This implies t ω t d m β t for every t and hence p

  • No School
  • AA 1
  • 306

This preview shows page 247 - 250 out of 306 pages.

This implies t ω ( t ) = - d M β ( t ) for every t and hence p ( F 1 ) - p ( F 0 ) = ω (1) - ω (0) = Z 1 0 t om ( t ) dt = - d M Z 1 0 β ( t ) dt. Thus p ( F 1 ) - p ( F 0 ) is exact and this proves (ii). We prove (iii). In Section 8.1.5 we have seen that the curvature of the pullback connection f * is in the local trivializations f * ψ α given by the 2-forms F f * α = f * F α Ω 1 ( f - 1 ( U α ) , g ) . Hence it follows directly from the definitions that p ( F f * ) = f * p ( F ). This proves (iii) and Theorem 8.3.2.
Image of page 247
240 CHAPTER 8. CONNECTIONS AND CURVATURE 8.3.3 The Euler Class of an Oriented Rank- 2 Bundle Let π : E M be an oriented Riemannian real rank-2 bundle over a smooth manifold. By Example 8.1.13 E is a vector bundle with structure group SO(2) = g = a - c c a a, c R , a 2 + c 2 = 1 . Its Lie algebra consists of all skew-symmetric real 2 × 2-matrices: so (2) = ξ = 0 - λ λ 0 λ R . The linear map e : so (2) R defined by e ( ξ ) := - λ 2 π is invariant under conjugation. (However, e ( g - 1 ξg ) = - e ( ξ ) whenever g O( n ) has determinant - 1. Thus we must assume that E is oriented.) Hence there is a characteristic class e ( E ) := [ e ( F )] H 2 ( M ) , (8.3.5) where is Riemannian connection on E . If we change the Riemannian structure on E then there is an orientation preserving automorphism of E intertwining the two inner products. (Prove this!) Thus the characteristic class e ( E ) is independent of the choice of the Riemannian metric. We prove below that (8.3.5) is the Euler class of E whenever M is a compact oriented manifold without boundary. Thus we have extended the definition of the Euler class of an oriented real rank-2 bundle to arbitrary base manifolds. Theorem 8.3.3. If E is an oriented real rank- 2 bundle over a compact oriented manifold M without boundary then (8.3.5) is the Euler class of E . Proof. Choose a smooth section s : M E that is transverse to the zero section and denote Q := s - 1 (0) . Choose a Riemannian metric on M and let exp : TQ ε U ε be the tubular neighborhood diffeomorphism in (7.2.11). Multiplying s by a suitable positive function on M we may assume that p M \ U ε/ 3 = | s ( p ) | = 1 .
Image of page 248
8.3. CHERN–WEIL THEORY 241 Next we claim that there is a Riemannian connection on E such that s = 0 on M \ U ε/ 2 . (8.3.6) To see this, we choose on open cover { U α } of M such that one of the sets is U α 0 = M \ U ε/ 3 and E admits a trivialization over each set U α . In particular, we can use s to trivialize E over U α 0 . Next we choose a partition of unity where ρ α 0 = 1 on M \ U ε/ 2 . Then the formula (8.1.6) in Step 6 of the proof of Proposition 8.1.3 defines a Riemannian connection that satisfies (8.3.6). By (8.3.6) we have F s = d s = 0 on M \ U ε/ 2 . Since F is a 2-form with values in the skew-symmetric endomorphisms of E we deduce that F = 0 on M \ U ε/ 2 . (8.3.7) The key observation is that, under this assumption, the 2-form τ ε := exp * e ( F ) Ω 2 c ( TQ ε ) is a Thom form on the normal bundle of Q . With this understood we obtain from Lemma 7.2.17 with τ Q = e ( F ) that Z M ω e ( F ) = Z Q ω = Z M ω s * τ for every closed form ω Ω m - 2 ( M ) and every Thom form τ Ω 2 c ( E ), where the last equation follows from Theorem 7.3.15. By Poincar´
Image of page 249
Image of page 250

You've reached the end of your free preview.

Want to read all 306 pages?

  • Fall '19
  • Manifold,

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

Stuck? We have tutors online 24/7 who can help you get unstuck.
A+ icon
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes