# How can i realize a real rank 2 vector bundle on $S^2$ of euler class $k\in\mathbb{Z}$ as a submersion of $S^3\times\mathbb{R}^2$?

Hi!

As the title says i'm wondering how can i realize a real vector bundle rank 2 on $S^2$ of euler class $k\in\mathbb{Z}$ as a submersion of $S^3\times\mathbb{R}^2$. My initial idea was to realize it as a quotient of an action of $S^1$, more precisely i consider the following action of $S^1$ on $S^3\times\mathbb{R}^2$: i consider $S^3$ as a subset of $\mathbb{C}^2$ and i identify $\mathbb{R}$ with $\mathbb{C}$ $$\rho:S^1\times S^3\times\mathbb{R}^2\rightarrow S^3\times\mathbb{R}^2$$ $$\rho(\theta,z_1,z_2,re^{i\alpha})=(z_1e^{i\theta},z_2e^{i\theta},re^{i(\alpha-k\theta)})$$ I can't see if $S^3\times\mathbb{R}^2/S^1$ is what i'm looking for because i can't se if this object has a vector bundle structure on $S^2$. Please can anyone help?

Thank you in advance and sorry if it is a silly question.

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This is a homework level question in an introductory course on vector bundles, and not really appropriate for MO. Have you considered reading something like Steenrod's book on fibre bundles, or Hatcher's vector bundle notes (available on-line)? –  Ryan Budney Nov 17 '10 at 20:16
I will do it, if necessary i'll delete the question. –  Italo Nov 17 '10 at 20:26
For future reference, for a question like this the math.stackexchange site (listed in the FAQ) is a more appropriate venue. –  Ryan Budney Nov 17 '10 at 20:36
I am not going to explain why $(S^3\times\mathbb R^2)/S^1$ is a vector bundle (this is indeed a a homework problem) but computing the Euler class is less trivial. I suggest you look at the associated circle bundles. Namely, $S^2$ can be described $(S^3\times S^1)/T^2$, and replacing $T^2$ with the circle $S^1_{1,k}\subset T^2$ gives the circle bundle $(S^3\times S^1)/S^1_{1,k}\to (S^3\times S^1)/T^2$ whose "$k$th root" is the Hopf bundle with Euler number $1$, or perhaps $-1$. Now you need tos how that taking "$k$th power" is multiplicative for the Euler number. –  Igor Belegradek Nov 17 '10 at 20:55