} ℝ^{n}

This leads to a description of G(k, n) as a quotient (by GL(k)) of the space of all k-frames. The latter is called the Stiefel manifold of k−frames and is denoted by F (k, n). We can identify

F (k, n) = {n × k matrices of rankk}
= {f : ℝ^{k} → ℝ^{n} : f is linear and injective}

Typeset by AMS-TEX

Then, G(k, n) = F (k, n)/ GL(n) where GL(k) acts on F (k, n) in the manner described above.

Exercise 1. Show that the GL(k) action on F (k, n) is a free right action.

Exercise 2. Show that F (k, n) is a smooth manifold of dimension kn.

Suppose that [v⃗_{1}, …, v⃗_{k}] ∈ G(k, n) has the property that the first k by k minor has non-zero determinant (since the matrix has rank k, at least one of the minors has non-zero determinant). Then, there is some A ∈ GL(k) such that

[v⃗_{1}, …, v⃗_{k}] · A = [\frac{Id}{B}]

That is, the first k rows have been reduced to the identity, and B is an (n − k) by k block. We define a map,

Mat(n − k, k) → U_{1,...,k} ⊂ GL(k, n)

by

B ↦ [\frac{Id}{B}]

Exercise 3. If I = {i_{1}, …, i_{k}} ⊆ {1, …, n}, then U_{I} defines a coordinate patch for G(k, n). Show that G(k, n) is covered by the set of all such U_{I} and that the coordinate transformations are smooth.

Note that the story gets better. Over ℂ^{n}, the coordinate transformations for G_{ℂ}(k, n) are holomorphic. Thus, G_{ℂ}(k, n) is a complex manifold of dimension k(n − k).

We now move to the second property. We first look at the projection π : F (k, n) → G(k, n). This clearly has fiber GL(k). In fact, it is a GL(k) bundle. We need to establish the local triviality over U_{I}. Over U_{{1,...,k}}, describe

[v⃗_{1}, …, v⃗_{k}] = [\frac{Id}{B}] · A

Define a map [v⃗_{1}, … v⃗_{k}] ↦ ([v⃗_{1}, …, v⃗_{k}], A). This is the trivializing map. So, π : F (k, n) → G(k, n) is a principal GL(k)-bundle. How does this bundle relate to γ_{k}^{n}?

Claim.

γ_{k}^{n} = F (k, n) ×_{GL(k)} ℝ^{k}

Exercise 4. Establish this for the case where k = 1. the tautological bundle over ℝℙ^{n−1}.

Remark. Given a k−plane, V^{k} ⊆ ℝ^{n}, we obtain an n − k plane given by ℝ^{n}/V^{k} (if we have a metric, then we can identify this with V^{⊥}m the othogonal complement of V).

We thus get a map q : G(k, n) → G(n − k, n). We can use this map to pull back Q → G(n − k, n) to G(k, n). We then have the following diagram:

[figure: γ_{k}^{n}  q^{∗}(Q)  Q; G(k, n); q; G(n − k, n)]

A natural question to ask is: how are γ_{k}^{n} and q^{∗}(Q) realted. It turns out that γ_{k}^{n} is the same as the dual bundle to q^{∗}(Q).

Exercise 5. Prove that γ_{k}^{n} ≅ (q^{∗}(Q))^{∗}.

273 Altgeld Hall, 1409 W. Green Street, Urbana, IL 61801 E-mail address: bradlow@math.uiuc.edu