Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the principal bundle $S^7$ with base space $\mathbb{C}P^3$ (3-dimensional complex projective space) and fiber $S^1\cong U(1)$. Can someone write to me the bundle projection $\pi:S^7\rightarrow\mathbb{C}P^3$ explicitly?

share|cite|improve this question
up vote 5 down vote accepted

Regard $S^7$ as the set of points in $\mathbb C^4$ of length $1$. If $p:\mathbb C^4\to\mathbb CP^3$ is the usual quotient map, then the restriction $p|_{S^7}:S^7\to\mathbb CP^3$ is the projection you are after.

share|cite|improve this answer
This, of course, works for other odd values of $7$. – Mariano Suárez-Alvarez Mar 14 '11 at 22:10
ok, I'll do that, thanks – Dar Far Mar 14 '11 at 22:12
+1 for "other odd values of 7" :) – t.b. Mar 14 '11 at 22:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.