Stably trivial bundle is trivial

Question

I have a smooth embedding $f:S^2\to \mathbb{R}^4$ and would like to show that the normal bundle $\nu\to K$ of the image $K:=f(S^2)$ is trivial.
I have already shown that it is stably trivial, i.e. $ \epsilon^3\oplus\nu\cong \epsilon^5$ for the trivial vector bundle $\epsilon\to K$ and that the Euler number $\chi(\nu)$ vanishes.

I have heard that there is an argument using classifying spaces of vector bundles to conclude that $\nu$ is actually trivial.
Does someone know a reference for this or an easier proof of my initial question which does not involve too much algebraic topology?

Answer 1

$2$-dimensional vector bundles over $S^2$ are classified by their Euler number or equivalently the homological self-intersection of their zero-section (see Milnor-Stasheff for instance).

The proof of this is probably easy enough for you to work out on your own. Consider trivializations of the North and South hemispheres of $S^2$. The vector bundle is determined by how these trivializations are glued together along the equator which in turn is determined by an element of $\pi_1(SO(2))\cong \Bbb Z$. Now all you have to check is that either generator of $\pi_1(SO(2))$ corresponds to a vector bundle with $e(\nu)=1$.

If $e(\nu)=[f(S^2)]^2\ne 0$ (here $[f(S^2)]^2$ denotes the intersection pairing on $H_2$), then $f(S^2)$ would be represent an element of $H_2(\Bbb R^4, \Bbb Z)$ with non-trivial intersection pairing. Since $H_2(\Bbb R^4,\Bbb Z)=0$, no such homology class exists.

$3$ and $4$ dimensional vector bundles over $S^2$ are in fact determined by their second Stiefel-Whitney class (which is a similar argument to the above using that $\pi_1(SO(3))\cong \pi_1(SO(4) \cong \Bbb Z_2$). In the case of a Whitney sum of a trivial bundle and a $2$-dimensional vector bundle, the second Stiefel-Whitney class is just the mod $2$ reduction of the Euler class. This implies that any vector bundle with even Euler class (e.g. $TS^2$) is stably trivial but not trivial.