- #1
zhentil
- 489
- 0
This is a question from Hirsch's Differential Topology book: show that there is a bijective correspondence between
[tex]K^k(S^n) \leftrightarrow \pi_{n-1}(GL(k)) [/tex],
where [tex]K^k(S^n) [/tex] denotes the isomorphism classes of rank k vector bundles over the sphere. The basic idea is that any vector bundle over the sphere has a trivializing cover consisting of two open sets diffeomorphic to [tex]\textbf{R}^n[/tex]. The transition function of such a cover restricts to a map from the equator [tex]S^{n-1} \rightarrow GL(k)[/tex]. Moreover, vector bundles with homotopic classifying maps are isomorphic. However, the reverse inclusion is eluding me. Why must two isomorphic bundles over the sphere have homotopic classifying maps?
I've seen a proof of this involving writing the Grassmanian as a fiber bundle of orthogonal groups, and using the exact homotopy sequence to conclude that [tex]\pi_n(G_{3k,k}) \cong \pi_{n-1}(GL(k)) [/tex], but since Hirsch never mentioned that, it strikes me that there must be an elementary way of seeing this. Any ideas?
[tex]K^k(S^n) \leftrightarrow \pi_{n-1}(GL(k)) [/tex],
where [tex]K^k(S^n) [/tex] denotes the isomorphism classes of rank k vector bundles over the sphere. The basic idea is that any vector bundle over the sphere has a trivializing cover consisting of two open sets diffeomorphic to [tex]\textbf{R}^n[/tex]. The transition function of such a cover restricts to a map from the equator [tex]S^{n-1} \rightarrow GL(k)[/tex]. Moreover, vector bundles with homotopic classifying maps are isomorphic. However, the reverse inclusion is eluding me. Why must two isomorphic bundles over the sphere have homotopic classifying maps?
I've seen a proof of this involving writing the Grassmanian as a fiber bundle of orthogonal groups, and using the exact homotopy sequence to conclude that [tex]\pi_n(G_{3k,k}) \cong \pi_{n-1}(GL(k)) [/tex], but since Hirsch never mentioned that, it strikes me that there must be an elementary way of seeing this. Any ideas?