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Let me first explain the background of my question.

As is well known, the group $SO(n+1)$ acts transitively on the sphere $S^n$, and the stabilizer is the group $SO(n)$, so that we get a fibration sequence $ SO(n) \to SO(n+1) \to S^n.$

Indeed this fibration is a principal-$SO(n)$-bundle and can actually be identified with the frame bundle of the tangent bundle of $S^n$, i.e., the tangent bundle $TS^n$ is the associated bundle to this principal-$SO(n)$-bundle along the tautological representation of $SO(n)$ on $\mathbb{R}^n$.

Now let us consider the special case $n = 4$. Here we get a principal-$SO(3)$-bundle $ SO(3) \to SO(4) \to S^3.$ But we can identify the $S^3$ with $Sp(1)$, the symplectic group acting on the quaternions. Via this identification $S^3$ sits canonically in $SO(4)$, i.e. we have an embedding of topological groups $S^3 \cong Sp(1) \to SO(4)$.

The standard theory of principal-bundles hence tells us that the above bundle splits (topologically) which means that there is a homeomorphism $ SO(4) \cong SO(3)\times S^3 $ in particular there is a continuous map $SO(4) \to SO(3)$ (which is not a morphism of lie groups).

It is well known that the homotopy type $BSO(n)$ represents the functor of $n$-dimensional oriented vector bundles, and via the clutching function construction the group $SO(n)$ itself represents the functor of $n$-dimensional vector bundles over suspensions of spaces. Hence the above map $SO(4) \to SO(3)$ tells us that there is a natural transformation between the functor of rank $4$ vector bundles over suspensions to the functor of rank $3$ vector bundles over suspensions.

Does anybody know a geometric construction of this transformation which I only understand homotopy theoretically?

Any ideas or references are greatly appreciated.

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    @DavidSpeyer: Yes, SU(2), sadly too late to edit.2012-12-22

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The double cover $\pi:S^3\times S^3\rightarrow SO(4)$ actually provides two homotopoically distinct maps from $S^3 = Sp(1)$ to $SO(4)$, given by restricting $\pi$ to either factor. (One can easily see that these two maps induce different maps on $\pi_3$, so are not homotopic).

The following theorem is contained in

K.Grove-W.Ziller, Lifting group actions and nonnegative curvature, Trans. Amer. Math. Soc. 363 (2011) 2865-2890.

Ziller has a freely accessible version here - see the middle of page $8$.

Theorem If $E\rightarrow M$ is a rank $4$ vector bundle over $M$ (where $M$ is a compact simply connected manifold.), then $\Lambda^2(E) = \Lambda^2(E)_+\oplus\Lambda^2(E)_-$ decomposes into self-dual and antiself dual forms. Then $\Lambda^2(E)_{\pm}$ are the two rank $3$ vector bundles over $M$ corresponding to the two sections above.

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    Well, assuming the usual space is connected (and I'm not exactly sure where this enters), it's suspension is simply connected, which is one of the hypothesis of their theorem. Also, I've been thinking about it and I know longer thing the decomposition comes (canonically) from the Hodge star. The issue is that the Hodge star requires a metric to define. On the other hand, I *think* that given the standard rep $V$ of $SU(2)\times SU(2)$, then $\Lambda^2 V$ has a unique nontrivial $SU(2)\times \{e\}$ invt. subspace as does $\{e\}\times SU(2)$ and the sum of these is $\Lambda^2 V$.2012-12-23