Since $\pi_4 (PU(2)) = \pi_4 (SO(3)) = {\mathbb Z}_2$, the twoelement group, we know that half of the twosphere bundles over the 5sphere $S^5$ are trivial and the other half are nontrivial and all isomorphic. Can you write an explicit concrete realization for this nontrivial bundle? I have in mind something along the lines of the Hirzebruch surface (see eg. http://en.wikipedia.org/wiki/Hirzebruch_surface) $\Sigma_1$ (or $\Sigma_k$, $k$ odd) which realizes the unique topologically nontrivial $S^2$ bundle over $S^2$. (Since $\pi_1 (SO(3) = {\mathbb Z}_2$ again ``half' the twosphere bundles over the 2sphere $S^2$ are trivial and the other half are nontrivial and all isomorphic.)

$\begingroup$ I'm trying to pull one back from $S^5/S^1 = \mathbb{CP}^2$, but it seems like the nontrivial bundle there comes from $\pi_3(Diff(S^2))$, not from $\pi_4$, grr. $\endgroup$ – Allen Knutson Aug 21 '13 at 4:22
Is this concrete enough? Recall that $\mathrm{SU}(3)$ fibers over $S^5$, with fibers equal to $\mathrm{SU}(2)$ and that this fibration is nontrivial. Let $S^1\subset \mathrm{SU}(2)$ be (any) subgroup and let $B = \mathrm{SU}(3)/S^1$. Then $B$ fibers over $S^5$ with fibers $S^2$. If $B$ were trivial, then $\mathrm{SU}(3)\to S^5$ would be trivial as well, but it is not, since $S^5$ is not parallelizable.
In more detail: Regard $\mathrm{SU}(3)$ as the set of triples $(e_1,e_2,e_3)$ of special unitary bases of $\mathbb{C}^3$. Define a map $\pi:\mathrm{SU}(3)\to S^5\subset\mathbb{C}^3$ by $$ \pi(e_1,e_2,e_3) = e_1\ . $$ This is a smooth submersion with fibers isomorphic to $\mathrm{SU}(2)$. Let $B$ be the set of pairs $(v,L)$, where $v\in S^5\subset\mathbb{C}^3$ and $L\in\mathbb{CP}^2$ is a line that is Hermitian orthogonal to the line spanned by $v$, i.e., $L$ is a line in $v^\perp\simeq\mathbb{C}^2$. Then $B\to S^5$ given by $(v,L)\mapsto v$ is a smooth $S^2$ bundle over $S^5$. If $B$ were trivial, there would be a section of $B$ over $S^5$ and hence a smooth mapping $\lambda:S^5\to\mathbb{CP}^2$ such that $\bigl(v,\lambda(v)\bigr)\in B$ for all $v\in S^5$. This would define a smooth complex line bundle $\Lambda$ over $S^5$, and, since every complex line bundle over $S^5$ is trivial, there would be a nonvanishing section of this line bundle, i.e., a mapping $\sigma:S^5\to S^5$ such that $\lambda(v) = \mathbb{C}\cdot\sigma(v)$ for all $v\in S^5$. However, then there would exist a unique mapping $\tau:S^5\to S^5$ such that $\zeta(v) = \bigl(v,\sigma(v),\tau(v)\bigr)$ is a special unitary frame for all $v\in S^5$, i.e., $\zeta:S^5\to \mathrm{SU}(3)$ would be a section of the nontrivial bundle $\mathrm{SU}(3)\to S^5$.

3$\begingroup$ A more succinct description of $B$ is that it is the pullback to $S^5$ of the projectivized tangent bundle of $\mathbb{CP}^2$. $\endgroup$ – Eric Wofsey Aug 21 '13 at 9:47

1$\begingroup$ @Eric: Yes, but that doesn't make it obvious that $B$ is nontrivial as a bundle over $S^5$. $\endgroup$ – Robert Bryant Aug 21 '13 at 11:17

$\begingroup$ Thanks as always Robert. You nailed it, and explained it very clear! Even better, the bundle you gave (in Eric W's incarnation) is exactly the one we wanted to understand (non)triviality ofR $\endgroup$ – Richard Montgomery Aug 21 '13 at 16:30