If $S^3=\{ (z_1,z_2)\in\mathbb{C}^2\mid \vert z_1\vert^2+\vert z_2\vert^2=1\}$ and $\pi:S³\rightarrow\mathbb{C}P^1$ for $(z_1,z_2)\mapsto [(z_1,z_2)]$ since $[(z_1,z_2)]=\{ (w_1,w_2)\in\mathbb{C}^2-\bar{0}\mid (z_1,z_2)=\lambda(w_1,w_2),\lambda\in\mathbb{C}-\{0 \}\}$.

I need to proof that $\pi$ is a submersion surjective and $\pi$ induces a fibre bundle with fibre $S^1$.

The surjectivity is almost immediate since the equivalence relation can restrictes a $S^3$ and for the fibre for the fact that $S^1$ act on $S^3$, $i.e.$ $(w,(z_1,z_2))\mapsto w\cdot(z_1,z_2)\in S^3$ and then $\mathbb{C}P^1=S^3/S^1$. For Ehresmann's Lemma $\pi$ induces a fibre bundle since the $S^3$ is compact (and then $\pi$ is proper function). But I had a trouble to prove $\pi$ is a submersion because I dont know how to express the derivative $D\pi$ since that is a projection for the equivalece relation. How am I supposed to derive the function $\pi$?


For $i=1,2$ let $V_i\subset \mathbb P^1$ be the subset defined by $z_i\neq 0$ and $U_i=\pi^{-1}(V_i)\subset S^3$.
We can write down explicitly a diffeomorphism (commuting with the projections onto $V_i$) between $U_i$ and the trivial bundle $V_i\times S^1$ as follows:

Define $$f_i:U_i\stackrel \cong\to V_i\times S^1: (z_1,z_2)\mapsto ([z_1:z_2],\frac {z_i}{ |z_i|})$$ and $$g_i:V_i\times S^1 \stackrel \cong\to U_i : ([w_1:w_2],s)\mapsto \frac {1}{||w||}(w_1\frac { |w_i|}{w_i}s,w_2\frac { |w_i|}{w_i}s)$$ Then $f_i,g_i$ are mutually inverse diffeomorphisms proving that $\pi$ is a fiber bundle with fiber $S^1$.

I have set up the notation in a way that makes it trivial to generalize the result to obtain a fibration $ S^{2n+1}\to \mathbb P^n(\mathbb C)$ with fiber $S^1$ just by allowing the index $i$ to run from $1$ to $n$.

  • $\begingroup$ How does it follow (if it does at all) that $\pi$ is a submersion? $\endgroup$ – user500094 Feb 7 '18 at 19:10
  • $\begingroup$ I see, every fiber bundle is a submersion. But what is $w$ and $||w||$ in your proof? $\endgroup$ – user500094 Feb 8 '18 at 3:57
  • $\begingroup$ @user500094: $w$ is a point of $\mathbb C^2$ satisfying $w_i\neq0$ and $\vert\vert w\vert\vert$ is its euclidean norm. $\endgroup$ – Georges Elencwajg Feb 8 '18 at 11:51

You could treat all of $CP^1$ except $z_2 = 0$ as being "parametrized" by $C$ via $z \mapsto (z, 1)$. Then the map you've defined just takes $(z_1, z_2) \mapsto z_1/z_z \in \mathbb C$. Now it's not so hard to compute the derivative.

You can then make a second coordinate chart for where $z_1 \ne 0$, via $z \mapsto (1, z)$, and do more or less the same thing.

The map you're looking at is called the Hopf Map, and is described in many topology books.

  • $\begingroup$ Dear John, you are avoiding the difficulty because you don't take any account of the equation of the sphere. $\endgroup$ – Georges Elencwajg Nov 11 '14 at 18:30
  • $\begingroup$ If $(g,\mathbb{C})$ ($g[(z_1,z_2)]=z_1/z_2$) is the chart for $z_2\not= 0$ then $D\pi$ is surjetive if the derivative of $g\circ\pi$ is surjective. It is right? $\endgroup$ – Donyarley Nov 11 '14 at 19:22
  • $\begingroup$ I don't think so, @Georges. He still needs to prove that the derivative of map $K: S^3 \to \mathbb C : (z_1, z_2) \mapsto z_1/z_2$ has maximal rank on each tangent space, rather than on all of $\mathbb R^4$. I was just giving him a way to avoid thinking about equivalence classes in the codomain. $\endgroup$ – John Hughes Nov 11 '14 at 21:35
  • $\begingroup$ Yes, Donyarley, that's mostly correct...as long as you regard the domain of $D(g \circ \pi)$ as the tangent space to $S^3$ at $(z_1, z_2)$ rather than the tangent space to $\mathbb R^4$ at $(z_1, z_2)$. (The first is 3-dimensional; the second is 4-dimensional.) $\endgroup$ – John Hughes Nov 11 '14 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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