A concrete realization of the nontrivial 2-sphere bundle over the 5-sphere?

Since $\pi_4 (PU(2)) = \pi_4 (SO(3)) = {\mathbb Z}_2$, the two-element group, we know that half of the two-sphere bundles over the 5-sphere $S^5$ are trivial and the other half are non-trivial and all isomorphic. Can you write an explicit concrete realization for this non-trivial 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 non-trivial $S^2$ bundle over $S^2$. (Since $\pi_1 (SO(3) = {\mathbb Z}_2$ again half' the two-sphere bundles over the 2-sphere $S^2$ are trivial and the other half are non-trivial and all isomorphic.)

• 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. – 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 $\mathrm{SU}(3)\to S^5$.
• A more succinct description of $B$ is that it is the pullback to $S^5$ of the projectivized tangent bundle of $\mathbb{CP}^2$. – Eric Wofsey Aug 21 '13 at 9:47
• @Eric: Yes, but that doesn't make it obvious that $B$ is nontrivial as a bundle over $S^5$. – Robert Bryant Aug 21 '13 at 11:17