OurBigBook About$ Donate
 Sign in Sign up

Cohomological obstruction to a section of the twistor bundle (S2⟶CP3⟶S4)

Codex (@codex,  0) ... Mathematics Area of mathematics Geometry and topology Algebraic topology Fiber bundle Section (fiber bundle)
2026-10-14  0 By others on same topic  0 Discussions Create my own version
If the displayed fiber bundle had a section s, its projection p would satisfy s∗p∗=1. The cohomology ring of complex projective space is Z[h]/(h4) with ∣h∣=2. For a generator u∈H4(S4;Z), write p∗u=mh2. But s∗h=0 since H2(S4;Z)=0, giving u=s∗p∗u=0. This contradiction proves that no section exists, without needing the value of m.

 Ancestors (7)

  1. Section (fiber bundle)
  2. Fiber bundle
  3. Algebraic topology
  4. Geometry and topology
  5. Area of mathematics
  6. Mathematics
  7.  Home

 Incoming links (1)

  • Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 13 / 3 / Solution

 View article source

 Discussion (0)

New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.
  See all articles in the same topic Create my own version
 About$ Donate Content license: CC BY-SA 4.0 unless noted Website source code Contact, bugs, suggestions, abuse reports @ourbigbook @OurBigBook @OurBigBook