A question on the non trivial rank three bundle on $S^2$ We know that number of rank $3$ vector bundles on $S^2$ is just the number of equivalence classes of maps from $S^1$ to $SO(3)$. This implies that there is one and only one non trivial vector bundle on $S^2$. Also we know that this rank $3$ vector bundle can be written as a whitney sum of a rank $2$ and trivial line bundle. Now I was wondering which rank $2$ vector bundles on $S^2$ when added to a trivial line bundle yield the non trivial bundle. 
My try: There are an integer number of rank $2$ vector bundles. We know the trivial bundle and the tangent bundle do not give a non trivial bundle. Also I calculated the clutching function for the tangent bundle, I got it as $z \to z^2$ after doing the identification of $SO(2)$ with $S^1$. So I am guessing that all odd degree clutching functions give a non trivial bundle and the even ones give a trivial bundle. I am confused as to how one proves this. Thanks.
 A: You're right: the "stabilising map" $\textrm{-} \oplus \epsilon^1$ from $r$-dimensional bundles to $(r+1)$-dimensional bundles gives the map $\pi_1(SO(r)) \to \pi_1(SO(r+1))$ induced by the inclusion $\mathrm{diag}(\textrm{–},1) : SO(r) \to SO(r+1)$.
If $r=2$, that's the  mod. 2 reduction $\mathbb Z \to \mathbb Z/2$. For $r \geq 3$, it's the identity of $\mathbb Z/2$. So the 2-dimensional bundles you're after are the one corresponding to the elements of $\pi_1(SO(2))$ which are not squares, that is, to bundles whose clutching function has an odd degree.
A: Consider the mapping $f_n : \mathbb S^1 \to SO(3)$, where $f_n(z) = z^n$ (treated as an element in $SO(2) \subset SO(3)$. The mapping is homotopic to $f_0$ if $n$ is even, and $f_1$ when $n$ is odd. This shows that the bundle $L_n \oplus \mathbb R$, where $L_n$ is defined by the clutching function $z\mapsto z^n$, is isomorphic to either the trivial bundle or $L_1 \oplus \mathbb R$. 
There is an explicit way to construct the homopoty. The two to one covering $f :\mathbb S^3 \to SO(3)$ is given here. Since $f_n$ are rotations which fixes $(0,0,1)$, we know that 
$$f\big( \cos (n\theta/2) + \vec k \sin (n\theta/2)\big) = f_n(\theta).$$
Thus $f_n$ lifts to $\mathbb S^3$ if and only if $n$ is even. In that case, pick any homotopy $\gamma$ from $\theta \mapsto \cos (n\theta/2) + \vec k \sin (n\theta/2$ to a constant map, then $f\circ \gamma$ is a homotopy from $f_n$ to a constant map. 
