THE GEOMETRY OF VECTOR BUNDLES AND

AN INTRODUCTION TO GAUGE THEORY

LECTURE 35

Professor Steven Bradlow Class Notes From Math 433

University of Illinois at Urbana-Champaign

April 24, 1998

Suppose E → B is a rank n = 2m bundle and that P is a SO(n) invariant polynomial on {A + A^{t} = 0} = Lie(O(n)). Then P (A) = P_{0}(A) + P_{1}(A) where P_{0} is a fully O(n) invariant polynomial and P_{1} is a SO(n) invariant polynomial. We saw that det(g)P_{1}(A) = P_{1}(gAg^{−1}) for g ∈ O(n).

Example 1. Define e(A) as follows. Fix an oriented, orthonormal basis for ℝ^{2n}, say {e_{i}}. Let Ae_{i} = A_{ji}e_{j}. Define

α(A) = ∑_{i<j} A_{ij}e_{i} ∧ e_{j} ∈ ⋀^{2}(ℝ^{2m})

[If A is a matrix of standard type I, then a direct calculation shows that

α(A) = ∑_{i=1}^{m} x_{i}e_{2i−1} ∧ e_{2i}]

Set

e(A) = \frac{1}{m!}(α(A)^{m}, e_{1} ∧ ⋯ ∧ e_{2m})

where ( , ) is the inner product on ⋀^{2m}(ℝ^{2m}). [For A as above we get that

α(A)^{m} = m!(x_{1}x_{2} … x_{m}e_{1} ∧ ⋯ ∧ e_{2m})

So, e(A) = x_{1} … x_{m}.]

Exercise 1. Show that (1) e(gAg^{−1}) = e(A)det(g) for g ∈ O(n).

(2) e^{2}(A) = det(A).

Fact: Any SO(n) invariant polynomial can be written as P (A) = e(A)P̂(A) where P̂ is a O(n) invariant polynomial.

Typeset by AMS-TEX

Definition 1. Given an oriented real vector bundle E → B of rank n = 2m, we define the Euler class by e(\frac{1}{2π} F) where F is the curvature of any orthogonal connection with respect to an oriented, orthonormal frame.

Note.

(1) If the rank of E is n = 2m, then e(E) ∈ H^{2m}(B; ℝ). In fact, e(E) ∈ H^{2m}(B; ℤ).

(2) The Euler class distributes nicely across sums. That is, e(E_{1} ⊕ E_{2}) = e(E_{1})e(E_{2}).

Special Case: Suppose that E = E_{r} is the underlying real bundle of a complex bundle E_{c}. From previous lectures, E_{r} is orientable and if E_{c} has complex dimension m, then E_{r} has real dimension 2m.

Question: How is e(E_{r}) related related to the Chern classes of E_{c}?

Clue: e^{2}(A) = det(A). Also, det(I + A) = 1 + ⋯ + det(A). So, e^{2}(E_{r}) = P_{m}(E_{r}). Also,

∑_{k=0}^{m} P_{k}(E_{r})(−1)^{k} = (1 + c_{1}(E_{c}) + ⋯ + c_{m}(E_{c}))(1 − c_{1}(E_{c}) + ⋯ + (−1)^{m}c_{m}(E_{c}))

This implies that P_{m}(E_{r})(−1)^{m} = (−1)^{m}c_{m}^{2}(E_{c}). So, e^{2}(E_{r}) = c_{m}^{2}(E_{c}). In fact, we shall see that e(E_{r}) = c_{m}(E_{c}), where E_{r} has a standard orientation induced by E_{c}.

The orientation on ℝ^{2m} induced by an orientation from ℂ^{m} is obtained as follows. Choose a basis {e_{a}}_{a=1}^{m} for ℂ^{m} over ℂ. Then {e_{a}, ie_{a}}_{a=1}^{m} is a basis for ℝ^{2m}. The orientation on ℝ^{2m} is obtained by declaring that this basis is positively oriented.

Under this choice of frames, the inclusion GL(m, ℂ) → GL(2m, ℝ) yields

( …_{\frac{iλ_{m}}{2π}}^{\frac{iλ_{1}}{2π}} ) ↦ ( \begin{matrix}0 & −\frac{λ_{1}}{2π} \\ \frac{λ_{1}}{2π} & 0 \\ & & … \\ & & & 0 & −\frac{λ_{m}}{2π} \\ & & & \frac{λ_{m}}{2π} & 0 \end{matrix} )

It follows from this that

c_{m}(E_{c}) = ∏_{i=1}^{m} −\frac{λ_{i}}{2π} = e(E_{r})

Note. The signs have been chosen so that this holds. If we change the orientation of E, then e(E) will change signs.

273 Altgeld Hall, 1409 W. Green Street, Urbana, IL 61801

E-mail address: bradlow@math.uiuc.edu